notes on algebra - buffalo
TRANSCRIPT
Notes on AlgebraMTH 619/620
2015.09.01
Contents
1 Monoids and groups 7
2 Subgroups 10
3 Homomorphisms of groups 12
4 The kernel and the image of a homomorphism 15
5 Normal subgroups, cosets and quotient groups 17
6 Isomorphism theorems 22
7 Index of a subgroup and order of an element 26
8 Free groups and presentations of groups 29
9 Direct products, direct sums, and free abelian groups 33
10 Categories and functors 36
11 Adjoint functors 41
12 Categorical products and coproducts 44
13 More on free abelian groups 48
14 Finitely generated abelian groups 53
15 Permutation representations and G-sets 60
16 Some applications of G-sets 65
17 The Sylow theorems 68
18 Application: groups of order pq 72
19 Group extensions and composition series 82
20 Simple groups 86
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21 Symmetric and alternating groups 89
22 Simplicity of alternating groups 95
23 Solvable groups 98
24 Nilpotent groups 101
25 Rings 108
26 Ring homomorphisms and ideals 112
27 Principal ideal domains and Euclidean rings 116
28 Prime ideals and maximal ideals 120
29 Zorn’s Lemma and maximal ideals 123
30 Unique factorization domains 126
31 Prime elements 130
32 PIDs and UFDs 133
33 Application: sums of two squares 136
34 Application: Fermat’s Last Theorem 140
35 Greatest common divisor 141
36 Rings of fractions 143
37 Factorization in rings of polynomials 149
38 Irreducibility criteria in rings of polynomials 156
39 Modules 162
40 Basic operations on modules 165
41 Free modules and vector spaces 167
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42 Invariant basis number 171
43 Projective modules 176
44 Projective modules over PIDs 180
45 The Grothendieck group 182
46 Injective modules 190
47 Exact functors 197
48 Tensor products 202
49 Tensor products of homomorphisms 207
50 Tensor products and adjoint functors 209
51 Tensor products for non-commutative rings 211
52 Restriction and extension of scalars 214
53 Application: rings with IBN 217
54 Algebras 218
55 Tensor product of algebras 219
56 Fields 220
57 Field extensions 222
58 Prime subfield and field characteristic 223
59 Algebraic and transcendental elements 225
60 Algebraic extensions 231
61 Separable elements 234
62 Derivatives and separable elements 236
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63 Separable extensions 241
64 Simple extensions 245
65 Simple extensions and intermediate fields 250
66 Construction of extensions 253
67 Algebraically closed fields 256
68 Roots of unity 262
69 Finite fields 265
70 Galois theory - motivation 268
71 Normal extensions 271
72 Galois extensions 277
73 Application: rational symmetric functions 282
74 The fundamental theorem of Galois theory 284
75 The fundamental theorem of algebra 290
76 Infinite Galois extensions 292
77 Abelian and cyclic extensions 295
78 Radical extensions 301
79 Solvable extensions 303
80 Solvability of polynomials by radicals 307
81 Straightedge and compass constructions 312
82 Construction of regular polygons 318
83 Transcendental extensions 321
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84 Algebraic sets 327
85 Hilbert basis theorem 329
86 Radical ideals 334
87 Integral extensions of rings 336
88 Noether Normalization Lemma 341
89 Hilbert Nullstellensatz 345
90 Zariski topology 350
91 Algebraic varieties 353
92 Regular functions 358
93 Suggested further reading 361
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1 Monoids and groups
1.1 Definition. A monoid is a set M together with a map
M ×M →M, (x, y) 7→ x · y
such that
(i) (x · y) · z = x · (y · z) ∀x, y, z ∈M (associativity);
(ii) ∃e ∈M such thatx · e = e · x = x
for all x ∈M (e = the identity element of M).
1.2 Examples.
1) Z with addition of integers (e = 0)
2) Z with multiplication of integers (e = 1)
3) Mn(R) = {the set of all n× n matrices with coefficients in R} with ma-trix multiplication (e = I = the identity matrix)
4) U = any setP (U) := {the set of all subsets of U}
P (U) is a monoid with A ·B := A ∪B and e = ∅.
5) Let U = any set
F (U) := {the set of all functions f : U → U}
F (U) is a monoid with multiplication given by composition of functions(e = idU = the identity function).
1.3 Definition. A monoid is commutative if x · y = y · x for all x, y ∈M .
1.4 Example. Monoids 1), 2), 4) in 1.2 are commutative; 3), 5) are not.
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1.5 Note. Associativity implies that for x1, . . . , xk ∈M the expression
x1 · x2 · · · · · xkhas the same value regardless how we place parentheses within it; e.g.:
(x1 · x2) · (x3 · x4) = ((x1 · x2) · x3) · x4 = x1 · ((x2 · x3) · x4)etc.
1.6 Note. A monoid has only one identity element: if e, e′ ∈ M are identityelements then
e = e · e′ = e′
1.7 Definition. A group is a monoid G such that for any x ∈ G there is y ∈ Gsatistying x · y = e = y · x.
The element y is called the inverse of x and it is denoted by x−1 (or by −x inthe additive notation).
A group G is commutative (or abelian) if x · y = y · x for all x, y ∈ G.
1.8 Examples.
1) Z,Q,R,C with addition
2) Q∗ = Q− {0}, R∗ = R− {0}, C∗ = C− {0} with multiplication
3) GLn(R) = {A ∈Mn(R) | det(A) = 0} with matrix multiplication(the n× n general linear group)
4) SLn(R) = {A ∈Mn(R) | det(A) = 1} with matrix multiplication(the n× n special linear group)
5) Let U = be any set and let
Perm(U) := {f : U → U | f is a bijection}Perm(U) with composition of functions is a group (the group of permu-tations of U)
Note. If U = {1, 2, . . . , n} then Perm(U) is called the symmetric groupon n letters and it is denoted by Sn.
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7) Let T = an equilateral triangle
GT = {I, R1, R2, S1, S2, S3}
1S2S 3S
IR1
R2
GT = the group of symmetries of T .
1.9 Proposition (Cancellation Law). If G is a group, x, y, x ∈ G and
xy = xz
then y = z.
Proof.
xy = xz
x−1xy = x−1xz
y = z
1.10 Note. The cancellation law does not hold for monoids. E.g. in M2(R)take
A =
(1 00 0
), B =
(0 00 1
), C =
(0 00 0
)Then AB = AC but A = C.
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2 Subgroups
2.1 Definition. If G is a group then a subgroup of G is a subset H ⊆ G suchthat
(i) e ∈ H;
(ii) if x, y ∈ H then xy ∈ H;
(iii) if x ∈ H then x−1 ∈ H.
2.2 Note. A subgroup of a group is by itself a group.
2.3 Examples.
1) If G is a group then G, {e} are subgroups of G
2) Z is a subgroup of Q, which is a subgroup of R, which is a subgroup of C.
3) SLn(R) is a subgroup of GLn(R)
4) H = {I, R1, R2} is a subgroup of GT
2.4 Note. If {Hi}i∈I is a family of subgroups of G then⋂i∈I Hi is also a
subgroup of G.
2.5 Definition. If G is a group and S is a subset of G then denote
⟨S⟩ = the smallest subgroup of G that contains S
⟨S⟩ is the subgroup of G generated by the set S.
2.6 Proposition. If S ⊆ G then ⟨S⟩ consists of all elements of the form
x±11 x±1
2 · · · · · x±1k
where x1, . . . , xk ∈ S.
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Proof. Exercise.
2.7 Definition. A set S ⊆ G generates G if ⟨S⟩ = G.
2.8 Example. S = {S1, S2} generates GT .
2.9 Definition. A group G is finitely generated if it is generated by some finitesubset S ⊆ G.
2.10 Note.
• Every finite group is finitely generated.
• Some infinite groups are finitely generated; e.g. Z = ⟨1⟩.
2.11 Definition. A group G is cyclic if G = ⟨a⟩ for some a ∈ G
2.12 Note. If G is cyclic, G = ⟨a⟩ then every element g ∈ G is of the form
g = an
for some n ∈ Z (where a−n := (a−1)n, a0 = e).
2.13 Examples.
1) Z = ⟨1⟩ is cyclic.
2) H := {I, R1, R2} ⊆ GT is cyclic:
H = ⟨R1⟩ and H = ⟨R2⟩
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3 Homomorphisms of groups
3.1 Definition. Let G, H be groups. A function f : G→ H is a group homo-morphism if for any a, b ∈ G we have
f(ab) = f(a)f(b)
3.2 Proposition. If f : G→ H is a homomorphism of groups and eG, eH denoteidentity elements in, respectively, G and H then
(i) f(eG) = eH
(ii) f(a−1) = f(a)−1 for any a ∈ G.
Proof. (i) We have
f(eG) = f(eG · eG) = f(eG) · f(eG)
Multiplying this equation by f(eG)−1 we obtain eH = f(eG).
(ii) Since by (i) we have f(eG) = eH therefore
f(a) · f(a−1) = f(a · a−1) = f(eG) = eH
It is now enough to multiply this equation from the left by f(a)−1.
3.3 Definition. A homomorphism f : G → H is an isomorphism if there is ahomomorphism g : H → G such that g ◦ f = idG and f ◦ g = idH .
3.4 Proposition. A map f : G → H is an isomorphism of groups iff f is ahomomorphism and a bijection.
Proof. Exercise.
3.5 Definition. If there exists an isomorphism f : G→ H then we say that thegroups G and H are isomorphic and we write G ∼= H.
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3.6 Definition. A homomorphism f : G→ G is called an endomorphism of G.An isomorphism f : G→ G is called an automorphism of G.
3.7 Examples.
1) idG : G→ G is an automorphism of G.
2) f : G→ G, f(g) = e ∀g∈G is an endomorphism of G.
3) If f : G→ H, g : H → K are homomorphisms then so is g ◦ f : G→ K.
4) For g ∈ G definecg : G→ G, cg(a) := gag−1
Check: cg is an automorphism of G. Automorphisms of this form are calledinner automorphisms of G.
Note. If G is an abelian group then cg = idG for all g ∈ G.
5) Recall: GLn(R) = {A ∈Mn | det(A) = 0}, R∗ = R− {0}We have the determinant function:
det : GLn(R)→ R∗
Since det(AB) = det(A) · det(B) this function is a homomorphism.
6) Let G ⊆ GL2(R)
G :=
{(1 r0 1
) ∣∣∣∣ r ∈ R}
G is a subgroup of GL2(R):(1 r0 1
)·(1 s0 1
)=
(1 r + s0 1
)(1 r0 1
)−1
=
(1 −r0 1
)We have homomorphisms:
f : R→ G and g : R→ G
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where
f(r) =
(1 r0 1
), g
((1 r0 1
))= r
Since g ◦ f = idG, f ◦ g = R we get G ∼= R.
3.8 Definition. If G is a group then
|G| := the number of elements of G
|G| is called the order of G.
3.9 Example. |GT | = 6, |Z| =∞.
3.10 Note. If G ∼= H then |G| = |H|.
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4 The kernel and the image of a homomorphism
4.1 Proposition. Let f : G→ H be a homomorphism.
1) If G′ is a subgroup of G then f(G′) is a subgroup of H.
2) If H ′ is a subgroup of H then f−1(H ′) is a subgroup of G.
Proof. Exercise.
4.2 Definition. If f : G→ H is a homomorphism then
• the image of f is the subgroup
Im(f) := f(G) ⊆ H
• the kernel of f is the subgroup
Ker(f) := f−1(eH) ⊆ G
4.3 Note. f : G→ H is an epimorphism (onto) iff Im(f) = H.
4.4 Proposition. f : G→ H is a monomorphism (1-1) iff Ker(f) = {eG}
Proof. (⇒) We have f(eG) = eH . Thus if f is 1-1 then f(g) = eH only ifg = eH . In other words we have then Ker(f) = {eH}.
(⇐) Assume that Ker(f) = {eG} and let f(a) = f(b) for some a, b ∈ G. Wehave:
f(ab−1) = f(a)f(b)−1 = eH
so ab−1 ∈ Ker(f). Therefore ab−1 = eG, and so a = b.
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4.5 Problem. Let G be a group, and let H be a subgroup of G. Is there ahomomorphism
f : G→ K
such that Ker(f) = H?
4.6 Note. The dual problem is trivial: if H is a subgroup of G then we havethe inclusion homomorphism
i : H ↪→ G
and Im(i) = H. It follows that any subgroup of G is an image of some homo-morphism.
4.7 Definition. A subgroup H ⊆ G is a normal subgroup if for every h ∈ H wehave
aha−1 ∈ H ∀a ∈ G
4.8 Notation. If H is a normal subgroup of G then we write H ◁G
4.9 Proposition. If f : G → H is a homomorphism then Ker(f) is a normalsubgroup of G.
Proof. If a ∈ G, h ∈ Ker(f) then
f(aha−1) = f(a)f(h)f(a)−1 = f(a) · e · f(a)−1 = f(a)f(a)−1 = e
so aha−1 ∈ Ker(f).
4.10 Examples.
1) Any subgroup of an abelian group is normal.
2) H := {I, R1, R2} is a normal subgroup of GT (check!).
3) K := {I, S1} is not a normal subgroup of GT (check!). As a consequenceK cannot be the kernel of any homomorphism GT → G.
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5 Normal subgroups, cosets and quotient groups
Recall. If f : G→ K is a homomorphism then Ker(f) is a normal subgroup ofG.
Next goal: If H is a normal subgroup of G then there is a homomorphismf : G→ K such that H = Ker(f).
5.1 Definition. If H is a subgroup of G then a left coset of H in G is a subsetof G of the form
aH := {ah | h ∈ H}for some a ∈ G.
A right coset of H in G is a subset of G of the form
Ha := {ha | h ∈ H}
for some a ∈ G.
5.2 Example.
Recall: GT = {I, R1, R2, S1, S2, S3}. Take H := {I, S1}. We have:
IH = {I · I, I · S1} = {I, S1} = HS1H = {S1 · I, S1 · S1} = {S1, I}S2H = {S2 · I, S2 · S1} = {S2, R2}S3H = {S3 · I, S3 · S1} = {S3, R1}R1H = {R1 · I, R1 · S1} = {R1, S3}R2H = {R2 · I, R2 · S1} = {R2, S2}
Note: IH = S1H, S2H = R2H, S3H = R1H
5.3 Lemma. If G is a group and H is a subgroup of G then
aH = bH iff a−1b ∈ H
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Proof. (⇒) Let aH = bH. Since e ∈ H thus
b = be ∈ bH = aH
so b = ah for some h ∈ H. Therefore a−1b = h ∈ H.
(⇐) Assume that a−1b ∈ H. For any h ∈ H we have
ah = a(a−1b)(a−1b)−1h = b((a−1b)−1)h ∈ bH
This gives: aH ⊆ bH. Also for any h ∈ H we have:
bh = (aa−1)bh = a(a−1b)h ∈ aH
so bH ⊆ aH. Therefore aH = bH.
5.4 Proposition. If H is a subgroup of G then for any a, b ∈ G either
aH = bH or aH ∩ bH = ∅
Proof. Let aH ∩ bH = ∅ and let c ∈ aH ∩ bH. Then
ah1 = c = bh2
for some h1, h2 ∈ H. This gives
a−1b = h1h−12 ∈ H
and so aH = bH by (5.3).
5.5 Corollary. If H is a subgroup of G then every element of G belongs to oneand only left coset of H.
5.6 Note. In general aH = Ha. For example, If H ⊆ GT , H = {I, S1} then
S2H = {S2, R2}, HS2 = {S2, R1}
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5.7 Proposition. A subgroup H of G is normal iff
aH = Ha ∀a ∈ G
Proof. Exercise.
5.8 Notation. If H is a subgroup of G then
G/H := the set of all left cosets of H in G
5.9. Multiplication of cosets. Let H ⊆ G, aH, bH ∈ G/H. Define
aH · bH := (ab)H
5.10 Note. In general this is not well defined, i.e. we may have aH = a′H,bH = b′H but (ab)H = (a′b′)H.
For example, take H = {I, S1} ⊆ GT . Recall:
S2H = R2H = {S2, R2}, S3H = R1H = {S3, R1}
However:(S2S3)H = R1H = {R1, S2}(R2R1)H = IH = {I, S1}
5.11 Proposition. If H is a normal subgroup of G then the multiplication ofcosets given in (5.9) is well defined.
Proof. If H ◁G then by (5.7) we have aH = Ha ∀a∈G.
Let aH = a′H, bH = b′H. Then
(ab)H = a(bH) = a(b′H) = a(Hb′) = (aH)b′ = (a′H)b′ = a′(b′H) = (a′b′)H
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5.12 Corollary/Definition. If H ◁G then G/H is a group with multiplicationdefined by (5.9). The identity elements in G/H is the coset eH = H ∈ G/H.The inverse of a coset aH is the coset a−1H.
The group G/H is called the quotient group (or the factor group) of G by H.
5.13 Example.
Take Z, the additive group of integers. Since Z is abelian every its subgroup isnormal. For n ∈ Z, n ≥ 2 define
nZ = {na | a ∈ Z}
e.g. 2Z = {· · · − 4,−2, 0, 2, 4, . . . }, 5Z = {. . . ,−10,−5, 0, 5, 10, . . . }
Note: nZ is a subgroup of Z.
Cosets of nZ in Z:k + nZ = {k + na | a ∈ Z}
e.g. 1 + 5Z = {· · · − 9,−4, 1, 6, 11, . . . }, 3 + 5Z = {. . . ,−7,−2, 3, 8, 13, . . . }
Note: k + nZ = l + nZ iff (k − l) ∈ nZ i.e. iff k = l + na for some a ∈ Z.
E.g.:1 + 5Z = 6 + 5Z = 11 + 5Z = −4 + 5Z
Recall: if n, k ∈ Z then there is a unique number l ∈ {0, 1, . . . , n− 1} such that
k = l + na
for some a ∈ Z. Thus every coset of nZ can be uniquely written as l+nZ wherel ∈ {0, 1, . . . , n− 1}.
Denote l := l + nZ. Then Z/nZ = {0, 1, . . . , n− 1}
The addition table in Z/5Z:
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+ 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
Recall: A group G is cyclic if it is generated by a single element: G = ⟨a⟩ forsome a ∈ G.
Note: For every n the group Z/nZ is cyclic: Z/nZ = ⟨1⟩.
5.14 Note. If H ◁G then we have a homomorphism
π : G→ G/H, π(a) := aH
This is the canonical epimorphism of G onto G/H.
We have:
Ker(π) = {a ∈ G | π(a) = eH}= {a ∈ G | aH = eH}= {a ∈ G | e−1a ∈ H}= {a ∈ G | a ∈ H}= H
5.15 Corollary. A subgroup H ⊆ G is the kernel of some homomorphismf : G→ K iff H is a normal subgroup
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6 Isomorphism theorems
6.1 Theorem. If f : G→ H is a homomorphism then there is a unique homo-morphism
f : G/Ker(f)→ H
such that the following diagram commutes:
Gf //
π
��
H
G/Ker(f)
f
;;
Moreover, f is a monomorphism and Im(f) = Im(f).
Proof. Denote: K := Ker(f). Define
f : G/K → H, f(aK) := f(a)
We have:
1) f is well defined:
If aK = bK then a−1b ∈ K, so f(a−1b) = e. Thus
f(b) = f(aa−1b) = f(a)f(a−1b) = f(a)
2) f is a homomorphism (check).
3) f is a unique homomorphism satisfying f ◦ π = f
Indeed, if g : G/K → H is some other homomorphism and g ◦ π = f then
f(a) = g ◦ π(a) = g(aK)
and so g(aK) = f(aK) for all aK ∈ G/K.
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4) f is 1-1:
We need: Ker(f) = {eK}.We have: if f(aK) = e then f(a) = e, so a ∈ K and so aK = eK.
5) Im(f) = Im(f) (obvious).
6.2 First Isomorphism Theorem. If f : G→ H is an epimorphism then
G/Ker(f) ∼= H
Proof. Take the map f : G/Ker(f) → H. Then Im(f) = Im(f) = H, so f isan epimorphism. Also, f is 1-1. Therefore f is a bijective homomorphism andthus it is an isomorphism.
6.3 Example.
Recall:GLn(R) = {A ∈Mn(R) | det(A) = 0}SLn(R) = {A ∈Mn(R) | det(A) = 1}
SLn(R) is a normal subgroup of GLn(R).
We have the homomorphism
det : GLn(R)→ R∗
Since this is an epimorphism and Ker(det) = SLn(R) we get
GLn(R)/SLn(R) ∼= R∗
6.4 Theorem. If G is a cyclic group then G ∼= {e} or G ∼= Z or G ∼= Z/nZ forsome n ≥ 2.
Proof. Let G = ⟨a⟩ for some a ∈ G. Definef : Z→ G, f(n) := an
Notice that
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1) f is a homomorphism
2) f is onto.
Thus by the First Isomorphism Theorem G ∼= Z/Ker(f).
Check: all subgroups H ⊆ Z are of the form H = nZ for some n ≥ 0.
It follows that G ∼= Z/nZ for some n ≥ 0. Also:
• if n = 0 then nZ = 0Z = {0} and G ∼= Z/{0} ∼= Z
• if n = 1 then nZ = 1Z = Z and G ∼= Z/Z ∼= {e}
• if n ≥ 2 then G ∼= Z/nZ.
6.5 Notation. If H,K are subgroups of G then
HK := {hk ∈ G | h ∈ H, k ∈ K}
6.6 Lemma. If H,K are subgroups of G then HK is a subgroup of G iff
HK = KH
Proof. Exercise.
6.7 Second Isomorphism Theorem. If H,K are subgroups of G and H ◁Gthen KH is a subgroup of G, (H ∩K)◁K and
K/(H ∩K) ∼= KH/H
Proof. Exercise.
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6.8 Third Isomorphism Theorem. Let K ⊆ H ⊆ G. If K,H are normalsubgroups of G then K ◁H, H/K ◁G/K and
(G/K)/(H/K) ∼= G/H
Proof. Exercise.
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7 Index of a subgroup and order of an element
7.1 Definition. Let H be a subgroup of G. Then
[G : H] := the number of distinct left cosets of H in G
This number is called the index of H in G.
7.2 Note.
1) The number of left cosets of H in G is the same as the number of right cosets(check!), so also
[G : H] = the number of distinct right cosets of H in G
2) If H is a normal subgroup of G then [G : H] = |G/H|.
7.3 Lemma. If H is a subgroup of G then every left (and right) coset of H inG has the same number of elements as H.
Proof. If a ∈ G then the map of sets
f : H → aH, f(h) = ah
is a bijection (check!).
7.4 Theorem (Lagrange). If H is a subgroup of G then
|G| = [G : H]|H|
Proof. Recall:
1) G is a disjoint union of left cosets of H (5.5)
2) each left coset has as many elements as H (7.3)
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This gives:
|G| = (number of left cosets of H) · (number of elements of H) = [G : H] · |H|
7.5 Definition. If a ∈ G then the order |a| of a is the order of the subgroup⟨a⟩ ⊆ G generated by a.
7.6 Proposition. |a| = n if n is the smallest positive integer such that an = e,and |a| =∞ if such n does not exist.
Proof. Take the homomorphism
f : Z→ ⟨a⟩, f(k) = ak
Note that f is onto. If an = e for all n > 0 then Ker(f) = {0}. Then f is anisomorphism and so |⟨a⟩| = |Z| =∞.
If an = e for some n > 0 and n is the smallest positive number with this propertythen Ker(f) = nZ (check!). Therefore ⟨a⟩ ∼= Z/nZ and so |⟨a⟩| = |Z/nZ| = n.
7.7 Proposition. If G is a finite group and a ∈ G then |a| divides |G|.
Proof. Follows from Lagrange’s theorem (7.4).
7.8 Note. If a number n divides |G| then G need not contain an element oforder n. E.g. |GT | = 6 but GT does not have an element of order 6:
|S1| = |S2| = |S3| = 2, |R1| = |R2| = 3, |I| = 1
We will see later that if p is a prime number that divides |G| then G contains anelement of order p.
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7.9 Definition. An element a ∈ G is a torsion element if |a| <∞. A group Gis a torsion group if all its elements are torsion elements. A group is torsion freeif it has not torsion elements.
7.10 Examples.
1) Every finite group is a torsion group.
2) Q/Z is also a torsion group.
3) Z,Q,R,C are torsion free.
4) Q∗ is neither torsion nor torsion free: 2 ∈ Q∗ has infinite order, −1 ∈ Q∗
has order 2.
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8 Free groups and presentations of groups
8.1. Let S be a set. A word in S is a finite sequence of the form
w = xλ11 xλ22 · · · · · xλkk
where xi ∈ S and λi = ±1 for i = 1, 2, . . . , k. Also, take
e := “the empty word”
(i.e. the word corresponding to the sequence of length 0).
We identify two words if one can be obtained from the other by a series of“cancellations” and “insertions” of subwords of the form xx−1 and x−1x, e.g.:
x1x2x3x−13 ∼ x1x2 ∼ x1x
−14 x4x2 ∼ x1x
−14 x1x
−11 x4x2
x1x−11 ∼ e
Let F (S) be the set of equivalence classes of words under this equivalence rela-tion.
Note. A word is reduced if it does not contain any subwords of the form xx−1
or x−1x. Every equivalence class in F (S) is represented by a unique reducedword.
Define multiplication in F (S) by concatenation of words, e.g.:
(x1x2x−11 ) · (x2x−1
3 ) = x1x2x−11 x2x
−13
F(S) with this multiplication becomes a group:
• the identity element in F (S): e
• inverses in F (S): (x1x2x−13 x2)
−1 = x−12 x3x
−12 x−1
1
29
8.2 Definition. F (S) is called the free group generated by the set S.
In general, a group G is free if G ∼= F (S) for some set S.
8.3 Note.
• If S = ∅ then F (S) = {e} is the trivial group.
• If S consists of a single element, S = {x} then F (S) is an infinite cyclicgroup, so F (S) ∼= Z.
8.4 Note. We have a map of sets
i : S → F (S), i(x) = x
8.5 Theorem (The universal property of free groups).Let S be a set and G be a group. For any map of sets f : S → G thereexists a unique homomorphism f : F (S) → G such that the following diagramcommutes:
Sf //
i
��
G
F (S)
f
==
Proof. f is defined by
f(xλ11 xλ22 · · · · · xλkk ) := f(x1)
λ1f(x2)λ2 · · · · · f(xk)λk
8.6 Corollary. Every group is the homeomorphic image of a free group.
Proof. Let G be a group. Take the set
S := {xg | g ∈ G}
30
We have a map of sets
f : S → G, f(xg) := g
This gives a homomorphism f : F (S)→ G. Since f is onto thus also f is onto,i.e. G = f(F (S)).
8.7 Note. By Corollary 8.6 and the First Isomorphism Theorem (6.2) we have
G ∼= F (S)/Ker(f)
One can show that any subgroup of a free group is free. In particular Ker(f) isfree. This shows that any group is isomorphic to a quotient of two free groups.
8.8 Definition. Let S be a set and let R be a subset of F (S). Then
⟨S | R⟩ := F (S)/H
where H is the smallest normal subgroup of F (S) such that R ⊆ H. We saythen that
• elements of S are generators of ⟨S | R⟩
• elements of R are relations (or relators) in ⟨S | R⟩
8.9 Definition. If G is a group and G ∼= ⟨S | R⟩ for some set S and somesubset R ⊆ F (S) then we say that ⟨S | R⟩ is a presentation of G.
We say that a group G is finitely presentable if it has a presentation such thatboth S and R are finite sets.
8.10 Examples.
1) F (S) ∼= ⟨S | ∅⟩
31
2) Z/nZ ∼= ⟨x | xn⟩Note: in particular Z/6Z ∼= ⟨x | x6⟩. Here is a different presentation ofZ/6Z:
Z/6Z ∼= ⟨x, y | x2, y3, xyxy2⟩
3) GT∼= ⟨x, y | x2, y3, xyxy⟩ (isomorphism: x 7→ S1, y 7→ R1)
4) Recall: Sn = the symmetric group on n letters (1.8)
Sn ∼= ⟨x1, . . . xn−1 | x2i , (xixi+1)3, (xixj)
2 for |i− j| > 1⟩
(isomorphism: xi 7→ σi where σi : {1, . . . , n} → {1, . . . , n}, σi(i) = i+1,σi(i+ 1) = i and σi(j) = j for j = i, i+ 1).
32
9 Direct products, direct sums, and free abeliangroups
9.1 Definition. A direct product of a family of groups {Gi}i∈I is a group∏i∈I Gi defined as follows. As a set
∏i∈I Gi is the cartesian product of the
groups Gi. Given elements (ai)i∈I , (bi)i∈I ∈∏
i∈I Gi we set
(ai)i∈I · (bi)i∈I := (aibi)i∈I
9.2 Definition. A weak direct product of a family of groups {Gi}i∈I is thesubgroup of
∏i∈I Gi given by∏
i∈I
wGi := {(ai)i∈I | ai = ei ∈ Gi for finitely many i only}
If all groups Gi are abelian then∏w
i∈IGi is denoted⊕
i∈I Gi and it is called thedirect sum of {Gi}i∈I .
9.3 Note. If I is a finite set then∏
i∈I Gi =∏w
i∈IGi.
9.4 Example.
Z/2Z× Z/2Z = Z/2Z⊕ Z/2Z = {(0, 0), (0, 1), (1, 0), (1, 1)}
Note. Z/2Z ⊕ Z/2Z is a the smallest non-cyclic group. It is called the Kleinfour group.
9.5 Example.
Z/2Z⊕ Z/3Z = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}
Note. Z/2Z⊕ Z/3Z is a cyclic group ((1, 1) is a generator), thus
Z/2Z⊕ Z/3Z ∼= Z/6Z
33
9.6. Let S be a set. Denote by Fab(S) the set of all expressions of the form∑x∈S
kxx
where kx ∈ Z and kx = 0 for finitely many x ∈ X only.
Fab(S) is an abelian group with addition defined by∑x∈S
kxx+∑x∈S
lxx :=∑x∈S
(kx + lx)x
9.7 Definition. The group Fab(S) is called the free abelian group generated bythe set S.
In general a group G is free abelian if G ∼= Fab(S) for some set S.
9.8 Proposition. If S is a set then
Fab(S) ∼=⊕x∈S
Z
Proof. The isomorphism is given by
f : Fab(S)→⊕x∈S
Z, f(∑x∈S
kxx) = (kx)x∈S
9.9 Note. We have a map of sets
i : S → Fab(S), i(x) = 1 · x
34
9.10 Theorem (The universal property of free abelian groups).Let S be a set and G be an abelian group. For any map of sets f : S → G thereexists a unique homomorphism f : F (S) → G such that the following diagramcommutes:
Sf //
i
��
G
Fab(S)
f
==
Proof. Define f by
f
(∑x∈S
kxx
):=∑x∈S
kxf(x)
Note: this is well defined since kx = 0 for almost all x ∈ S.
35
10 Categories and functors
10.1 Definition. A category C consists of
1) a collection of objects Ob(C)
2) for any a, b ∈ Ob(C) a set HomC(a, b) of morphisms from a to b
3) for any a, b, c ∈ Ob(C) a function (“composition law”)
HomC(a, b)× HomC(b, c)→ HomC(a, c)
(f , g) 7→ g ◦ f
such that the following conditions are satisfied:
• Associativity.f ◦ (g ◦ h) = (f ◦ g) ◦ h
for any morphisms f, g, h for which these compositions are defined.
• Identity. For any c ∈ Ob(C) there is a morphism idc ∈ HomC(c, c) suchthat
f ◦ idc = f, idc ◦ g = g
for any f ∈ HomC(c, d), g ∈ HomC(b, c).
10.2 Examples.
1) Set = the category of all sets.
• Ob(Set) = the collection of all sets
• HomSet(A,B) = { all maps of sets f : A→ B }
2) Gr = the category of all groups
• Ob(Gr) = the collection of all groups
• HomGr(G,H) = { all homomorphisms f : G→ H }
36
3) Ab = the category of all abelian groups
• Ob(Ab) = the collection of all abelian groups
• HomAb(G,H) = { all homomorphisms f : G→ H }
4) Top = the category of all topological spaces
• Ob(Top) = the collection of all topological spaces
• HomTop(X, Y ) = { all continuous maps f : X → Y }
5) Let G be a group. Define a category CG as follows:
• Ob(CG) = {∗}• HomCG
(∗, ∗) = { elements of G }• composition of morphisms = multiplication in G
6) A very small category C:
cf // d
• Ob(C) = {c, d}• HomC(c, d) = {f}HomC(d, c) = ∅HomC(c, c) = idcHomC(d, d) = idd
10.3 Definition. A morphism f : c → d in a category C is an isomorphism ifthere exists a morphism g : d→ c such that gf = idc and fg = idd.
If for some c, d ∈ C there exist an isomorphism f : c → d then we say that theobjects c and d are isomorphic and we write c ∼= d.
37
10.4 Note. For an object c ∈ C define
Aut(c) := { all isomorphisms f : c→ c }
Aut(c) with composition of morphisms is a group.
10.5 Definition. Let C,D be categories. A (covariant) functor F : C → D
consists of
1) an assignmentOb(C)→ Ob(D), c 7→ F (c)
2) for every c, c′ ∈ C a function
HomC(c, c′)→ HomD(F (c), F (c
′)), f 7→ F (f)
such that F (gf) = F (g)F (f) and F (idc) = idF (c).
10.6 Note. If F : C → D is a functor and f : c → c′ is an isomorphism in C
then F (f) : F (c)→ F (c′) is an isomorphism in D.
In particular if c ∼= c′ in C then F (c) ∼= F (c′) in D.
10.7 Examples.
1) U : Gr → Set
If G ∈ Gr then U(G) = { the set of elements of G }If f : G→ H is a homomorphism then U(f) : U(G)→ U(H) is the mapof sets underlying this homomorphism.
2) U : Ab→ Set,
defined the same way as in 1).Note. The functors U in 1), 2) are called forgetful functors.
38
3) Let G be a group. The commutator of a, b ∈ G is the element
[a, b] := aba−1b−1
Note: [a, b] = e iff ab = ba.
The commutator subgroup of G is the subgroup [G,G] ⊆ G generated bythe set S = {[a, b] | a, b ∈ G}.Note.
(a) [G,G] = {e} iff G is an abelian group.
(b) [G,G] is a normal subgroup of G (check!).
(c) G/[G,G] is an abelian group (check!).
(d) If f : G→ H is a homomorphism then f([G,G]) ⊆ [H,H].
(e) If f : G→ H is a homomorphism then f induces a homomorphism
fab : G/[G,G]→ H/[H,H]
given by fab(a[G,G]) = f(a)[H,H].
The abelianization functor Ab: Gr → Ab is given by
Ab(G) := G/[G,G], Ab(f) := fab
4) Recall: if S is a set then F (S) is the free group generated by S.
A map of sets f : S → T defines a homomorphism
f : F (S)→ F (T )
given by f(xλ11 xλ22 · · · · · xλkk ) = f(x1)
λ1f(x2)λ2 · · · · · f(xk)λk .
Check: the assignment
S 7→ F (S), (f : S → T ) 7→ (f : F (S)→ F (T ))
Defines a functor F : Set→ Gr. This is the free group functor.
39
5) Similarly we have the free abelian group functor
Fab : Set→ Ab
where
• Fab(S) = the free abelian group generated by the set S
• if f : S → T then Fab(f) : Fab(S)→ Fab(T ) is given by
Fab(f)
(∑x∈S
kxx
)=∑x∈S
kxf(x)
40
11 Adjoint functors
11.1 Definition. Given two functors
L : C→ D and R : D→ C
we say that L is the left adjoint functor of R and that R is the right adjointfunctor of L if for any object c ∈ C we have a morphism ηc : c → RL(c) suchthat:
1) for any morphism f : c→ c′ in C the following diagram commutes:
cf //
ηc
��
c′
ηc′
��RL(c)
RL(f)// RL(c′)
2) for any c ∈ C and d ∈ D the map of sets
HomD(L(c), d) −→ HomC(c, R(d))
(L(c)f→ d) 7−→ (c
ηc→ RL(c)R(f)→ R(d))
is a bijection.In such situation we say that (L,R) is an adjoint pair of functors.
11.2 Note.
1) The collection of morphisms {ηc}c∈C is called the unit of adjunction of (L,R).
2) For any adjoint pair (L,R) we also have morphisms {εd : LR(d) → d}d∈Dsatisfying analogous conditions as {ηc}c∈C. This collection of morphisms is calledthe counit of the adjunction.
11.3 Note. The morphism ηc is universal in the following sense. For any d ∈ D
and any morphism f : c → R(d) in C there is a unique morphism f : L(c) → d
41
in D such that the following diagram commutes:
cf //
ηc
��
R(d)
RL(c)
R(f)
<<
This property is equivalent to part 2) of Definition 11.1.
11.4 Examples.
1) Recall that we have functors:
F : Set→ Gr, Gr ← Set : U
where F = free group functor, U = forgetful functor.
The pair (F,U) is an adjoint pair. For S ∈ Set the unit of adjunction isgiven by the function
iS : S → UF (S), iS(x) = x
The universal property of free groups (9.10) says that for any G ∈ Grand any map of sets f : S → U(G) there is a unique homomorphismf : F (S)→ G such that we have a commutative diagram
Sf //
iS
��
U(G)
UF (S)
U(f)
<<
2) We have functors
Fab : Set→ Ab, Set← Ab : U
where Fab = free abelian group functor, U = forgetful functor. Similarlyas in 1) one can check that (Fab, U) is an adjoint pair.
42
3) Recall that we have the abelianization functor
Ab: Gr → Ab, Ab(G) = G/[G,G]
This functor is left adjoint to the inclusion functor
J : Ab→ Gr, J(G) = G
(check!).
11.5 Note. It is not true that every functor has a left or right adjoint.
43
12 Categorical products and coproducts
12.1 Definition. Let {ci}i∈I be a family of objects in a category C. A (categor-ical) product of the family {ci}i∈I is an object p ∈ C equipped with morphismsπi : p → ci for all i ∈ I that satisfies the following universal property. For anyobject d ∈ C and a family of morphisms {fi : d → ci}i∈I there exists a uniquemorphism f : d→ p such that πif = fi for all i ∈ I.
df
��
f1
��
f2
%%
p π1//
π2
��
c1
c2
12.2 Note. If a categorical product of {ci}i∈I exists then it is defined uniquelyup to isomorphism. We then write:
p =∏i∈I
ci
12.3 Examples.
1) In the category of groups Gr the categorical product of a family {Gi}i∈Iis the direct product of groups
∏i∈I Gi.
Indeed, we have projection homomorphisms:
πi0 :∏i∈I
Gi → Gi0 , πi0((gi)i∈I) = gi0
Also, if for some group H we have homomorphisms fi : H → Gi then thisdefines a homomorphism
f : H →∏i∈I
Gi, f(h) = (fi(h))i∈I
Moreover, f is the unique homomorphism such that we have πif = fi.
44
2) By a similar argument if {Gi}i∈I is a family of abelian groups then thedirect product
∏i∈I Gi is the categorical product of the family {Gi}i∈I in
the category Ab.
3) In the category Set the categorical product of a family of sets {Ai}i∈I isthe cartesian product of sets
∏i∈I Ai.
12.4 Definition. Let {ci}i∈I be a family of objects in a category C. A (categori-cal) coproduct of the family {ci}i∈I is an object d ∈ C equipped with morphismsεi : ci → d for all i ∈ I that satisfies the following universal property. For anyobject b ∈ C and a family of morphisms {fi : ci → b}i∈I there exists a uniquemorphism f : d→ b such that fεi = fi for all i ∈ I.
c1
ε1
��
f1
c2
f2
44
ε2 // df
��b
12.5 Note. If a categorical coproduct of {ci}i∈I exists then it is defined uniquelyup to isomorphism. We then write:
d =∐i∈I
ci
12.6 Examples.
1) In the category of sets Set the categorical coproduct of a family of sets{Ai}i∈I is the disjoint union of sets
∐i∈I Ai.
45
2) In the category of abelian groups Ab the categorical coproduct of a familyof abelian groups {Gi}i∈I is the direct sum
⊕i∈I Gi.
The homomorphisms εi0 : Gi0 →⊕
i∈I Gi are given by g 7→ (gi)i∈I where
gi =
{g if i = i0eGi
otherwise
Given an abelian group H and homomorphisms fi : Gi → H we have ahomomorphism
f :⊕i∈I
Gi → H, f((gi)i∈I) =∑i∈I
fi(gi)
This is the unique homomorphism satisfying fεi = fi for all i ∈ I.
3) If {Gi}i∈I is a family of groups then∏w
i∈I Gi is not, in general, a coproductof {Gi}i∈I . Take e.g. G1 = Z/2Z, G2 = Z/3Z. We have homomorphisms
f1 : Z/2Z→ GT , f(1) = S1
f2 : Z/3Z→ GT , f(1) = R1
However, there is no homomorphism f : Z/2Z ⊕ Z/3Z → GT such thatfεi = f for i = 1, 2.
12.7. Construction of coproducts in Gr.
Let {Gi}i∈I be a family of groups, and let S =∐
i∈I Gi be the disjoint union ofsets of elements of these groups. A word in S is a sequence
a1a2 . . . ak
where k ≥ 0 and a1, a2, . . . , ak ∈ S. Consider the equivalence relation of wordsgenerated by the following conditions:
1) if eGiis the trivial element in Gi for some i ∈ I then
a1 . . . ajaj+1 . . . ak ∼ ai . . . ajeGiaj+1 . . . ak
46
2) if aj, aj+1 belong to the same group Gi for some i ∈ I then
a1 . . . ajaj+1 . . . ak ∼ a1 . . . (ajaj+1)︸ ︷︷ ︸product in Gi
. . . ak
Denote ∏i∈I
∗Gi := { equivalence classes of words }
This set is a group with multiplication defined by concatenation of words.
12.8 Definition. The group∏∗
i∈I Gi is called the free product of the family{Gi}i∈I
12.9 Proposition. If {Gi}i∈I is a family of groups then∏∗
i∈I Gi is the coproductof the family {Gi}i∈I in the category of groups.
12.10 Note. The free product∏∗
i∈I Z is isomorphic to the free group generatedby the set I.
47
13 More on free abelian groups
Recall. G is a free abelian group if
G ∼=⊕i∈I
Z
for some set I.
13.1 Definition. Let G be an abelian group. A set B ⊆ G is a basis of G if
• B generates G
• if for some x1, . . . xk ∈ B and n1, . . . , nk ∈ Z we have
n1x1 + · · ·+ nkxk = 0
then n1 = · · · = nk = 0.
13.2 Theorem. An abelian group G has a basis iff G is a free abelian group.
Proof. If B is a basis of G then the map
f :⊕x∈B
Z→ G, f((nx)x∈B) =∑x∈B
nxx
is an isomorphism (check!).
Conversely, if we have an isomorphism
f :⊕i∈I
Z→ G
then for j ∈ I take δj ∈⊕
i∈I Z where δj = (ni)i∈I such that
ni =
{1 if i = j
0 otherwise
Then B = {f(δj)}i∈I is a basis of G (check!).
48
13.3 Proposition. If G is a free abelian group then any two bases of G havethe same cardinality.
Notation. If X is a set then |X| = cardinality of X.
13.4 Lemma. If {Gi}i∈I is a family of abelian groups and Hi is a subgroup ofGi for i ∈ I then ⊕
i∈I
G1 /⊕i∈I
Hi∼=
⊕i∈I
(Gi/Hi)
Proof. Exercise.
Proof of Proposition 13.3.
Let B, B′ be two bases of G. We want to show: |B| = |B′|.
We have isomorphisms ⊕x∈B
Z ∼= G ∼=⊕y∈B′
Z
Case 1. B,B′ - finite sets , |B| = m , |B′| = n.
Take 2G := {2a | a ∈ G}. This is a subgroup G. Since G ∼=⊕
x∈B Z, usingLemma 13.4 we obtain
G/2G ∼=⊕x∈B
Z/2Z
Similarly, since G ∼=⊕
x∈B′ Z we have
G/2G ∼=⊕y∈B′
Z/2Z
This gives:
2m = |⊕x∈B
Z/2Z| = |G/2G| = |⊕y∈B′
Z/2Z| = 2n
so m = n.
49
Case 2. B - finite set, B′ - infinite set.
As before this would give:⊕x∈B
Z/2Z ∼= G/2G ∼=⊕x∈B′
Z/2Z
This is however impossible since⊕
x∈B Z/2Z is a finite group and⊕
y∈B′ Z/2Zis an infinite group.
Case 3. B,B′ - infinite sets.
Check: if B is an infinite set then
|B| = |⊕x∈B
Z|
It follows that we have
|B| = |⊕x∈B
Z| = |G| = |⊕y∈B′
Z| = |B′|
13.5 Definition. If G is a free abelian group then the rank of G is the cardinalityof a basis of G.
Note. By (13.3) rank of G does not depend on the choice of basis of G.
13.6 Theorem. Let G be a free abelian group of a finite rank n and let H bea subgroup of G. Then H is a free abelian group and
rankH ≤ rankG
Note. This theorem is true also if G is a free abelian group of an infinite rank.
13.7 Lemma. If f : G→ H is an epimorphism of abelian groups and H is freeabelian group then
G ∼= H ⊕Ker(f)
50
Proof. Recall (from homework): if we have homomorphisms
f : G→ H, g : H → G
such that fg = idH then G ∼= H ⊕ Ker(f). If follows that we only need toconstruct the homomorphism g.
Take a basis B of H. Since f is onto, for every x ∈ B there is ax ∈ G such thatf(ax) = x. Let g : H → G be the unique homomorphism satisfying
g(x) = ax
for all x ∈ B. Then fg(x) = x for all x ∈ B and so fg = idH .
Proof of Theorem 13.6.
We can assume that G = Zn.
We want to show: if H ⊆ Zn then H is a free abelian group and rankH ≤ n.
Induction with respect to n:
If n = 1 then H = kZ for some k ≥ 0 so H = {0} or H ∼= Z.
Next, assume that for some n every subgroup of Zn is a free abelian group ofrank ≤ n, and let H ⊆ Zn+1. Take the homomorphism
f : Zn+1 → Z, f(m1, . . . ,mn+1) = mn+1
We have:Ker(f) = {(m1, . . . ,mn, 0) | mi ∈ Z} ∼= Zn
We have an epimorphism:f |H : H → Im(f)
Since Im(f |H) ⊆ Z, thus Im(f |H) is a free abelian group and so by Lemma 13.7we have
G ∼= Im(f |H)⊕Ker(f |H)We also have:
Ker(f |H) = Ker(f) ∩H
51
It follows that that Ker(f |H) is a subgroup of Ker(f), and since Ker(f) is a freeabelian group of rank n by the inductive assumption we get that Ker(f |H) is afree abelian group of rank ≤ n. Therefore
H ∼= Im(f |H)︸ ︷︷ ︸free abelianrank ≤ 1
⊕ Ker(f |H)︸ ︷︷ ︸free abelianrank ≤ n
and so H is a free abelian group of rank ≤ n+ 1.
52
14 Finitely generated abelian groups
Goal. Describe all isomorphism types of finitely generated abelian groups.
14.1 Proposition. If G is an abelian group generated by n elements then
G ∼= F/H
where F is a free abelian group of rank n and H is some subgroup of F .
Proof. We have G = ⟨a1, . . . an⟩ for some a1, . . . , an ∈ G. Let {x1, . . . , xn} bea basis of F . We have a homomorphism
f : F → G, f(xi) = ai
Take H = Ker(f). Since f is an epimorphism by the First Isomorphism Theorem(6.2) we have
G ∼= F/H
Recall. If rankF = n and H ⊆ F then H is a free abelian group of rank ≤ n.
14.2 Theorem. Let F be a free abelian group of rank n and let H be a subgroupof F There exists a basis {x1, . . . , xn} of F and integers d1, . . . , dr > 0 suchthat
• di|di+1 for i = 1, . . . , r
• {d1x1, . . . , drxr} is a basis of H.
14.3 Theorem. If G is a finitely generated abelian group then
G ∼= (Z/d1Z)⊕ . . .⊕ (Z/drZ)⊕ Zk
for some k ≥ 0 and d1, . . . , dr > 0 such that di|di+1 for i = 1, . . . , r.
53
Proof. By (14.1) we haveG ∼= F/H
for some free abelian group F of finite rank and H ⊆ F . Let rankF = n. ByTheorem 14.2 there is a basis {x1, . . . , xn} of F such that {d1x1, . . . , drxr} isa basis of H for some d1, . . . , dr > 0, di|di+1.
We have an isomorphism
f : F → Z⊕ · · · ⊕ Z︸ ︷︷ ︸n times
where f(x1) = (1, 0, . . . , 0), f(x2) = (0, 1, . . . , 0), . . . , f(xn) = (0, . . . , 0, 1)Notice that
f(H) = d1Z⊕ . . .⊕ drZ ⊕ {0} ⊕ . . .⊕ {0}. This gives
G ∼= (Z⊕ · · · ⊕ Z)/(d1Z⊕ . . .⊕ drZ ⊕ {0} ⊕ . . .⊕ {0})Using (13.4) we obtain
G ∼= (Z/d1Z)⊕ . . .⊕ (Z/drZ)⊕ Zk
where k = n− r.
Proof of Theorem 14.2.
Let {y1, . . . , yn} be any basis of F and let {h1, . . . , hm} ⊆ H be any set gener-ating H. We have
hi = ai1y1 + ai2y2 + ainyn
for some aij ∈ Z. Consider the matrix
A =
a11 . . . a1n...
...am1 . . . amn
Note: columns of A correspond to basis elements of F and rows of A correspondto generators of H.
Consider the following operations on matrices:
54
1) interchange of two rows
2) multiplication of a row by (−1)3) addition of a multiple of one row to another row.
These operations are called elementary row operations for matrices of integers.Elementary column operations are defined analogously.
Notice that:
• application of an elementary row operation to the matrix A corresponds toreplacing of the set of generators of H by another set of generators of H;
• application of an elementary column operation to the matrix A correspondsto passing to a new basis of F and rewriting the generators of H in termsof this new basis.
Key step. Starting with any matrix of integers A and applying a sequence ofelementary row and column operations we can obtain a matrix of the form
B =
(D 0
0 0
)where 0’s denote zero matrices (of appropriate dimensions) and D is a squarediagonal matrix
D =
d1 0 . . . 00 d2 . . . 0...
.... . .
...0 0 . . . dr
for some d1, . . . , dr > 0 such that di|di+1 for all i.
Note. The matrix B is called the Smith normal form of the matrix A.
Let {x1, . . . , xn} be the basis of F corresponding to columns of the matrix B. Interms of this basis a set of generators of H is given by {d1x1, . . . , drxr, 0, . . . , 0}.It follows that {d1x1, . . . drxr} is a basis of H.
55
Key step. How to compute the Smith normal form of a matrix A.
Step 1. Produce a matrix of the form
A1 =
a11 0 . . . 0
0... A′
1
0
This can be done as follows.
(1a) By interchanging rows and columns if necessary we can make sure thata11 = 0. Also, by multiplying the first row by (−1) we can get a11 > 0.
(1b) By adding multiples of the first row to the other rows (and multiples ofthe first column to the other columns) we can make all other entries in thefirst row and the first column positive and smaller than a11.
(1c) If all these other entries of the first row and column are 0 we are done. Ifsome entry is non-zero then by replacing rows (or columns) we can movethat entry to the (1, 1)-position. Then we go back to (1b).
(1d) After a finite number of iterations we get a matrix of the form of A1.
Step 2. Given a matrix A1 as above make sure that the entry a11 divides allentries of A′
1.
This can be done as follows.
(2a) If all entries of A′1 are already divisible by a11 we are done.
(2b) If some entry aij is not divisible by a11 add the i-th row to the first row.Then go back to (1b).
(2c) After a finite number of iterations we get a matrix of the form of A1 witha11 dividing all entries of A′.
Step 3. We are done with the first row and and the first column. Next we applythe same steps recursively to reduce A′
1.
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14.4 Lemma. Let G be a group and let a1, . . . , ak ∈ G be elements such that|ai| = ni. Assume that for i = 1, . . . , k we have
gcd(ni,∏j =i
ni) = 1
Then |a1 · . . .·ak| = n1 · . . . · nk.
Proof. Exercise.
14.5 Corollary. If m > 0 is an integer and m = pn11 · . . . · pnk
k where p1, . . . , pkare distinct primes then
Z/mZ ∼= (Z/pn11 Z) ⊕ . . .⊕ (Z/pnk
k Z)
Proof. It is enough to show that the group (Z/pn11 Z) ⊕ . . .⊕ (Z/pnk
k Z) containsan element of order n. This follows however from Lemma 14.4.
14.6 Theorem. If G is a finitely generated abelian group then G is isomorphicto a finite direct sum groups Z and Z/pnZ where p is a prime.
Proof. This follows directly from (14.3) and (14.5).
14.7 Theorem. For any finitely generated abelian group G there are unique(up to order) integers pn1
1 , . . . , pnss > 1 where p1, . . . , ps are primes, and a unique
integer k ≥ 0 such that
G ∼= (Z/pn11 Z)⊕ . . .⊕ (Z/pns
k Z)⊕ Zk
14.8 Note. The integers pn11 , . . . , p
nss are called the elementary divisors of the
group G.
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14.9 Lemma. If p is a prime number and
0 < n1 ≤ · · · ≤ ns 0 < m1 ≤ · · · ≤ mr
be integers such that
(Z/pn1Z)⊕ · · · ⊕ (Z/pnsZ) ∼= (Z/pm1Z)⊕ · · · ⊕ (Z/pmrZ)
Then s = r and ni = mi for all i.
Proof. Exercise.
Proof of Theorem 14.7.
Assume thatG ∼= (Z/pn1
1 Z)⊕ . . .⊕ (Z/pnss Z)⊕ Zk (1)
andG ∼= (Z/qm1
1 Z)⊕ . . .⊕ (Z/qmrr Z)⊕ Zl (2)
where k, l ≥ 0, pi, qj are primes and and n1,mj > 0. We want to show thatk = l, r = s and (after reordering) pni
i = qmii for all i.
DefineGt := {a ∈ G | |a| <∞}
This is the torsion subgroup of G. From the isomorphism (1) we obtain:
Gt∼= (Z/pn1
1 Z)⊕ . . .⊕ (Z/pnss Z)
while from the isomorphism (2) we get
Gt∼= (Z/qm1
1 Z)⊕ . . .⊕ (Z/qmrr Z)
If follows thatG/Gt
∼= Zk and G/Gt∼= Zl
Using Proposition 13.3 we obtain from here that k = l.
Next, we want to show that the family of integers {pnii | i = 1, . . . , s} is the
same as the family {qmj
j | j = 1, . . . , r}.
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For a prime p define
Gp := {a ∈ G | |a| = pl for some l > 0}
From the isomorphisms (1) and (2) we get⊕pi=p
(Z/pnii Z) ∼= Gp
∼=⊕qj=p
(Z/qmj
j Z)
Using Lemma 14.9 we obtain that the family {pnii | pi = p} is the same as
{qnj
j | qj = p}. Since this holds for all primes p we are done.
59
15 Permutation representations and G-sets
Recall. If C is a category and c ∈ C then
Aut(c) = the group of automorphisms of c
15.1 Definition. A representation of a group G in a category C is a homomor-phism
ϱ : G→ Aut(c)
Special types of representations:
• linear representations = representations in the category of vector spaces
• permutation representations = representations in the category of sets.
15.2 Note. Let S be a set and let
ϱ : G→ Aut(S)
be a permutation representation of G. The homomorphism ϱ defines a map
G× S → S, (a, x) 7→ ϱ(a)(x)
Denote a · x := ϱ(a)(x). We have:
1) (ab) · x = a · (b · x) for a, b ∈ G, x ∈ S2) e · x = x for all x ∈ S
15.3 Definition. An action of a group G on a set S is a map
G× S → S, (a, x) 7→ a · x
satisfying conditions 1) - 2) above.
A G-set is a set S equipped with an action of the group G.
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15.4 Note. A permutation representation G → Aut(S) determines a G-setstructure on the set S. Conversely, the structure of a G-set on S determines apermutation representation G→ Aut(S).
15.5 Examples.
1) Take S = G. The map
G×G→ G, (a, b) 7→ ab
defines a G-set structure on G. We say that the group G acts on itself byleft translations.
2) Let H be a subgroup of G. Take S = G/H. The map
G×G/H → G/H, (a, bH) 7→ (ab)H
defines a G-set structure on G/H.
3) Take again S = G. The map
G×G→ G, (a, b) 7→ aba−1
defines another G-set structure on G. We say that the group G acts onitself by conjugations.
4) R is a Z/2Z-set with the action
Z/2Z× R→ R
given by 0 · x = x, 1 · x = −x.
15.6 Definition. If S, T are G-sets then a G-equivariant map is a function
f : S → T
such that f(a · x) = a · f(x) for all a ∈ G, x ∈ S.
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15.7 Note. For a given group G the collection of G-sets and G-equivariant mapforms a category SetG.
15.8 Proposition. A G-equivariant map
f : S → T
is an isomorphism of G-sets iff f is a bijection.
15.9. Proposition/Definition. Let S be a G-set and let x ∈ G. The set
Gx := {a ∈ G | a · x = x}
is a subgroup of G. It is called the stabilizer of x (or the isotropy group of x).
15.10 Definition. Let S be a G-set. The action of the groupG on S is transitiveif for any x, y ∈ S there is a ∈ G such that a · x = y.
15.11 Proposition. For any transitive G-set S we have an isomorphism of G-sets:
S ∼= G/Gx
where x is any element of S.
Proof. We have a map
f : G/Gx → S, f(aGx) = a · x
Check that
• f is well defined
• f is G-equivariant
• f is a bijection.
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It follows that f an isomorphism of G-sets.
15.12 Proposition. If S is a transitive G-set and x, y ∈ S then
Gx = a−1Gya
where y = a · x
Proof. For b ∈ Gy we have
(a−1ba) · x = (a−1b) · y = a−1 · y = x
It follows that a−1Gya ⊆ Gx. On the other hand, if c ∈ Gx then
(aca−1) · y = (ac) · x = a · x = y
Thus aca−1 ∈ Gy and so c = a−1(aca−1)a ∈ a−1Gya. This shows that Gx ⊆a−1Gya.
15.13 Definition. If S is a G-set then the orbit of an element x ∈ S is the set
Orb(x) := {y ∈ S | y = a · x for some a ∈ G}
15.14 Proposition. Let S be a G-set.
1) If x ∈ S then Orb(x) is the unique subset of S such that
• x ∈ Orb(x)
• G acts transitively on Orb(x).
2) If x, y ∈ S then either Orb(x) ∩Orb(y) = ∅ or Orb(x) = Orb(y).
3) If x, y ∈ S then Orb(x) = Orb(y) iff x = a · y for some a ∈ G.
Proof. Exercise.
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15.15 Corollary. Every G-set S is a disjoint union of its orbits. For an elementx ∈ S we have an isomorphism of G-sets
Orb(x) ∼= G/Gx
Proof. This follows directly from Proposition 15.14 and Proposition 15.11.
15.16 Corollary. Let G be a finite group and let S be a finite G-set .
1) For any x ∈ S we have
|Orb(x)| · |Gx| = |G|
2) If Orb(x1), . . .Orb(xn) are all distinct orbits of S then
|S| =n∑i=1
|Orb(xi)|
Proof.
1) By Corollary 15.15 we have |Orb(x)| = [G : Gx], so by Lagrange’s Theorem(7.4)
|G| = [G : Gx] · |Gx| = |Orb(x)| · |Gx|.
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16 Some applications of G-sets
Recall. If G is a finite group and a ∈ G then |a| divides |G|.16.1 Theorem (Cauchy). Let G be a finite group. If |G| is divisible by a primenumber p then there exists a ∈ G such that |a| = p.
Proof. LetS = {(a1, . . . , ap) | ai ∈ G, a1 · . . . · ap = e}
Notice that (a1, . . . , ap) ∈ S iff a1, . . . , ap−1 are arbitrary elements of G andap = (a1 · . . . · ap−1)
−1. If follows that
|S| = |G|p−1
In particular p divides S. Define an action of Z/pZ on S by cyclic permutations:
k · (a1, . . . , ap) := (ak+1 , . . . , ap, a1, . . . , ak)
for k ∈ Z/pZ.
Notice that:
1) The number of elements of each orbit of this action divides |Z/pZ| = p.Therefore for any (a1, . . . , ap) ∈ S we have
|Orb((a1, . . . , ap))| = p or |Orb((a1, . . . , ap))| = 1
2) |Orb((a1, . . . , ap))| = 1 iff a1 = · · · = ap. We have then
ap1 = a1 · . . . · ap = e
3) |Orb((e, . . . , e))| = 1
Assume that Orb((e, . . . , e)) is the only orbit consisting of only one element.Then every other orbit of S has p elements. From part 2) of Corollary 15.16 weobtain then
|S| = kp+ 1
for some k. This is however impossible since p divides |S|. Therefore thereis some element (a1, . . . , ap) ∈ S such that (a1, . . . , ap) = (e, . . . , e) and|Orb((a1, . . . , ap))| = 1. It follows that a1 = e and ap1 = e. Thus |a1| = p.
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16.2 Definition. Let p be a prime number. A group G is a p-group if order ofevery element of G is a power of p.
16.3 Proposition. If G is a finite group then G is a p-group iff |G| = pn forsome n.
Proof. If |G| = pn and a ∈ G then by (7.7) |a| divides pn, so |a| = pr for somer ≥ 0.
Conversely, assume that |G| = qm for some prime q = p. Then by Cauchy’sTheorem (16.1) there is an element a ∈ G such that |a| = q. Therefore G isnot a p-group.
Recall. If G is a group then the center of G is the subgroup
Z(G) = {a ∈ G | ab = ba for all b ∈ G}
Note. In general it may happen that Z(G) = {e} (take e.g. G = GT ).
16.4 Theorem. If G is a finite p-group then Z(G) = {e}.
Proof. Consider the action of G on itself by conjugations:
G×G→ G, a · b := aba−1
Let |G| = pn. Notice that:
1) If b ∈ G then |Orb(b)| divides pn, so |Orb(b)| = pr for some r ≥ 0. Inparticular either p divides |Orb(b)| or |Orb(b)| = 1.
2) |Orb(b)| = 1 iff b ∈ Z(G).
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As a consequence if Z(G) = {e} then |Orb(e)| = 1 and p divides the numberof elements in every other orbit. From part 2) of Corollary 15.16 we obtain then
pn = |G| = kp+ 1
which is impossible. Therefore Z(G) = {e}.
16.5 Note. Not every p-group is abelian. For example take the group of quater-nions Q8 (see Hungerford p. 33) and the group of symmetries of a square D∗
4
(see Hungerford p. 25). These group are non-abelian, but |Q8| = |D∗4| = 8, so
they are 2-groups.
16.6 Definition. Let S be a G-set. A element x ∈ S is a fixed point of theaction of G on S if a · x = x for all a ∈ G.
Note. If S is a G-set then x is a fixed point of the action of G iff |Orb(x)| = 1.
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17 The Sylow theorems
Recall. If G is a finite group and H ⊆ G is a subgroup then |H| divides |G|.
Note.
1) If G is an abelian group and n divides |G| then there is a subgroup H ⊆ Gsuch that |H| = n (homework).
2) This is not true in general when G is a non-abelian group (homework).
17.1 Definition. Let G be a finite group such that |G| = prm where p is aprime and p ∤ m. We say that a subgroup P ⊆ G is a Sylow p-subgroup of G if|P | = pr.
17.2 First Sylow Theorem. If G is a finite group and p is a prime numberdividing |G| then G contains a Sylow p-subgroup.
Proof.
Assume that |G| = prm where p ∤ m.
Let S be the set of all subsets A ⊆ G such that |A| = pr. Define an action ofG on S as follows. If A ∈ S, A = {a1, . . . , apr} and b ∈ G then
b · A := {ba1, . . . , bapr}
Notice that
|S| =(prm
pr
)=
pr∏j=1
pr(m− 1) + j
j
so p does not divide |S|. A a consequence there is an orbit Orb(A0) in S suchthat p ∤ |Orb(A0)|. Take the stabilizer GA0 . We want to show that |GA0 | = pr.We have
prm = |G| = |Orb(A0)| · |GA0 |
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so pr divides |GA0|. On the other hand, notice that for any a ∈ G there is someelement b · A0 ∈ Orb(A0) such that a ∈ b · A0. Ineed, if A0 = {a1, . . . , apr}then
(aa−11 ) · A0 = {a, . . . }
Since G has prm elements and each set b · A0 contains pr elements, thus wemust have |Orb(A0)| ≥ m, and consequently |GA0| ≤ pr. It follows that |GA0| =pr.
17.3 Second Sylow Theorem. If P is a Sylow p-subgroup of a finite group Gthen for any p-subgroup H ⊆ G we have
aHa−1 ⊆ P
for some a ∈ G.
In particular if P , P ′ are two Sylow p-subgroups of G then P = aP ′a−1 for somea ∈ G.
Proof. Assume that |G| = prm where p ∤ m and let |H| = ps.
Take G/P , the set of left cosets of P . The group H acts on G/P by lefttranslations:
H ×G/P → G/P, a · (bP ) := (ab)P
Since |H| = ps, for every bP ∈ G/P we have |Orb(bP )| = pk for some k ≤ s.So, either p divides |Orb(bP )| or |Orb(bP )| = 1.
On the other hand |G/P | = [G : P ] = m, and p ∤ m, so there must be anorbit whose number of elements is not divisible by p. It follows that there isb0P ∈ G/P such that
|Orb(b0P )| = 1
As a consequence ab0P = b0P for all a ∈ H, i.e. Hb0P ⊆ b0P . Since Hb0 ⊆Hb0P we obtain
Hb0 ⊆ b0P
or equivalently: b−10 Hb0 ⊆ P .
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17.4 Example. Take the group GT . We have: |GT | = 6 = 3 · 2.
• Sylow 3-subgroup of GT : {I, R1, R2}• Sylow 2-subgroups of GT : P = {I, S1}, P ′ = {I, S2}, P ′′ = {I, S3}.
We have: P = R1P′R−1
1 and P = R2P′′R−1
2 .
17.5 Third Sylow Theorem. Let G be a finite group such that |G| = prmwhere r > 0 and p ∤ m. If s is the number of Sylow p-subgroups of G then
1) s | m2) s ≡ 1 (mod p)
17.6 Definition. Let G be a group and H be a subgroup of G. The normalizerof H in G is the subgroup NG(H) ⊆ G given by
NG(H) = {a ∈ G | aHa−1 = H}
17.7 Note.
1) H ◁NG(H). In fact, NG(H) is the biggest subgroup of G that containsH a its normal subgroup.
2) H ◁G iff NG(H) = G.
3) Let S be the set of all subgroups of G. The group G acts on S byconjugations:
G× S → S, a ·H := aHa−1
If H is a subgroup of G then
Orb(H) = { all subgroups of G conjugate to H }NG(H) = stabilizer of H
By (15.16) this gives:
(number of subgroups conjugate to H) = |Orb(H)| = [G : NG(H)]
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Proof of Theorem 17.5. Let P be a Sylow p-subgroup of G. By (17.3) all otherSylow p-subgroups of G are conjugate to P , so we get
s = (number of subgroups conjugate to P ) = [G : NG(P )]
Since P ⊆ NG by Lagrange’s Theorem (7.4) we have
|G|︸︷︷︸q
prm
= [G : NG(P )] · |NG(P )| = [G : NG(P )]︸ ︷︷ ︸qs
·[NG(P ) : P ] · |P |︸︷︷︸qpr
It follows that s | m.
Next, let S = {P1, . . . , Ps} be the set of all Sylow p-subgroups of G. The groupP1 acts on S by conjugations:
P1 × S → S, a · Pi := aPia−1
Notice that:
• |Orb(Pi)| divides |P1| = pr, so for i = 1, . . . , s either p divides |Orb(Pi)|or |Orb(Pi)| = 1.
• |Orb(P1)| = 1
• If i > 1 then |Orb(Pi)| > 1.
Indeed, if |Orb(Pi)| = 1 for some i > 1 then aPia−1 = Pi for all a ∈ P1.
Check: in this case P1Pi is a p-subgroup of G. Since P1 = Pi we wouldalso have |P1Pi| > pr which is impossible.
As a consequence we get
s = |S| = |Orb(P1)|︸ ︷︷ ︸q1
+∑|( all other orbits )|︸ ︷︷ ︸
qmultiples of p
Therefore s ≡ 1 (mod p).
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18 Application: groups of order pq
Recall. If G is a group, |G| = p where p is a prime then G ∼= Z/pZ.
Goal. Classify all groups of order pq where p, q are prime numbers.
18.1 Proposition. If G is a group of order p2 for some prime p then eitherG ∼= Z/p2Z or G ∼= Z/pZ⊕ Z/pZ.
Proof. It is enough to show that G is abelian since then the statement followsfrom the classification of finitely generated abelian groups (14.7).
Since G is a p-group by Theorem 16.4 we have Z(G) = {e}. If Z(G) = G thenG is abelian. If Z(G) = G then G/Z(G) is a group of order p and thus it is anon-trivial cyclic group. This is however impossible by Problem 8 of HW 1.
18.2 Proposition. If G is a group of order pq for some primes p, q such thatp > q and q ∤ (p− 1) then
G ∼= Z/pqZ
Proof. If is enough to show that G contains an element of order pq.
Let sp denote the number of Sylow p-subgroups of G. By the Third SylowTheorem (17.5) we have
sp | q and sp = 1 + kp
Since q is a prime the first condition gives sp = 1 or sp = q. Since p > q thesecond condition implies then that sp = 1.
Similarly, let sq be the number of Sylow q-subgroups of G. We have
sq | p and sq = 1 + kq
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The first condition gives sq = 1 or sq = p. If sq = p then the second conditiongives p = 1 + kq, or p − 1 = kq. This is however impossible since q ∤ (p − 1).Therefore we have sq = 1.
We obtain that G has exactly one Sylow p-subgroup P (of order p) and exactlyone Sylow q-subgroup Q (of order q). By the Second Sylow Theorem (17.3)every element of G of order p belongs to the subgroup P and every element oforder q belongs to the subgroup Q. It follows that G contains exactly p − 1elements of order p, exactly q − 1 elements of order q, and one trivial element(of order 1). Since for all p, q we have
pq > (p− 1) + (q − 1) + 1
there are elements of G of order not equal to 1, p, or q. Any such element musthave order pq.
Note. If |G| = pq, and q | (p − 1) then G need not be isomorphic to Z/pqZ(take e.g. G = GT ).
18.3. Defintition. Let N , K be groups and let
φ : K → Aut(N)
be a homomorphism. The semidirect product of N and K with respect to φ isthe group N ⋊φ K such that
N ⋊φ K = N ×K
as sets. Multiplication in N ⋊φ K is given by
(a1, b1) · (a2, b2) := (a1(φ(b1)(a2)), b1b2)
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18.4 Note.
1) If φ is the trivial homomorphism then N ⋊φ K = N ×K.
2) If (a, b) ∈ N ⋊φ K then
(a, b)−1 = (φ(b−1)(a−1), b−1)
3) N ⋊φ K contains subgroups
N ∼= {(a, e) | a ∈ N} and K ∼= {(e, b) | b ∈ K}
We have N ◁N ⋊φ K, and N ∩K = {(e, e)}.
18.5 Examples.
1) Notice that Aut(Z/3Z) ∼= Z/2Z. Let
φ : Z/2→ Aut(Z/3Z)
be the isomorphism. Check:
Z/3Z ⋊φ Z/2Z ∼= GT
2) In general if p is a prime then we have
Aut(Z/pZ) ∼= Z/(p− 1)Z
For any n | (p− 1) there is a unique cyclic subgroup H ⊆ Aut(Z/pZ) oforder n. Let
φ : Z/nZ→ Aut(Z/pZ)
be any homomorphism such that Im(φ) = H. Then Z/pZ⋊φ Z/nZ. is anon-abelian group of order pn.
3) If N is any abelian group then the map
inv : N → N, inv(a) = a−1
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is an automorphism of N . This gives a homomorphism
φ : Z/2Z→ Aut(N)
such that φ(1) = inv. We obtain in this way a group N ⋊φ Z/2Z of order2|N |.Special cases:
• If N = Z/nZ then this group is called the dihedral group of order 2n andit is denoted Dn. The group Dn is isomorphic to the group of all isometriesof a regular polygon with n sides (exercise). In particular D3
∼= GT .
• If N = Z then this group is the infinite dihedral group and it is denotedby D∞. The group D∞ is isomorphic to the free product Z/2Z ∗ Z/2Z(exercise).
Recall. From HW 1: If G,H are abelian groups and f : G → H, g : H → Gare homomorphisms such that fg = idH then G ∼= Ker(f)⊕H.
18.6 Proposition. If G, H are groups and f : G → H, g : H → G are homo-morphisms such that fg = idH then
G ∼= Ker(f)⋊φ H
where φ : H → Aut(Ker(f)) is given by
φ(b)(a) := g(b)ag(b)−1
for a ∈ Ker(f), b ∈ H.
Proof. Exercise.
18.7 Proposition. Let p, q be prime numbers such that p > q and q | (p− 1).Then, up to isomorphism, there are only two groups of order pq:
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– the abelian group Z/pqZ ∼= Z/pZ× Z/qZ
– the non-abelian group Z/pZ ⋊φ Z/qZ where
φ : Z/qZ→ Aut(Z/pZ)
is any non-trivial homomorphism.
Proof. By the same argument as in the proof of Proposition 18.2 we get that Ghas only one Sylow p-subgroup. Call this subgroup P . We have P ◁G.
Let Q be any Sylow q-subgroup. Consider the quotient map
f : G→ G/P
Take the restriction f |Q : Q → G/P . Notice that Ker(f |Q) = Ker(f) ∩ Q =P ∩Q = {e}, so f |Q is a monomorphism. In addition |Q| = q = |G/P |, so f |Qis an isomorphism. As a consequence we have a homomorphism
g : G/P(f |Q)−1
// Q ↪→ G
Since fg = idG/P by Proposition 18.6 we obtain
G ∼= P ⋊φ G/P
for some homomorphism φ : G/P → Aut(P ). Also, since P ∼= Z/pZ andG/P ∼= Z/qZ we get
G ∼= Z/pZ ⋊φ Z/qZfor some φ : Z/qZ→ Aut(Z/pZ).
If φ is the trivial homomorphism then G ∼= Z/pZ × Z/qZ ∼= Z/pqZ. If φ isnon-trivial then G is a non-abelian group. Notice that since q | (p − 1) suchnon-trivial homomorphism exists by (18.5).
It remains to show that for any two non-trivial homomorphisms
φ, ψ : Z/qZ→ Aut(Z/pZ)
we have an isomorphism
Z/pZ ⋊φ Z/qZ ∼= Z/pZ ⋊ψ Z/qZ
(exercise).
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18.8 Example. For any odd prime p there are two non-isomorphic groups oforder 2p:
– the cyclic group Z/2pZ– the dihedral group Dp.
77
H. U. Besche, B. Eick, E. A. O’BrienA millennium project: constructing small groupsInternational Journal of Algebra and Computation 12(5) (2002) 623-644.
19 Group extensions and composition series
19.1 Definition. Let
. . . −→ Gifi−→ Gi+1
fi+1−→ Gi+2 −→ . . .
be a sequence of groups and group homomorphisms. This sequence is exact ifIm(fi) = Ker(fi+1) for all i.
19.2 Definition. A short exact sequence is an exact sequence of the form
1 −→ Nf−→ G
g−→ K −→ 1
(where 1 is the trivial group).
19.3 Note.
1) A sequence 1→ Nf−→ G
g−→ K → 1 is a short exact sequence iff
• f is a monomorphism
• g is an epimorphism
• Im(f) = Ker(g).
2) If H ◁G then we have a short exact sequence
1 −→ H −→ G −→ G/H −→ 1
Morever, up to an isomorphism, every short exact sequence is of this form:
1 // Nf //
∼=��
Gg //
=
��
K //
∼=��
1
1 // Ker(g) // G // G/Ker(g) // 1
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19.4 Definition. If a group G fits into a short exact sequence
1 −→ Nf−→ G
g−→ K −→ 1
then we say that G is an extension of K by N .
19.5 Example. For any n > 1 the dihedral group Dn and the cyclic groupZ/2nZ are non-isomorphic extensions of Z/nZ by Z/2Z.
19.6 Definition. A group G is a simple group if G = {e} and the only normalsubgroups of G are G and {e}.
19.7 Example. Z/pZ is a simple group for every prime p.
Note. A group G is simple iff it is not a non-trivial extension of any group.
19.8 Definition. If G is a group then a normal series of G is a sequence ofsubgroups
{e} = G0 ⊆ G1 ⊆ G2 ⊆ . . . ⊆ Gk = G
such that Gi−1 ◁Gi for all i.
A composition series ofG is a normal series such that all quotient groupsGi/Gi−1
are simple.
19.9 Example. Take the dihedral group D4 = Z/4Z ⋊ Z/2Z. We have acomposition series
{0} ⊆ Z/2Z ⊆ Z/4Z ⊆ D4
Another composition series of D4:
{0} ⊆ Z/2Z ⊆ Z/2Z× Z/2Z ⊆ D4
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19.10 Theorem (Jordan – Holder). If G = {e} is a finite group then
1) G has a composition series.
2) All composition series of G are equivalent in the following sense. If wehave composition series
{e} = G0 ⊆ . . . ⊆ Gk = G and {e} = H0 ⊆ . . . ⊆ Hl = G
then k = l and there is a bijection σ : {1, . . . , k} → {1, . . . , k} such thatfor i = 1, . . . , k we have an isomorphism
Gi/Gi−1∼= Hσ(i)/Hσ(i)−1
Proof. Exercise (or see Hungerford p. 111).
Upshot. If G = {e} is a finite group then G can be obtained by taking successiveextensions of simple groups as follows.
1) Take a composition series
{e} = G0 ⊆ G1 ⊆ G2 ⊆ . . . ⊆ Gk = G
2) For every i = 1, . . . , k we have a short exact sequence
1 −→ Gi−1 −→ Gi −→ Gi/Gi−1 −→ 1
where Gi/Gi−1 is a simple group. Therefore
G1 is an extension of G1/G0 by G0
G2 is an extension of G2/G1 by G1
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
G = Gk is an extension of Gk/Gk−1 by Gk−1
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The grand plan for classifying all finite groups (The Holder Program)
1) Classify all finite simple groups.
2) For any two groups N , K describe all possible extensions of K by N .
Good news: part 1) is done. See
R. Solomon, A brief history of the classification of the finite simple groups,Bulletin AMS 38 (3) (2001), 315-352.
M. Aschbacher, The status of the classification of the finite simple groups, No-tices AMS 51(7) (2004), 736-740.
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20 Simple groups
Recall. A group G = {e} is a simple group if the only normal subgroups of Gare G and {e}.
Note. If G is a simple group then any non-trivial homomorphism f : G→ H isa monomorphism.
20.1 Proposition. If G is an abelian group then G is simple iff G ∼= Z/pZ forsome prime p.
Proof. Exercise.
20.2 Proposition. If G is a simple p-group then G ∼= Z/pZ.
Proof. If G is a p-group then Z(G) = {e} by (16.4). We have Z(G)◁G, so ifG is simple we must have G = Z(G). Therefore G is a simple abelian p-group,and so G ∼= Z/pZ.
20.3 Lemma. There are no non-abelian simple groups of order prm where p isa prime, r ≥ 1, p ∤ m and prm ∤ m!.
Proof. Assume that G is a simple, non-abelian group of such order. We musthave m > 1 (since if m = 1 then G is a p-group). Let P be a Sylow p-subgroupof G. Consider the action of G on the left cosets G/P :
G×G/P → G/P, a · bP = (ab)P
This action defines a homomorphism
ϱ : G→ Perm(G/P )
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where Perm(G/P ) is the group of all permutations of the set G/P . SinceG = P this homomorphism is non-trivial, and so, since G is a simple group, ϱ isa monomorphism. Therefore G can be identified with a subgroup of Perm(G/P ).By Lagrange’s Theorem (7.4) we obtain that |G| divides |Perm(G/P )|. Since|Perm(G/P )| = m! this gives
prm | m!
which contradicts assumptions of the lemma.
20.4 Theorem. There are no non-abelian simple groups of order < 60.
Proof. Check: If 1 ≤ n < 60, and n = 30, 40, 56 then n is of the form prm forsome prime p, and r,m ≥ 1 such that p ∤ m and prm ∤ m!.
By Lemma 20.3 we obtain then that a non-abelian group G of order n < 60 maybe simple only if n = 30, 40 and 56.
Assume that |G| = 30 = 2 ·3 ·5. We will show that G cannot be a simple group.
We argue by contradiction. Assume that G is simple and let s3 be the numberof Sylow 3-subgroups of G. We have
s3 | 10 and s3 ≡ 1 (mod 3)
It follows that either s3 = 1 or s3 = 10. Since G is simple s3 = 1, so s3 = 10.
Notice that if P , P ′ are two distinct Sylow 3-subgroups of G then P ∩P ′ = {e}.We obtain:
– G contains 10 Sylow 3-subgroups.
– each Sylow 3-subgroup contains 2 elements of order 3.
It follows that G contains 20 elements of order 3.
By a similar argument we obtain that
– G must contain 6 Sylow 5-subgroups.
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– each Sylow 5-subgroup contains 4 elements of order 3.
so we have 24 elements of order 5 in G. This is however impossible, since20 + 24 > 30 = |G|.
In a similar way one can show that if |G| = 40 or |G| = 56 then G is not asimple group (exercise)
Next goal: there are infinitely many non-abelian simple finite groups. In partic-ular there is a simple group of order 60.
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21 Symmetric and alternating groups
Recall. The symmetric group on n letters is the group
Sn = Perm({1, . . . , n})
21.1 Theorem (Cayley). If G is a group of order n then G is isomorphic to asubgroup of Sn.
Proof. Let S be the set of all elements of G. Consider the action of G on S
G× S → S, a · b := ab
This action defines a homomorphism ϱ : G→ Perm(S). Check: this homomor-phism is 1-1. It follows that G is isomorphic to a subgroup of Perm(S). Finally,since |S| = n we have Perm(S) ∼= Sn.
21.2 Notation. Denote[n] := {1, . . . , n}
If σ ∈ Sn, σ : [n]→ [n] then we write
σ =
(1 2 3 . . . n
σ(1) σ(2) σ(3) . . . σ(n)
)
21.3 Definition. A permutation σ ∈ Sn is a cycle of length r (or r-cycle) ifthere are distinct integers i1, . . . , ir ∈ [n] such that
σ(i1) = i2, σ(i2) = i3, . . . , σ(ir) = i1
and σ(j) = j for j = i1, . . . , ir.
A cycle of length 2 is called a transposition.
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Note. The only cycle of length 1 is the identity element in Sn.
21.4 Notation. If σ is a cycle as above then we write
σ = (i1 i2 . . . ir)
21.5 Example. In S5 we have(1 2 3 4 5
1 4 2 5 3
)= (2 4 5 3)
Note: (2 4 5 3) = (4 5 3 2) = (5 3 2 4) = (3 2 4 5).
21.6 Definition. Permutations σ, τ ∈ Sn are disjoint if
{i ∈ [n] | σ(i) = i} ∩ {j ∈ [n] | τ(j) = j} = ∅
21.7 Proposition. If σ, τ are disjoint permutations then στ = τσ.
Proof. Exercise.
21.8 Proposition. Every non-identity permutation σ ∈ Sn is a product of dis-joint cycles of length ≥ 2. Moreover, this decomposition into cycles is unique upto the order of factors.
21.9 Example. Let σ ∈ S9
σ =
(1 2 3 4 5 6 7 8 9
4 7 1 3 2 6 5 9 8
)Then σ = (1 4 3)(2 7 5)(8 9).
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Proof of proposition 21.8. Consider the action of Z on the set [n] given by
k · i = σk(i)
for k ∈ Z, i ∈ [n]. Notice that
Orb(i) = {σk(i) | k ∈ Z}
Define σi : [n]→ [n]
σi(j) =
{σ(j) if j ∈ Orb(i)
j otherwise
Notice that σi is a bijection since σ(Orb(i)) = Orb(i). Thus σi ∈ Sn. Check:
1) σi is a cycle of length |Orb(i)|.
2) if Orb(i1), . . . ,Orb(ir) are all distinct orbits of [n] containing more thanone element then σi1 , . . . , σir are non-trivial, disjoint cycles and
σ = σi1 · . . . · σir
Uniqueness of decomposition - easy.
21.10 Proposition. Every permutation σ ∈ Sn is a product of (not necessarilydisjoint) transpositions.
Proof. By Proposition 21.8 it is enough to show that every cycle is a product oftranspositions. We have:
(i1 i2 i3 . . . ir) = (i1 ir)(i1 ir−1) · . . . · (i1 i3)(i1 i2)
Note. For σ ∈ Sn we have a bijection
σ × σ : [n]× [n]→ [n]× [n]
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given by σ × σ(i, j) = (σ(i), σ(j)). Define
Sσ := {(i, j) ∈ [n]× [n] | i > j and σ(i) < σ(j)}
21.11 Definition. A permutation σ ∈ Sn is even (resp. odd) if the number ofelements of Sσ is even (resp. odd).
21.12 Theorem. 1) The map sgn: Sn → Z/2Z defined by
sgn(σ) =
{0 if σ is even
1 if σ is odd
is a homomorphism.
2) If σ is a transposition then sgn(σ) = 1, so this homomorphism is non-trivial.
Proof. 1) Let σ, τ ∈ Sn. Denote sσ = |Sσ|. We want to show
sτσ ≡ sτ + sσ (mod 2)
Let [n]+ := {(i, j) ∈ [n] × [n] | i > j}. Define subsets Pσ, Rσ, Pτ , Rτ ⊆ [n]+
as follows:
Pσ :={(i, j) | σ−1(i) > σ−1(j)}Rσ :={(i, j) | σ−1(i) < σ−1(j)}Pτ :={(i, j) | τ(i) > τ(j)}Rτ :={(i, j) | τ(i) < τ(j)}
Notice that sσ = |Rσ| and sτ = |Rτ |. Notice also that (i, j) ∈ Sτσ iff either(τ(i), τ(j)) ∈ Pσ ∩Rτ or (τ(j), τ(i)) ∈ Rσ ∩ Pτ . This gives
sτσ = |Pσ ∩Rτ |+ |Rσ ∩ Pτ |
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On the other hand we have:
sσ = |Rσ| = |Rσ ∩ Pτ |+ |Rσ ∩Rτ |
sτ = |Rτ | = |Pσ ∩Rτ |+ |Rσ ∩Rτ |Therefore
sσ + sτ = |Rσ ∩ Pτ |+ |Pσ ∩Rτ |+ 2|Rσ ∩Rτ | = sτσ + 2|Rσ ∩Rτ |
and so sτ + sσ ≡ sτσ (mod 2).
2) Exercise.
21.13 Definition/Proposition. The set
An = {σ ∈ Sn | σ is even}
is a normal subgroup of Sn. It is called the alternating group on n letters.
Proof. It is enough to notice that An = Ker(sgn).
Note. We haveSn/An ∼= Z/2Z
Since |Sn| = n! thus |An| = n!2.
21.14 Proposition. If σ ∈ Sn then σ is even (resp. odd) iff σ is a product ofan even (resp. odd) number of transpositions.
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Proof. If σ = τ1. . .τm where τ1, . . . , τm are transpositions then
sgn(σ) = sgn(τ1. . .τm) =m∑i=1
sgn(τi) =m∑i=1
1
Thus sgn(σ) = 0 iff m is even and sgn(σ) = 1 iff m is odd.
Note. If follows that if a permutation σ ∈ Sn is a product of an even numberof transpositions then it cannot be written as a product of an odd number oftranspositions (and vice versa).
21.15 Corollary. A permutation σ ∈ Sn is even iff
σ = σ1σ2. . .σr
where σi is a cycle of length mi and∑r
i=1(mi − 1) is even.
Proof. It is enough to notice that by the proof of Proposition 21.10 a cycle oflength m is a product of m− 1 transpositions.
Note. The usual notation for the sign of a permutation is
sgn(σ) =
{1 if σ is even
−1 if σ is odd
where {−1, 1} ∼= Z/2Z is the multiplicative group of units in Z.
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22 Simplicity of alternating groups
22.1 Theorem. The alternating group An is simple for n ≥ 5.
22.2 Lemma. For n ≥ 3 every element of An is a product of 3-cycles.
Proof. It is enough to show that if n ≥ 3 and τ , σ are transpositions in Sn thenτσ is a product of 3-cycles.
Case 1) τ , σ are disjoint transpositions: τ = (i j), σ = (k l) for distinctelements i, j, k, l ∈ [n]. Then we have
τσ = (i j k)(j k l)
Case 2) τ , σ are not disjoint: τ = (i j), σ = (j k). Then
τσ = (i j k)
22.3 Lemma. If n ≥ 5 and σ, σ′ are 3-cycles in Sn then
σ′ = τστ−1
for some τ ∈ An
Proof. Check: if (i1 i2 . . . ir) is a cycle in Sn then for any ω ∈ Sn we have
ω(i1 i2 . . . ir)ω−1 = (ω(i1) ω(i2) . . . ω( ir))
If σ = (i1 i2 i3), σ′ = (j1 j2 j3) then take ω ∈ Sn such that ω(ik) = jk for
k = 1, 2, 3. We haveσ′ = ωσω−1
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If ω ∈ An we can then take τ := ω.
Assume then that ω ∈ An. Since n ≥ 5 there are r, s ∈ [n] such that (r s) andσ = (i1 i2 i3) are disjoint cycles. Take τ = ω(r s). Then τ ∈ An. Moreover,since (r s) commutes with σ we have
τστ−1 = ω(r s)σ(r s)−1ω−1 = ωσω−1 = σ′
22.4 Corollary. If n ≥ 5 andH is a normal subgroup of An such thatH containssome 3-cycle then H = An.
Proof. By Lemma 22.3H contains all 3-cycles, and so by Lemma 22.2 it containsall elements of An.
Proof of Theorem 22.1. Let n ≥ 5, H ◁ An and H = {(1)}. We need to showthat H = An. By Corollary 22.4 it will suffice to show that H contains some3-cycle.
Let (1) = σ be an element of H with the maximal number of fixed points in [n].We will show that σ is 3-cycle. Take the decompositon of σ into dosjoint cycles:
σ = σ1σ2 · . . . · σm
Case 1) σ1, . . . , σm are transpositions.
Since σ ∈ An we must then have m ≥ 2. Say, σ1 = (i j), σ2 = (k l). Takes = i, j, k, l and let τ = (k l s) ∈ An. Since H is normal in An we have
τστ−1σ−1 ∈ H
Check:
1) τστ−1σ−1 = (1) since τστ−1σ−1(k) = k
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2) τστ−1σ−1 fixes every element of [n] fixed by σ
3) τστ−1σ−1 fixes i, j.
Thus τστ−1σ−1 has more fixed points than σ which is impossible by the definitionof σ.
Case 2) σr is a cycle of length ≥ 3 for some 1 ≤ r ≤ m.
We can assume r = 1: σ1 = (i j k . . . ). If σ = σ1 and σ1 is a 3-cycle we aredone.
Otherwise σ must move at least two more elements, say p, q. In such case takeτ = (k p q). We have
τστ−1σ−1 ∈ HCheck:
1) τστ−1σ−1 = (1) since τστ−1σ−1(k) = k
2) τστ−1σ−1 fixes every element of [n] fixed by σ
3) τστ−1σ−1 fixes j.
Thus τστ−1σ−1 has more fixed points than σ which is again impossible by thedefinition of σ.
As a consequence σ must be a 3-cycle.
22.5. Classification of simple finite groups.
1) cyclic groups Z/pZ, p – prime
2) alternating groups An, n ≥ 5
3) finite simple groups of Lie type, e.g. projective special linear groups
PSLn(F) := SLn(F)/Z(SLn(F))F -finite field, n ≥ 2 (and n > 2 if F = F2 or F = F3).
4) 26 sporadic groups (the smallest: Mathieu group M11, |M11| = 7920, thebiggest: the Monster M , |M | ≈ 8 · 1053).
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23 Solvable groups
Recall. Every finite group G has a composition series:
{e} = G0 ⊆ . . . ⊆ Gk = G
where Gi−1 ◁Gi and Gi/Gi−1 is a simple group.
23.1 Definition. A group G is solvable if it has a composition series
{e} = G0 ⊆ . . . ⊆ Gk = G
such that for every i the group Gi/Gi−1 is a simple abelian group (i.e. Gi/Gi−1∼=
Z/piZ for some prime pi).
23.2 Example.
1) Every finite abelian group is solvable.
2) For n ≥ 5 the symmetric group Sn has a composition series
{(1)} ⊆ An ⊆ Sn
and so Sn is not solvable.
23.3 Proposition. A finite group G is solvable iff it has a normal series
{e} = H0 ⊆ . . . ⊆ Hl = G
such that Hj/Hj−1 is an abelian group for all j.
Proof. Exercise.
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Recall.
1) If G is a group the [G,G] is the commutator subgroup of G
[G,G] = ⟨{aba−1b−1 | a, b ∈ G}⟩
2) [G,G] is the smallest normal subgroup of G such that G/[G,G] is abelian:if G/H for some H ◁G then [G,G] ⊆ H.
23.4 Definition. For a group G the derived series of G is the normal series
· · · ⊆ G(2) ⊆ G(1) ⊆ G(0) = G
where Gi+1 = [G(i), G(i)] for i ≥ 1. The group G(i) is called the i-th derivedgroup of G.
23.5 Theorem. A group G is solvable iff G(n) = {e} for some n ≥ 0.
Proof. Exercise.
23.6 Theorem.
1) Every subgroup of a solvable group is solvable.
2) Ever quotient group of a solvable group is solvable.
3) If H◁G, and both H and G/H are solvable groups then G is also solvable.
Proof.
1) If H ⊆ G then H(i) ⊆ G(i). Thus if G(n) = {e} then H(n) = {e}.
2) For H ◁G take the canonical epimorphism f : G→ G/H. We have
f(G(i)) = (G/H)(i)
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Therefore if G(n) = {e} then (G/H)(n) = {e}.
3) Assume that H ◁ G, and that H(m), (G/H)(n) are trivial groups. Considerthe canonical epimorphism f : G→ G/H. We have
f(G(n)) = (G/H)(n) = {e}
Therefore G(n) ⊆ Ker(f) = H. As a consequence we obtain
G(n+m) =(G(n)
)(m) ⊆ H(m) = {e}
23.7 Theorem (Feit-Thompson). Every finite group of odd order is solvable.
Proof. See: W. Feit, J.G. Thompson, Solvability of groups of odd order, PacificJournal of Mathematics 13(3) (1963), 775-1029.
23.8 Corollary. There are no non-abelian finite simple groups of odd order.
Proof. Let G = {e} be a simple group of odd order. By Theorem 23.7 G issolvable so [G,G] = G. Since [G,G] ◁ G, by simplicity of G we must have[G,G] = {e}, and so G is an abelian group.
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24 Nilpotent groups
24.1. Recall that if G is a group then
Z(G) = {a ∈ G | ab = ba for all b ∈ G}
Note that Z(G)◁G. Take the canonical epimorphism π : G→ G/Z(G). SinceZ (G/Z(G))◁G/Z(G) we have:
π−1 (Z (G/Z(G)))◁G
Define:Z1(G) :=Z(G)
Zi(G) :=π−1i (Z (G/Zi−1(G))) for i > 1
where πi : G→ G/Zi−1(G). We have Zi(G)◁G for all i.
24.2 Definition. The upper central series of a group G is a sequence of normalsubgroups of G:
{e} = Z0(G) ⊆ Z1(G) ⊆ Z2(G) ⊆ . . .
24.3 Definition. A group G is nilpotent if Zi(G) = G for some i.
If G is a nilpotent group then the nilpotency class of G is the smallest n ≥ 0such that Zn(G) = G.
24.4 Proposition. Every nilpotent group is solvable.
Proof. If G is nilpotent group then the upper central series of G
{e} = Z0(G) ⊆ Z1(G) ⊆ . . . ⊆ Zn(G) = G
is a normal series.
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Moreover, for every i we have
Zi(G)/Zi−1(G) = Z(G/Zi−1(G))
so all quotients of the upper central series are abelian.
24.5 Note. Not every solvable group is nilpotent. Take e.g. GT . We haveZ(GT ) = {I}, and so
Zi(GT ) = {I}for all i. Thus GT is not nilpotent. On the other hand GT is solvable with acomposition series
{I} ⊆ {I, R1, R2} ⊆ GT
24.6 Proposition.
1) Every abelian group is nilpotent.
2) Every finite p-group is nilpotent.
Proof.
1) If G is abelian then Z1(G) = G.
2) If G is a p-group then so is G/Zi(G) for every i. By Theorem 16.4 if G/Zi(G)is non-trivial then its center Z(G/Zi(G)) a non-trivial group. This means thatif Zi(G) = G then Zi(G) ⊆ Zi+1(G) and Zi(G) = Zi+1(G). Since G is finitewe must have Zn(G) = G for some G.
24.7 Definition. A central series of a group G is a normal series
{e} = G0 ⊆ . . . ⊆ Gk = G
such Gi ◁G and Gi+1/Gi ⊆ Z(G/Gi) for all i.
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24.8 Proposition. If {e} = G0 ⊆ . . . ⊆ Gk = G is a central series of G then
Gi ⊆ Zi(G)
Proof. Exercise.
24.9 Corollary. A group G is nilpotent iff it has a central series.
Proof. If G is nilpotent then
{e} = Z0(G) ⊆ Z1(G) ⊆ . . . ⊆ Zn(G) = G
is a central series of G.
Conversely, if{e} = G0 ⊆ . . . ⊆ Gk = G
is a central series of G then by (24.9) we have G = Gk ⊆ Zk(G), so G = Zk(G),and so G is nilpotent.
24.10 Note. Given a group G define
Γ0(G) :=G
Γi(G) :=[G,Γi−1(G)] for i > 0.
We have. . . ⊆ Γ1(G) ⊆ Γ0(G) = G
24.11 Proposition. If G is a group then
1) Γi(G)◁G for all i
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2) Γi(G)/Γi+1(G) ⊆ Z(G/Γi+1(G)) for all i
Proof. Exercise.
24.12 Definition. If Γn(G) = {e} then
{e} = Γn(G) ⊆ . . . ⊆ Γ0(G) = G
is a central series of G. It is called the lower central series of G.
24.13 Proposition. A group G is nilpotent iff Γn(G) = {e}
Proof. Exercise.
24.14 Theorem.
1) Every subgroup of a nilpotent group is nilpotent.
2) Ever quotient group of a nilpotent group is nilpotent.
Proof. Exercise.
24.15 Note. The properties of nilpotent group given in Theorem 24.14 areanalogous to the first two properties of solvable groups from Theorem 23.6. Thethird part of that theorem (if H, G/H are solvable then so is G) is not true fornilpotent groups. For example, take G = GT and H = {I, R1, R2}. Both Hand G/H are nilpotent, but by (24.5) G is not.
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24.16 Proposition. If G1, . . . , Gk are nilpotent groups then the direct productG1 × · · · ×Gk is also nilpotent.
Proof. It will be enough to show that the statement holds for k = 2, then thegeneral case will follow by induction with respect to k. Notice for any groupsG1, G2 we have Γi(G1 × G2) = Γi(G1) × Γi(G2). If G1 and G2 are nilpotentthen by (24.13) there exists n ≥ 0 such that Γn(G1) = {e} and Γn(G2) = {e}.This implies that Γn(G1 × G2) is trivial, and so using (24.13) again we obtainthat G1 ×G2 is nilpotent.
24.17 Corollary. If p1, . . . , pk are primes and Pi is a pi-group then P1× . . .×Pkis a nilpotent group.
Proof. Follows from (24.6) and (24.16).
24.18 Theorem. Let G be a finite group. The following conditions are equiva-lent.
1) G is nilpotent.
2) Every Sylow subgroup of G is a normal subgroup.
3) G isomorphic to the direct product of its Sylow subgroups.
24.19 Lemma. If G is a finite group and P is a Sylow p-subgroup of G then
NG(NG(P )) = NG(P )
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Proof. Since P ⊆ NG(P ) ⊆ G and P is a Sylow p-subgroup of G therefore P isa Sylow p-subgroup of NG(P ). Moreover, P ◁ NG(P ), so P is the only Sylowp-subgroup of G.
Take a ∈ NG(NG(P )). We will show that a ∈ NG(P ). We have
aPa−1 ⊆ aNG(P )a−1 = NG(P )
As a consequence aPa−1 is a Sylow p-subgroup of NG(P ), and thus aPa−1 = P .By the definitions of normalizer this gives a ∈ NG(P ).
24.20 Lemma. If H is a proper subgroup of a nilpotent group G (i.e. H ⊆ G,and H = G), then H is a proper subgroup of NG(H).
Proof. Let k ≥ 0 be the biggest integer such that Zk(G) ⊆ H. Take a ∈Zk+1(G) such that a ∈ H. We will show that a ∈ NG(H).
We have
H/Zk(G) ⊆ G/Zk(G) and Zk+1(G)/Zk(G) = Z(G/Zk(G))
If follows that for every h ∈ H we have
ahZk(G) = (aZk(G))(hZk(G)) = (hZk(G))(aZk(G)) = haZk(G)
Therefore ha = ahh′ for some h′ ∈ Zk(G) ⊆ H, and so a−1ha = hh′ ∈ H. Asa consequence a−1Ha = H, so a−1 ∈ NG(H), and so also a ∈ NG(H).
Proof of Theorem 24.18.
1)⇒ 2) Let P be a Sylow p-subgroup of G. It suffices to show that NG(P ) = G.
Assume that this is not true. Then NG(P ) is a proper subgroup G, and so byLemma 24.20 it is also a proper subgroup of NG(NG(P )). On the other handby Lemma 24.19 we have NG(NG(P )) = NG(P ), so we obtain a contradiction.
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2) ⇒ 3) Exercise.
3) ⇒ 1) Follows from Corollary 24.17.
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25 Rings
25.1 Definition. A ring is a set R together with two binary operations: addition(+) and multiplication (·) satisfying the following conditions:
1) R with addtion is an abelian group.
2) multiplication is associative: (ab)c = a(bc)
3) addition is distributive with respect to multiplication:
a(b+ c) = ab+ ac (a+ b)c = ac+ bc
The ring R is commutative if ab = ba for all a, b ∈ R.
The ring R is a ring with identity if there is and element 1 ∈ R such thata1 = 1a = a for all a ∈ R. (Note: if such identity element exists then it isunique)
25.2 Examples.
1) Z, Q, R, C are commutative rings with identity.
2) Z/nZ is a ring with multiplication given by
k(nZ) · l(nZ) := kl(nZ)
3) If R is a ring then
R[x] = {a0 + a1x+ . . .+ anxn | ai ∈ R, n ≥ 0}
is the ring of polynomials with coefficients in R and
R[[x]] = {a0 + a1x+ . . . | ai ∈ R}
is the ring of formal power series with coefficients in R.
If R is a commutative ring then so are R[x], R[[x]]. If R has identity thenR[x], R[[x]] also have identity.
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4) If R is a ring then Mn(R) is the ring of n × n matrices with coefficientsin R.
5) The set 2Z of even integers with the usual addition and multiplication isa commutative ring without identity.
6) If G is an abelian group then the set Hom(G,G) of all homomorphismsf : G→ G is a ring with multiplication given by composition of homomor-phisms and addition defined by
(f + g)(a) := f(a) + g(a)
7) If R is a ring and G is a group then define
R[G] := {∑g∈G
agg | ag ∈ R, ag = 0 for finitely many g only }
addition in R[G]: ∑g∈G
agg +∑g∈G
bgg =∑g∈G
(ag + bg)g
multiplication in R[G]:(∑g∈G
agg
)(∑g∈G
bgg
)=∑g∈G
(∑hh′=g
ahah′
)g
The ring R[G] is called the group ring of G with coefficients in R.
25.3 Definition. Let R be a ring. An element 0 = a ∈ R is a left (resp. right)zero divisor in R if there exists 0 = b ∈ R such that ab = 0 (resp. ba = 0).
An element 0 = a ∈ R is a zero divisor if it is both left and right zero divisor.
25.4 Example. In Z/6Z we have 2 · 3 = 0, so 2 and 3 are zero divisors.
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25.5 Definition. An integral domain is a commutative ring with identity 1 = 0that has no zero divisors.
25.6 Proposition. Let R be an integral domain. If a, b, c ∈ R are non-zeroelements such that
ac = bc
then a = b.
Proof. We have (a− b)c = 0. Since c = 0 and R has no zero divisors this givesa− b = 0, and so a = b.
25.7 Definition. Let R be a ring with identity. An element a has a left (resp.right) inverse if there exists b ∈ R such that ba = 1 (resp. there exists c ∈ Rsuch that cb = 1).
An element a ∈ R is a unit if it has both a left and a right inverse.
25.8 Proposition. If a is a unit of R then the left inverse and the right inverseof a coincide.
Proof. If ba = 1 = ac then
b = b · 1 = b(ac) = (ba)c = 1 · c = c
25.9 Note. The set of all units of a ring R forms a group R∗ (with multiplica-tion). E.g.:
Z∗ = {−1, 1} ∼= Z/2ZR∗ = R− {0}(Z/14Z)∗ = {1, 3, 5, 9, 11, 13} ∼= Z/6Z
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25.10 Definition. A division ring is a ring R with identity 1 = 0 such that everynon-zero element of R is a unit.
A field is a commutative division ring.
25.11 Examples.
1) R, Q, C are fields.
2) Z is an integral domain but it is not a field.
3) The ring of real quaternions is defined by
H := {a+ bi+ cj + dk | a, b, c, d ∈ R}Addition in H is coordinatewise. Multiplication is defined by the identities:
i2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j
The ring H is a (non-commutative) division ring with the identity
1 = 1 + 0i+ 0j + 0k
The inverse of an element z = a+ bi+ cj + dk is given by
z−1 = (a/∥z∥)− (b/∥z∥)i− (c/∥z∥)j − (d/∥z∥)kwhere ∥z∥ =
√a2 + b2 + c2 + d2
25.12 Proposition. The following conditions are equivalent.
1) Z/nZ is a field.
2) Z/nZ is an integral domain.
3) n is a prime number.
Proof. Exercise.
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26 Ring homomorphisms and ideals
26.1 Definition. Let R, S be rings. A ring homomorphism is a map
f : R→ S
such that
1) f(a+ b) = f(a) + f(b)
2) f(ab) = f(a)f(b)
26.2 Note. If R, S are rings with identity then these conditions do not guaranteethat f(1R) = 1S.
Take e.g. rings with identity R1, R2 and define
R1 ⊕R2 = {(r1, r2) | r1 ∈ R1, R2}
with addition and multiplication defined coordinatewise. Then R1⊕R2 is a ringwith identity (1R1 , 1R2). The map
f : R1 → R1 ⊕R2, f(r1) = (r1, 0)
is a ring homomorphism, but f(1R1) = (1R1 , 1R2).
26.3 Note. Rings and ring homomorphisms form a category Ring.
26.4 Proposition. A ring homomorphism f : R→ S is an isomorphism of ringsiff f is a bijection.
Proof. Exercise.
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26.5 Definition. If f : R→ S is a ring homomorphism then
Ker(f) = {a ∈ R | f(a) = 0}
26.6 Proposition. A ring homomorphism is 1-1 iff Ker(f) = {0}
Proof. The same as for groups (4.4).
26.7 Definition. A subring of a ring R is a subset S ⊆ R such that S is anadditive subgroup of R and it is closed under the multiplication.
A left ideal of R is a subring I ⊆ R such that for every a ∈ I and b ∈ R wehave ab ∈ I. A right ideal of R is defined analogously.
A ideal of R is a subring I ⊆ R such that I is both left and right ideal.
26.8 Notation. If I is an ideal of R then we write I ◁R.
26.9 Proposition. If f : R → S is a ring homomorphism then Ker(f) is anideal of R.
Proof. Exercise.
26.10 Definition. If I is an ideal of a ring R then the quotient ring R/I isdefined as follows.
R/I := the set of left cosets of I in R
Addition: (a+I)+(b+I) = (a+b)+I, multiplication: (a+I)(b+I) = ab+I.
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26.11 Note. If I ◁R then the map
π : R→ R/I, π(a) = a+ I
is a ring homomorphism. It is called the canonical epimorphism of R onto R/I.
26.12 Theorem. If f : R → S is a homomorphism of rings then there is aunique homomorphism
f : R/Ker(f)→ S
such that the following diagram commutes:
Rf //
π
��
S
R/Ker(f)
f
;;
Moreover, f is a monomorphism and Im(f) = Im(f).
Proof. Similar to the proof of Theorem 6.1 for groups.
26.13 First Isomorphism Theorem. If f : R→ S is a homomorphism of ringsthat is an epimorphism then
R/Ker(f) ∼= S
Proof. Take the map f : R/Ker(f) → S. Then Im(f) = Im(f) = S, so f isan epimorphism. Also, f is 1-1. Therefore f is a bijective homomorphism andthus it is an isomorphism.
26.14 Note. Let I, J ◁R. Check:
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1) I ∩ J ◁R
2) I + J ◁R where I + J = {a+ b | a ∈ I, b ∈ J}
26.15 Second Isomorphism Theorem. If I, J are ideals of R then
I/(I ∩ J) ∼= (I + J)/J
Proof. Exercise.
26.16 Third Isomorphism Theorem. If I, J are ideals of R and J ⊆ I thenI/J is a ideal of R/J and
(R/J)/(I/J) ∼= R/I
Proof. Exercise.
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Note. From now until Section 39 all rings are commutative rings withidentity 1 = 0 unless stated otherwise. Also, all ring homomorphismspreserve the identity.
27 Principal ideal domains and Euclidean rings
27.1 Definition. If R is a ring and S is a subset of R then denote
⟨S⟩ = the smallest ideal of R that contains S
We say that ⟨S⟩ is the ideal of R generated by the set S.
27.2 Note. We have
⟨S⟩ = {b1a1 + . . . bkak | ai ∈ I, bi ∈ R, k ≥ 0}
27.3 Definition. An ideal I ◁ R is finitely generated if I = ⟨a1, . . . , an⟩ forsome a1, . . . , an ∈ R.
An ideal I ◁R is a principal ideal if I = ⟨a⟩ for some a ∈ R.
27.4 Definition. A ring R is a principal ideal domain (PID) if it is an integraldomain (25.5) such that every ideal of R is a principal ideal.
27.5 Proposition. The ring of integers Z is a PID.
Proof. Let I ◁ Z. If I = {0} then I = ⟨0⟩, so I is a principal ideal. If I = {0}then let a be the smallest integer such that a > 0 and a ∈ I. We will show thatI = ⟨a⟩.
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Since a ∈ I we have ⟨a⟩ ⊆ I. Conversely, if b ∈ I then we have
b = qa+ r
for some q, r ∈ Z, 0 ≤ r ≤ a − 1. This gives r = b − qa, so r ∈ I. Since ais the smallest positive element of I, this implies that r = 0. Therefore b = qa,and so b ∈ ⟨a⟩.
27.6 Proposition. If F is a field then F is a PID.
Proof. If I ◁ F and 0 = a ∈ I then for every b ∈ F we have
b = (ba−1)a ∈ I
And so I = F. As a consequence the only ideals of F are {0} = ⟨0⟩ andF = ⟨1⟩.
27.7 Proposition. If F is a field then the ring of polynomials F[x] is a PID.
Proof. Let I ∈ R. If I = {0} then I = ⟨0⟩. Otherwise let 0 = p(x) be apolynomial such that p(x) ∈ I and deg p(x) ≤ deg q(x) for all q(x) ∈ I − {0}.Check that I = ⟨p(x)⟩.
27.8 Note. Z[x] is not a PID. E.g. the ideal ⟨2, x⟩ is not a principal ideal ofZ[x] (check!).
27.9 Definition. A Euclidean domain is an integral domain R equipped with afunction
N : R− {0} −→ N = {0, 1, . . . }such that
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1) N(ab) ≥ N(a) for all a, b ∈ R− {0}2) for any a, b ∈ R, a = 0 there exist q, r ∈ R such that
b = qa+ r
and either r = 0 or N(r) < N(a).
The function N is called the norm function on R.
27.10 Examples.
1) Z is a Euclidean domain with the norm function given by the absolutevalue.
2) Any field F is a Euclidean domain with N(a) = 0 for all a ∈ F− {0}.
3) If F is a field then F[x] is a Euclidean domain with N(p(x)) = deg p(x)for p(x) ∈ F[x], p(x) = 0.
Note: Z[x] is not a Euclidean domain with N(p(x)) = deg p(x). E. g.there are no q(x), r(x) ∈ Z[x] such that either r(x) = 0 or deg r(x) < 0and that
x = 2q(x) + r(x)
4) The ring of Gaussian integers is the subring Z[i] ⊆ C given by
Z[i] := {a+ bi ∈ C | a, b ∈ Z}
Z[i] is a Euclidean domain with
N(a+ bi) = a2 + b2 = (a+ bi)(a+ bi)
(exercise).
5) DefineZ[√−5] := {a+ b
√5i | a, b ∈ Z}
Exercise: Z[√−5] is not a Euclidean domain.
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27.11 Theorem. If R is a Euclidean domain then R is a PID.
Proof. Let I◁R, I = {0}. Choose a ∈ I such that a = 0 and thatN(a) ≤ N(b)for all b ∈ I − {0}. Check that ⟨a⟩ = I.
27.12 Note. It is not true that every PID is a Euclidean domain. Take e.g.α = 1
2+
√192i and let
Z[α] = {a+ bα | a, b ∈ Z}Then Z[α] is a PID, but it is not a Euclidean domain. See J. C. Wilson, Aprincipal ideal ring that is not a euclidean ring, Mathematics Magazine, 46 (1)(1973), 34-38. (Note: this link requires a JSTOR access)
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28 Prime ideals and maximal ideals
28.1 Definition. Let R be a ring.
1) An ideal I ◁R is a prime ideal if I = R and for any a, b ∈ R we have
ab ∈ I iff either a ∈ I or b ∈ I
2) An ideal I ◁ R is a maximal ideal if I = R and for any J ◁ R such thatI ⊆ J ⊆ R we have either J = I or J = R.
28.2 Examples.
1) The zero ideal {0} ∈ R is a prime ideal iff R is an integral domain, and itis a maximal ideal iff R is a field.
2) Recall that if I ◁ Z then I = nZ for some n ≥ 0. Check:
(nZ is a prime ideal) iff (nZ is a maximal ideal) iff (n is a prime number)
3) ⟨x⟩◁ Z[x] is prime ideal (check!) but it is not a maximal ideal:
⟨x⟩ ⊆ ⟨2, x⟩◁ Z[x]
28.3 Proposition. Let I ◁R
1) The ideal I is a prime ideal iff R/I is an integral domain.
2) The ideal I is a maximal ideal iff R/I is a field.
28.4 Corollary. Every maximal ideal is a prime ideal.
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Proof. If I ◁ R is a maximal ideal then R/I is a field. In particular R/I is anintegral domain, and so I is a prime ideal.
Proof of Proposition 28.3.
1) Assume that I ◁R is a prime ideal. For a+ I, b+ I ∈ R/I we have
(a+ I)(b+ I) = ab+ I
Thus if (a + I)(b + I) = 0 + I then ab + I = 0 + I i.e. ab ∈ I. Since I is aprime ideal we get that either a ∈ I (and so a + I = 0 + I) or b ∈ I (and sob+ I = 0 + I). Therefore R/I is an integral domain.
The other implication follows from a similar argument.
2) Assume that I ◁R is a maximal ideal. Let a+ I ∈ R/I, a+ I = 0+ I. Wewant to show that there exists b+ I ∈ R/I such that
(a+ I)(b+ I) = 1 + I
Take the ideal J = ⟨a⟩+ I. We have
I ⊆ J ⊆ R
and I = J (since a ∈ I). Since I is a maximal ideal we must have J = R. Inparticular 1 ∈ J , so
1 = ab+ c
for some b ∈ R, c ∈ I. This gives
1 + I = (ab+ c) + I = ab+ I = (a+ I)(b+ I)
Conversely, assume that R/I is a field, and let J ◁R be an ideal such that
I ⊆ J ⊆ R
121
We will show that either J = I or J = R.
Take the canonical epimorphism π : R → R/I. Since π(J) is an ideal of R/I(check!) and R/I is a field we have either π(J) = {0 + I} or π(J) = R/I.Also, since I ⊆ J , we have π−1(π(J)) = J . If follows that either
J = π−1({0 + I}) = I or J = π−1(R/I) = R
28.5 Examples.
1) Since an ideal nZ of Z is prime (and maximal) iff n is a prime numbertherefore Z/nZ is an integral domain (and in fact a field) iff n is a primenumber.
2) Take x2 + 1 ∈ R[x]. We have an epimorphism of rings
f : R[x] −→ C f(p(x)) = p(i)
Check: Ker(f) = ⟨x2 + 1⟩. By the First Isomorphism Theorem (26.13)we get R[x]/⟨x2 + 1⟩ ∼= C. Since C is a field this shows that ⟨x2 + 1⟩ isa maximal ideal of R[x].
28.6 Note. For I, J ◁R define
IJ := {a1b1 + . . .+ akbk | ai ∈ I, bi ∈ J, k ≥ 0}Check: IJ is an ideal of R.
28.7 Proposition. Let I ◁ R, I = R. The ideal I is a prime ideal iff for anyideals J1, J2 such that
J1J2 ⊆ I
we have either J1 ⊆ I or J2 ⊆ I.
Proof. Exercise.
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29 Zorn’s Lemma and maximal ideals
Goal:
29.1 Theorem. If R is a ring, I ◁ R, and I = R then there exists a maximalideal J ◁R such that I ⊆ J .
29.2 Definition. A partially ordered set (or poset) is a set S equipped with abinary relation ≤ satisfying
1) x ≤ x for all x ∈ S (reflexivity)
2) if x ≤ y and y ≤ z then x ≤ z (transitivity)
3) if x ≤ y and y ≤ x then y = x (antisymmetry).
29.3 Example. If A is a set and S is the set of all subsets of A then S is aposet with ordering given by the inclusion of subsets.
29.4 Definition. A linearly ordered set is a poset (S,≤) such that for anyx, y ∈ S we have either x ≤ y or y ≤ x.
29.5 Definition. If (S,≤) is a poset then an element x ∈ S is a maximalelement of S if we have x ≤ y only for y = x.
29.6 Example. Let S be the set of all proper subsets of a set A ordered withrespect to the inclusion. For every a ∈ A the set A− {a} is a maximal elementof S.
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29.7 Note. If (S,≤) is a poset and T ⊆ S then T is also a poset with orderinginherited from S.
29.8 Definition. Let (S,≤) is a poset and let T ⊆ S. An upper bound of T isan element x ∈ S such that y ≤ x for all y ∈ T .
29.9 Definition. If (S,≤) is a poset. A chain in S is a subset T ⊆ S such thatT is linearly ordered.
29.10 Zorn’s Lemma. If (S,≤) is a non-empty poset such that every chain inS has an upper bound in S then S contains a maximal element.
Proof of Theorem 29.1. Let I = R be an ideal of R, and let S be the set of allideals J ◁R such that I ⊆ J and J = R. Notice that S = ∅ since I ∈ S.
The set S is a poset ordered with respect to inclusion of ideals. We will showthat every chain in S has an upper bound in S. Let
T = {Ji}i∈Abe a chain in S. Check: JT :=
⋃i∈A Ji is an ideal of R. Moreover, I ⊆ JT .
Finally, we have JT = R. Indeed, otherwise 1 ∈ JT , and so 1 ∈ Ji for somei ∈ A. This would give Ji = R, which contradicts our assumptions.
It follows that JT ∈ S. Since Ji ⊆ JT for all i ∈ A, we obtain that JT is anupper bound of the chain T .
By Zorn’s Lemma (29.10) there is a maximal element J ∈ S. This means, inparticular, that J is an ideal of R such I ⊆ J . Moreover, let K◁R be any idealsuch that K = R and J ⊆ K. We have
I ⊆ J ⊆ K
which means that K ∈ S. Maximality of J is S implies then that J = K. Thisshows that J is a maximal ideal of R.
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29.11 Corollary. Every ring contains a maximal ideal.
Proof. If R is a ring then by Theorem 29.1 there exists a maximal ideal I ◁ Rsuch that I contains the zero ideal {0}◁R.
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30 Unique factorization domains
Motivation:
30.1 Fundamental Theorem of Arithmetic.
If n ∈ Z, n > 1 thenn = p1p2 · . . . · pk
where p1, . . . , pk are primes. Moreover, this decomposition is unique up to re-ordering of factors.
Goal. Extend this to other rings.
30.2 Definition. Let R be an integral domain. An element a ∈ R is irreducibleif a = 0, a is not a unit and if a = bc for some b, c ∈ R then either b or c is aunit.
30.3 Examples.
1) n ∈ Z is irreducible iff n = ±p where p is a prime number.
2) A field has no irreducible elements.
3) Take p(x) ∈ R[x], p(x) = x2 + 1. Then p(x) is irreducible in R[x].
4) Take p(x) ∈ C[x], p(x) = x2 + 1. Then p(x) is not irreducible in C[x]:
p(x) = (x− i)(x+ i)
30.4 Note. If a ∈ R is an irreducible element and u ∈ R is a unit then ua isirreducible.
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30.5 Definition. Let R be an integral domain. Elements a, b ∈ R are associatesif a = ub for some unit u ∈ R. We write: a ∼ b.
30.6 Examples.
1) If m,n ∈ Z then m ∼ n iff m = ±n.2) Check: units in R[x] are non-zero polynomials of degree 0. It follows that if
p(x), q(x) ∈ R[x] then p(x) ∼ q(x) iff p(x) = aq(x) for some a ∈ R−{0}.
30.7 Definition. A unique factorization domain (UFD) is an integral domain Rthat satisfies the following conditions:
1) if a ∈ R is a non-zero, non-unit element then
a = b1 · . . . · bkfor some irreducible elements b1, . . . , bk ∈ R
2) if b1, . . . , bk, c1, . . . , cl are irreducible elements such that
b1 · . . . · bk = c1 · . . . · clthen k = l and for some permutation σ : {1, . . . , k} → {1, . . . , k} we haveb1 ∼ cσ(1), . . . , bk ∼ cσ(k).
30.8 Examples.
1) Z is a UFD by the Fundamental Theorem of Arithmetic (30.1).
2) If F is a field then F is a UFD since all non-zero elements of F are units.
Recall. Z[√−5] is the subring of C given by
Z[√−5] = {a+ b
√5i | a, b ∈ Z}
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30.9 Proposition. Z[√−5] is not a UFD.
Proof. For a+ b√5i ∈ Z[
√−5] define
N(a+ b√5i) = (a+ b
√5i)(a+ b
√5i) = a2 + 5b2 ∈ N
Notice that
1) N(α) = 1 iff α = ±12) N(α) = 0 iff α = 0
3) N(α) = 3 for all α ∈ Z[√−5]
4) N(αβ) = N(α)N(β)
Observation 1. The only units in Z[√−5] are 1 and −1.
As a consequence for any α, β we have α ∼ β iff α = ±β.
Indeed, if α ∈ Z[√−5] is a unit then
N(α)N(α−1) = N(αα−1) = N(1) = 1
Therefore N(α) = 1, and so α = ±1.
Observation 2. If α ∈ Z[√−5] is an element such that N(α) = 9 then α is
irreducible.
Indeed, if α = ββ′ then
N(β)N(β′) = N(α) = 9
Therefore N(β) must be either 1 (and so β is a unit), 3 (impossible), or 9 (andthen N(β′) = 1, i.e. β′ is a unit) .
128
Take 9 ∈ Z[√−5]. We have
3 · 3 = 9 = (2 +√5i)(2−
√5i)
By Observation 2 the elements 3, 2 +√5i , 2−
√5i are irreducible in Z[
√−5].
On the other hand by Observation 1 we obtain
3 ∼ (2 +√5i), 3 ∼ (2−
√5i)
As a consequence 9 does not have a unique factorization in Z[√−5]
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31 Prime elements
31.1 Definition. Let R be an integral domain, and let a, b ∈ R. We say that adivides b if b = ac for some c ∈ R. We then write: a | b.
31.2 Proposition. If R is an integral domain and a, b ∈ R then a ∼ b iff a | band b | a.
Proof. Exercise.
31.3 Definition. Let R is an integral domain. An element a ∈ R is a primeelement if a = 0, a is a non-unit and if a | bc then either a | b or a | c.
31.4 Example.
1) In Z we have
{prime elements} = {± prime numbers} = {irreducible elements}
2) By the proof of Proposition 30.9 in Z[√−5] the element α = 2 +
√5i is
irreducible. On the other hand α is not a prime element:
α | (3 · 3) but α ∤ 3
31.5 Proposition. If R is an integral domain and a ∈ R is a prime elementthen a is irreducible.
Proof. Let a ∈ R be a prime element and let a = bc. We want to show thateither b or c must be a unit in R.
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We have a | (bc), and since a is a prime element it implies that a | b or a | c.
We can assume that a | b. Since also b | a, thus by (31.2) we obtain that a ∼ b,i.e. a = bu for some unit u ∈ R. Therefore we have
bc = a = bu
By (25.6) this gives u = c, and so c is s unit.
31.6 Proposition. If R is a UFD and a ∈ R then a is an irreducible element iffa is a prime element.
Proof.
(⇐) Follows from Proposition 31.5.
(⇒) Assume that a ∈ R is irreducible and that a | (bc). We want to show thateither a | b or a | c.
If b = 0 then b = a · 0 so a | 0. If b is a unit then c = b−1bc so a | c.
As a consequence we can assume that b, c are non-zero, non-units.
Since a | (bc) there is d ∈ R such that bc = ad. Assume that d is not a unit.Since R is a UFD we have decompositions:
b = b1 · . . . · bm, c = c1 · . . . · cn, d = d1 · . . . · dp
where bi, cj, dk are irreducible. This gives
b1 · . . . ·bm · c1 · . . . ·cn = a · d1 · . . . ·dp
By the uniqueness of decomposition in UFDs this implies that either a ∼ bi forsome i or a ∼ cj for some j. In the first case we get a | b, and in the secondcase a | c.
If d is a unit the argument is similar
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31.7 Theorem. An integral domain R is a UFD iff
1) every non-zero, non-unit element of R is a product of irreducible elements
2) every irreducible element in R is a prime element.
Proof.
(⇒) Follows from the definition of UFD (30.7) and Propositon 31.6.
(⇐) Assume that R satisfies conditions 1)-2) of the theorem. We only need toshow that if b1, . . . , bk, c1, . . . , cl are irreducible elements in R such that
b1 · . . . · bk = c1 · . . . · clthen k = l, and after reordering factors we have b1 ∼ c1, . . . , bk ∼ ck.
We argue by induction with respect to k.
If k = 1 then we have b1 = c1 · . . . · cl. Since b1 is irreducible this implies thatl = 1, and so b1 = c1.
Next, assume that the uniqueness property holds for some k and that we have
b1 · . . . · bk · bk+1 = c1 · . . . · clwhere bi, cj are irreducible elements. This implies that bk+1 | (c1 · . . . · cl). Bycondition 2) of the theorem bk+1 is a prime element. It follows that bk | cj forsome 1 ≤ j ≤ l. We can assume that bk+1 | cl. Then cl = abk+1 for somea ∈ R. Also, since cl, bk+1 are irreducible a must be a unit. This shows thatbk+1 ∼ cl. Furthermore, we obtain from here that
b1 · . . . · bk · bk+1 = c1 · . . . · cl−1 · abk+1
Since R is an integral domain we get
b1 · . . . · bk = c1 · . . . · cl−1a
Since bk is irreducible and a is a unit the product cl−1a is an irreducible element.Therefore by the inductive assumption k = l− 1, and after reordering of factorswe have
b1 ∼ c1, . . . , bk−1 ∼ ck−1, bk ∼ cl−1a ∼ cl−1
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32 PIDs and UFDs
32.1 Theorem. If R is a PID then it is a UFD.
32.2 Lemma. If R is a PID and I1, I2, . . . are ideals of R such that
I1 ⊆ I2 ⊆ . . .
then there exists n ≥ 1 such that In = In+1 = . . . .
Proof. Take I =⋃∞i=1 Ii. Check: I is an ideal of R. Since R is a PID we have
I = ⟨a⟩ for some a ∈ I. Take n ≥ 1 such that a ∈ In. Then we get
I ⊆ In ⊆ In+1 ⊆ . . . ⊆ I
It follows that In = In+1 = . . . = I.
32.3 Lemma. Let R be a PID. An element a ∈ R is irreducible iff ⟨a⟩ is amaximal ideal of R.
Proof. Exercise.
32.4 Lemma. If R is an integral domain then a ∈ R is a prime element iff ⟨a⟩is a non-zero prime ideal of R.
Proof. Exercise.
Proof of Theorem 32.1. Let R be a PID. By Theorem 31.7 it suffices to showthat
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1) every non-zero, non-unit element of R is a product of irreducible elements
2) every irreducible element in R is a prime element.
1) We argue by contradiction. Assume that a0 ∈ R is a non-zero, non-unitelement that is not a product of irreducibles. This implies that
a0 = a1b1
for some non-zero, non-unit elements a1, b1 ∈ R.
Next, if both a1 and b1 were products of irreducibles, then a0 would be also aproduct of irreducibles, contradicting our assumption. We can then assume thata1 is not a product of irreducibles, and so in particular we have
a1 = a2b2
for some non-zero, non-unit elements a2, b2 ∈ R.
By induction we obtain that for i = 1, 2, . . . there exists non-zero, non-unitelements ai, bi ∈ R such that ai = ai+1bi+1 for all i ≥ 0.
Consider the chain of ideals
⟨a0⟩ ⊆ ⟨a1⟩ ⊆ . . .
By Lemma 32.2 we obtain that ⟨an⟩ = ⟨an+1⟩ for some n ≥ 0. This means thatan = an+1u for some unit u ∈ R (check!). As a consequence we obtain
an+1bn+1 = an = an+1u
and so bn+1 = u. This is a contradiction, since bn+1 is not a unit.
2) Let a ∈ R be an irreducible element and let a | (bc). We need to show thateither a | b or a | c.
Assume that a ∤ b. This implies that b ∈ ⟨a⟩ and so
⟨a⟩ = ⟨a⟩+ ⟨b⟩
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Since by Lemma 32.3 the ideal ⟨a⟩ is a maximal ideal we obtain then that⟨a⟩+ ⟨b⟩ = R, and so in particular 1 ∈ ⟨a⟩+ ⟨b⟩. Therefore
1 = ar + bs
for some r, s ∈ R, and soc = a(rc) + (bc)s
Since a | a(rc) and a | (bc)s we obtain from here that a | c.
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33 Application: sums of two squares
Recall. The ring of Gaussian integers:
Z[i] = {a+ bi | a, b ∈ Z}
Z[i] is a Euclidean domain with the norm function N(a + bi) = a2 + b2. Inparticular it follows that Z[i] is a UFD.
33.1 Note. For α, β ∈ Z[i] we have
N(αβ) = N(α)N(β)
33.2 Proposition. An element α ∈ Z[i] is a unit iff N(α) = 1.
Proof. Exercise.
33.3 Corollary. The only units in Z[i] are ±1 and ±i.
33.4 Proposition. If p ∈ Z is a prime number and for some α ∈ Z[i] we haveN(α) = p then α is an irreducible element in Z[i].
Proof. Assume that α = βγ. Then we have
p = N(α) = N(β)N(γ)
Since p is a prime number we get that either N(β) = 1 or N(γ) = 1, and soeither β or γ is a unit in Z[i].
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33.5 Theorem. Let p be an odd prime. The following conditions are equivalent.
1) p = a2 + b2 for some a, b ∈ Z2) p ≡ 1 (mod 4)
33.6 Lemma. If p ∈ Z is a prime number and p ≡ 1 (mod 4) then there ism ∈ Z such that
m2 ≡ −1 (mod p)
33.7 Proposition. If p ∈ Z is a prime number then the groups of units (Z/pZ)∗of the ring Z/pZ is a cyclic group of order (p− 1).
Proof. See Propositon 38.8.
Proof of Lemma 33.6. By Proposition 33.7 we have
(Z/pZ)∗ ∼= Z/(p− 1)Z
Since p ≡ 1 (mod 4), thus 4 | (p − 1), and so the group (Z/pZ)∗ has a cyclicsubgroup of order 4. Let a ∈ Z/pZ be a generator of this subgroup. Then a2 isan element of order 2 in (Z/pZ)∗, so a2 = −1.
Take the canonical epimorphism π : Z→ Z/pZ and let m ∈ π−1(a). Then
π(m2) = a2 = π(−1)
which gives m2 ≡ −1 (mod p)
Proof of Theorem 33.5.
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1) ⇒ 2) Assume that p = a2 + b2. Since p is odd we can assume that a is even,a = 2m, and b is odd, b = 2n+ 1. Then we have
p = (2m)2 + (2n+ 1)2 = 4m2 + 4n2 + 4n+ 1
and so p ≡ 1 (mod 4).
2) ⇒ 1) Since p ≡ 1 (mod 4) by Lemma 33.6 there exists m ∈ Z such thatp | (m2 + 1). In Z[i] we have
m2 + 1 = (m+ i)(m− i)
Therefore in Z[i] we have p | (m + i)(m − i). On the other hand p ∤ (m ± i)since otherwise for some a, b ∈ Z we would have
m± i = p(a+ bi) = pa+ pbi
and so pb = ±1 which is impossible.
As a consequence we obtain that p is not a prime element in Z[i]. Since Z[i] isa UFD, prime elements in Z[i] coincide with irreducible elements and so p is notirreducible. Therefore
p = αβ
for some non-units α, β ∈ Z[i]. We have
p2 = N(p) = N(α)N(β)
By Lemma 33.2 we have N(α) = 1 = N(β) so N(α) = N(β) = p. Therefore,if α = a+ bi then p = a2 + b2.
33.8 Note.
1) One can use factorization in Z[i] to show that, in general, a positive integern is a sum of two squares iff n is of the form
n = 2kpmi1 . . .pmr
r q2n11 . . .q2ns
s
where k,mi, nj ≥ 0, and pi, qj are prime numbers such that pi ≡ 1 (mod 4)and qj ≡ 3 (mod 4).
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2) The ring Z[i] can be also used to describe all Pythagorean triples, e.i.triples of positive integers (x, y, z) that satisfy the equation x2 + y2 = z2.
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34 Application: Fermat’s Last Theorem
34.1 Fermat’s Last Theorem.
For n ≥ 3 there are no integers x, y, z > 0 such that xn + yn = zn.
Kummer’s idea of the proof.
1) It is enough to show that xn + yn = zn has no integral solutions for n = 4and for n = p where p is an odd prime. Indeed, if n ≥ 3 is any integer thenn = mk where either k = 4 or k is an odd prime, and if x = a, y = b, z = cwould be a solution of
xn + yn = zn
then x = am, y = bm, z = cm would be a solution of
xk + yk = zk
2) The case m = 4 was proved by Fermat.
3) Take an odd prime p. Let ζ = e2πi/p be a pth primitive root of 1, and let Z[ζ]be the smallest subring of C that contains ζ.
For any x, y ∈ Z we have a factorization in Z[ζ]:
xp + yp =
p−1∏i=0
(x+ ζ iy)
Therefore, if xp + yp = zp then in Z[ζ] we have
zp =
p−1∏i=0
(x+ ζ iy)
Assuming that Z[ζ] is a UFD we can compare these factorizations and try to geta contradiction.
Problem. For some values of p (e.g. p = 23) the ring Z[ζ] is not a UFD.
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35 Greatest common divisor
35.1 Definition. Let R be a ring and let a1, . . . an ∈ R. We say that b ∈ R isa greatest common divisor of a1, . . . , an if
1) b | ai for i = 1, . . . , n
2) if c | ai for i = 1, . . . , n then c | b.In such case we write b ∼ gcd(a1, . . . , an).
35.2 Note.
1) If b is a greatest common divisor of a1, . . . , an and b′ ∼ b then b′ is also agreatest common divisor of a1, . . . , an.
2) In general gcd(a1, . . . , an) need not exists. E.g. gcd(9, 3(2 +√5i)) does
not exist in Z[√−5].
35.3 Theorem. If R is a UFD then gcd(a1, . . . , an) exists for any a1, . . . , an.
35.4 Lemma. If R is a ring such that gcd(a1, a2) exists for any a1, a2 ∈ R thengcd(a1, . . . , an) exists for any a1, . . . , an ∈ R.
Proof. Check:
gcd(a1, . . . , an) ∼ gcd(gcd(a1, . . . , an−1), an)
Proof of Theorem 35.3. Let a, b ∈ R. By Lemma 35.4 it is enough to show thatgcd(a, b) exists.
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If a = 0 then gcd(a, b) ∼ b (check!).
If a is a unit then gcd(a, b) ∼ a (check!).
Therefore we can assume that a, b are non-zero, non-unit elements of R. SinceR is a UFD in such case we have
a = uck11 ck22 . . .c
kmm , b = vcl11 c
l22 . . .c
lmm
where u, v are units, c1, . . . , cm are distinct irreducible elements, and ki, li ≥ 0.Check: we have
gcd(a, b) ∼ cn11 c
n22 . . .c
nmm
where ni = min{ki, li} for i = 1, . . .m.
35.5 Definition. Let R be a ring. Elements a1, . . . , an ∈ R are relatively primeif gcd(a1, . . . , an) ∼ 1.
35.6 Note.
1) The elements a1, . . . , an are relatively prime iff every common divisor ofa1, . . . , an is a unit (check!).
2) If R is a UFD, a, b, c ∈ R, gcd(a, b) ∼ 1 and a | bc then a | c (check!).
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36 Rings of fractions
Recall. If R is a PID then R is a UFD.
In particular
• Z is a UFD
• if F is a field then F[x] is a UFD.
Goal. If R is a UFD then so is R[x].
Idea of proof.
1) Find an embedding R ↪→ F where F is a field.
2) If p(x) ∈ R[x] then p(x) ∈ F[x] and since F[x] is a UFD thus p(x) has aunique factorization into irreducibles in F[x].
3) Use the factorization in F[x] and the fact that R is a UFD to obtain afactorization of p(x) in R[x].
36.1 Definition. If R is a ring then a subset S ⊆ R is a multiplicative subset if
1) 1 ∈ S2) if a, b ∈ S then ab ∈ S
36.2 Example.
Multiplicative subsets of Z:
1) S = Z
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2) S0 = Z− {0}
3) Sp = {n ∈ Z | p ∤ n} where p is a prime number.
Note: Sp = Z− ⟨p⟩.
36.3 Proposition. If R is a ring, and I is a prime ideal of R then S = R − Iis a multiplicative subset of R.
Proof. Exercise.
36.4. Construction of a ring of fractions.
Goal. For a ring R and a multiplicative subset S ⊆ R construct a ring S−1Rsuch that every element of S becomes a unit in S−1R.
Consider a relation on the set R× S:
(a, s) ∼ (a′, s′) if s0(as′ − a′s) = 0 for some s0 ∈ S
Check: ∼ is an equivalence relation.
Elements of S−1R = (equivalence classes of R× S under the relation ∼).
Notation. a/s := the equivalence class represented by (a, s)
Note.
1) If 0 ∈ S then (a, s) ∼ (a′, s′) for all (a, s), (a′, s′) ∈ R × S, and so S−1Rconsists of only one element.
2) If R is an integral domain and 0 ∈ S then (a, s) ∼ (a′, s′) iff as′ = a′s.
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Multiplication in S−1R:
(a/s)(a′/s′) = (aa′)/ss′
Addition in S−1R:a/s+ a′/s′ = (as′ + a′s)/ss′
Check:
1) The operations of addition and multiplication in S−1R are well defined.
2) These operations define a commutative ring structure on S−1R.
3) The mapφS : R→ S−1R, φS(r) = r/1
is a ring homomorphism.
Note. If s ∈ S then φS(s) = s/1 is a unit in S−1R:
(s/1)(1/s) = s/s = 1/1 = 1S−1R
36.5 Definition. If S is a multiplicative subset of a ring R then S−1R is calledthe ring of fractions of R with respect to S.
36.6 Theorem. If S is a multiplicative subset of a ring R and f : R → R′ is aring homomorphism such for ever s ∈ S the element f(s) ∈ R′ is a unit, thenthere exists a unique homomorphism
f : S−1R→ R′
such that the following diagram commutes:
Rf //
φS
��
R′
S−1R
f
==
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Proof. Define f(r/s) = f(r)f(s)−1. Check that
1) f is a well defined ring homomorphism
2) f is the only homomorphism such that f = fφS.
36.7 Examples.
1) If S ⊆ Z, S = Z then S−1Z ∼= {0}.
2) If S0 ⊆ Z, S0 = Z− {0} then S−10 Z ∼= Q.
3) If Sp ⊆ Z, Sp = Z − ⟨p⟩ then S−1p Z is isomorphic to the subring of Q
consisting of all fractions mnsuch that p ∤ n.
4) If R = Z/6Z, S = {1, 3} ⊆ R then S−1R ∼= Z/2Z (check!).
36.8 Proposition. If S is a multiplicative subset of a ring R then
Ker(φS) = {a ∈ R | sa = 0 for some s ∈ S}
Proof. We have φS(r) = 0/1 iff s(r · 1− 0 · 1) = 0 for some s ∈ S.
36.9 Corollary. If R is an integral domain and S ⊆ R is a multiplicative subsetsuch that 0 ∈ S then the homomorphism
φS : R→ S−1R
is 1-1. As a consequence in this case we can identify R with a subring of S−1R.
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36.10 Note.
1) If R is an integral domain and S = R−{0} then the ring S−1R is a field.In this case S−1R is called the field of fractions of R.
2) If I is a prime ideal of a ring R then the set S = R− I is a multiplicativesubset of R. In this case the ring S−1R is called the localization of R atI and it is denoted RI .
36.11 Definition. A ring R is a local ring if R has exactly one maximal ideal.
36.12 Examples.
1) If F is a field then it is a local ring with the maximal ideal I = {0}.
2) Check: if F is a field then F[[x]] is a local ring with the maximal ideal ⟨x⟩.
3) Check: if F is a field then F[x] is not a local ring since for every a ∈ F theideal ⟨x− a⟩ is a maximal ideal in F[x].
4) Z is not a local ring.
36.13 Proposition. If R is a local ring then the maximal ideal J ◁ R consistsof all non-units of R.
Proof. Since J = R thus J does not contain any units.
Conversely, if a ∈ R is a non-unit then 1 ∈ ⟨a⟩ so ⟨a⟩ = R. Therefore byTheorem 29.1 ⟨a⟩ is contained in some maximal ideal of R. Since J is the onlymaximal ideal we obtain ⟨a⟩ ⊆ J , and so a ∈ J .
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36.14 Proposition. If R is a ring and I ◁ R is a prime ideal then the ring RI
is a local ring, and the maximal ideal J ◁RI is given by
J := {a/s | a ∈ I, s ∈ I}
Proof. Exercise.
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37 Factorization in rings of polynomials
Goal:
37.1 Theorem. If R is a UFD then so is R[x].
37.2 Lemma. If R is an integral domain then p(x) ∈ R[x] is a unit in R[x] iffdeg p(x) = 0 and p(x) = a0 where a0 is a unit in R.
Proof. Exercise.
37.3 Lemma. If R is an integral domain and p(x) = a0 is a polynomial ofdegree 0 in R[x] then p(x) is irreducible in R[x] iff a0 is irreducible in R.
Proof. Exercise.
37.4 Definition. If R is a UFD and p(x) = a0 + a1x + . . . anxn ∈ R[x], then
p(x) is a primitive polynomial if gcd(a0, . . . , an) ∼ 1.
37.5 Lemma. If R is a UFD and p(x) ∈ R[x] is an irreducible polynomial suchthat deg p(x) > 0 then p(x) is a primitive polynomial.
Proof. If p(x) = a0 + a1x+ . . . anxn is not primitive then
p(x) = gcd(a0, . . . an)q(x)
for some q(x) ∈ R[x], deg q(x) > 0. Since both gcd(a0, . . . , an) and q(x) arenon-units in R[x] the polynomial p(x) is not irreducible.
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37.6 Lemma (Gauss). If R is a UFD and p(x), q(x) ∈ R[x] are primitive poly-nomials then p(x)q(x) is also primitive.
Proof. Assume that r(x) = p(x)q(x) is not primitive. Then we have
r(x) = c · r(x)
where c ∈ R is an irreducible element, and r(x) ∈ R[x]. Take the canonicalepimorphism
π : R −→ R/⟨c⟩This defines a homomorphism of rings of polynomials
π : R[x] −→ (R/⟨c⟩)[x]
given by
π(a0 + a1x+ . . .+ anxm) := π(a0) + π1(a)x+ . . .+ π(an)x
n
Note:
1) Since c is irreducible element thus ⟨c⟩ is a prime ideal and so R/⟨c⟩ is adomain. As a consequence (R/⟨c⟩)[x] is a domain.
2) We have
π(p(x))π(q(x)) = π(c · r(x)) = 0 · π(r(x)) = 0
so either π(p(x)) = 0 or π(q(x)) = 0.
We can assume that π(p(x)) = 0. Then p(x) = c · p(x) for some p(x) ∈ R[x].Since p(x) is primitive we get a contradiction.
37.7 Lemma. Let R be a UFD and K be the field of fractions of R.
1) For any non-zero polynomial p(x) ∈ K[x] there is an element c(p) ∈ Kand a primitive polynomial p∗(x) ∈ R[x] such that
p(x) = c(p)p∗(x)
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2) If p(x) = c(p)p∗(x) and p(x) = c(p)p∗(x) for some c(p), c(p) ∈ K, andsome primitive polynomials p∗(x), p∗(x) ∈ R[x] then there exists a unitu ∈ R such that
c(p) = uc(p), p∗(x) = u−1p∗(x)
3) If p(x), q(x) ∈ K[x] are non-zero polynomials then for some unit u ∈ Rwe have
c(pq) = uc(p)c(q) (p(x)q(x))∗ = u−1p∗(x)q∗(x)
37.8 Definition. If p(x) ∈ K[x] then the element c(p) ∈ K is called the contentof p(x).
Proof of Lemma 37.7.
1) If p(x) ∈ K[x] then there is there exists 0 = c ∈ R such that cp(x) ∈ R[x].Let b ∈ R be a greatest common divisor of coefficients of cp(x). Take
p∗(x) = c/b · p(x), c(p) = b/c
Check that p∗(x) ∈ R[x] and p∗(x) is a primitive polynomial.
2) We havec(p)p∗(x) = p(x) = c(p)p∗(x)
where c(p), c(p) ∈ K are non-zero elements and p∗(x), p∗(x) ∈ R[x] are primitivepolynomials. Let p∗(x) = a0 + a1x+ . . .+ anx
n.
We havep∗(x) = c(p)c(p)−1p∗(x)
Since c(p)c(p)−1 ∈ K there b, d ∈ R such that c(p)c(p)−1 = b/d. We canassume that gcd(b, d) ∼ 1. We have
dp∗(x) = bp∗(x) = ba0 + ba1x+ . . .+ banxn
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This gives: d | bai for i = 1, . . . , n. By (35.6) this gives that d | ai for all i.Since p∗(x) is a primitive polynomial it implies that d is a unit in R. By ananalogous argument we obtain that b is also a unit in R.
As a consequence u = db−1 is a unit in R and we have
c(p) = uc(p), p∗(x) = u−1p∗(x)
3) For p(x), q(x) ∈ K[x] we have
c(p)c(q)p∗(x)q∗(x) = p(x)q(x) = c(pq)(p(x)q(x))∗
We have c(p)c(q), c(pq) ∈ K, (p(x)q(x))∗ ∈ R[x] is primitive by constructionand p∗(x)q∗(x) ∈ R[x] is primitive by Lemma 37.6.
Applying part 2) we obtain that there is a unit u ∈ R such that
c(pq) = uc(p)c(q) (p(x)q(x))∗ = u−1p∗(x)q∗(x)
37.9 Note. Check:
1) Let p(x) ∈ K[x]. Then p(x) ∈ R[x] iff c(p) ∈ R.
2) p(x) ∈ R[x] is primitive iff p(x) = up∗(x) for some unit u ∈ R.
37.10 Proposition. Let R be a UFD and K be the ring of fractions of R.
1) A polynomial p(x) ∈ K[x] of non-zero degree is irreducible in K[x] iffp∗(x) is irreducible in R[x].
2) A polynomial p(x) ∈ R[x] of non-zero degree is irreducible in R[x] iff p(x)is primitive and it is irreducible in K[x].
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Proof.
1) (⇐) If p(x) is not irreducible in K[x] then p(x) = q(x)r(x) for someq(x), r(x) ∈ K[x], deg q(x), deg r(x) > 0. By Lemma 37.7 we have
p∗(x) = uq∗(x)r∗(x)
for some unit u ∈ R. Therefore p∗(x) is not irreducible in R[x].
(⇒) If p∗(x) is not irreducible in R[x] then
p∗(x) = q(x)r(x)
where q(x), r(x) are non-units in R[x]. Since p∗(x) is primitive we must havedeg q(x), deg r(x) > 0. Then
p(x) = c(p)q(x)r(x)
and so p(x) is not irreducible in K[x].
2) (⇒) If p(x) ∈ R[x] is irreducible then it must be a primitive polynomial.Therefore p(x) = up∗(x) for some unit u ∈ R. Since p(x) is irreducible in R[x],thus p∗(x) is also irreducible in R[x], and so by part 1) p(x) is irreducible inK[x].
(⇐) Since p(x) is irreducible in K[x] by part 1) we get that p∗(x) is irreducible inR[x]. Also, since p(x) is primitive we have p(x) = up∗(x) for some unit u ∈ R.It follows that p(x) is irreducible in R[x].
Proof of Theorem 37.1. By Theorem 31.7 we need to show that:
1) Every non-zero, non-unit p(x) ∈ R[x] is a product of irreducible polyno-mials.
2) Every irreducible polynomial p(x) ∈ R[x] is a prime element of R[x].
153
1) Recall that irreducible polynomials of degree 0 inR[x] correspond to irreducibleelements of R. Since R is a UFD it follows that if p(x) ∈ R[x] has degree 0then p(x) is a product of irreducibles in R[x].
If deg p(x) > 0 then p(x) is a non-zero non-unit element of K[x]. Since K[x] isa UFD we have
p(x) = q1(x)q2(x). . .qk(x)
where q1(x), . . . , qk(x) are irreducible polynomials in K[x]. We have
p(x) = (c(q1)c(q2). . .c(qk))q∗1(x)q
∗2(x). . .q
∗k(x)
By Proposition 37.10 q∗i (x) are irreducible polynomials in R[x]. Moreover, byLemma 37.7 we have
c(q1)c(q2). . .c(qk) = uc(p)
for some unit u ∈ R, and also since p(x) ∈ R[x] we have c(p) ∈ R. Thereforec(q1)c(q2). . .c(qk) ∈ R, and since R is a UFD we have
c(q1)c(q2). . .c(qk) = a1a2. . .al
where a1, . . . , al ∈ R are irreducible elements in R[x]. As a consequence weobtain a factorization of p(x):
p(x) = a1a2. . .alq∗1(x)q
∗2(x). . .q
∗k(x)
where a1, . . . , al, q∗1(x), . . . , q
∗k(x) are irreducible elements in R[x].
2) Let p(x) ∈ R[x] be an irreducible polynomial. We need to show that if forsome q(x), r(x) ∈ R[x] we have p(x) | q(x)r(x) then either p(x) | q(x) orp(x) | r(x).
Exercise: this holds if deg p(x) = 0.
If deg p(x) > 0, then since p(x) is irreducible in R[x] by Proposition 37.10 weobtain that p(x) is primitive and it is irreducible in K[x]. Since K[x] is a UFDwe obtain that p(x) is a prime element of K[x], and so p(x) | q(x) or p(x) | r(x)in K[x].
We can assume that p(x) | q(x) in K[x]. Then there exists h(x) ∈ K[x] suchthat
p(x)h(x) = q(x)
154
We have c(p)c(h) = uc(q) for some unit u ∈ R. Also, since q(x) ∈ R[x] wehave that c(q) ∈ R and therefore c(p)c(h) ∈ R. Finally, since p(x) is primitivethus c(p) is a unit in R. We obtain:
c(h) = c(p)−1uc(q) ∈ R
Therefore h(x) ∈ R[x], and so p(x) | q(x) in R[x].
37.11 Corollary. Z[x] is a UFD
37.12 Corollary. If R is a UFD then the ring of polynomials of n variablesR[x1, . . . , xn] is also a UFD.
Proof. It is enough to notice that
R[x1, . . . , xn] ∼= R[x1, . . . , xn−1][xn]
(check!).
155
38 Irreducibility criteria in rings of polynomials
38.1 Theorem. Let p(x), q(x) ∈ R[x] be polynomials such that
p(x) = a0 + a1x+ . . .+ anxn, q(x) = b0 + b1x+ . . .+ bmx
m
and an, bm = 0. If bm is a unit in R then there exist unique polynomialsr(x), s(x) ∈ R[x] such that
p(x) = s(x)q(x) + r(x)
and either deg r(x) < deg q(x) or r(x) = 0.
Proof. Exercise (or see Hungerford p.158).
38.2 Definition. If R is a ring and p(x) ∈ R[x] then p(x) defines a function
p : R −→ R, a 7→ p(a)
A function of this form is called a polynomial function.
38.3 Note. Different polynomials may define the same polynomial function.
E.g. if p(x) = x+ 1, q(x) = x2 + 1 are polynomials in Z/2Z[x] then p(x), q(x)define the same function Z/2Z→ Z/2Z:
p(0) = q(0) = 1, p(1) = q(1) = 0
38.4 Definition. Let p(x) ∈ R[x]. An element a ∈ R is a root of p(x) iff(a) = 0.
156
38.5 Proposition. An element a ∈ R is a root of p(x) ∈ R[x] iff (x−a) | p(x).
Proof.
(⇐) If (x − a) | p(x) then p(x) = q(x)(x − a) for some q(x) ∈ R[x] sop(a) = q(a)(a− a) = 0.
(⇒) Assume that p(a) = 0. By Theorem 38.1 we have
p(x) = s(x)(x− a) + r(x)
where deg r(x) = 0, so r(x) = b for some b ∈ R. This gives
0 = p(a) = s(a)(a− a) + b
Thus b = 0, and so p(x) = s(x)(x− a).
38.6 Corollary. If R is an integral domain and 0 = p(x) ∈ R[x] is a polynomialof degree n then R[x] has at most n distinct roots in R.
Proof. Let a1, . . . , ak ∈ R be all distinct roots of p(x). By (38.5) we have
p(x) = (x− a1)q1(x)
for some q1(x) ∈ R[x]. Also we have
0 = p(a2) = (a2 − a1)q1(a2)
Since R is an integral domain and a2 − a1 = 0 we obtain q1(a2), and so
q2(x) = (x− a2)q3(x)
for some q3(x) ∈ R[x]. This gives
p(x) = (x− a1)(x− a2)q3(x)
157
By induction we obtain
p(x) = (x− a1) · . . .·(x− ak)qk(x)
for some 0 = qk(x) ∈ R[x]. This gives
deg p(x) = deg((x− a1) · . . .·(x− ak)qk(x)) ≥ k
38.7 Note.
1) Corollary 38.6 in not true if R is not an integral domain. E.g. if R = Z/6Zand p(x) = x2 + x then 0, 2, 3 ∈ Z/6Z are roots of p(x).
2) Corollary 38.6 is not true is R is a non-commutative ring (even if R has nozero divisors). For example, if R = H and p(x) = x2+1 then ±i,±j,±k ∈H are roots of p(x).
38.8 Proposition. Let R is an integral domain. If G is a finite subgroup of themultiplicative group of units of R then G is a cyclic group.
Proof. Let a ∈ G be an elements of a maximal order in G. We will show that⟨a⟩ = G. We argue by contradiction. Assume that there exists an elementb ∈ G− ⟨a⟩, and let |b| = m. By assumption m ≤ |a|. If m ∤ |a| then |ab| > |a|(check!) which is impossible by the choice of a. Therefore |a| = km for somek > 0. This shows that there are k + 1 distinct elements of order m in G:b, ak, a2k, . . . , a(m−1)k. This is however impossible since each of these elementswould be a root of the polynomial f(x) = xm − 1 ∈ R[x] and by Corollary 38.6this polynomial has at most m roots in R.
38.9 Proposition. If F is a field and p(x) ∈ F[x] is a polynomial such thatdeg p(x) > 1 and p(x) has a root in F then p(x) is not irreducible in F[x].
158
Proof. By (38.6) we have p(x) = q(x)(x − a) for some q(x) ∈ F[x]. Sincedeg p(x) > 1 we have deg q(x) > 0, so q(x) and (x − a) are not units inR[x].
38.10 Corollary. Let R be a UFD and let K be the field of fractions of R. Ifp(x) ∈ R[x] is a polynomial such that deg p(x) > 1 and p(x) has a root in Kthen p(x) is not irreducible in R[x].
Proof. By (38.6) p(x) is not irreducible in K[x], so by (37.10) it is also notirreducible in R[x].
38.11 Proposition (Integral root test). Let R be a UFD, let K be the field offractions of R and let p(x) ∈ R[x] be a polynomial
p(x) = a0 + a1x+ . . .+ anxn
where an = 0. If a ∈ K is a root of p(x) then a is of the form a = b/s whereb, s ∈ R, gcd(b, s) ∼ 1, b | a0 and s | an.
In particular, in an = 1 then a ∈ R and a | a0.
Proof. Exercise.
38.12 Theorem (Eisenstein Irreducibility Criterion). Let R be a UFD. If
p(x) = a0 + a1x+ . . .+ anxn
is a primitive polynomial in R[x] such that deg p(x) > 0, and b ∈ R is anirreducible element b ∈ R such that
1) b ∤ an2) b | ai for all i < n
159
3) b2 ∤ a0then p(x) is irreducible in R[x].
Proof. Assume that p(x) is not irreducible in R[x]. Then we have
p(x) = q(x)r(x)
for some non-units q(x), r(x) ∈ R[x]. Since p(x) is primitive we must havedeg q(x), deg r(x) > 0. Let
q(x) = c0 + c1x+ . . .+ ckxk, r(x) = d0 + d1x+ . . .+ dlx
l
Notice that since b is irreducible it is a prime element of R and so by (32.4)the ideal ⟨b⟩ is a prime ideal of R. As a consequence R/⟨b⟩ is an integraldomain. Consider the canonical epimorphism π : R → R/⟨b⟩ and the inducedhomomorphism of rings of polynomials
π : R[x] −→ R/⟨b⟩[x]By assumption on p(x) we have
π(p(x)) = π(an)xn
On the other hand we have
π(p(x)) = π(q(x))π(r(x))
Check: since R/⟨b⟩ is an integral domain we must have
π(q(x)) = π(ck)xk, π(r(x)) = π(dl)x
l
In particular π(c0) = π(d0) = 0, so b | c0 and b | d0. On the other handa0 = c0d0, so b
2 | a0 which contradicts the assumption on a0.
38.13 Example. If p ∈ Z is a prime number then q(x) = xn−p is an irreduciblepolynomial in Z[x].
Note: by (37.10) q(x) is also irreducible in Q[x]. This shows in particular thatq(x) has no roots in Q, and so that n
√p is an irrational number for all primes p
and all n > 1.
160
38.14 Proposition. Let R be an integral domain and let c ∈ R. A polynomialp(x) =
∑ni=0 aix
i is irreducible iff the polynomial p(x − c) =∑n
i=0(x − c)i isirreducible.
Proof. It is enough to notice that the map
f : R[x]→ R[x], f(p(x)) = p(x− c)
is an isomorphism of rings.
38.15 Example. Let p ∈ Z be a prime number, and let
q(x) = xp−1 + xp−2 + . . .+ x+ 1
We will show that q(x) is irreducible in Z[x]. We have
q(x) =xp − 1
x− 1
This gives
q(x+ 1) =(x+ 1)p − 1
(x+ 1)− 1
=(x+ 1)p − 1
x
= xp−1 +
(p
1
)xp−2 +
(p
2
)xp−3 + . . .+
(p
p− 1
)
Since p |(pk
)for k = 1, . . . , p − 1 and p2 ∤
(pp−1
)the polynomial q(x + 1) is
irreducible in Z[x], and so q(x) is also irreducible.
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39 Modules
39.1 Definition. Let R be a (possibly non-commutative) ring. A left R-moduleis an abelian group M together with a map
R×M →M, (r,m) 7→ rm
satisfying the following conditions:
1) r(m+ n) = rm+ rn
2) (r + s)m = rm+ sm
3) (rs)m = r(sm)
4) If R is a ring with identity 1 ∈ R then 1m = m for all m ∈M .
A right R-module is defined analogously.
39.2 Definition. If M,N are left R-modules then a map
f : M −→ N
is a left R-modules homomorphism if f is a homomorphism of abelian groupsand f(rm) = rf(m) for all r ∈ R, m ∈M .
39.3 Note. Left R-modules and their homomorphisms form a category R-Mod.Analogously, right R-modules form a category Mod-R.
39.4 Examples.
1) If I is a left ideal of R then I is a left module of R. In particular R is aleft R-module.
2) If F is a field then left (or right ) F-modules are vector spaces over F, andhomomorphisms of F-modules are F-linear maps.
162
3) The category of left (or right) Z-modules is isomorphic to the category ofabelian groups: if G is an abelian group then G has a natural Z-modulestructure such that for n ∈ Z and g ∈ G we have
ng =
n times︷ ︸︸ ︷g + · · ·+ g for n > 0
0 for n = 0
(−g) + · · ·+ (−g)︸ ︷︷ ︸|n| times
for n < 0
4) Let G be an abelian group, and let R = Hom(G,G) be the ring of homo-morphisms of G (with the usual addition of homomorphims and multipli-cation given by composition of homomorphisms). We have a map
R×G→ G, φ · g = φ(g)
Check: this defines a left R-module structure on G.
Note: the multiplication
G×R→ G, g · φ = φ(g)
does not define a right module structure on G. Indeed, for φ, ψ ∈ R wehave:
(g · ψ) · φ = (ψ(g)) · φ = φ(ψ(g))
One the other hand g · (ψ · φ) = ψ(φ(g)). Since in general ψ(φ(g)) =φ(ψ(g)) we get that
(g · ψ) · φ = g · (ψ · φ)
39.5 Note. For a ring R define a ring Rop as follows:
• Rop = R as abelian group
• r ·Rop s := sr
163
We have:(left R-modules) = (right Rop-modules)
(check!). In particular, if R is a commutative ring then R = Rop, and so
(left R-modules) = (right R-modules)
Note. From now on by an R-module we will understand a left R-module.
164
40 Basic operations on modules
40.1 Definition. If M in an R-module then a submodule of M is an additivesubgroup N ⊆M such that if r ∈ R and n ∈ N then rn ∈ N .
40.2 Note. If f : M → N is a homomorphism of R-modules then Ker(f) :=f−1(0) is a submodule of M and Im(f) := f(M) is a submodule of N .
40.3 Definition. If M is an R-module and S is a subset of M then the modulegenerated by S is the submodule ⟨S⟩ ⊆M that is the smallest submodule of Mcontaining S.
If ⟨S⟩ = M then we say that the set S generates M . A module M is finitelygenerated if M = ⟨S⟩ for some finite set S.
40.4 Note. If M is an R-module and S ⊆M then
⟨S⟩ = {r1m1 + . . .+ rkmk | ri ∈ R, mi ∈ S}
40.5 Definition. If M is an R-module and N ⊆ M is a submodule then thequotient module M/N is the quotient abelian group with multiplication definedby
r(m+N) := rm+N
for r ∈ R, m+N ∈M/N .
40.6 First Isomorphism Theorem. If f : M → N is a homomorphism ofR-modules that is onto then
M/Ker(f) ∼= N
165
Proof. Similar to the proof of Theorem 6.1 for groups.
40.7 Definition. If {Mi}i∈I is a family of R-modules then the direct productof {Mi}i∈I is the module∏
i∈I
Mi = {(mi)i∈I | mi ∈Mi }
with addition and multiplication by R defined coordinatewise.
The direct sum of {Mi}i∈I is the submodule⊕
i∈IMi of∏
i∈IMi given by⊕i∈I
Mi := {(mi)i∈I | mi = 0 for finitely many i only }
40.8 Note. Recall the notions of categorical products and copruducts (Section12). Check:∏
i∈IMi is the categorical product of the family {Mi}i∈I in the category R-Mod.⊕i∈IMi is the categorical coproduct of {Mi}i∈I in the category R-Mod.
166
41 Free modules and vector spaces
41.1 Definition. Let M be an R-module. A set S ⊆M is linearly independentif for any distinct m1, . . . ,mk ∈ S we have
r1m1 + . . .+ rkmk = 0
only if r1 = . . . = rk = 0.
41.2 Definition. Let M be an R-module. A set B ⊆ M is a basis of M if Bis linearly independent and B generates M .
41.3 Definition. An R-module M is a free module if M has a basis.
41.4 Theorem. Let R be a ring with identity 1 = 0 and let F be an R-module.The following conditions are equivalent.
1) F is a free module.
2) F ∼=⊕
i∈I R for some set I.
3) There is a non-empty subset B ⊆ F satisfying the following universalproperty. For any R-module M and any map of sets f : B → M thereis a unique R-module homomorphism f : F → N such that the followingdiagram commutes:
Bf //
i
��
M
F
f
>>
Here i : B ↪→ F is the inclusion map.
Proof. Exercise.
167
41.5 Note. Let R be a ring with a an identity and let U : R-Mod→ Set be theforgetful functor. Check: U has a left adjoint functor
FRMod : Set→ R-Mod
One can show that an R-module M is free iff M ∼= FRMod(S) for some set S(exercise).
41.6 Note. We have:
(free Z-modules) = (free abelian groups)
41.7 Theorem. If R is a division ring then every R-module is free.
Proof. Let M be an R-module. It is enough to show that M has a basis.
Let S be the set of all linearly independent subsets of M ordered with respectto inclusion of subsets.
Claim 1. S has a maximal element.
Indeed, by Zorn’s Lemma (29.10) it is enough to show that every chain in S hasan upper bound. Let then
T = {Bi}i∈Ibe a chain in S. Take B :=
⋃i∈I Bi. We have Bi ⊆ B for all i ∈ I, so it suffices
check that B is a linearly independent. Let then b1, . . . , bk ∈ B, and assumethat
r1b1 + . . .+ rkbk = 0
We need to show that r1 = . . . = rk = 0.
We have b1 ∈ Bi1 , . . . , bk ∈ Bik for some i1, . . . , ik ∈ I. Since {Bi}i∈I is a chainwe can assume that
Bi1 ⊆ Bi2 ⊆ . . . ⊆ Bik
168
As a consequence b1, . . ., bk ∈ Bik , and since Bik is a linearly independent setwe get that r1 = . . . = rk = 0.
Claim 2. If B is a maximal element in S then B is a basis of M .
Indeed, by the definition of S the set B is linearly independent so it is enough toshow that ⟨B⟩ =M . Assume that this is not true, and let m ∈M − ⟨B⟩. Takethe set
B′ = B ∪ {m}Notice that the set B′ is linearly independent. To see this, assume that for someb1, . . . , bk ∈ B and r1, . . . rk, s ∈ R we have
r1b1 + . . .+ rkbk + sm = 0
If s = 0 then s is a unit (since R is a division ring) and so
m = (−s−1r1)b1 + . . .+ (−s−1rk)bk
This is however impossible since m ∈ ⟨B⟩. Therefore s = 0, and so
r1b1 + . . .+ rkbk = 0
Linear independence of B gives then s = r1 = . . . = rk = 0
As a consequence we get that B′ ∈ S and B ⊊ B′. This is however a contra-diction since B is a maximal element of S.
41.8 Note. Let R be a division ring and let M be an R-module. By a similarargument as in the proof of Theorem 41.7 we can show that:
1) if V ⊆ M is a linearly independent set then there is a basis B of M suchthat V ⊆ B;
2) if V ⊆ M is a set generating M then then there is a basis B of M suchthat B ⊆ V .
169
41.9 Note.
1) For a general ring R it is not true that a linearly independent subset V ofa free R-module F can be always extended to a basis. Take e.g.
R = Z, F = Z, V = {2}
2) It is also not true in general that if V is a set generating a free R-modulethen V contains a basis of F . Take e.g.
R = Z, F = Z, V = {2, 3}
170
42 Invariant basis number
42.1 Definition. A ring R has the invariant basis number (IBN) property if forany free R-module F and for any bases two B, B′ of F we have |B| = |B′|.
42.2 Definition. If a ring R has IBN then for a free R-module F the rank ofF is the cardinality of a basis of F .
42.3 Example. Since free Z-modules correspond to free abelian groups byProposition 13.3 the ring of integers Z has IBN.
42.4 Notation. For a ring R and n > 0 denote Rn :=⊕n
i=1R.
42.5 Example. Let F be a field and let V be an F-vector space with an infinite,countable basis. Let R be the ring of all linear maps V → V :
R = HomF(V, V )
with pointwise addition and with multiplication given by composition. We have
Rn ∼= Rm
for every m,n ≥ 0 (exercise). Thus R does not have IBN.
42.6 Theorem. Let R be a ring with identity and let F be a free R-module. IfF has an infinite basis B then for any other basis B′ of F we have |B| = |B′|.
42.7 Corollary. Let R be a ring with identity. The following conditions areequivalent.
171
1) R has IBN
2) If F is a free R module with two finite bases B and B′ then |B| = |B′|.
3) For any m,n > 0 if Rm ∼= Rn then m = n.
Proof. Follows directly from Theorem 42.6.
Proof of Theorem 42.6. Let F be a free R module with an infinite basis B. LetB′ be any other basis of F .
Claim 1. The basis B′ is infinite.
Indeed, assume that B′ is finite. Since F = ⟨B⟩ thus every element of B′ is alinear combination of a finite number of elements of B and so B′ ⊆ {b1, . . . , bn}where {b1, . . . , bn} is some finite subset of B. This gives
⟨b1, . . . , bn⟩ ⊇ ⟨B′⟩ = F
so ⟨b1, . . . , bn⟩ = F . Since B is an infinite set there is b ∈ B such that b ∈{b1, . . . , bn}. On the other hand b ∈ F = ⟨b1, . . . , bn⟩. This is a contradictionsince B is a linearly independent set.
Next, assume that B, B′ are two infinite bases of F . We can also assume that|B′| ≤ |B|.
Claim 2. Let T = {b′1, . . . , b′k} be a finite subset of B′ and let
BT := {b ∈ B | b ∈ ⟨T ⟩}
Then BT is a finite subset of B.
Indeed, each b′i is a linear combination of a finite number of elements of B andso T ⊆ ⟨b1, . . . , bn⟩ where {b1, . . . , bn} is some finite subset of B. This gives
BT ⊆ ⟨T ⟩ ⊆ ⟨b1, . . . , bn⟩
172
By linear independence of B we must then have BT ⊆ {b1, . . . , bn}
Claim 3. |B| ≤ |B′|.
Indeed, let SB′ be the set of all finite subsets of B′. Note that since B′ is aninfinite set we have |SB′ | = |B′|.
We have a map of setsf : B → SB′
such that f(b) = {b′1, . . . , b′k} if we have
b = r1b′1 + · · ·+ rkb
′k
for some non-zero elements r1, . . . , rk ∈ R. Since B′ is a basis of F this map iswell defined.
Notice that for T ∈ SB′ we have
b ∈ f−1(T ) iff b ∈ ⟨T ⟩
and so by Claim 2 the set f−1(T ) is finite for all T ∈ SB′ . As a consequence weobtain
|B| = |⋃
T∈SB′
f−1(T )| ≤ |⋃
T∈SB′
N| = |SB′| · ℵ0 = |B′| · ℵ0
Since B′ is an infinite set we have |B′| · ℵ0 = |B′|, and so |B| ≤ |B′|.
Since by assumption we had |B| ≤ |B′| and by Claim 3 we have |B| ≤ |B′| weobtain that |B| = |B′|.
42.8 Theorem. If R is a division ring then R has IBN.
173
Proof. By Corollary 42.7 it is enough to show that if F is a free R-module withtwo finite bases B = {b1, . . . , bn} and B′ = {b′1, . . . , b′m} then n = m.
We will argue by induction with respect to n
Assume that n = 1, and so B = {b1}. If B′ = {b′1, . . . , b′m} for some m > 1then we have
b′1 = r1b1 and b′2 = r2b1
for some r1, r2 ∈ R, r1, r2 = 0. Therefore r−11 b′1 − r−1
2 b′2 = 0 which contradictsthe assumption that B′ is a inearly independent set. As a consequence we musthave m = 1.
Next, assume that n ≥ 1 is a number such that if a free R-module has abasis consisting of n elements then every other basis of that module also has nelements.
Let F be a free R-module with a basis B = {b1, . . . , bn+1} consisting of n + 1elements and let B′ = {b′1, . . . , b′m} be another basis of F . Since ⟨B′⟩ = F wehave
bn+1 = r1b′1 + . . .+ rmb
′m
for some r1, . . . , rm ∈ R. Also, since bn+1 = 0 we have ri = 0 for some i. Wecan assume that rm = 0. Let B′′ := {b′1, . . . , b′m−1, bn+1}. Check: B′′ is a basisof F .
Take the canonical epimorphism
π : F → F/⟨bn+1⟩
Check: since F is a free module with basis B := {b1, . . . , bn, bn+1}, thusF/⟨bn+1⟩ is a free module with basis {π(b1), . . . , π(bn)}. On the other hand,since F has a basis B′′ := {b′1, . . . , b′m−1, bn+1} therefore {π(b′1), . . . , π(b′m−1)}is a basis of F/⟨bn+1⟩.
By the inductive assumption we obtain than n = m− 1, and so n+ 1 = m
174
42.9 Note. Let I be an ideal of R and let M be an R-module. Define:
IM := {rm | r ∈ I, m ∈M}
Check:
1) IM a submodule of M .
2) M/IM has a structure of a R/I-module with the multiplication given by
(r + I)(m+ IM) = rm+ IM
42.10 Theorem. Let R be a ring with identity and let I = R be an ideal of R.If R/I has IBN then R also has IBN.
Proof. Let F be a free R-module and let B = {b1, . . . , bn} be a basis of F .Check: F/IF is a free R/I-module with basis {b1 + IF, . . . , bn + IF}.
Since R/I has IBN any basis of F/IF has n elements. As a consequence anybasis of F also has n elements.
42.11 Corollary. If R is a commutative ring with identity 1 = 0 then R hasIBN.
Proof. Let I be a maximal ideal in R. Then R/I is a field and we have thecanonical homomorphism
π : R→ R/I
By Theorem 42.8 R/I has IBN, so by Theorem 42.10 R also has IBN.
42.12 Note. Corollary 42.11 can be generalized as follows. If
f : R→ S
is an epimorphism of rings of identity such that S is a division ring then R hasIBN.
175
43 Projective modules
43.1 Note. If F is a free R-module and P ⊆ F is a submodule then P neednot be free even if P is a direct summand of F .
Take e.g. R = Z/6Z. Notice that Z/2Z and Z/3Z are Z/6Z-modules and wehave an isomorphism of Z/6Z-modules:
Z/6Z ∼= Z/2Z⊕ Z/3Z
Thus Z/2Z and Z/3Z are non-free modules isomorphic to direct summands ofthe free module Z/6Z.
43.2 Definition. An R-module P is a projective module if there exists an R-module Q such that P ⊕Q is a free R-module.
43.3 Examples.
1) If R is a ring with identity then every free R-module is projective.
2) Z/2Z and Z/3Z are non-free projective Z/6Z-modules.
43.4 Definition. Let
. . . −→Mifi−→Mi+1
fi+1−→Mi+2 −→ . . .
be a sequence of R-modules and R-module homomorphisms. This sequence isexact if Im(fi) = Ker(fi+1) for all i.
43.5 Definition. A short exact sequence is an exact sequence of R-modules theform
0 −→ Nf−→M
g−→ K −→ 0
(where 0 is the trivial R-module).
176
43.6 Note.
1) A sequence 0→ Nf−→M
g−→ K → 0 is short exact iff
• f is a monomorphism
• g is an epimorphism
• Im(f) = Ker(g).
2) If M ′ is a submodule of M then we have a short exact sequence
0 −→M ′ −→M −→M/M ′ −→ 0
Morever, up to an isomorphism, every short exact sequence is of this form:
0 // Nf //
∼=��
Mg //
=
��
K //
∼=��
0
0 // Ker(g) //M //M/Ker(g) // 0
43.7 Definition. A short exact sequence
0 −→ Nf−→M
g−→ K −→ 0
is split exact if there is an isomorphism φ : M∼=−→ N⊕K such that the following
diagram commutes:
0 // Nf //
=��
Mg //
φ ∼=��
K //
=��
0
0 // N // N ⊕K // K // 0
43.8 Proposition. Let R be a ring and let 0 → Nf−→ M
g−→ K → 0 be ashort exact sequence of R-modules. The following conditions are equivalent.
1) The sequence is split exact.
177
2) There exists a homomorphism h : K →M such that gh = idK .
3) There exists a homomorphism k : M → N such that kf = idN
Proof. Exercise.
43.9 Theorem. Let R be a ring with identity and let P be an R-module. Thefollowing conditions are equivalent.
1) P is a projective module.
2) For any homomorphism f : P → N and an epimorphism g : M → N thereis a homomorphism h : P →M such that the following diagram commutes:
P
h
~~
f
��M
g // N
3) Every short exact sequence 0→ Nf−→M
g−→ P → 0 splits.
Proof.
(1) ⇒ (2) Let Q be a module such that P ⊕ Q is a free module, and letB = {bi}i∈I be a basis of P ⊕Q. Since g is an epimorphism for every i ∈ I wecan find mi ∈M such that g(mi) = f(bi). Define
h : P ⊕Q −→M
by
h
(∑i
ribi
):=∑i
rimi
178
Check: since B is a basis of P ⊕ Q the map h is a well defined R-modulehomomorphism and gh = f . Then we can take h = h|P .
(2) ⇒ (3) We have a diagram
P
idP
��M
g // P
Since g is an epimorphism there is h : P → M such that gh = idP . Therefore
by (43.8) the sequence 0→ Nf−→M
g−→ P → 0 splits.
(3) ⇒ (1) We have the canonical epimorphism of R-modules:
f :⊕p∈P
R→ P
This gives a short exact sequence
0 −→ Ker(f)−→⊕p∈P
Rf−→ P −→ 0
By assumption on P this sequence splits. so we obtain
P ⊕Ker(f) ∼=⊕p∈P
R
and thus P is a projective module.
43.10 Corollary. If R is a ring with identity, P is a projective R-module andf : M → P is an epimorphism of R-modules then M ∼= P ⊕Ker(f).
Proof. We have a short exact sequence
0 −→ Ker(f)−→M f−→ P −→ 0
which splits by Theorem 43.9.
179
44 Projective modules over PIDs
44.1 Theorem. If R is a PID, F is a free R-module of a finite rank, andM ⊆ Fis a submodule then M is a free module and rankM ≤ rankF .
44.2 Corollary. If R is a PID then every finitely generated projective R-moduleis free.
Proof. If P is a finitely generated projective R-module then we have an epimor-phism f : Rn → P for some n > 0. By Corollary 43.10 we have an isomorphism
P ⊕Ker(f) ∼= Rn
Therefore we can identify P with a submodule of Rn, and thus by Theorem 44.1P is a free module.
44.3 Note. Theorem 44.1 is true also for infinitely generated free modules overPIDs. As a consequence Corollary 44.2 is true for all (non necessarily finitellygenerated) projective modules over PIDs.
Proof of Theorem 44.1 (compare with the proof of Theorem 13.6).
We can assume that F = Rn.
We want to show: if M ⊆ Rn then M is a free R-module and rankM ≤ n.
Induction with respect to n:
If n = 1 thenM ⊆ R, soM is an ideal of R. Since R is a PID we haveM = ⟨a⟩for some a ∈ R. If a = 0 then M = {0} is a free module of rank 0. Otherwisewe have an isomorphism of R-modules
f : R∼=−→M, f(r) = ra
180
and so M is a free module of rank 1.
Next, assume that for some n every submodule of Rn is a free R-module of rank≤ n, and let M ⊆ Rn+1. Take the homomorphism of R-modules
f : Rn+1 → R, f(r1, . . . , rn+1) = rn+1
We have:Ker(f) = {(r1, . . . , rn, 0) | ri ∈ R} ∼= Rn
We have an epimorphism:
f |M : M → Im(f |M)
Since Im(f |M) ⊆ R, thus Im(f |M) is a free R-module, and so by Corollary 43.10we have
M ∼= Im(f |M)⊕Ker(f |M)
We also have:Ker(f |M) = Ker(f) ∩M
It follows that that Ker(f |M) is a submodule of Ker(f), and since Ker(f) is afree R-module of rank n by the inductive assumption we get that Ker(f |M) is afree R-module of rank ≤ n. Therefore
M ∼= Im(f |M)︸ ︷︷ ︸free
rank ≤ 1
⊕ Ker(f |M)︸ ︷︷ ︸free
rank ≤ n
and so M is a free R-module of rank ≤ n+ 1.
181
45 The Grothendieck group
Recall. If R is a ring with IBN and F is a free, finitely generated R-module then
rankF = number of elements of a basis of F
Goal. Extend the notion of rank to finitely generated projective modules.
Idea.
1) Rank should be additive: rank(P ⊕Q) = rankP + rankQ.
2) Rank of a module need not be an integer. Each ring determines a groupK0(R) such that for each finitely generated projective module rank of P is anelement [P ] ∈ K0(R).
Recall. A commutative monoid is a set M together with addition
M ×M →M, (x, y) 7→ x+ y
and with a trivial element 0 ∈ M such that the addition is associative, commu-tative and 0 + x = x for all x ∈M .
45.1 Example. Let ProjfgR be the set of isomorphism classes of finitely generatedprojective R-modules. For a projective finitely generated R-module P denote
[P ] = the isomorphism class of P
The set ProjfgR is a commutative monoid with addition given by
[P ] + [Q] := [P ⊕Q]
The identity element in ProjfgR is [0], the isomorphism class of the zero module.
182
45.2 Theorem. Let M be a commutative monoid. There exists an abeliangroup Gr(M) and a homomorphism of monoids
αM : M → Gr(M)
that satisfies the following universal property. If G is any abelian group andf : M → G is a homomorphism of monoids then there exists a unique homomor-phism of groups f : Gr(M)→ G such that the following diagram commutes:
Mf //
αM
��
G
Gr(M)
f
==
Moreover, such group Gr(M) is unique up to isomorphism.
45.3 Note. Let CMono denote the category of commutative monoids. We havethe forgetful functor
U : Ab −→ CMono
Theorem 45.2 is equivalent to the statement that this functor has a left adjoint
Gr: CMono→ Ab, M 7→ Gr(M)
45.4 Definition. LetM be a commutative monoid. The group Gr(M) is calledthe group completion or the Grothendieck group of the monoid M .
Proof of Theorem 45.2.
Construction of Gr(M). Let M be a commutative monoid. Define
Gr(M) :=M ×M/ ∼
183
where
(x, y) ∼ (x′, y′) iff x+ y′ + t = x′ + y + t for some t ∈M
Check: ∼ is an equivalence relation on M ×M . Notation:
[x, y] := the equivalence class of (x, y)
(Intuitively: [x, y] = x− y)
Note: for any x ∈M we have [x, x] = [0, 0] since x+ 0 = 0 + x.
Addition in Gr(M):
[x, y] + [x′, y′] = [x+ x′, y + y′]
Check: this operation is well defined, it is associative, and it has [0, 0] as theidentity element.
Additive inverses in Gr(M):
−[x, y] = [y, x]
Indeed: [x, y] + [y, x] = [x+ y, y + x] = [0, 0]
Construction of the homomorphism αM : M → Gr(M).
DefineαM : M → Gr(M), x 7→ [x, 0]
The universal property of Gr(M).
Let G be an abelian group and let f : M → G be a homomorphism of commu-tative monoids. Define
f : Gr(M)→ G, f([x, y]) := f(x)− f(y)Check:
184
1) f is a well defined group homomorphism.
2) fαM = f
3) f is the only homomorphism Gr(M)→ G satisfying condition 2).
Uniqueness of Gr(M) follows from the universal property.
45.5 Examples.
1) Gr(N) ∼= Z
2) Let M = N ∪ {∞} with n+∞ =∞ for all n ∈ M . Then Gr(M) is thetrivial group. Indeed, for any m,n ∈M we have
[m,n] = [∞,∞]
since m+∞ =∞+ n.
3) If G is an abelian group then Gr(G) ∼= G.
45.6 Definition. If R is a ring then K0(R) := Gr(ProjfgR )
45.7 Notation. For [P ], [Q] ∈ ProjfgR denote
[P ]− [Q] :=[P,Q] ∈ K0(R)
[P ] :=[P, 0]
−[Q] :=[0, Q]
45.8 Proposition. Let R be a ring with identity. If P , Q are finitely generatedprojective R-modules then [P ] = [Q] in K0(R) iff there exists n ≥ 0 such thatP ⊕Rn ∼= Q⊕Rn.
185
Proof.
(⇒) If P ⊕Rn ∼= Q⊕Rn then in K0(R) we have
[P ] + [Rn] = [P ⊕Rn] = [Q⊕Rn] = [Q] + [Rn]
and so [P ] = [Q]
(⇒) If [P ] = [Q] in K0(R) then
P ⊕ S ∼= Q⊕ S
for some finitely generated projective R-module S.
Exercise: If S is a finitely generated projective R-module then there is a finitelygenerated projective R-module T such that S ⊕ T ∼= Rn for some n ≥ 0.
We obtainP ⊕Rn ∼= P ⊕ S ⊕ T ∼= Q⊕ S ⊕ T ∼= Q⊕Rn
45.9 Definition. Let R be a ring with identity. We say that R-modules M , Nare stably isomorphic if M ⊕Rn ∼= N ⊕Rn for some n ≥ 0.
45.10 Definition. Let R be a ring with identity. We say an R-module M isstably free if M ⊕Rn ∼= Rm for some m,n ≥ 0.
45.11 Note. Let R be a ring with identity. We have a homomorphism φ : Z→K0(R) given by
φ(n) :=
{[Rn] for n ≥ 0
−[R−n] for n < 0
186
45.12 Proposition. Let R be a ring with identity, and let φ : Z → K0(R) bethe homomorphism as in (45.11).
1) φ is 1-1 iff R has IBN.
2) φ is an epimorphism iff every finitely generated projective R-module isstably free.
Proof.
1) By Proposition 45.8 for n ≥ 0 we have n ∈ Ker(φ) iff Rn⊕Rm ∼= 0⊕Rm forsome m ≥ 0. If R has IBN this is possible only if n = 0, and so Ker(φ) = {0}.
Conversely, assume that R does not have IBN. Then Rn ∼= Rm for some n > m.This gives Rn−m ⊕Rm ∼= 0⊕Rm, and so φ(n−m) ∈ Ker(φ).
2) (⇒) Assume that φ is an epimorphism. Then for every finitely generatedprojective R-module P we have [P ] = [Rn] for some n ≥ 0 or [P ] = −[Rn] forsome n ≥ 0.
If [P ] = [Rn] then by Proposition 45.8 we have P ⊕ Rm ∼= Rn ⊕ Rm and so Pis a stably free module.
If [P ] = −[Rn] then[0] = [P ] + [Rn] = [P ⊕Rn]
Again by Proposition 45.8 this gives 0 ⊕ Rm ∼= P ⊕ Rn ⊕ Rm, and again weobtain that P is stably free.
(⇐) The group K0(R) is generated by elements [P ] where P is a finitely gener-ated projective R-module, so it is enough to show that for any such P we have[P ] = φ(k) for some k ∈ Z.
Since P is stably free we have P ⊕Rn ∼= Rm for some n,m ≥ 0. This gives
[P ] + [Rn] = [P ⊕Rn] = [Rm]
Therefore [P ] = [Rm]− [Rn] = φ(m)− φ(n) = φ(m− n).
187
45.13 Example.
Here is an example of a stable free module that is not free. For details see:
R. G. Swan, Vector bundles and projective modules, Transactions AMS 105 (2)(1962), 264-277.
Let B be a compact, normal, topological space and let p : E → B be a realvector bundle over B. Define:
C(B) = {f : B → R | f - continuous }
C(B) is a ring (with pointwise addition and multiplication). Let Γ(p) be the setof all continuous sections of p:
Γ(p) = {s : B → E | ps = idB}
Note: Γ(p) is an C(B)-module with poinwise addition and pointwise multiplica-tion by elements of C(B).
Fact 1. The module Γ(p) is free iff p is a trivial vector bundle.
Fact 2. If p : E → B and q : E ′ → B are real vector bundles over B then wehave an isomorphism of C(B)-modules:
Γ(p⊕ q) ∼= Γ(p)⊕ Γ(q)
Upshot. If p : E → B, q : E ′ → B are bundles such that p is non-trivial, butboth q and p ⊕ q are trivial bundles then Γ(p) is stably free C(B)-module thatis not free.
Indeed, in such case we have:
Γ(p)⊕ Γ(q)︸︷︷︸free
∼= Γ(p⊕ q)︸ ︷︷ ︸free
188
Fact 3. It is possible to find vector bundles as above. Take e.g. p : TS2 → S2
to be the tangent bundle of the 2-dimensional sphere, and q : S2 ×R1 → S2 tobe the 1-dimensional trivial bundle over S2.
Note: one can also show that a C(B)-module M is finitely generated projectivemodule iff M ∼= Γ(p) for some vector bundle p : E → B.
189
46 Injective modules
Recall. If R is a ring with identity then an R-module P is projective iff one ofthe following equivalent conditions holds:
1) For any homomorphism f : P → N and an epimorphism g : M → N thereis a homomorphism h : P →M such that the following diagram commutes:
P
h
~~
f
��M
g // N
2) Every short exact sequence 0→ Nf−→M
g−→ P → 0 splits.
46.1 Proposition. Let R be a ring and let J be an R-module. The followingconditions are equivalent.
1) For any homomorphism f : N → J and an monomorphism g : M → Nthere is a homomorphism h : M → J such that the following diagramcommutes:
J
M g//
f
OO
N
h
``
2) Every short exact sequence 0→ Jf−→M
g−→ N → 0 splits.
Proof. Exercise.
190
46.2 Definition. An R-module J is an injective module if J satisfies one of theequivalent conditions of Proposition 46.1.
46.3 Theorem (Baer’s Criterion).Let R be a ring with identity and let J be an R-module. The following conditionsare equivalent.
1) J is an injective module.
2) For every left ideal I ◁ R and for every homomorphisms of R-modulesf : I → J there is a homomorphism f : R→ J such f |I = f .
Proof.1) ⇒ 2) Given a homomorphism f : I → J we have a diagram
J
I �� i //
f
OO
R
where i : I ↪→ R is the inclusion homomorphism. By the definition of an injectivemodule there is a homomorphism f : R→ J such that f = f i = f |I
2) ⇐ 1) Assume that J is an R-module satisfying 2). It is enough to showthat if M is an R-module, N is a submodule of M , and f : N → J is an R-module homomorphism then there exists a homomorphism f : M → J such thatf |N = f
Let S be a set of all pairs (K, fK) such that
(i) K is a submodule of M such that N ⊆ K ⊆M
(ii) fK : K → J is a homomorphism such that fK |N = f
191
Define partial ordering on S as follows:
(K, fK) ≤ (K ′, fK′) if K ⊆ K ′ and fK′|K = fK
Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S containsa maximal element (K0, fK0).
It will suffice to show that K0 = M . Assume, by contradiction, that K0 = M ,and let m0 ∈M −K0. Define
I := {r ∈ R | rm0 ∈ K0}
Check: I is an ideal of R and the map
g : I → J, g(r) = fK0(rm0)
is a homomorphism of R-modules. By the assumptions on J we have a homo-morphism g : R→ J such that g|I = g. Define
K0 +Rm0 := {k + rm0 | k ∈ K, r ∈ R}
Check: K0 +Rm0 is a submodule of M and the map
f ′ : K0 +Rm0 → J, f ′(k + rm0) = fK0(k) + g(r)
is a well defined homomorphism of R-modules such that f ′|N = f . This showsthat (K0 +Rm0, f
′) ∈ S. We also have
(K0, fK0) < (K0 +Rm0, f′)
This is impossible since by assumption (K0, fK0) is a maximal element in S.
46.4 Corollary. Let R be an integral domain and let K the field of fractions ofR. Then K is an injective R-module.
Proof. Let I be an ideal of R and let f : I → K be a homomorphism of R-modules. For 0 = r, s ∈ I we have
rf(s) = f(rs) = sf(r)
192
As consequence in K we have f(r)/r = f(s)/s for any 0 = r, s ∈ I. Denotethis element by a. Define
f : R→ K, f(r) := ra
Check: f is a homomorphism of R-modules and f |I = f .
By Baer’s Criterion (46.3) it follows that K is an injective R-module.
46.5 Example. Q is an injective Z-module.
46.6 Definition. Let R be an integral domain. An R-module M is divisible iffor every r ∈ R− {0} and for every m ∈M there is n ∈M such that rn = m.
46.7 Theorem. If R is a PID then an R-module J is injective iff J is divisible.
Proof. Exercise.
46.8 Example.
Since Z is a PID injective Z-modules are divisible Z-modules (i.e. divisible abeliangroups).
Exercise: an abelian group G is divisible iff G is isomorphic to a direct sum ofcopies of Q and Z(p∞) for various primes p.
46.9 Corollary. If R is a PID, J is an injective R-module and K is a submoduleof J then J/K is injective.
193
Proof. Since J is divisible thus so is J/K (check!).
46.10 Note. If R is not a PID then a quotient of an injective R-module neednot be injective.
Note. If R is a ring with identity then for any R-module M there exists anepimorphism of R-modules:
f : P −→M
where P is a projective module (take e.g. P =⊕
m∈M R).
46.11 Theorem. If R is a ring with identity then for any R-module M thereexist a monomorphism
j : M −→ J
where J is an injective R-module.
46.12 Lemma. For any abelian group G there exists a monomorphism
i : G −→ D
where H is a divisible abelian group.
Proof. We have an epimorphism f :⊕
g∈G Z→ G which gives an isomorphism
φ : G∼=−→⊕g∈G
Z/Ker(f)
Moreover, the monomorphism⊕
g∈G Z→⊕g∈GQ induces a monomorphism
ψ :⊕g∈G
Z/Ker(f) −→⊕g∈G
Q/Ker(f)
We can take D :=⊕
g∈GQ/Ker(f) and i := ψφ.
194
46.13 Note. Let G is an abelian group, let R let be a ring, and let HomZ(R,G)be the set of all homomorphisms of abelian groups φ : R→ G.
Check: HomZ(R,G) is an R-module with pointwise addition and with multipli-cation by elements of R given by
(r · φ)(s) := φ(sr)
for r, s ∈ R.
46.14 Lemma. If D is a divisible abelian group and R is a ring with identitythen HomZ(R,D) in an injective R-module.
Proof. Exercise.
Proof of Theorem 46.11. Let M be an R-module. Consider M as an abeliangroup. By Lemma 46.12 we have a monomorphism of abelian groups
i : M −→ D
where D is a divisible abelian group. Consider the induced map
i∗ : HomZ(R,M)→ HomZ(R,D), i∗(φ) = i ◦ φ
Check:
1) i∗ is a monomorphism.
2) i∗ is a homomorphism of R-modules.
Since M is an R-module we also have a map
f : M → HomZ(R,M), f(m)(r) = rm
Check:
1) f is a monomorphism.
195
2) f is a homomorphism of R-modules.
As a consequence we obtain a monomorphism of R-modules
i∗f : M −→ HomZ(R,D)
Moreover, by Lemma 46.12 HomZ(R,D) is an injective R-module.
196
47 Exact functors
47.1 Definition. A chain complex of R-modules is a sequence of R-modulesand R-homomorphisms
. . . −→Mi+1di+1−→Mi
di−→Mi−1di−1−→Mi−2 −→ . . .
such that didi+1 = 0 for all i.
47.2 Note. If M∗ = (Mi, di) is a chain complex then Im(di+1) ⊆ Ker(di).
47.3 Definition. If M∗ = (Mi, di) is a chain complex of R-modules then thei-th homology module of M∗ is the module
Hi(M∗) = Ker(di)/ Im(di+1)
Recall. A sequence
M∗ = (. . . −→Mi+1di+1−→Mi
di−→Mi−1di−1−→Mi−2 −→ . . .)
is exact if Ker(di) = Im(di+1) for all i. Therefore M∗ is exact iff Hi(M∗) = 0for all i.
47.4 Definition. Let R, S be rings. A functor F : R-Mod → S-Mod is exactif for every short exact sequence of R-modules
0 −→ Nf−→M
g−→ K −→ 0
the sequence
0 −→ F (N)F (f)−→ F (M)
F (g)−→ F (K) −→ 0
is short exact.
197
47.5 Note. If F : R-Mod→ S-Mod is a functor such that F (0) = 0 and
M∗ = (. . . −→Mi+1di+1−→Mi
di−→Mi−1 −→ . . .)
is a chain complex of R-modules then
F (M∗) = (. . . −→ F (Mi+1)F (di+1)−→ F (Mi)
F (di)−→ F (Mi−1) −→ . . .)
is a chain complex of S-modules.
Moreover, the functor F is exact iff for every chain complex M∗ we have isomor-phisms
F (Hi(M∗)) ∼= Hi(F (M∗))
for all i.
47.6 Note. For a ring R and R-modules L,M let HomR(L,M) be the set of allR-module homomorphisms φ : L→M . Notice that HomR(L,M) is an abeliangroup (with respect to the pointwise addition of homomorphisms). Moreover,for any homomorphism of R-modules f : M → N the map
f∗ : HomR(L,M)→ HomR(L,N), f∗(φ) = f ◦ φ
is a homomorphism of abelian groups. This defines a functor
HomR(L,−) : R-Mod −→ Ab
This functor is in general not exact. Take e.g. R = Z, L = Z/2Z. We have ashort exact sequence of abelian groups:
0→ Z ·2−→ Z −→ Z/2Z→ 0
On the other hand the sequence
0→ HomZ(Z/2Z,Z) −→ HomZ(Z/2Z,Z) −→ HomZ(Z/2Z,Z/2Z)→ 0
is not exact since HomZ(Z/2Z,Z) ∼= 0 and HomZ(Z/2Z,Z/2Z) ∼= Z/2Z.
198
47.7 Proposition. Let R be a ring and let L be an R-module. If
0 −→ Nf−→M
g−→ K −→ 0
is a short exact sequence of R-modules then
0 −→ HomR(L,N)f∗−→ HomR(L,M)
g∗−→ HomR(L,K)
is an exact sequence of abelian groups.
Proof. Exercise.
47.8 Theorem. Let R be a ring with identity and let P be an R-module. Thefunctor HomR(P,−) is exact iff P is a projective module.
Proof. By Proposition 47.7 is suffices to show that P is a projective module ifffor every epimorphism of R-modules g : M → K the map
g∗ : HomR(P,M) −→ HomR(P,K)
is an epimorphism. This follows directly from Theorem 43.9.
47.9 Definition. Let C,D be categories. A contravariant functor F : C → D
consists of
1) an assignmentOb(C)→ Ob(D), c 7→ F (c)
2) for every c, c′ ∈ C a function
HomC(c, c′)→ HomD(F (c
′), F (c)), f 7→ F (f)
such that F (idc) = idF (c) and F (gf) = F (f)F (g).
199
47.10 Example. Let R be a ring and let L be an R-module. For any homo-morphism of R-modules f : M → N we have a map
f ∗ : HomR(N,L) −→ HomR(M,L), f ∗(φ) = φ ◦ f
Moreover, f ∗ is a homomorphism of abelian groups.
This defines a contravariant functor
HomR(−, L) : R-Mod −→ Ab
47.11 Note. The functor HomR(−, L) is in general not exact. Take e.g. R = Z,L = Z. We have a short exact sequence of abelian groups:
0→ Z ·2−→ Z −→ Z/2Z→ 0
On the other hand the sequence
0→ HomZ(Z/2Z,Z) −→ HomZ(Z,Z) −→ HomZ(Z,Z)→ 0
is not exact.
47.12 Proposition. Let R be a ring and let L be an R-module. If
0 −→ Nf−→M
g−→ K −→ 0
is a short exact sequence of R-modules then
0 −→ HomR(K,L)g∗−→ Hom(M,L)
f∗−→ Hom(M,L)
is an exact sequence of abelian groups.
Proof. Exercise.
200
47.13 Theorem. Let R be a ring with identity and let J be an R-module. Thefunctor HomR(−, J) is exact iff J is an injective module.
Proof. By Proposition 47.12 is suffices to show that J is an injective module ifffor every monomorphism of R-modules f : N →M the map
f ∗ : HomR(M,J) −→ HomR(N, J)
is an epimorphism. This follows directly from Proposition 46.1.
47.14 Note. Let
M∗ = (. . . −→Midi−→Mi−1 −→ . . .)
be a chain complex of R-modules. For any R-module L we have the inducedchain complex of abelian groups
HomR(M∗, L) = (. . . −→ HomR(Mi−1, L)d∗i−→ HomR(Mi, L) −→ . . .)
Homology groups of the complex HomR(M∗, L) are called cohomology groupsof M∗ with coefficients in L. We denote:
H i(M∗, L) := Hi(HomR(M∗, L))
If L is an injective module then by Theorem 47.13 the functor HomR(−, L) isexact. By (47.5) in such case we have
H i(M∗, L) ∼= HomR(Hi(M∗), L)
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48 Tensor products
Note. For now all rings are commutative rings with identity.
48.1 Definition. Let M,N,K be R-modules. A function
f : M ×N −→ K
is bilinear if
1) f(m1 +m2, n) = f(m1, n) + f(m2, n)f(m,n1 + n2) = f(m,n1) + f(m,n2)
2) f(rm, n) = rf(m,n) = f(m, rn)
48.2 Theorem. Let M , N be R-modules. There exists an R-module M ⊗R Nand a bilinear map
η : M ×N →M ⊗R Nthat satisfies the following universal property. For any R-module K and a bilin-ear map f : M × N → K there exists a unique homomorphism of R-modulesf : M ⊗R N → K such that the following diagram commutes:
M ×N f //
η
��
K
M ⊗R N
f
<<
Moreover, such R-module M ⊗R N is unique up to isomorphism.
48.3 Definition. If M,N are R-modules then the module M ⊗R N is calledthe tensor product of M and N .
202
Proof of Theorem 48.2.
Construction of M ⊗R N .
Let F (M ×N) denote the free R-module with basis {(m,n) | m ∈M,n ∈ N}Note: every element of F (M ×N) is of the form∑
i
ri(mi, ni)
for some ri ∈ R.
Let S be the submodule of F (M×N) generated by all elements of the followingforms:
(i) (m1 +m2, n)− (m1, n)− (m2, n)(m,n1 + n2)− (m,n1)− (m,n2)
(ii) (rm, n)− r(m,n)(m, rn)− r(m,n)
Define M ⊗R N := F (M ×N)/S
Denote:m⊗ n = (the equvalence class of (m,n) in M ⊗R N)
Note:
1) Every element of M ⊗R N can be written (non-uniquely!) as a linearcombination ∑
i
ri(mi ⊗ ni)
2) In M ⊗R N we have
(i) (m1 +m2)⊗ n = m1 ⊗ n+m2 ⊗ nm⊗ (n1 + n2) = m⊗ n1 +m⊗ n2
(ii) (rm)⊗ n = r(m⊗ n) = m⊗ (rn)
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Construction of the map η.
Define:η : M ×N →M ⊗R N, η(m,n) = m⊗ n
By (i)-(ii) above η is bilinear.
The universal property of M ⊗R N .
If f : M ×N → K is a bilinear map define f : M ⊗R N → K by
f
(∑i
ri(mi ⊗ ni))
=∑i
rif(mi, ni)
Check:
1) f is a well defined R-module homomorphism
2) fη = f
3) f is the only homomorphism M ⊗R N → K satisfying condition 2).
Uniqueness of M ⊗R N up to isomorphism follows directly from the universalproperty.
48.4 Example. Let m,n > 0, gcd(m,n) = 1
Z/mZ⊗Z Z/nZ = ?
Note: if f : Z/mZ× Z/nZ→ K is a bilinear map then
f(k, l) = f(k · 1, l · 1) = (kl) · f(1, 1)
Therefore the map f is determined by the value of f(1, 1).
204
Moreover, since gcd(m,n) = 1 we have am+ bn = 1 for some m,n ∈ Z. Thisgives:
f(1, 1) = f(1, (am+ bn) · 1)= f(1, (am) · 1) + f(1, (bn) · 1)= f((am) · 1, 1) + f(1, (bn) · 1)= f(0, 1) + f(1, 0)
= f(0 · 1, 1) + f(1, 0 · 1)= 0 · f(1, 1) + 0 · f(1, 1)= 0
Therefore for any bilinear map f we have f(1, 1) = 0, and so f = 0. It followsthat
Z/mZ⊗Z Z/nZ = 0
since the zero module satisfies the universal property:
Z/mZ× Z/nZ f=0 //
��
K
0
f=0
::
48.5 Proposition. If M is an R-module then
R⊗RM ∼= M
Proof. We have a bilinear map
η : R×M →M, η(r,m) = rm
(note: this works only if R is commutative!).
Moreover, if f : R×M → K is any bilinear map then define
f : M → K, f(m) = f(1,m)
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Check: f is the unique R-module homomorphism such that fη = f .
By the universal property of tensor products this gives M ∼= R⊗RM .
48.6 Proposition.
1) If M , N are R-modules then we have an isomorphism
φ : M ⊗R N → N ⊗RM, φ(m⊗ n) = n⊗m
2) If M , N , K are R-modules then
M ⊗R (N ⊗R K) ∼= (M ⊗R N)⊗R K
3) If Mα, N are R-modules then
(⊕α
Mα)⊗R N ∼=⊕α
(Mα ⊗R N)
Proof. Exercise.
206
49 Tensor products of homomorphisms
Let f : M → K, g : N → L be homomorphisms of R-modules. Notice that wehave a bilinear map
f × g : N : M ×N → K ⊗R L, f × g(m,n) = f(m)⊗ g(n)
By the unniversal property of tensor products f × g induces a unique homomor-phism of R-modules
f ⊗ g : M ⊗R N → K ⊗R L, f ⊗ g(m⊗ n) = f(m)⊗ g(n)
49.1 Proposition.
1) Tensor product preserves composition of homomorphisms: if we have ho-momorphisms of R-modules
Mf−→ K
f ′−→ K ′
Ng−→ L
g′−→ L′
then(f ′ ◦ f)⊗ (g′ ◦ g) = (f ′ ⊗ g′) ◦ (f ⊗ g)
2) Tensor product preserves identity maps:
idM ⊗ idN = idM⊗RN
Proof. Exercise.
49.2 Corollary. If f : M → K, g : N → L are isomorphisms of R-modules then
f ⊗ g : M ⊗R N → K ⊗R L
is also an isomorphism.
207
Proof. Exercise.
49.3 Proposition. If F , F ′ are free R-modules then F ⊗R F ′ is also a freemodule.
Proof. We haveF ∼=
⊕i∈I
R, F ′ ∼=⊕j∈J
R
Therefore
F ⊗R F ′ ∼= (⊕i∈I
R)⊗R (⊕j∈J
R)
∼=⊕i∈I
⊕j∈J
(R⊕R R)
∼=⊕
(i,j)∈I×J
R
49.4 Note. If F , F ′ are free modules, {bi}i∈I is a basis of F and {b′j}j∈J is abasis of F ′ then {bi ⊗ bj}(i,j)∈I×J is a basis of F ⊗R F ′ (exercise).
49.5 Corollary. If P , Q are projective R-modules then P⊗RQ is also projective.
Proof. Since P , Q are projective R-modules there exists modules P ′, Q′ suchthat P ⊕ P ′, Q⊕Q′ are free modules. We have
(P ⊕ P ′)⊗R (Q⊕Q′)︸ ︷︷ ︸free
∼= (P ⊗R Q)⊕ (P ⊗R Q′)⊕ (P ′ ⊗R Q)⊕ (P ′ ⊗R Q′)
Therefore P ⊗R Q is a direct summand of a free module, and so it is projective.
208
50 Tensor products and adjoint functors
Recall. Let R be a commutative ring. If M , N are R-modules then
HomR(M,N) = (the set of all R-module homomorphisms M → N)
Note.
1) HomR(M,N) is an abelian group with addition given by
(f + g)(m) := f(m) + g(m)
2) HomR(M,N) is an R-module with multiplication by elements of R givenby
(rf)(m) := r · f(m)
(Note: rf need not be an R-module homomorphism if R is a non-commu-tative ring.)
3) If g : N → N ′ is a homomorphism of R-modules then the map
g∗ : HomR(M,N)→ HomR(M,N ′), g∗(f) = g ◦ f
is a homomorphism of R-modules.
Upshot. For any R-module M we have a functor
Hom(M,−) : R-Mod −→ R-Mod
50.1 Proposition. For any R-module M the functor
−⊗RM : R-Mod −→ R-Mod
is left adjoint to the functor HomR(M,−).
209
50.2 Note. Proposition 50.1 says that for any R-modules N , K we have anatural bijection of sets
HomR(N ⊗RM,K)∼=−→ HomR(N,HomR(M,K))
Compare this with the exponential law for in set theory: if A, B, C are sets thenwe have a bijection
Map(A×B,C) ∼=−→ Map(A,Map(B,C))
Sketch of proof of Proposition 50.1.
Denote:
Hom2R(N,M ;K) = (set of all bilinear maps N ×M → K)
By the universal property of tensor ptoducts we have a bijection
HomR(N ⊗RM,K) ∼= Hom2R(N,M ;K)
Therefore we only need a bijection
Φ: Hom2R(N,M ;K) −→ HomR(N,HomR(M,K))
For f ∈ Hom2R(N,M ;K) define:
Φ(f) : N → HomR(M,K), Φ(f)(n) = f(n,−)
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51 Tensor products for non-commutative rings
Recall. If R is a commutative ring and M , N , K are R-modules then we havea bijection
(bilinear mapsM ×N → K
)∼=
(R-linear mapsM ⊗R N → K
)
Note. If R is a non-commutative ring then there are few bilinear map since forany such map f : M ×N → K we have
(sr) · f(m,n) = s · f(rm, n) = f(rm, sn) = r · f(m, sn) = (rs) · f(m,n)
A more appropriate notion in this setting is a middle linear map.
51.1 Definition. Let R be a ring with identity. Let M be a right R-module, Nbe a left R-module, G be an abelian group. A map
f : M ×N → G
is middle linear if
1) f(m1 +m2, n) = f(m1, n) + f(m2, n)f(m,n1 + n2) = f(m,n1) + f(m,n2)
2) f(mr, n) = f(m, rn)
51.2 Examples.
1) If R is a ring then the map
µ : R×R→ R, µ(r, s) = rs
is middle linear
211
2) In general, if N is a left R-module then the map
µ : R×N → N, µ(r, n) = rn
is middle linear.
51.3 Definition. If R is a ring with identity, M is a right R-module and N isa left R-module then M ⊗R N is the abelian group given by
M ⊗R N := Fab(M ×N)/S
Here Fab(M ×N) is the free abelian group generated by the set M ×N and Sis the subgroup of Fab(M ×N) generated by all elements of the following forms:
(i) (m1 +m2, n)− (m1, n)− (m2, n)(m,n1 + n2)− (m,n1)− (m,n2)
(ii) (mr, n)− (m, rn)
51.4 Note. Let m⊗ n denote the element of M ⊗R N represented by (m,n).We have a middle linear map
η : M ×N →M ⊗R N, η(m,n) = m⊗ n
51.5 Theorem. Let R be a ring with identity, let M be a right R-module, Nbe a left R-module. If G be an abelian group and and f : M × N → G is amiddle linear map then there exists a unique homomorphism of abelian groupsf : M ⊗R N → G such that the following diagram commutes:
M ×N f //
η
��
G
M ⊗R N
f
<<
Moreover, M ⊗R N is unique up to isomorphism abelian group satisfying thisproperty.
212
Proof. Similar to the proof of Theorem 48.2.
213
52 Restriction and extension of scalars
52.1 Definition. Let f : R → S be a homomorphism of rings, and let M be aleft S-module. Let f ∗M be the left R-module given by
f ∗M :=M as abelian group
r ·m :=f(r)m for r ∈ R,m ∈M
We say that f ∗M is obtained from M by restriction of scalars.
52.2 Example. Let V be an R-vector space and let i : Q→ R be the inclusion.Then i∗V is a Q-vector space.
52.3 Note. For a ring homomorphism f : R → S restriction of scalars definesa functor
f ∗ : S-Mod −→ R-Mod
52.4 Note. Let f : R→ S be a ring homomorphism. Then:
1) S is a right R-module with multiplication given by
s · r := sf(r) for s ∈ S, r ∈ R
Therefore for any left R-module N we have an abelian group S ⊗R N
2) S ⊗R N has a structure of a left S-module with multiplication given by
s · (s′ ⊗ n) := (ss′)⊗ n
(check!).
214
52.5 Definition. If f : R → S is a homomorphism of rings and N is a leftR-module then we say that the left S-module S ⊗R N is obtained from N byextension of scalars. We denote:
f∗N := S ⊗R N
52.6 Example. If V be an R-vector space and i : R→ C be the inclusion thenThen i∗V = C⊗RV is a C-vector space. We say that i∗V is the complexificationof V .
52.7 Note. For a ring homomorphism f : R→ S extension of scalars defines afunctor
f∗ : R-Mod −→ S-Mod
52.8 Note. The functor f∗ is left adjoint to the functor f ∗. That is, for any leftR-module N and any left S-module M we have a natural bijection of sets
HomS(f∗M,N)∼=−→ HomR(M, f ∗N)
(exercise).
52.9 Proposition. Let f : R→ S be a homomorphism of rings.
1) If M , M ′ are left R-modules and M ∼= M ′ then f∗M ∼= f∗M′.
2) f∗R ∼= S
3) f∗(⊕
αMα) ∼=⊕
α f∗Mα
Proof. Exercise.
215
52.10 Corollary. Let f : R→ S be a homomorphism of rings. If F is a free R-module with basis {bα}α∈A then f∗F is a free S-module with basis {1⊗ bα}α∈A.
Sketch of proof. We have F ∼=⊕
αR, so
f∗F ∼= f∗(⊕α
R) ∼=⊕α
f∗R ∼=⊕α
S
52.11 Corollary. Let f : R→ S be a homomorphism of rings. If P is a projectiveR-module then f∗P is a projective S-module.
Proof. Since P is projective there exists an R-module Q such that P ⊕ Q is afree R-module. Then we have
f∗(P ⊕Q) ∼= f∗P ⊕ f∗Q
By Corollary 52.10 f∗(P ⊕ Q) is a free S-module, so f∗P is a projective S-module.
52.12 Note. Restriction of scalars does not preserve free and projective modules.E.g, let i : Z→ Q be the inclusion. Then Q is a free Q-module, but i∗Q is notfree (or projective) Z-module.
216
53 Application: rings with IBN
Recall. From Section 42:
A ring R has the invariant basis number (IBN) property if
Rm ∼= Rn iff m = n
53.1 Proposition. If f : R→ S is a homomorphisms of rings with identity suchthat S has IBN then R also has IBN.
Proof. If Rm ∼= Rn then f∗Rm ∼= f∗R
n. By Corollary 52.10 we have f∗Rm ∼=
Sm, f∗Rn ∼= Sn, and since S has IBN we have m = n.
Recall:
42.11 Corollary. If R is a commutative ring with identity 1 = 0 then R hasIBN.
New proof. Let I ◁R be a maximal ideal. We have a ring homomorphism
π : R→ R/I
Since R/I is a field it has IBN, and so by Proposition 53.1 the ring R also hasIBN.
217
54 Algebras
54.1 Definition. LetR be a commutative ring with identity. A is an (associative,unital) R-algebra if
1) A is a ring with identity
2) A is an R-module
3) for any r ∈ R, a, b ∈ A we have
r · (ab) = (ra)b = a(rb)
A homomorphism of R-algebras is a ring homomorphism
f : A→ B
such that f(ra) = rf(a) for all a ∈ A, r ∈ R.
54.2 Examples.
1) Every ring is a Z-algebra.
2) If R is a commutative ring then R[x1, . . . , xn] is an R-algebra.
3) If R is a commutative ring and Mn(R) is the ring of all n × n matriceswith coefficients in R then Mn(R) is an R-algebra.
218
55 Tensor product of algebras
55.1 Note.
1) If A, B are R-algebras then they are R-modules, so A ⊗R B is also anR-module.
2) Check: A⊗R B has a structure of an R-algebra with multiplication
(A⊗R B)× (A⊗R B) −→ A⊗R B
given by:(∑i
ri(ai ⊗ bi))·(∑
j
sj(a′j ⊗ b′j)
):=∑i,j
risj(aia′j ⊗ bib′j)
55.2 Example. Let R be a commutative ring. We have an isomorphism ofR-algebras:
R[x]⊗R R[y] ∼= R[x, y]
(exercise).
219
56 Fields
Recall. A field is a commutative ring K with identity 1 = 0 such that allnon-zero elements of K are invertible.
56.1 Examples.
1) Q, R, C, Fp (p - prime number).
2) R/I where R is a commutative ring with identity 1 = 0 and I ◁ R is amaximal ideal.
E.g. R = K[x] where K is a field, I = ⟨p(x)⟩ where p(x) ∈ K[x] is anirreducible polynomial.
3) A field of fractions: S−1R where R is an integral domain and S = R−{0}.Note. If K is a field and R = K[x1, . . . , xn] then
K(x1, . . . , xn) := S−1R
is the field of rational functions in n variables with coefficients in K.
56.2 Proposition. If K, L are fields and
f : K −→ L
is a homomorphism then f is a monomorphism.
Proof. We have Ker(f) ◁ K. Since K is a field we must have Ker(f) = {0}(and so f is a monomorphism), or Ker(f) = K (which is impossible sincef(1) = 1 = 0.
220
Problems.
1) How do embeddings of fields K ↪→ L look like? (Field extensions)
2) If L is a field what are automorphisms L → L? If φ : L → L is suchautomorphism and K is a subfield of L what is φ(K)? (Galois Theory)
221
57 Field extensions
57.1 Definition. If K, L are fields and K ⊆ L then we say that L is a fieldextension of K.
57.2 Note. If K ⊆ L then L is a vector space over K.
57.3 Notation.
• L/K := extension L of a field K.
• [L : K] := dimK L. The number [L : K] is called the degree of theextension L/K.
57.4 Examples.
1) [R : Q] =∞ since |R| > |Q|Exercise: show that the subset
{√p | p – prime } ⊆ R
is linearly independent over Q.
2) [C : R] = 2 since {1, i} is a basis of C over R.
57.5 Note. [L : K] = 1 iff L = K.
222
58 Prime subfield and field characteristic
58.1 Proposition. If L is a field and {Ki}i∈I is a family of subfields of L then⋂i∈I Ki is a subfield of L.
Proof. Exercise.
58.2 Definition. The prime subfield of a field L is the intersection of all subfieldsof L.
58.3 Definition. If L is a field then the characteristic of L is the smallestpositive integer χ(L) such that
1L + · · ·+ 1L︸ ︷︷ ︸χ(L) times
= 0
If 1L + · · ·+ 1L︸ ︷︷ ︸n times
= 0 for all n > 0 then χ(L) = 0.
58.4 Example.
χ(Q) = χ(R) = χ(C) = 0.
χ(Fp) = p.
58.5 Proposition. If χ(L) = 0 then it is a prime number.
Proof. Exercise.
223
58.6 Theorem. Let L be a field and let K be the prime subfield of L.
1) If χ(L) = 0 then K ∼= Q.
2) if χ(L) = p > 0 then K ∼= Fp.
Proof. Exercise.
224
59 Algebraic and transcendental elements
59.1 Notation. If L/K is a field extension, and S is a subset of L then K(S)is the smallest subfield L such that K ⊆ K(S) and S ⊆ K(S).
Note.
1) If a ∈ L then
K(a) =
{r0 + r1a+ · · ·+ rna
n
s0 + s1a+ · · ·+ smam
∣∣∣∣ ri, sj ∈ K, s0 + · · ·+ smam = 0
}2) If a1, . . . , an ∈ L then
K(a1, . . . , an) = K(a1, . . . , an−1)(an)
59.2 Definition. Let L/K be a field extension. An element a ∈ L is algebraicover K if there exists r0, . . . , rn ∈ K, rn = 0 such that
r0 + r1a+ · · ·+ rnan = 0
Equivalently, a ∈ L is algebraic over K if there exists a non-zero polynomialp(x) = r0 + r1x+ · · ·+ rnx
n ∈ K[x] such that p(a) = 0.
An element a ∈ L is transcendental over K if it is not algebraic over K.
59.3 Example.
1)√2 ∈ R is a root of p(x) = x2 − 2 ∈ Q[x], so
√2 is algebraic over Q.
2) e, π ∈ R are transcendental over Q (harder to show).
225
59.4 Proposition. If L/K is a field extension and a ∈ L is a transcendentalelement over K then
K(a) ∼= K(x)
where K(x) is the field of rational functions of variable x with coefficients in K.
Proof. We have a homomorphism of rings
φa : K[x]→ L, φa(f) = f(a)
Since a is transcendental over K this map is a monomorphism, so φa(f) isan invertible element for all f = 0. Therefore, by the universal property oflocalizations of rings (36.6) there exists a homomorphism
φa : K(x)→ L, φa(f/g) = f(a)/g(a)
We have Im(φa) = K(a), so K(a) ∼= K(x).
59.5 Example.Q(π) ∼= Q(x) ∼= Q(e)
59.6 Proposition. Let L/K be a field extension and let a ∈ L be an algebraicelement over K.
1) There exists an irreducible polynomial p(x) ∈ K[x] such that p(a) = 0and if q(a) = 0 for some q(x) ∈ K[x] then p(x) | q(x).
2) We have an isomorphism K(a) ∼= K[x]/⟨p(x)⟩.
Proof.
1) Take the homomorphism of rings
φa : K[x]→ L, φa(f) = f(a)
226
Since a is algebraic over K we have Ker(φa) = {0}. Also, since K[x] is a PIDthus Ker(φa) = ⟨p(x)⟩ for some p(x) ∈ K[x].
If q(a) = 0 for some q(x) ∈ K[x] then q(x) ∈ ⟨p(x)⟩, so p(x) | q(x).
In order to see that p(x) is irreducible assume that p(x) = g(x)h(x) for someg(x), h(x) ∈ K[x]. Then
0 = p(a) = g(a)h(a)
so either g(a) = 0 or h(a) = 0. We can assume that g(a) = 0. Then p(x) | g(x).Since also g(x) | p(x) we obtain that h(x) must be a unit in K[x].
2) We haveK[x]/⟨p(x)⟩ ∼= Im(φa)
It is then enough to show that Im(φa) = K(a).
It is clear that Im(φa) ⊆ K(a). On the other hand, since p(x) is an irre-ducible element of K[x] then by (32.3) ⟨p(x)⟩ is a maximal ideal of K[x] andso K[x]/⟨p(x)⟩ is a field. As a consequence Im(φa) is a subfield of L, andK ∪ {a} ⊆ Im(φa). Since K(a) is the smallest subfield containing K ∪ {a} weget that K(a) = Im(φa).
59.7 Notation. Let L/K be a field extension and let a ∈ L be an algebraicelement over K. By Proposition 59.6 there is a unique irreducible polynomialp(x) ∈ K[x] of the form
p(x) = xn + rn−1xn−1 + . . .+ r1x+ r0
such that p(a) = 0. Denote this polynomial by irrKa (x).
Note. Polynomials of the form
f(x) = xn + rn−1xn−1 + . . .+ r1x+ r0
(with the highest degree coefficient equal to 1) are called monic polynomials.
227
59.8 Proposition. Let L/K, M/K be extensions of a field K and let a ∈ L,b ∈M be algebraic elements over K. The following conditions are equivalent.
1) irrKa (x) = irrKb (x)
2) there exists an isomorphism
φ : K(a)→ K(b)
such that φ|K = idK and φ(a) = φ(b).
Proof.
(1) ⇒ (2) If irrKa (x) = irrKb (x) = p(x) then we have isomorphisms
K(a)φa←− K[x]/⟨p(x)⟩ φb−→ K(b)
Take φ = φb ◦ φ−1a .
(2)⇒ (1) It is enough to check that irrKa (b) = 0. If irrKa (x) = xn+rn−1xn−1+
. . .+ r1x+ r0 Then we have
irrKa (b) = bn + rn−1bn−1 + . . .+ r1b+ r0
= φ(a)n + rn−1φ(a)n−1 + . . .+ r1φ(a) + r0
Since ri ∈ K we have ri = φ(ri), so
irrKa (b) = φ(a)n + φ(rn−1)φ(a)n−1 + . . .+ φ(r1)φ(a) + φ(r0)
= φ(an + rn−1an−1 + . . .+ r1a+ r0)
= φ(0)
= 0
59.9 Note. If L/K is a field extension, a, b ∈ L and irrKa (x) = irrKb (x) then we
may have K(a) = K(b). Take e.g. K/L = C/Q, a = 3√2, b = 3
√2(1
2+
√32i).
ThenirrQa (x) = irrQb (x) = x3 − 2
but Q(a) ⊆ R, Q(b) ⊆ R, so Q(a) = Q(b).
228
59.10 Proposition. Let L/K be a field extension and let a ∈ L be an algebraicelement over K, and let p(x) = irrKa (x). If deg p(x) = n then
1) K(a) = {s0 + s1a+ . . .+ sn−1an−1 | si ∈ K}
2) the set {1, a, . . . , an−1} is a basis of K(a) over K
3) [K(a) : K] = n
Proof.
1) We have the isomorphism
φa : K[x]/⟨p(x)⟩ −→ K(a), φa(f) = f(a)
Also, for f(x) ∈ K[x] we have
f(x) = q(x)p(x) + r(x)
for some q(x), r(x) ∈ K[x], deg r(x) < n. This gives
f(a) = q(a)p(a) + r(a) = 0 + r(a) = r(a)
Ii follows that K(a) = {r(a) | r(x) ∈ K[x], deg r(x) < n}.
2) By part 1) we have
K(a) = SpanK(1, a, . . . an−1)
so it suffices to show that the set {1, a, . . . , an−1} is linearly independent overK. Assume that
s0 · 1 + s1a+ . . .+ sn−1an−1 = 0
for some si ∈ K. Then a is a root of the polynomial g(x) = s0 + s1x + . . . +sn−1x
n−1. Since deg g(x) < deg p(x) we must have g(x) = 0, so s0 = s1 =. . . = sn−1 = 0.
3) Obvious by part 2).
229
59.11 Example.
Take√2 ∈ R. Since irrQ√
2(x) = x2 − 2, thus
Q(√2) = {a+ b
√2 | a, b ∈ Q}
Notice that
(a+ b√2)−1 =
1
a2 − 2b2(a− b
√2)
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60 Algebraic extensions
60.1 Notation. If L/K is a field extension and a ∈ L then
degK(a) := [K(a) : K]
Note.
1) If a is a transcendental element over K then K(a) ∼= K(x), so degK(a) =∞.
2) If a is an algebraic element over K then degK(a) = deg irrKa (x) <∞.
Upshot. If L/K is a field extension then an element a ∈ L is algebraic over Kiff degK(a) <∞.
60.2 Definition. A field extension L/K is an algebraic extension if all elementsof L are algebraic over K.
60.3 Proposition. If [L : K] <∞ then L/K is an algebraic extension.
Note. L = Q(√2,√3,√5,√7, . . . ) is an algebraic extension of Q even though
[L : Q] =∞.
60.4 Lemma. Let K ⊆ L ⊆ M be field extensions. If {ai}i∈I is a basis of Mover L and {bj}j∈J is a basis of L over K then {aibj}(i,j)∈I×J is a basis of Mover K. As a consequence
[M : K] = [M : L] · [L : K]
231
Proof.
Claim 1. The set {aibj}(i,j)∈I×J spans M over K.
Indeed, if m ∈M thenm =
∑i
liai
for some li ∈ L. Also, for each li we have
li =∑j
kijbj
for some kij ∈ K. Therefore we obtain
m =∑i
(∑j
kijbj
)ai =
∑ij
kij(aibj)
Claim 2. The set {aibj}(i,j)∈I×J is linearly independent over K.
Indeed, if∑
ij kij(aibj) = 0 for some kij ∈ K then
0 =∑i
(∑j
kijai
)bj
Since∑
j kijai ∈ L and the set {bj}j∈J is linearly independent over L we have∑j kijai = 0 for each j. Then, since the set {ai}i∈I is linearly independent over
K we have kij = 0 for all i, j.
Proof of Proposition 60.3.
Let a ∈ L. We have K ⊆ K(a) ⊆ L, so by Lemma 60.4 we get
[L : K(a)] · [K(a) : K] = [L : K] <∞Therefore [K(a) : K] <∞, and so a is an algebraic element over K.
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60.5 Corollary. If L/K is a field extension and a1, . . . , an ∈ L then
[K(a1, . . . , an) : K] ≤ degK(a1) · . . . · degK(a1)In particular, if a1, . . . , an are algebraic elements over K then [K(a1, . . . , an) :K] <∞ and so K(a1, . . . , an) is an algebraic extension of K.
Proof. Exercise.
60.6 Corollary. If M/L, L/K are algebraic extensions then M/K is also analgebraic extension.
Proof. Let a ∈M . We need to show that a is algebraic over K.
Since a is algebraic over L there is a polynomial f(x) = l0 + l1x+ . . .+ lnxn in
L[x] such that f(a) = 0.
Let N = K(l0, . . . , ln). Notice that f(x) ∈ N [x], so a is an algebraic elementover N . In particular we have [N(a) : N ] < ∞. Moreover, since l0, . . . , ln arealgebraic elements over K by Corollary 60.5 we have [N : K] <∞. This gives
[N(a) : K] = [N(a) : N ] · [N : K] <∞Finally, since K(a) ⊆ N(a) we have
[K(a) : K] ≤ [N(a) : K] <∞so a is an algebraic element over K.
60.7 Proposition. Let L/K be a field extension and let
Kalg(L) = {a ∈ L | a is an algebraic element over K}Then Kalg(L) is a subfield of L.
Proof. We need to show that if a, b ∈ L are algebraic elements over K, thena± b, ab, a/b are also algebraic over K. This holds since by (60.5) K(a, b)/Kis an algebraic extension and a± b, ab, a/b ∈ K(a, b).
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61 Separable elements
61.1 Example.
Note: since√2,√3 ∈ R are algebraic elements over Q, thus Q(
√2,√3)/Q is
an algebraic extension. We have:
Q ⊆ Q(√2 +√2) ⊆ Q(
√2,√3)
Claim: Q(√2 +√2) = Q(
√2,√3)
Indeed, it is enough to show that [Q(√2,√3) : Q(
√2 +√3)] = 1
We have irrQ√2(x) = x2 − 2 and irrQ√
2(x) = x2 − 3, so by (60.5)
[Q(√2,√3) : Q] ≤ degQ√
2· degQ√
3= 4
On the other hand√2 +√3 is a root of p(x) = x4 − 10x2 + 1, and since p(x)
is irreducible in Q[x] (check!) we have
[Q(√2 +√3) : Q] = deg p(x) = 4
This gives
[Q(√2,√3) : Q(
√2 +√3)] · [Q(
√2 +√3) : Q]︸ ︷︷ ︸
= 4
= [Q(√2,√3) : Q]︸ ︷︷ ︸
≤ 4
so [Q(√2,√3) : Q(
√2 +√3)] = 1.
Goal. If L/K is a field extension and a1, . . . , an ∈ L are algebraic elements overK then there exists an element c ∈ L such that
K(a1, . . . , an) = K(c)
provided that a1, . . . , an are separable over K.
234
Recall. Let L/K be a field extension. An element a ∈ L is a root of f(x) ∈ K[x]iff
f(x) = (x− a)g(x)for some g(x) ∈ L[x].
61.2 Definition. If L/K is a field extension and a ∈ L then a is a multiple rootof f(x) ∈ K[x] if
f(x) = (x− a)kg(x)for some k > 1 and g(x) ∈ L[x].
61.3 Definition. A polynomial f(x) ∈ K[x] is separable if it has no multipleroots in any extension of K.
61.4 Definition. Let L/K be a field extension and let a ∈ L be an algebraicelement over K. The element a is separable over K if irrKa (x) ∈ K[x] is aseparable polynomial.
235
62 Derivatives and separable elements
62.1 Definition. If f(x) ∈ K[x], f(x) = a0 + a1x + . . . + anxn then the
derivative of f(x) is the polynomial
f ′(x) = a1 + 2a2x+ . . .+ nanxn−1
62.2. Properties of derivatives.
1) (f + g)′ = f ′ + g′
2) (cf)′ = c · f ′ for c ∈ K
3) (fg)′ = f ′g + fg′
4) (fn)′ = nfn−1 · f ′
5) (f ◦ g)′ = (f ′ ◦ g) · g′
62.3 Proposition. A polynomial f(x) ∈ K[x] is separable iff f and f ′ have nocommon roots in any extension of K.
Proof.
(⇒) Assume that there exists an extension L/K such that f(a) = f ′(a) = 0for some a ∈ L. We have
f(x) = (x− a)g(x)
for some g(x) ∈ L[x]. It follows that
f ′(x) = 1 · g(x) + (x− a)g′(x)
and sog(x) = f ′(x)− (x− a)g′(x)
236
This givesg(a) = f ′(a)− (a− a)g′(a) = 0
Therefore g(x) = (x− a)h(x) for some h(x) ∈ L[x], and
f(x) = (x− a)2h(x)
Thus a is a multiple root of f(x).
(⇐) Assume that f(x) is not a separable polynomial. Then there exists anextension L/K and a ∈ L such that
f(x) = (x− a)kg(x)
for some k > 1 and g(x) ∈ L[x]. This gives
f ′(x) = k(x− a)k−1g(x) + (x− a)kg′(x)
so f ′(a) = f(a) = 0.
62.4 Proposition. Let L/K be a field extension, let a ∈ L be an algebraicelement over K and let p(x) = irrKa (x). The element a is separable over K iffp′(x) = 0.
Proof. (⇐) Assume that p′(x) = 0. Then
p(a) = p′(a) = 0
and by Proposition 62.3 p(x) is not a separable polynomial.
(⇒) Assume that a is not separable over K. Then p(x) is not a separablepolynomial, so by Proposition 62.3 there exists an extension M/K and b ∈ Msuch that
p(b) = p′(b) = 0
237
Since p(x) is an irreducible polynomial in K[x] we have p(x) = irrKb (x). Itfollows that p(x) | p′(x). However, deg p(x) > deg p′(x), so we must havep′(x) = 0.
62.5 Corollary. LetK be a field such that χ(K) = 0. If L/K is a field extensionthen every algebraic element a ∈ L is separable over K.
Proof. If p(x) = irrKa (x) then deg p(x) ≥ 1, so p′(x) = 0.
62.6 Example.
Let p be a prime number, and let
K = Fp(t)
be the field of rational functions of variable t with coefficients in Fp. Check:
p(x) = xp − t
is an irreducible polynomial inK[x]. Let L/K be an extension such that p(a) = 0for some a ∈ L. Then p(x) = irrKa (x), and since p′(x) = 0 then element a isnot separable over K.
62.7 Proposition. If K is a field, χ(K) = p = 0 and f(x) ∈ K[x] thenf ′(x) = 0 iff f(x) ∈ K[xp].
Proof.
(⇒) If f(x) ∈ K[xp] then
f(x) = a0 + apxp + a2px
2p + . . .+ anpxnp
238
for some akp ∈ K. Then
f ′(x) = papxp−1 + 2pa2px
2p−1 + . . .+ npanpxnp−1 = 0
(⇐) If f(x) = a0+a1x+. . .+anxn and f ′(x) = 0 then kak = 0 for k = 0, . . . , n.
Therefore either ak = 0 or p | k. It follows that f(x) ∈ K[xp].
62.8 Corollary. Let L/K be a field extension, let χ(K) = p = 0 and let a ∈ Lbe an algebraic element over K. Then a is separable over K iff irrKa (x) ∈ K[xp].
Proof. Follows from (62.4) and (62.7).
62.9 Proposition. Let L/K be a field extension such that χ(K) = p = 0, andlet a ∈ L be an algebraic element over K. The element a is separable over Kiff K(a) = K(ap).
Proof. (⇐) If a is separable over K then it is separable over K(ap) (check!).Also, a is a root of g(x) = xp − ap ∈ K(ap)[x]. Let f(x) = irrK(ap)
a (x). Thenf(x) | g(x).
On the other hand we have g(x) = (x − a)p, so f(x) = (x − a)k for some1 ≤ k ≤ p. Since f(x) is a separable polynomial we must have f(x) = x−a, andsince f(x) ∈ K(ap)[x] this gives that a ∈ K(ap). It follows that K(a) = K(ap).
(⇒) Assume that a is not separable overK, and let f(x) = irrKa (x). By Corollary62.8 we have f(x) ∈ K[xp], so
f(x) = r0 + rpxp + . . .+ rnpx
np
239
for some r0, . . . , rnp ∈ K. Let g(x) = r0 + rpx + . . . + rnpxn. We have
g(ap) = f(a) = 0, so
degK(ap) ≤ deg g(x) < deg f(x) = degK(a)
This gives
[K(ap) : K] = degK(ap) < degK(a) = [K(a) : K]
so K(ap) = K(a).
62.10 Corollary. Let K be a field such that χ(K) = p = 0. If L/K is a fieldextension and a1, . . . , an ∈ L are elements separable over K then
K(a1, . . . , an) = K(ap1, . . . , apn)
62.11 Notation. If K,L ⊆M are field extensions then
KL :=
(the smallest subfield of L
containing K ∪ L
)
Note: if L/K is a field extension and a1, . . . , an ∈ L then
K(a1, . . . , an) = K(a1)K(a2) · . . . ·K(an)
Proof of Corollary 62.10.
Using Proposition 62.9 we obtain:
K(a1, . . . , an) = K(a1) · . . . ·K(an) = K(ap1) · . . . ·K(apn) = K(ap1, . . . , apn)
240
63 Separable extensions
63.1 Definition. A field extension L/K is separable if every element a ∈ L isseparable over K.
63.2 Note. By (62.5) if χ(K) = 0 then every algebraic extension of K isseparable.
63.3 Notation. If L is a field and χ(L) = p = 0 then
Lp := {ap | a ∈ L}
Note.
1) Lp is a subfield of L since if ap, bp ∈ Lp then apbp = (ab)p ∈ Lp andap + bp = (a+ b)p ∈ Lp.
2) The mapφ : L→ Lp, φ(a) = ap
is a field isomorphism.
63.4 Proposition. Let K be a field such that χ(K) = p = 0. If L/K is a fieldextension and [L : K] <∞ then L/K is separable iff KLp = L.
63.5 Lemma. If K ⊆M ⊆ L and N ⊆ L are field extensions then
[NM : NK] ≤ [M : K]
Proof. Exercise.
241
Proof of Proposition 63.4.
(⇐) If L/K is separable and a ∈ L then by Proposition 62.9 we have
a ∈ K(a) = K(ap) ⊆ KLp
This gives L ⊆ KLp, and since also KLp ⊆ L we obtain KLp = L.
(⇒) Assume that L = KLp, and let b ∈ L. We need to show that b is a separableelement over K. By Proposition 62.9 it suffices to show that K(b) = K(bp).
Take the isomorphismφ : L→ L, φ(a) = ap
We have[φ(L) : φ(K(b))] = [L : K(b)] ≤ [L : K] <∞
Consider the extensions φ(K(b)) ⊆ φ(L) ⊆ L and K ⊆ L. By Lemma 63.5 wehave
[Kφ(L) : Kφ(K(b))] ≤ [φ(L) : φ(K(b))]
Notice that Kφ(L) = KLp = L and Kφ(K(b)) = KKp(bp) = K(bp), so thisgives
[L : K(bp)] ≤ [φ(L) : φ(K(b))] = [L : K(b)]
One the other hand we have
[L : K(bp)] = [L : K(b)] · [K(b) : K(bp)]
Therefore we must have [L : K(bp)] = [L : K(b)] and [K(b) : K(bp)] = 1. Thisgives
K(bp) = K(b)
63.6 Theorem. If L/K is a field extension, χ(K) = p = 0 and a1, . . . , an ∈ Lare elements separable over K then K(a1, . . . , an) is a separable extension of K.
242
Proof. Denote M := K(a1, . . . , an). Since [M : K] < ∞ by Proposition 63.4it suffices to show that KMp =M . We have
KMp = K(K(a1, . . . , an))p = K(Kp(ap1, . . . , a
pn)) = K(ap1, . . . , a
pn)
By (62.10) we also have
K(ap1, . . . , apn) = K(a1, . . . , an) =M
Therefore KMp =M .
63.7 Theorem. If M/L and L/K are separable extensions then M/K is alsoa separable extension.
Proof. Let a ∈M we need to show that a is separable over K.
Let f(x) = irrLa (x), f(x) = xn + rn−1xn−1 + . . .+ r1x+ r0. Define
N := K(r0, . . . , rn−1) ⊆ L
Notice that f(x) = irrNa (x), and since f(x) is a separable polynomial the elementa is separable over K. Moreover, we have
1) N/K is a separable extension (since L/K is separable and N ⊆ L)
2) [N : K] <∞Therefore by Proposition 63.4 we have KNp = N .
Similarly, since
1) N(a)/N is a separable extension (by (63.6), since a is separable over N)
2) [N(a) : N ] <∞thus N(N(a))p = N(a).
Finally we have:
1) [N(a) : K] = [N(a) : N ] · [N : K] <∞
2) K(N(a))p = KNp(N(a))p = N(N(a))p = N
243
so by Proposition 63.4 N(a)/K is a separable extension. Since a ∈ N(a) it is aseparable element over K.
63.8 Proposition. Let L/K be a field extension and let
Ksep(L) = {a ∈ L | a is separable over K}
Then Ksep(L) is a subfield of L.
Proof. We need to show that if a, b ∈ L are separable elements over K, thena± b, ab, a/b are also separable over K. This holds since by (63.6) K(a, b)/Kis an separable extension and a± b, ab, a/b ∈ K(a, b).
63.9 Definition. The separability degree of an extension L/K is the number
[L : K]sep := [Ksep(L) : K]
244
64 Simple extensions
64.1 Definition. A field extension L/K is simple if L = K(a) for some a ∈ L.
64.2 Theorem. If K is an infinite field and L/K is a separable extension suchthat [L : K] <∞ then L/K is a simple extension.
64.3 Note. Theorem 64.2 is true also if K is a finite field (later, see (69.5)).
Recall. If K is a field then K[x] is a Euclidean domain, so
1) for every f(x), g(x) ∈ K[x] there exists gcd(f(x), g(x)) ∈ K[x]
2) there exist λ(x), µ(x) ∈ K[x] such that
gcd(f(x), g(x)) = λ(x)f(x) + µ(x)g(x)
64.4 Lemma. Let L be a field, let f(x), g(x), h(x) ∈ L[x] be polynomials suchthat
h(x) ∼ gcd(f(x), g(x))
If K ⊆ L is a subfield such that f(x), g(x) ∈ K[x] then
h(x) = ah1(x)
for some 0 = a ∈ L and some h1(x). In particular is h(x) is a monic polynomialthen h(x) ∈ K[x].
Proof. Exercise.
245
64.5 Lemma. Let K be an infinite field, let L/K be a field extension and letf(x), g(x) ∈ L[x]. If f(0) = 0 and g(x) = 0 then there exists 0 = d ∈ K suchthat
gcd(f(x), g(dx)) ∼ 1
Proof. If deg f(x) = 0 or deg g(x) = 0 the statement is obvious. Otherwise wehave
f(x) = a · f1(x) · . . . · fn(x)g(x) = b · g1(x) · . . . · gn(x)
where 0 = a, b ∈ L and fi(x), gj(x) ∈ L[x] are irreducible monic polynomials:
fi(x) = ai,0 + ai,1x+ . . .+ xri
gj(x) = bj,0 + bj,1x+ . . .+ xsj
Note:
1) Since f(0) = 0 thus ai,0 = 0 for all i.
2) For any 0 = d ∈ K we have
g(dx) = b · g1(dx) · . . . · gn(dx)
Also, g1(dx), . . . , gn(dx) are irreducible polynomials in L[x] since gj(dx) =φ(gj(x)) where φ is the ring isomorphism
φ : L[x]→ L[x], φ(q(x)) := q(dx)
Let d ∈ K be an element such that gcd(f(x), g(dx)) ∼ 1. Then for some i, jwe have fi(x) ∼ gj(dx), i.e.
fi(x) = cgj(dx)
foe some 0 = c ∈ L. This gives:
ai,0 + ai,1x+ · · ·+ xri = cbj,0 + cbj,1dx+ · · ·+ cdsjxsj
As a consequence we obtain:
246
1) ri = sj
2) ai,0 = cbj,0
3) 1 = cdsj
Thereforedsj = c−1 = bj,0a
−1i,0
Since K is an infinite field we can find d ∈ K such that dsj = bj,0a−1i,0 for all i, j.
Then fi(x) ∼ gj(dx) for all i, j, and so gcd(f(x), g(dx)) ∼ 1.
64.6 Lemma. Let K be an infinite field and let L/K be a separable extensionsuch that L = K(a, b) let a, b ∈ L. There exists c ∈ L such that
K(a, b) = K(c)
Proof. Let f(x) = irrKa (x), g(x) = irrKb (x). Since f(a) = 0 we have
f(x) = (x− a)f1(x)
for some f1(x) ∈ L[x]. Moreover, since a is separable over K we have f1(a) = 0.
Denote:F (x) := f1(x+ a), G(x) := g(x+ b)
We have F (0) = f1(a) = 0 and G(x) = 0, so by Lemma 64.5 there exists0 = d ∈ K such that
gcd(F (x), G(dx)) ∼ 1
Take c := b− da. We will show that K(a, b) = K(c).
It is enough to show that a ∈ K(c). Indeed, we have K(c) ⊆ K(a, b), and ifa ∈ K(c) then also b = c+ da ∈ K(c), so in such case K(a, b) ⊆ K(c).
To see that a ∈ K(x) consider the ring isomorphism
φ : L[x]→ L[x], φ(q(x)) = q(x− a)
247
Since gcd(F (x), G(dx)) ∼ 1, thus also gcd(φ(F (x)), φ(G(dx))) ∼ 1. Moreoverwe have
φ(F (x)) = φ(f1(x+ a)) = f1((x− a) + a) = f1(x)
φ(G(dx)) = φ(g(dx+ b)) = g(d(x− a) + b) = g(dx− da+ b) = g(dx+ c)
Denote h(x) := g(dx+ c) ∈ K(c)[x]. We have:
h(a) = g(da− da+ b) = 0
soh(x) = (x− a)h1(x)
for some h1(x) ∈ L[x]. This gives:
gcd(f(x), h(x)) = gcd((x− a)f1(x), (x− a)h1(x))= (x− a) · gcd(f1(x), h1(x))
Since gcd(f1(x), h(x)) ∼ 1 and h1(x) | h(x) we have gcd(f1(x), h1(x)) ∼ 1.Therefore we get
gcd(f(x), h(x)) ∼ (x− a)Finally, notice that f(x), h(x) ∈ K(c)[x]. Thus, by Lemma 64.4, (x − a) ∈K(c)[x], and so a ∈ K(c).
Proof of Theorem 64.2.
Since [L : K] <∞ we have
L = K(a1, . . . , an)
for some a1, . . . , an ∈ L. We will show that L/K is a simple extension byinduction with respect to n.
If n = 1 then L = K(a1), so L/K is simple.
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Next, assume that for some n any extension K(a1, . . . , an)/K is simple, and letL = K(a1, . . . , an+1). Then we have
L = K(a1, . . . , an)(an+1)
By the inductive assumptionK(a1, . . . , an) = K(b) for some b ∈ L, so we obtain
L = K(b)(a) = K(b, a)
Finally, by Lemma 64.6 we have K(a, b) = K(c) for some c ∈ L, and so L/Kis simple.
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65 Simple extensions and intermediate fields
65.1 Definition. If L/K is a field extension then an intermediate field of L/Kis a field M such that
K ⊆M ⊆ L
65.2 Theorem. If L/K is a field extension, K is an infinite field and [L : K] <∞ then L/K is a simple extension iff L/K has finitely many intermediate fields.
65.3 Note. Theorem 65.2 is true also when K is a finite field. If K is finite and[L : K] <∞ then L is also a finite field, and so it has L finitely many subfields.Also, every finite extension of a finite field is simple (see Theorem 69.5).
65.4 Corollary. If L/K is a separable extension, K is an infinite field and[L : K] <∞ then L/K has finitely many intermediate fields.
Proof. By (64.2) L/K is a simple extension, so this follows from Theorem 65.2.
Proof of Theorem 65.2.
(⇒) Let L/K be a simple extension:
L = K(a)
for some a ∈ L, and let f(x) = irrKa (x). LetM be an intermediate field of L/Kand let g(x) = irrMa (x). Since K ⊆M we have g(x) | f(x) in L[x].
Claim. If g(x) = r0 + r1x+ · · ·+ rn−1xn−1 + xn then
M = K(r0, . . . , rn−1)
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Indeed, denote N := K(r0, . . . , rn−1). Notice that g(x) = irrNa (x). SinceM(a) = L = N(a) this gives
[L :M ] = degM(a) = deg g(x) = degN(a) = [L : N ]
On the other hand we have
[L : N ] = [L :M ] · [M : N ]
Therefore [M : N ] = 1, and so M = N .
We obtained that if K ⊆M ⊆ L then M = K(r0, . . . , rn−1) where
g(x) = r0 + r1x+ · · ·+ rn−1xn−1 + xn
is some monic polynomial in L[x] such that g(x) | f(x). Since there are onlyfinitely many such polynomials there are only finitely many such fields M .
(⇐) Assume that L/K has finitely many intermediate fields. We want to show:
L = K(c)
for some c ∈ L. Since [L : K] <∞ we have
L = K(a1, . . . , an)
for some a1, . . . , an ∈ L. We can assume that n = 2, since then we can argueby induction as in the proof of (64.2).
Thus, assume that we have L = K(a1, a2). For k ∈ K define
ck := a1 + ka2 ∈ L
We haveK ⊆ K(ck) ⊆ L
Notice that since K is an infinite field there are infinitely many elements of theform ck. On the other hand, by assumption L/K has finitely many intermediatefields, so there exist k, k′ ∈ K, k = k′ such that
K(ck) = K(ck′)
251
We will show K(ck) = K(a1, a2) = L.
We have K(ck) ⊆ K(a1, a2). Also, since ck, ck′ ∈ K(ck) we have
K(ck) ∋ ck − ck′ = (a1 + ka2)− (a1 + k′a2) = (k − k′)a2
Since 0 = k − k′ ∈ K this gives a2 ∈ K(ck), and then also
a1 ∈ ck − ka2 ∈ K(ck)
Therefore K(a1, a2) ⊆ K(ck).
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66 Construction of extensions
66.1 Proposition. Let K be a field and let p(x) ∈ K[x] be an irreduciblepolynomial. There exists a filed extension L/K such that p(x) has a root in L.
Proof. Let K[t] be the ring of polynomials of variable t. Since p(t) ∈ K[t] is anirreducible polynomial the ideal I = ⟨p(x)⟩ is a maximal ideal of K[t], so K[t]/Iis a field. Denote L := K[t]/I.
We have an embedding
i : K → L, i(a) = a+ I
so we can identify K with a subfield of L. Assume that
p(x) = a0 + a1x+ . . .+ anxn
Under the above identification we have
p(x) = (a0 + I) + (a1 + I)x+ . . .+ (an + I)xn
Let β := t+ I ∈ L. We have
p(β) = (a0 + I) + (a1 + I)(t+ I) + . . .+ (an + I)(t+ I)n
= (a0 + a1t+ . . .+ anTn) + I
= p(t) + I
= 0 + I
Therefore p(β) = 0 in L.
66.2 Definition. Let L/K be a field extension and let p(x) ∈ K[x]. We saythat p(x) splits over L if p(x) decomposes into a product of polynomials of degree1 in L[x]
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66.3 Proposition. Let K be a field and let p(x) ∈ K[x] be a polynomial suchthat deg p(x) > 0. There exists a field extension M/K such that p(x) splitsover M .
Proof. We argue by induction with respect to n = deg p(x).
If n = 1 then p(x) = ax+ b, and we can take M = K.
Next, assume that the statement holds for some n and let deg p(x) = n+1. Wehave
p(x) = p1(x) · . . . · pk(x)where p1(x), . . . , pk(x) ∈ K[x] are irreducible polynomials. By Proposition 66.1there exists an extension L/K such that p1(x) has a root a ∈ L. Then we have
p1(x) = (x− a)q1(x)
for some q1(x) ∈ L[x]. Take
p(x) := q1(x)p2(x) · . . . · pk(x) ∈ L[x]
We have p(x) = (x − a)p(x). Moreover, since deg p(x) = n by inductiveassumption there exists a field extension M/L such that p(x) splits over M . Iffollows that p(x) also splits over M .
66.4 Definition. If K is a field and p(x) ∈ K[x] then we say that a field L isa splitting field of p(x) if
1) K ⊆ L
2) p(x) splits over L
3) p(x) does not split over any proper subfield of L.
66.5 Proposition. Let K be a field. For any polynomial p(x) ∈ K[x], suchthat deg p(x) > 0 there exists a splitting field of p(x).
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Proof. By Proposition 66.3 there exists a field extension L/K such that p(x)splits over L:
p(x) = b(x− a1)(x− a2) · . . . · (x− an)for some b ∈ K, a1, . . . , an ∈ L. Then K(a1, . . . , an) is a splitting field ofp(x).
66.6 Example.
Take p(x) = x4 − 10x2 + 1 ∈ Q[x]. The polynomial p(x) splits over R[x]:
p(x) = (x−(√2+√3)) ·(x−(−
√2−√3)) ·(x−(−
√2+√3)) ·(x−(
√2−√3))
Therefore L = Q(±(√2 +√3), ±(
√2−√3)) is a splitting field of p(x).
Note: L = Q(√2,√3).
66.7 Proposition. Let K be a field, let p(x) ∈ K[x], deg p(x) > 0 and letL,L′ be splitting fields of p(x). There exists an isomorphism
φ : L→ L′
such φ|K = idK .
Proof. Exercise.
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67 Algebraically closed fields
67.1 Definition. A field is algebraically closed if the only irreducible polynomialsin K[x] are polynomials of degree 1.
67.2 Proposition. Let K be a field. The following conditions are equivalent.
(i) K is algebraically closed.
(ii) If p(x) ∈ K[x] and deg p(x) > 0 then p(x) splits over K.
(iii) If p(x) ∈ K[x] and deg p(x) > 0 then p(x) has a root in K.
(iv) K is the only algebraic extension of K.
Proof. Exercise.
67.3 Proposition. Every algebraically closed field is infinite.
Proof. Let K be a finite field, K = {a1, . . . , an}. Take the polynomial p(x) ∈K[x] given by
p(x) = (x− a1)(x− a2) · . . . · (x− an) + 1
We have p(ai) = 1 for all ai ∈ L, so p(x) has no roots in K. Therefore K isnot algebraically closed.
67.4 Definition. Let L/K be a field extension. The field L is an algebraicclosure ofK if L is an algebraically closed field and L/K is an algebraic extension.
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67.5 Proposition. Let L/K be an algebraic field extension. The followingconditions are equivalent
(i) L is an algebraic closure of K.
(ii) If p(x) ∈ K[x] is a polynomial such that deg p(x) > 0 then p(x) splitsover L.
Proof.
(i)⇒ (ii) If L is an algebraic closure of K, then L is an algebraically closed fieldand so by Proposition 67.2 every polynomial p(x) ∈ L[x] such that deg p(x) > 0splits over L. Since K[x] ⊆ L[x] the statement of (ii) follows.
(iI) ⇒ (i) Since L/K is an algebraic extension we only need to show that L isan algebraically closed field. By Proposition 67.2 it will suffice to show that forany algebraic extension M/L we have M = L.
Let then M/L be an algebraic extension. In such case the extension M/K isalso algebraic so if a ∈ M then a is a root of some polynomial p(x) ∈ K[x].Since by assumption p(x) splits over L we get that a ∈ L. Therefore M = L.
67.6 Theorem. For any field K there exists an extension K/K such that K isan algebraic closure of K.
67.7 Lemma. Let M/K be an algebraic extension.
1) If K is an infinite field then |K| = |M |.2) If K is a finite field then |M | ≤ ℵ0.
Proof. Exercise.
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Proof of Theorem 67.6. Let U be a set such that K ⊆ U and |U | > |K|. If Kis a finite field assume also that U is an uncountable set.
Let S be the set of all fields L such that L ⊆ U and L is an algebraic extensionof K. Define a partial ordering on S where L ≤ L′ if L′ is a field extension of L.Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S containsa maximal element K.
We will show that K is an algebraic closure of K. Since the extension K/K isalgebraic we only need to show that K is an algebraically closed field.
Assume, by contradiction, that K is not algebraically closed. By Proposition67.2 there is then an algebraic extension M/K where M = K. Then M/K isalso an algebraic extension and so by Lemma 67.7 we have |M | < |U |. As aconsequence there exists a monomorphism of sets
φ : M → U
such that φ|K = idK . We can introduce a field structure on the set φ(M) suchthat φ becomes a field isomorphism. Then φ(M) ∈ S and K < φ(M) which isimpossible by maximality of K.
67.8 Proposition. Let K/K be an algebraic closure and let L/K be any alge-braic extension of K. There exists an embedding
φ : L→ K
such that φK = idK .
67.9 Lemma. Let L/M be a field extension, let a ∈ L be an algebraic elementover M , and let
ψ : M → K
be a field homomorphism such that K is algebraically closed. Then there existsa homomorphism ψ : M(a)→ K such that ψ|M = ψ.
258
Proof. Exercise (compare with (59.8)).
Proof of Proposition 67.8. Let S be a set of all pairs (M,ψ) such that
(i) M is a field such that K ⊆M ⊆ L
(ii) ψ : M → K is a homomorphism such that ψ|K = idK
Define partial ordering on S as follows:
(M,ψ) ≤ (M ′, ψ′) if M ⊆M ′ and ψ′|M = ψ
Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S containsa maximal element (M0, ψ0).
Assume that M0 = L and let a ∈ L/M0. By Lemma 67.9 there exists ahomomorphism
ψ0 : M0(a)→ K
such that ψ0|M0 = ψ0. It follows that (M0, ψ0) < (M0(a), ψ0) which is impos-sible since (M0, ψ0) is a maximal element. Therefore M0 = L and we can takeφ = ψ0.
67.10 Note. Proposition 67.8 can be generalized as follows. Let K ⊆ L ⊆ Mbe algebraic extensions, let K be an algebraic closure of K, and let
φ : L→ K
be an embedding such that φ|K = idK . Then there exists an embedding
φ : M → K
such that φL = φ (exercise).
259
67.11 Corollary. If L/K, L′/K are algebraic closures of K then there exists anisomorphism
φ : L→ L′
such that φ|K = idK .
Proof. By Proposition 67.8 there exists a homomorphism
φ : L→ L′
such that φ|K = idK . It suffices to show that φ is an epimorphism. We have
K = φ(K) ⊆ φ(L) ⊆ L′
Since L is algebraically closed and φ(L) ∼= L we get that φ(L) is also algebraicallyclosed and so (by Proposition 67.2) the only algebraic extension of φ(L) is φ(L)itself. On the other hand, since L′/K is an algebraic extension and K ⊆ φ(L)the extension L′/φ(L) is algebraic. This gives that L′ = φ(L).
67.12 Fundamental Theorem of Algebra. The field C of complex numbersis algebraically closed.
Proof. Later.
Recall. If L/K is a field extension then
Kalg(L) = {a ∈ L | a is an algebraic element over K}
is a subfield of L.
260
67.13 Proposition. If L/K is a field extension and L is an algebraically closedfield then the field Kalg(L) is also algebraically closed. As a consequence Kalg(L)is an algebraic closure of K.
Proof. Let p(x) ∈ Kalg(L)[x], deg p(x) > 0. It suffices to show that p(x) has aroot in Kalg(L).
The field L is algebraically closed, so p(x) has a root a ∈ L. Consider theextensions
Kalg(L)(a)/Kalg(L) and Kalg(L)/K
These extensions are algebraic, so by Proposition 60.6 Kalg(L)(a)/K is also analgebraic extension. This means in particular that a ∈ L is an algebraic elementover K, so a ∈ Kalg(L).
67.14 Note. Since C is an algebraically closed field thus Q := Qalg(C) is analgebraic closure of Q.
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68 Roots of unity
68.1 Definition. Let K be a field. We say that a ∈ K is an n-th root of unityif an = 1, i.e. if a is a root of the polynomial fn(x) = xn − 1 ∈ K[x].
68.2 Proposition. Let K be a field and let
µn(K) := {a ∈ K | an = 1}
Then µn(K) is a subgroup of the multiplicative group K∗ = K − {0}.
Proof. Exercise.
68.3 Note.
1) If m|n then µm(K) is a subgroup of µn(K).
2) If χ(K) = p = 0 and n = pkm then µn(K) = µm(K). Indeed, we have
fn(x) = xpkm − 1p
k
= (xm − 1)pk
= (fm(x))pk
so roots of fn are the same as roots of fm.
68.4 Theorem. For any field K the group µn(K) is cyclic and |µn(K)| | n.
Proof. Since µn(K) is a finite subgroup of K∗ it is cyclic by 38.8. so µn(K) =⟨a⟩ for some a ∈ µn(K). This gives |µn(K)| = |a|. Also, since an = 1, we have|a| | n.
262
68.5 Corollary. Let K be an algebraically closed field. If χ(K) = p = 0 assumealso that p ∤ n. Then we have
µn(K) ∼= Z/nZ
Proof. By Theorem 68.4 it is enough to show that |µn(K)| = n.
Let fn(x) = xn − 1 ∈ K[x]. We have
f ′n(x) = nxn−1
Since fn(x) and f′n(x) have no common roots thus by Proposition 62.3 fn(x) is
a separable polynomial. As a consequence fn(x) has n distinct roots in K, andso |µn(K)| = n.
68.6 Definition. An element a ∈ K is a primitive n-th root of unity if an = 1and am = 1 for all 0 < m < n.
68.7 Note.
1) A primitive n-th root of unity exists in K iff |µn(K)| = n. In such casea ∈ K is primitive n-th roots of unity if a generates µn(K).
2) If |µn(K)| = m then µn(K) = µm(K) and µn(K) is generated by aprimitive m-th root of unity in K.
68.8 Examples.
1) In Q for any n ≥ 1 we have
µ2n−1(Q) = {1} = µ1(Q), µ2n(Q) = {1,−1} = µ2(Q)
Thus 1 is a primitive 1-st root of unity and −1 is a primitive 2-nd root ofunity. For n > 2 there are no primitive n-roots of unity in Q.
263
2) Since C is an algebraically closed field we have |µn(C)| = n for all n > 0,and so C contains a primitive n-th root of unity for every n. We have
µn(C) = {εk := cos(2πk/n) + i sin(2πk/n) | k = 0, . . . , n− 1}
and εk is a primitive n-th root of unity iff gcd(k, n) = 1.
264
69 Finite fields
69.1 Proposition. If K is a finite field and χ(K) = p then |K| = pn for somen ≥ 1.
Proof. Since χ(K) thus by Theorem 58.6 Fp can be identified with the primitivesubfield of K. Since K is a finite field we have [K : Fp] = n for some n ≥ 1.This means that K is an n dimensional vector space over Fp, and so K has pn
elements.
69.2 Proposition. If K is a finite field, such that χ(K) = p then |K| = pn iffK is a splitting field of the polynomial
f(x) = xpn − x ∈ Fp[x]
69.3 Lemma. If K be a field and φ : K → K be a field homomorphism then
L := {a ∈ K | φ(a) = a}
is a subfield of K.
Proof. Exercise.
Proof of Proposition 69.2.
(⇒) We will show that if |K| = pn then every element of K is a root of f(x).Since f(x) can have at most pn roots it will follow that K is a splitting field off(x).
265
Let a ∈ K. If a = 0 then a is a root of f(x). Assume then that a = 0. Then ais an element of the multiplicative group K∗ = K − {0}. Since |K∗| = pn − 1thus we have we have ap
n−1 = 1. This gives
f(a) = apn − a = a(ap
n−1 − 1) = 0
(⇐) Assume thatK is a splitting field of f(x). Consider the field homomorphism
φ : K → K, φ(a) = apn
By Lemma 69.3 the set
L = {a ∈ K | φ(a) = a}
is a subfield of K. On the other hand L consists of all roots of f(x) in K, andso since f(x) splits over K it also splits over L. Since K is a splitting field off(x) this implies that K = L, i.e. every element of K is a root of f(x).
It suffices to show then that f(x) has pn distinct roots, i.e. that it is a separablepolynomial. This follows from Proposition 62.3, since f ′(x) = −1, and so f(x)and f ′(x) have no roots in common.
69.4 Corollary. For every prime p and n ≥ 1 there exists a field K such thatχ(K) = p and |K| = pn. Moreover such field is unique up to isomorphism.
Proof. Take the polynomial f(x) = xpn − x ∈ Fp[x]. By Proposition 66.5 there
exists a splitting field K of f(x) and by Proposition 69.2 we have |K| = pn.
Assume now that K ′ is another field such that |K ′| = pn. By Proposition 69.2we have that K ′ is also a splitting field of f(x). Uniqueness of splitting fields(66.7) gives then that K ′ ∼= K.
266
Recall:
64.2 Theorem. If K is an infinite field and L/K is a separable extension suchthat [L : K] <∞ then L/K is a simple extension.
69.5 Theorem. If K is an finite field and L/K is a field extension such that[L : K] <∞ then L/K is a simple extension.
69.6 Note. If K is a finite field then any algebraic extension L/K is a separableextension (exercise).
Proof of Theorem 69.5.
Assume that χ(K) = p, and that Fp ⊆ K. We will show that a L = Fp(c) forsome c ∈ L. Since Fp(c) ⊆ K(c) this will imply that K(c) = L.
Since K is a finite field and [L : K] < ∞ the field L is finite, and so |L| = pn
for some n ≥ 1. Let L∗ = L − {0} be the multiplicative group of L. By theproof of Proposition 69.2 we have
L∗ = {a | apn−1 = 1}
i.e. L∗ is the group of (pn − 1)-st roots of unity in L:
L∗ = µpn−1(L)
By Theorem 68.4 µpn−1(L) is a cyclic group. Let c be a generator of µpn−1(L).Then we have L = Fp(c).
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70 Galois theory - motivation
70.1 Proposition. Let L/K be a field extension and let
φ : L→ L
be an automorphism such that φ|K = idK . If a ∈ L is a root of a polynomialf(x) ∈ K[x] then φ(a) is also a root of f(x).
Proof. If f(x) = r0 + r1x+ . . .+ rnxn then we have
f(φ(a)) = r0 + r1φ(a) + . . .+ rnφ(a)n
Also, since ri ∈ K we have ri = φ(ri), so we obtain
f(φ(a)) = φ(r0) + φ(r1)φ(a) + . . .+ φ(rn)φ(a)n
= φ(f(a))
= φ(0)
= 0
70.2 Corollary. Let L/K be an algebraic extension, let f(x) ∈ K[x], and letS = {a1, . . . , an} ⊆ L be the set of all roots of f(x) in L. If
φ : L→ L
is an automorphism such that φ|K = idK then φ|S is a permutation of S.
Moreover, if L is a splitting field of f(x) then the automorphism φ is uniquelydetermined by this permutation.
Proof. By Proposition 70.1 we have φ(S) ⊆ S. Also, since φ is a bijection andS is a finite set thus φ|S : S → S is also a bijection i.e. it is a permutation ofthe set S.
268
Recall that if L is a splitting field of f(x) we have
L = K(a1, . . . , an)
Thus, since φ|K = id, the automorphism φ is determined by the restrictionφ|{a1,...,an}.
Note. Let L be a splitting field of a polynomial f(x) ∈ K[x], and let S ⊆ L bethe set of all roots of f(x). It is not true that every permutation of S can beextended to an automorphism φ : L→ L such that φ|K = idK .
70.3 Example.
Let f(x) = x4 − 10x2 + 1 ∈ Q[x]. Roots of f(x) in C are:
√2 +√3,
√2−√3, −
√2 +√3, −
√2−√3
It follows that the field L := Q(√2,√3) is a splitting field of f(x).
Letφ : L→ L
be an automorphism such that φ|Q = idQ. Notice that we must have φ(√2) =
±√2 since ±
√2 are the only roots of g(x) = x2 − 2. Similarly we must have
φ(√3) = ±
√3.
Assume that φ(√2 +√3) =
√2−√3. Then we must have
φ(√2) =
√2, and φ(
√3) = −
√3
This gives that
φ(√2−√3) =
√2 +√3
φ(−√2 +√3) = −
√2−√3
φ(−√2−√3) = −
√2 +√3
269
By this argument one gets that there are only 4 permutations of the set ofroots f(x) that can extend to an automorphism of L, and that each of thesepermutations is determined by the value of φ(
√2 +√3).
Note. Let L be a splitting field of f(x) ∈ K[x], and let S ⊆ L be the set ofall roots of f(x). Permutations of S that extend to automorphisms of L form asubgroup of the group of permutations of S (check!). We will see that propertiesof this group provide information about the polynomial f(x) and its roots.
270
71 Normal extensions
71.1 Definition. Let L/K be an algebraic extension and let
Gal(L/K) := {φ : L→ L | φ is an automorphism and φ|K = idK}
Then Gal(L/K) is a group (with composition of automorphism), and it is calledthe Galois group of the extension L/K.
71.2 Proposition. Let L be a field and let Aut(L) be the group of all auto-morphisms of L. If H is a subgroup of Aut(L) then the set
LH = {a ∈ L | φ(a) = a for all φ ∈ H}
is a subfield of L.
Proof. Exercise.
71.3 Definition. If L is a field and H ⊆ Aut(L) is a subgroup then the subfieldLH ⊆ L is called the fixed field of H.
71.4 Note. Let L be a field. We have maps
Φ:
(subfields
of L
)⇆
(subgroups
of Aut(L)
): Ψ
where Φ(K) = Gal(L/K) and Ψ(H) = LH . In general these maps are notinverses of each other. Take e.g. L = Q(3
√2) and let K = Q ⊆ L. We have
Φ(K) = Gal(L/K) = {idL}
(check!). This gives
Ψ(Φ(K)) = LGal(L/K) = L = K
271
71.5 Definition. An algebraic extension L/K is called a Galois extension ifK = LGal(L/K).
Goal. An extension L/K is a Galois extension iff it is a normal and separableextension.
71.6 Definition. An algebraic extension L/K is a normal extension if for everya ∈ L the polynomial irrKa (x) splits over L.
71.7 Example.
1) If K is the algebraic closure of K then the extension K/K is normal.
2) The extension Q(3√2)/Q is not normal. Indeed, 3
√2 ∈ Q(3
√2), but the
polynomialirrQ3√2
(x) = x3 − 2
does not split over Q(3√2).
71.8 Proposition. Let K be an algebraic closure of a field K. An algebraicextension L/K is normal iff for any two homomorphisms
φ, ψ : L→ K
such that φ|K = ψ|K = idK we have φ(L) = ψ(L).
Proof.
(⇒) Let φ, ψ : L → K be homomorphisms satisfying φ|K = ψ|K = idK . It isenough to show that for any a ∈ L we have φ(a) ∈ ψ(L).
272
Let a ∈ L and let f(x) = irrKa (x). Since L/K is a normal extension f(x) splitsover L:
f(x) = (x− a1)(x− a2) · . . . · (x− an)where ai ∈ L and a1 = a. We have
φ(f(x)) = (x− φ(a1))(x− φ(a2)) · . . . · (x− φ(an))ψ(f(x)) = (x− ψ(a1))(x− ψ(a2)) · . . . · (x− ψ(an))
On the other hand, since f(x) ∈ K[x] and φ|K = ψ|K = idK we have
φ(f(x)) = f(x) = ψ(f(x))
Since K[x] is a UFD and x − φ(ai), x − ψ(ai) are irreducible polynomials inK[x] we must have x− φ(a1) = x− ψ(ai) for some i. Since a1 = a this gives
φ(a) = ψ(ai) ∈ ψ(L)
(⇐) Assume that L/K is an algebraic extension which is not normal. Thenthere exists a ∈ L such that the polynomial f(x) = irrKa (x) does not split overL. Let φ : L→ K be a homomorphism such that φ|K = idK . Since L ∼= φ(L)we have that φ(a) ∈ φ(L) is a root of f(x), but f(x) does not split over φ(L).On the other hand f(x) splits over K, so there exists b ∈ K − φ(L) such thatf(b) = 0. By Proposition 59.8 there exists an isomorphism
ψ : K(a)→ K(b)
such that ψ|K = idK . By (67.10) ψ can be extended to a homomorphism
ψ : L→ K
Since b ∈ ψ(L) and b ∈ φ(L) we get that φ(L) = ψ(L).
71.9 Proposition. Let L/K be a field extension such that [L : K] < ∞. Theextension L/K is normal iff L is a splitting field of some polynomial f(x) ∈ K[x].
273
Proof.
(⇒) Since [L : K] <∞ we have
L = K(a1, . . . , an)
for some a1, . . . , an ∈ L. Let fi(x) = irrKai(x). Since L/K is a normal extensionfi(x) splits over L for each i, so the polynomial
f(x) = f1(x) · . . . · fn(x)
also splits over L. Moreover, since a1, . . . , an are roots of f(x) and they generateL, thus L is a splitting field of f(x).
(⇐) Let L be a splitting field of some polynomial f(x) ∈ K[x]. Then we have
f(x) = (x− a1)(x− a2) · . . . · (x− an)
for some a1, . . . , an ∈ L, and L = K(a1, . . . , an).
Let K be an algebraic closure of K and let
φ, ψ : L→ K
be field homomorphisms such that φ|K = ψ|K = idK . We have
φ(f(x)) = (x− φ(a1))(x− φ(a2)) · . . . · (x− φ(an))ψ(f(x)) = (x− ψ(a1))(x− ψ(a2)) · . . . · (x− ψ(an))
On the other hand, since f(x) ∈ K[x] e have
φ(f(x)) = f(x) = ψ(f(x))
It follows that for every 1 ≤ i ≤ n there exists 1 ≤ j ≤ n such that φ(ai) =ψ(aj). This gives
φ(L) = K(φ(a1), . . . , φ(an)) ⊆ K(ψ(a1), . . . , ψ(an)) = ψ(L)
Similarly we get ψ(L) ⊂ φ(L). Therefore φ(L) = ψ(L), and so by Proposition71.8 L/K is a normal extension.
274
71.10 Proposition. Let K,L,M,N be fields such that K,L,M ⊆ N , N/K isan algebraic extension and L/K is a normal extension. Then LM/KM is alsoa normal extension.
71.11 Corollary. If K ⊆ L ⊆ M are field extensions and M/K is normal thenM/L is also normal.
Proof. We have M = ML, L = KL, and by Proposition 71.10 the extensionML/KL is normal.
Proof of Proposition 71.10. Let KM be an algebraic closure of KM , and let
φ, ψ : LM → KM
be two embeddings of LM such that φ|KM = ψ|KM = id|KM . By Proposition71.8 it is enough to show that φ(LM) = ψ(LM).
Notice that since KM/K and N/KM are algebraic extensions we have
KM = N = K
Since φ|K = ψ|K = id|K and L/K is a normal extension we have φ(L) = ψ(L).Moreover, φ|M = ψM = idM , so φ(M) = ψ(M) =M . This gives
φ(LM) = φ(L)φ(M) = ψ(L)ψ(M) = ψ(LM)
71.12 Note. It is not true that if K ⊆ L ⊆ M and M/L are L/K are normalextensions then M/K is also normal. Take e.g.
Q ⊆ Q(√2) ⊆ Q(4
√2)
275
71.13 Definition. Let L/K be an algebraic extension. An normal closure ofL/K is an extension M/K such that
1) L ⊆M
2) M/K is a normal extension
3) M does not have any proper subfields satisfying 1) and 2).
71.14 Proposition. For any L/K is any algebraic extension there exists a normalclosure M/K of L/K. Moreover, if M , M ′ are normal closures of L/K thenthere exists an isomorphism
φ : M →M ′
such that φ|L = idL.
Proof. Let L be an algebraic closure of L, and let S = {Mα} be the set of allsubfields of L such that L ⊆ Mα and that Mα/K is a normal extension. Theset S is non-empty since L ∈ S. Take M =
⋂αMα. Check: M/K is a normal
extension, and so it is a normal closure of L/K.
Let M ′/K be another normal closure of L/K. Since M ′/L is an algebraicextensions by (67.8) there exists an embedding
φ : M ′ → L
such that φ|L = id|L. Since L ⊆ φ(M ′) ⊆ L and φ(M)/K is a normalextension, thus by the construction of M we have M ⊆ φ(M ′). On the otherhand φ(M ′) is a normal closure of L/K. This implies that φ(M) =M ′.
276
72 Galois extensions
72.1 Theorem. An algebraic extension L/K is a Galois extension iff L/K isseparable and normal.
Proof.
(⇒) Let L/K be a Galois extension let a ∈ L, and let f(x) = irrKa (x). It isenough to show that f(x) splits over L and that it is a separable polynomial.
Let S = {a1, . . . , an} be the set of all distinct roots of f(x) in L where, say,a = a1. Let
g(x) = (x− a1)(x− a2) · . . . · (x− an) ∈ L[x]
Notice that g(x) is a separable polynomial that splits over L. Therefore it sufficesto show that f(x) = g(x).
Furthermore, it is enough to show that f(x)|g(x). Indeed since g(x)|f(x) andsince f(x) and g(x) are monic polynomials this will imply that f(x) = g(x).
Let φ ∈ Gal(L/K). By Corollary 70.2 φ permutes the set {a1, . . . , an}. Thisgives
φ(g(x)) = (x− φ(a1))(x− φ(a2)) · . . . · (x− φ(an))= (x− a1)(x− a2) · . . . · (x− an)= g(x)
Therefore, if g(x) = r0 + r1x + · · · + rnxn then we have φ(ri) = ri, i.e. ri ∈
LGal(L/K) for i = 0, . . . , n. Since L/K is a Galois extension we have LGal(L/K) =K, and so we obtain that g(x) ∈ K[x].
Since g(a) = g(a1) = 0 and f(x) = irrKa (x) this gives f(x)|g(x).
(⇐) Let L/K be a separable and normal extension and let a ∈ L−K. We needto show that there exists φ ∈ Gal(L/K) such that φ(a) = a.
277
Let K be an algebraic closure of K such that L ⊆ K. If f(x) = irrKa (x) ∈ K[x]then deg f(x) > 1 (since a ∈ K), and f(x) has no multiple roots (since a isseparable over K). It follows that there exists b ∈ K such that b = a andf(b) = 0. By Proposition 59.8 there exists an isomorphism
φ : K(a)→ K(b) ⊆ K
such that φ|K = idK . By (67.10) we can extend φ to an embedding φ : L→ K.
Let i : L ↪→ K be the inclusion homomorphism. Since L/K is a normal ex-tension thus by (71.8) we have φ(L) = i(L) = L. It follows that φ gives anautomorphism φ : L → L. Since φ|K = idK we have φ ∈ Gal(L/K) andφ(a) = b = a.
72.2 Corollary. If K ⊆ L ⊆ M are fields and M/K is a Galois extension thenM/L is also a Galois extension.
Proof. The extension M/K is normal and separable, so M/L is also normal (by(71.11)) and separable.
72.3 Theorem. Let L be a field, let Aut(L) be the group of automorphisms ofL, and let H be a finite subgroup of Aut(L). Then:
1) L/LH is a Galois extension
2) [L : LH ] = |H|
3) Gal(L/LH) = H
278
72.4 Note. Recall that for a field L we have maps of sets:
Φ:
(subfields
of L
)⇆
(subgroups
of Aut(L)
): Ψ
where Φ(K) = Gal(L/K) and Ψ(H) = LH . By part 3) of Theorem 72.3 weobtain that if H ⊆ Aut(L) is a finite subgroup then
Φ(Ψ(H)) = H
Proof of Theorem 72.3.
1) We need to show that L/LH is an algebraic extension and that
LH = LGal(L/LH)
Notice that if G,H are subgroups of Aut(L) and G ⊆ H then LH ⊆ LG. SinceH ⊆ Gal(L/LH) this gives LGal(L/LH) ⊆ LH . On the other hand, by definitionof Gal(L/LH), if φ ∈ Gal(L/LH) then φ|LH = idLH , so LH ⊆ LGal(L/LH).Thus we obtain LH = LGal(L/LH).
It remains to show that L/LH is an algebraic extension. Let a ∈ L and let
Ha := {φ(a) | φ ∈ H}
Since H is a finite group the set Ha is finite, say Ha = {a1, . . . , an} wherea1 = a. Let
g(x) = (x− a1)(x− a2) · . . . · (x− an) ∈ L[x]
Notice that if φ ∈ H then φ|Ha is a bijection:
φ|Ha : Ha→ Ha
As a consequence we have φ(g(x)) = g(x) for all φ ∈ H. This means thatg(x) ∈ LH [x]. Since g(a) = g(a1) = 0 we get that a is an algebraic elementover LH .
279
2) Let a ∈ L. The argument in part 1) shows that a is a root of a polynomialg(x) ∈ LH [x] such that deg g(x) ≤ |H|. It follows that for any a ∈ L we have
[LH(a) : LH ] ≤ |H| (∗)
Claim. [L : LH ] ≤ |H|.
Indeed, if [L : LH ] > |H| then we can find an intermediate field
LH ⊆ K ⊆ L
such that |H| < [K : LH ] < ∞. By part 1) L/LH is a Galois extension, so itis separable. This implies that K/LH is also a a separable extension. By (64.2)and (69.5) we obtain then that K/LH is a simple extension i.e.
K = LH(a)
for some a ∈ K. This gives
[LH(a) : LH ] = [K : LH ] > |H|
which contradicts the inequality (∗).
We obtain that L/LH is a separable extension, and [L : LH ] ≤ |H| < ∞. Itfollows that L/LH is a simple extention, L = LH(a) for some a ∈ L. Letf(x) = irrKa (x). We have
deg f(x) = [L : LH ] ≤ |H|
On the other hand each automorphism φ ∈ Gal(L/LH) is uniquely determinedby the value of φ(a). Since φ(a) is a root of f(x) we obtain that
|Gal(L/LH)| ≤ deg(f(x))
Since H ⊆ Gal(L/LH) this gives
|H| ≤ |Gal(L/LH)| ≤ deg f(x) ≤ |H|
Therefore |H| = |Gal(L/LH)| = [L : LH ].
280
3) We have H ⊆ Gal(L/LH), and by the proof of part 2), |H| = |Gal(L/LH)|.Since H is a finite group this implies that H = Gal(L/LH).
72.5 Corollary. Let L/K be a field extension such that [L : K] < ∞. Theextension L/K is a Galois extension iff [L : K] = |Gal(L/K)|.
Proof.
(⇐) If L/K is a Galois extension then K = LGal(L/K). Since [L : K] < ∞ bypart 2) of Theorem 72.3 we obtain [L : K] = |Gal(L/K)|.
(⇒) Assume that [L : K] = |Gal(L/K)|. We want to show thatK = LGal(L/K).
We have K ⊆ LGal(L/K) ⊆ L, so
|Gal(L/K)| = [L : K] = [L : LGal(L/K)] · [LGal(L/K) : K]
By part 2) of Theorem 72.3 we also have [L : LGal(L/K)] = |Gal(L/K)|. Itfollows that [LGal(L/K) : K] = 1, and so K = LGal(L/K).
281
73 Application: rational symmetric functions
Let L = K(t1, . . . , tn) be the field of rational functions of variables t1, . . . , tnwith coefficients in a field K, and let Sn be the symmetric group on n letters.
The group Sn can be identified with a subgroup of the group Aut(L) of auto-morphims of L as follows. If σ ∈ Sn then σ induces an automorphism
σ : L −→ L
such that σ|K = id|K and σ(ti) = tσ(i).
73.1 Definition. If f(t1,...,tn)g(t1,...,tn)
∈ LSn then we say that f(t1,...,tn)g(t1,...,tn)
is symmetricfunction.
73.2 Example. Let
f(x) = (x− t1)(x− t2) · · · · · (x− tn) ∈ L[x]For σ ∈ Sn we have
σ(f(x)) = (x− tσ(1))(x− tσ(2)) · · · · · (x− tσ(n)) = f(x)
Therefore f(x) ∈ LSn [x].
Notice that we have
f(x) = xn − σ1xn−1 + σ2xn−2 − · · ·+ (−1)n−1σn−1x+ (−1)nσn
where
σ1 =∑i
ti
σ2 =∑i<j
titj
σ3 =∑i<j<k
titjtk
...
σn = t1t2 · . . . · tn
If follows that σ1, . . . , σn ∈ LSn .
282
73.3 Definition. The polynomials σ1, . . . , σn are called elementary symmetricpolynomials in variables t1, . . . , tn.
73.4 Theorem.
K(t1, . . . , tn)Sn = K(σ1, . . . , σn) ⊆ K(t1, . . . , tn)
Proof. Denote
L := K(t1, . . . , tn), M := K(σ1, . . . , σn)
We have M ⊆ LSn so it suffices to show that [LSn :M ] = 1
Notice that L is a splitting field of the separable polynomial
f(x) = xn − σ1xn−1 + · · ·+ (−1)n−1σn−1 + (−1)nσn
where f(x) ∈ M [x], and so L/M is a Galois extension. By Corollary 70.2 weobtain that Gal(L/M) is a subgroup of the group of permutations of the set{t1, . . . , tn} of roots of f(x). From Corollary 72.3 it follows that
[L :M ] = |Gal(L/M)| ≤ |Sn|
On the other hand by Theorem 72.3 we have
[L : LSn ] = |Sn|
Since [L : M ] = [L : LSn ] · [LSn : M ] this implies that [L : M ] = |Sn| and that[LSn :M ] = 1.
283
74 The fundamental theorem of Galois theory
Let L/K be an algebraic extension. We have maps of sets
Φ:
(intermediate fields
K ⊆M ⊆ L
)⇆
(subgroups
H ⊆ Gal(L/K)
): Ψ
where Φ(M) = Gal(L/M) and Ψ(H) = LH .
74.1 Theorem. Let L/K be a Galois extension such that [L : K] <∞.
1) The maps Φ and Ψ are bijections and Ψ = Φ−1.
2) If M is an intermediate field of L/K then M/K is a Galois extension iffΦ(M) = Gal(L/M) is a normal subgroup of Gal(L/K). Moreover in suchcase we have
Gal(M/K) ∼= Gal(L/K)/Φ(M)
Proof.
1) We will show that Ψ(Φ(M)) =M and Φ(Ψ(H)) = H.
Let K ⊆M ⊆ L. We have
Ψ(Φ(M)) = Ψ(Gal(L/M)) = LGal(L/M)
Since L/K is a Galois extension thus L/M is also a Galois extension (72.2), soLGal(L/M) =M .
Next, if H is a subgroup of Gal(L/K) then
Ψ(Φ(H)) = Ψ(LH) = Gal(L/LH)
and by Theorem 72.3 we have Gal(L/LH) = H.
2) (⇒) Assume that M is an intermediate field of L/K such that M/K is aGalois extension.
284
Claim. If φ ∈ Gal(L/K) then φ(M) =M .
Indeed, let K be an algebraic closure of K such that L ⊆ K. We have inclusions
K ⊆M ⊆ L ⊆ K
Let i : M ↪→ K be the inclusion map. For φ ∈ Gal(L/K) consider the embed-ding
φ : M → L ↪→ K
Since M/K is a normal extension by Proposition 71.8 we have
φ(M) = i(M) =M
As a consequence we obtain a group homomorphism
f : Gal(L/K)→ Gal(M/K)
given by f(φ) = φ|M Notice that
Ker(f) = {φ ∈ Gal(L/K) | φ|M = idM} = Gal(L/M) = Φ(M)
Therefore Φ(M) is a normal subgroup of Gal(L/K).
(⇐) Assume that M is an intermediate field of L/K such that Φ(M) =Gal(L/M) is a normal subgroup of Gal(L/K).
Claim. If φ ∈ Gal(L/K) then φ(M) =M .
Indeed, let φ ∈ Gal(L/K) and let a ∈ M . For any ψ ∈ Gal(L/M) we haveφ−1ψφ ∈ Gal(L/M) since Gal(L/M) is a normal subgroup of Gal(L/K). Thisgives φ−1ψφ(a) = a, or equivalently:
ψ(φ(a)) = φ(a)
for all ψ ∈ Gal(L/M). Thus, φ(a) ∈ LGal(L/M). Since L/M is a Galoisextension we have LGal(L/M) = M . It follows that φ(a) ∈ M for any a ∈ Mand so φ(M) ⊆ M . Substituting φ−1 in place of φ we also obtain from hereφ−1(M) ⊆M . This gives
M = φ(φ−1(M)) ⊆ φ(M)
285
and so we obtain that φ(M) =M for all φ ∈ Gal(L/M).
As before we obtain then that there exists a group homomorphism
f : Gal(L/K)→ Gal(M/K)
where f(φ) = φ|M and Ker(f) = Gal(L/M). Moreover we have M Im(f) = K(since LGal(L/K) = K), and so by Theorem 72.3M/K is a Galois extension withGal(M/K) = Im(f). By the first isomorphism theorem for groups we also get
Im(f) ∼= Gal(L/K)/Ker(f) = Gal(L/K)/Gal(L/M) = Gal(L/K)/Φ(M)
74.2 Theorem. Let L/K be a Galois extension. Let H1, H2 be subgroups ofGal(L/K) and let Φ(H1) =M1, Φ(H2) =M2. Then:
1) Φ(H1 ∩H2) =M1M2
2) Φ(⟨H1, H2⟩) = M1 ∩M2 where ⟨H1, H2⟩ is the subgroup of Gal(L/K)generated by H1, H2.
Proof. Exercise.
74.3 Example.
Take the extension L := Q(√2,√3) of Q. This is a Galois extension since
Q(√2,√3) is a splitting field of a polynomial f(x) = (x2 − 2)(x2 − 3) ∈ Q[x].
We have [L : Q] = 4 so |Gal(L/Q)| = 4.
If σ ∈ Gal(L/Q) then σ is determined by the values of σ(√2) and σ(
√3). We
also have σ(√2) = ±
√2 and σ(
√3) = ±
√3. This gives 4 possible combinations
for the values of σ(√2), σ(
√3), and since Gal(L/Q) has 4 elements each of
these possibilities defines an element of Gal(L/Q).
286
There are 2 non-isomorphic groups of order 4: Z/4Z and Z/2Z. Notice that forany σ ∈ Gal(L/Q) we have σ2 = idL. Thus Gal(L/Q) does not contain anyelements of order 4, and so Gal(L/Q) ∼= Z/2Z⊕ Z/2Z.
Let φ, ψ ∈ Gal(L/Q) be the automorphism given by
φ(√2) = −
√2, φ(
√3) =
√3
ψ(√2) =
√2, ψ(
√3) = −
√3
Notice that Gal(L/Q) = {idL, φ, ψ, ψφ} The lattice of subgroups of Gal(L/Q)looks as follows:
Gal(L/Q)
�ϕ� �ψ� �ψϕ�
{idL}
We have:
L⟨φ⟩ = Q(√3), L⟨ψ⟩ = Q(
√2), L⟨ψφ⟩ = Q(
√2√3) = Q(
√6)
Therefore the corresponding lattice of intermediate fields of the extension L/Qlooks as follows:
Q
Q(√
2)Q(√
3) Q(√
6)
L
Notice that since Gal(L/Q) is abelian all of its subgroups are normal and so allextensions M/Q where M = Q(
√2), Q(
√3), Q(
√6) are Galois extensions.
287
74.4 Example.
Take the polynomial p(x) = x4− 2 ∈ Q[x], and let L ⊆ C be a splitting field off(x). The roots of p(x) in C are ±4
√2 and ±i4
√2, so we have
L = Q(4√2, i)
Since p(x) is irreducible in Q[x] we have [Q(4√2) : Q] = 4. Also, [L : Q(4
√2)] =
2, so we obtain
[L : Q] = [L : Q(4√2)] · [Q(4
√2) : Q] = 8
Thus |Gal(L/Q)| = 8.
If σ ∈ Gal(L/Q) then σ is determined by the values of σ(4√2) and σ(i). The
possible values of σ(4√2) are ±4
√2 and ±i4
√2, while the possible values of σ(i)
are ±i. This gives 8 possible combinations, and since Gal(L/Q) has 8 elementseach of these combinations defines an element of Gal(L/Q).
Let φ, ψ ∈ Gal(L/Q) be the automorphism given by
φ(4√2) = i4
√2, φ(i) = i
ψ(4√2) = 4
√2, ψ(i) = −i
Notice thatψφ(4√2) = −i4
√2 = i4
√2 = φψ(4
√2)
so Gal(L/Q) is a non-abelian group. We have a group homomorphism
f : Gal(L/Q)→ Z/2
where for σ ∈ Gal(L/Q) we set f(σ) = 0 if σ(i) = i and f(σ) = 1 if σ(i) = −i.Notice that Ker(f) is the subgroup generated by φ, and since |φ| = 4 we haveKer(f) ∼= Z/4Z. We also have a group homomorphism
g : Z/2Z→ Gal(L/Q)
such that g(1) = ψ. Since fg = idZ/2Z by Proposition 18.6 we obtain thatGal(L/Q) is isomorphic to the semidirect product:
Gal(L/Q) ∼= Z/4Z ⋊ Z/2Z ∼= ⟨φ⟩⋊ ⟨ψ⟩
288
which means that Gal(L/Q) is isomorphic to the dihedral group D4 (18.5).In particular every element of Gal(L/Q) is of the form φiψj for 0 ≤ i ≤ 3,0 ≤ j ≤ 1.
The lattice of subgroups of Gal(L/Q) looks as follows:
Gal(L/Q)
�ϕ�
�ψ�
{idL}
�ϕ2ψ� �ϕ3ψ�
�ϕ2,ψ� �ϕ2,ϕψ�
�ϕ2� �ϕψ�
This corresponds to the following lattice of intermediate fields of L/Q:
Q
L
Q(i)
Q(√
2, i)
Q(√
2)
Q(4√
2) Q(i4√
2)
Q(i√
2)
Q((1 + i)4√
2) Q((1− i)4√
2)
289
75 The fundamental theorem of algebra
Recall:
67.12 Fundamental Theorem of Algebra. The field C of complex numbersis algebraically closed.
75.1 Lemma. 1) If f(x) ∈ R[x] is a polynomial of an odd degree then f(x)has a root in R. In particular if f(x) ∈ R[x] is a polynomial of an odd degreethen f(x) is irreducible iff deg f(x) = 1.
2) If f(x) ∈ C[x] is a polynomial of degree 2 then f(x) splits over C. In particularthere are no extensions L/C such that [L : C] = 2.
Proof. Exercise.
75.2 Lemma. If L/R is an algebraic extension, [L : R] <∞, and [L : R] is anodd number then L = R.
Proof. Since L/R is a separable extension and [L : R] < ∞ thus by (64.2) wehave L = R(a) for some a ∈ L. Let f(x) = irrRa (x). Then deg f(x) = [L : R]is an odd number. Since f(x) is an irreducible polynomial by Lemma 75.1 weobtain [L : R] = deg f(x) = 1. Thus L = R.
75.3 Lemma. If L/R is an algebraic extension such that [L : R] < ∞ then[L : R] = 2n for some n ≥ 0.
Proof. Assume that there exists an odd prime p such that p | [L : R]. Let M bea normal closure of the extension L/K. We have
[M : R] = [M : L] · [L : R]
290
so p | [M : R].
Let P be the Sylow 2-subgroup of Gal(M/R). Since M/R is a Galois extensionwe have |Gal(M/R)| = [M : R], and so p | |Gal(M/R)|. On the other handp ∤ |P | so P = Gal(M/R). It follows that MP =MGal(M/R) = R. Moreover wehave
[M : R] = [M :MP ] · [MP : R]so:
[MP : R] = [M : R]/[M :MP ] = |Gal(M/R)|/|P |and so [MP : R] is an odd number.
As a consequence we obtain that MP/R is an extension of an odd degree andthat MP = R. This is however impossible by Lemma 75.2.
Proof of Theorem 67.12.
Let L/C be an algebraic extension, [L : C] <∞, and let M be a normal closureof L/C. Then M/C is a Galois extension. We have
[M : R] = [M : C] · [C : R] = 2[M : C]
By Lemma 75.3 we have [M : R] = 2n for some n, and thus [M : C] = 2n−1.
Assume that n− 1 > 0. Then Gal(M/C) is a non-trivial 2-group, and so it hasa normal subgroup H such that
Gal(M/C)/H ∼= Z/2Z
By Theorem 74.1 we obtain that MH/C is a Galois extension and
[MH : C] = |Gal(MH/C)| = |Z/2Z| = 2
which is impossible by Lemma 75.1.
Therefore [M : C] = 1, i.e. M = C. Since C ⊆ L ⊆ M this also gives thatL = C. As a consequence we obtain that there are no non-trivial algebraicextensions of C. By Proposition 67.2 this shows that C is an algebraically closedfield.
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76 Infinite Galois extensions
76.1 Note. If L/K is a Galois extension such that [L : K] =∞ then the maps
Φ:
(intermediate fields
K ⊆M ⊆ L
)⇆
(subgroups
H ⊆ Gal(L/K)
): Ψ
need not be bijections since not every subgroup of of Gal(L/K) will be of theform Gal(L/M) for some intermediate field M .
76.2 Example.
TakeL = Q(
√2,√3,√5, . . . )
Then L/Q is a Galois extension, [L : Q] =∞.
If σ ∈ Gal(L/Q) then σ(√p) = ±√p for each prime p. This gives σ2 = idL,
so all elements of Gal(L/Q) are of order 2, and so Gal(L/Q) is an infinitedimensions vector space over F2. If follows that Gal(L/Q) has uncountablymany subgroups H such that [Gal(L/Q) : H] = 2 (check).
Let H be such subgroup and assume that we have an intermediate field K ⊆M ⊆ L such that Gal(L/M) = H. Since H is a normal subgroup of Gal(L/Q)we obtain that M/Q is a Galois extension and
Gal(M/Q) = Gal(L/Q)/H ∼= Z/2Z
In particular [M/Q] = 2 < ∞, so by (64.2) we have M = K(a) for somea ∈ R, degQ(a) = 2. The set of such elements a is countable, so there areonly countably many such intermediate fields M . As a consequence there is asubgroup H ⊆ Gal(L/Q) such that H = Gal(L/M) for any intermediate fieldM .
In order to fix it we need to describe, given a Galois extension L/K, whichsubgroups of Gal(L/K) are of the form Gal(L/M) for some intermediate fieldK ⊆M ⊆ L.
292
Let L/K be an algebraic field extension and let A = {a1, . . . , an}, B ={b1, . . . , bn} be finite subsets of L. Denote
U(A,B) := {σ ∈ Gal(L/K) | σ(ai) = bi}
76.3 Definition/Proposition. For any algebraic field extension L/K the col-lection of subsets
U := {U(A,B) ⊆ Gal(L/K) | A,B ⊆ L, |A| = |B| <∞}
is a basis of some topology on Gal(L/K). This topology is called the Krulltopology.
76.4 Proposition. Let L/K be a algebraic field extension. If K ⊆ M ⊆ L isan intermediate field. then Gal(L/M) is a closed subset of Gal(L/K).
Proof. Let Gal(L/M) denote the closure of Gal(L/M) in the Krull topology onGal(L/K). We will show that Gal(L/M) = Gal(L/M).
Let ψ ∈ Gal(L/M). Take an element a ∈M and let
A := {a}, B := {ψ(a)}
Then U(A,B) is an open neighborhood of ψ inGal(L/K). Since ψ ∈ Gal(L/M)there exists φ ∈ U(A,B) such that φ ∈ Gal(L/M). We have
φ(a) = ψ(a)
On the other hand, since a ∈M we have φ(a) = a. Therefore we obtain ψ(a) =a. Since this holds for an arbitrary element a ∈M we get that ψ ∈ Gal(L/M).
293
76.5 Proposition. Let L/K be a Galois extension and let H ⊂ Gal(L/K) bea closed subgroup. Then H = Gal(L/LH).
Proof. We have H ⊆ Gal(L/LH). Conversely, let φ ∈ Gal(L/LH). We willshow that φ ∈ H.
We argue by contradiction. Assume that φ ∈ H. Since H is a closed sub-group there exists an open neighborhood U(A,B) ⊆ Gal(L/K) such thatφ ∈ U(A,B) and H ∩Gal(L/LH) = ∅.
Let A = {a1, . . . , an} and let N ⊆ L be the normal closure of LH(a1, . . . , an).Since N/LH is a normal extension for any ψ ∈ Gal(L/LH) we have ψ(N) = N .Define
H ′ := {ψ|N : N → N | ψ ∈ H} ⊆ Gal(N/LH)
Notice that φ|N ∈ H ′, so H ′ = Gal(N/LH). Since N/LH is a Galois extensionand [N : LH ] < ∞ by the Fundamental Theorem of Galois Theory (74.1) weobtain
NH′ = NGal(N/LH) = LH
On the other hand if a ∈ NH′then ψ(a) = a for all ψ ∈ H, so a ∈ LH . Thus
NH′ ⊆ LH . Since obviously we also have LH ⊆ NH′we get NH′
= LH whichis a contradiction.
76.6 Corollary. If L/K is a Galois extension then the maps
Φ:
(intermediate fields
K ⊆M ⊆ L
)⇆
(closed subgroups
H ⊆ Gal(L/K)
): Ψ
given by Φ(M) = Gal(L/M), Ψ(H) = LH are bijections and Ψ = Φ−1.
76.7 Note. If L/K is a Galois extension such that [L : K] < ∞ then Krulltopology on Gal(L/K) is the discrete topology (check!), and so any subgroupof Gal(L/K) is a closed subgroup.
294
77 Abelian and cyclic extensions
77.1 Definition. Let L/K be a Galois extension, [L : K] <∞. The extensionL/K is abelian (resp. cyclic) if the group Gal(L/K) is abelian (resp. cyclic).
77.2 Proposition. If L/K is an abelian (resp. cyclic) extension and M is anintermediate field then L/M andM/K are also abelian (resp. cyclic) extensions.
Proof. Exercise.
77.3 Definition. An extension L/K is a cyclotomic extension of degree n if Lis a splitting field of the polynomial f(x) = xn − 1 ∈ K[x].
77.4 Note. If χ(K) = p = 0 and n = mpk then xn − 1 = (xm − 1)pk, so
cyclotomic extensions of orders n and m coincide. Thus we can always assumethat p ∤ n.
77.5 Theorem. Let L/K be a cyclotomic extension of degree n. If χ(K) =p = 0 assume that p ∤ n. Then
1) L = K(ζ) where ζ is a primitive n-th root of unity;
2) Gal(L/K) is isomorphic to a subgroup of (Z/nZ)∗, the multiplicative ofunits of the ring Z/nZ.
295
Proof.
1) Since L is a splitting field of f(x) = xn− 1 thus L contains a primitive n-throot of unity. Moreover, every other root of f(x) = xn − 1 in L is of the formζk for some 1 ≤ k ≤ n, so L = K(ζ).
2) If φ ∈ Gal(L/K) then φ is uniquely determined determined by the value ofφ(ζ) ∈ L. Since ζ is a primitive n-th root of unity φ(ζ) also must be a primitiven-th root of unity. This means that φ(ζ) = ζk where 1 ≤ k ≤ n, gcd(n, k) = 1.
Define a mapΦ: Gal(L/K)→ (Z/nZ)∗
where Φ(φ) = k if φ(ζ) = ζk. Then Φ is a monomorphism and it is a homo-morphism of groups (check!). It follows that Gal(L/K) is isomorphic to thesubgroup Φ(Gal(L/K)) ⊆ (Z/nZ)∗.
77.6 Note. If L is the cyclotomic extension of Q of degree n then
Gal(L/Q) ∼= (Z/nZ)∗
(see Hungerford Ch. V, Proposition 8.3).
77.7 Corollary. If L/K is a cyclotomic extension of degree n then L/K is anabelian extension. Moreover, if n is a prime number and χ(K) ∤ n then L/K isa cyclic extension.
Proof. By Theorem 77.5 the group Gal(L/K) can be identified with a subgroupof (Z/nZ)∗. Since the group (Z/nZ)∗ is abelian thus so is Gal(L/K).
If n is prime then (Z/nZ)∗ is a cyclic group (see Proposition 38.8), so Gal(L/K)is cyclic.
296
77.8 Theorem (Kronecker-Weber).If L/Q is an abelian extension then L ⊆ M where M/Q is some cyclotomicextension of Q.
Proof. See e.g.:
M.J. Greenberg, An elementary proof of the Kronecker-Weber theorem, TheAmerican Mathematical Monthly 81(6) (1974), 601-607.
(with erratum in The American Mathematical Monthly 82(8) (1975), 803)
77.9 Theorem. Let K be a field such that K contains a primitive n-th root ofunity ζ and let L/K be a field extension.
1) If L/K is a cyclic extension and [L : K] = n then L = K(b) for some b ∈ Lsuch that bn ∈ K
2) If L = K(b) for some b ∈ L such that bn ∈ K then L/K is a cyclic extension.
77.10 Lemma. Let L be a field and let φ1, . . . , φn be distinct automorphismsof L. For c1, . . . , cn ∈ L consider the map
ψ : L→ L, ψ(a) =n∑i=1
ciφi(a)
If φ(a) = 0 for all a ∈ L then c1 = · · · = cn = 0
Proof. We argue by induction with respect to n.
If n = 1 the statement is obvious.
297
Assume that the statement of the lemma holds for some n, and let φ1, . . . , φn+1 ∈Aut(L), c1, . . . , cn+1 ∈ L be such that
n+1∑i=1
ciφi(a) = 0
for all a ∈ L. For any a, b ∈ L, a = 0 we have
0 =n+1∑i=1
ciφi(ab)
=n+1∑i=1
ciφi(a)φi(b)
Since a = 0 we have φn+1(a) = 0 so we have
0 =1
φn+1(a)
n+1∑i=1
ciφi(a)φi(b)
=n+1∑i=1
ciφi(a)
φn+1(a)φi(b)
Since also∑
i ciφi(b) = 0 we obtain
0 =n+1∑i=1
ciφi(a)
φn+1(a)φi(b)−
∑i
ciφi(b)
=n+1∑i=1
ci
(φi(a)
φn+1(a)− 1
)φi(b)
=n∑i=1
ci
(φi(a)
φn+1(a)− 1
)φi(b)
By the inductive assumption this gives
ci
(φi(a)
φn+1(a)− 1
)= 0
for all i = 1, . . . , n and all 0 = a ∈ L. If ci = 0 for some 1 ≤ i ≤ n this wouldgive
φi(a) = φn+1(a)
298
for all a ∈ L which is impossible since φ1, . . . , φn+1 are distinct automorphisms.Thus we obtain c1 = . . . = cn = 0, and so also cn+1 = 0.
Proof of Theorem 77.9.
1) Assume that L/K is a cyclic extension, [L : K] = n, and let φ be a generatorof Gal(L/K). Let ψ : L→ L be the map given by
ψ(a) =n∑i=1
ζ iφi(a)
By Lemma 77.10 there exists a ∈ L such that ψ(a) = 0.
Notice that we have
φ(ψ(a)) =n∑i=1
ζ iφi+1(a)
= ζ−1
n∑i=1
ζ i+1φi+1(a)
= ζ−1ψ(a)
By induction we obtain
φk(ψ(a)) = ζ−kψ(a)
for all k.
Claim 1. L = K(ψ(a)).
Indeed, it is enough to show that degK(ψ(a)) = [L : K] = n.
Let f(x) = irrKψ(a)(x). By (70.1) for any 1 ≤ k ≤ n the element φk(ψ(a)) is aroot of f(x). Moreover, if 1 ≤ k, l ≤ n and k = l then we have
φk(ψ(a)) = ζ−kψ(a) = ζ−lψ(a) = φl(ψ(a))
299
so f(x) has n distinct roots in L. It follows that deg f(x) ≥ n, and sodegK(ψ(a)) ≥ n. Also, since ψ(a) ∈ L we have degK(ψ(a)) ≤ [L : K] = n.Therefore degK(ψ(a)) = n.
Claim 2. ψ(a)n ∈ K.
Indeed, notice that since L/K is a Galois extension we have K = LGal(L/K).Also, since φ generates Gal(L/K) we have
K = LGal(L/K) = {c ∈ L | φ(c) = c}Thus it suffices to show that φ(ψ(a)n) = ψ(a)n. We have
φ(ψ(a)n) = φ(ψ(a))n = (ζ−1ψ(a))n = ζ−nψ(a)n = ψ(a)
From Claims 1 and 2 it follows that in the statement of the theorem we can takeb = ψ(a).
2) Assume that L = K(b) for some b ∈ L such that bn ∈ K. We need to showthat L/K is a Galois extension and that Gal(L/K) is a cyclic group.
Consider the polynomial f(x) = xn−bn ∈ K[x]. Notice that f(x) has n distinctroots b, ζb, . . . , ζn−1b ∈ L. It follows that L is a splitting field of f(x) and thatf(x) is a separable polynomial. As a consequence L/K is a Galois extension.
Next, notice that any automorphism φ ∈ Gal(L/K) is uniquely determined bythe value of φ(b), and that φ(b) = ζkb for some 0 ≤ k ≤ n− 1.
Let µn(K) denote the multiplicative group of n-th roots of unity in K (68.2).Define a map
Φ: Gal(L/K)→ µn(K)
by Φ(φ) = ζk if φ(b) = ζkb. Then Φ is a monomorphism, and it is a homomor-phism of groups (check!). It follows that Gal(L/K) is isomorphic to a subgroupof µn(K). By (68.4) the group µn(K) is cyclic, and so Gal(L/K) is also a cyclicgroup.
300
78 Radical extensions
78.1 Definition. A field extension L/K is a radical extension if there existelements a1, . . . , an ∈ L such that L = K(a1, . . . , an) and for every i = 1, . . . , nthere exists mi ≥ 1 such that
amii ∈ K(a1, . . . , ai−1)
78.2 Definition. Let K be an algebraic closure of a field K. An element a ∈ Kcan be expressed by radicals over K if a ∈ L for some radical extension L/K.
78.3 Note. An element a ∈ K can expressed by radicals over K iff a can beobtained from elements of K using arithmetic operators +, −, ·, / and rootextraction.
78.4 Definition. A polynomial f(x) ∈ K[x] can be solved by radicals if allroots of f(x) can be expressed by radicals over K.
78.5 Proposition. If f(x) ∈ K[x] and L is a splitting field of f(x) then f(x)can be solved by radicals iff L ⊆M where M/K is a radical extension.
78.6 Lemma. If K is an algebraic closure of K and L/K, M/K are radicalextensions such that L,M ⊆ K then LM/K is also a radical extension.
Proof. Exercise.
301
Proof of Proposition 78.5.
(⇒) Let L = K(a1, . . . , an) where a1, . . . , an are roots of f(x). Let K be analgebraic closure of K such that L ⊆ K. By assumption for every ai there existsLi ⊆ K such that ai ∈ Li and Li/K is a radical extension. Then we have
L = K(a1, . . . , an) ⊆ L1L2 · · ·Ln
and by Lemma 78.6 L1L2 · · ·Ln/K is a radical extension.
(⇐) Obvious.
78.7 Proposition. Let L/K be a radical extension. If N is a normal closure ofL/K then N/K is also a radical extension.
Proof. Let N be a normal closure of of L/K. We have:
N =∏
φ∈Gal(N/K)
φ(L)
(exercise). Since L/K is a radical extension thus for every φ ∈ Gal(L/K) theextension φ(L)/K is also a radical extension (check!). By Lemma 78.6 it follows
that that(∏
φ∈Gal(N/K) φ(L))/K is also a radical extension.
302
79 Solvable extensions
Recall.
• A group G is solvable if it has a normal series
{e} = G0 ⊆ . . . ⊆ Gk = G
such that for every i the group Gi/Gi−1 is abelian.
• Abelian groups and finite p-groups are solvable.
• The symmetric group Sn is solvable iff n ≤ 4.
• Subgroups and quotients groups of a solvable group are solvable.
• If H ◁G and H, G/H are solvable groups then G is solvable.
79.1 Definition. A Galois extension L/K is solvable if [L : K] < ∞ andGal(L/K) is a solvable group.
79.2 Theorem. Let L/K be a Galois extension such that [L : K] < ∞ andχ(K) = 0. The following conditions are equivalent:
(i) There exists a radical extension M/K such that L ⊆M .
(ii) L/K is a solvable extension.
79.3 Note. Theorem 79.2 holds also if χ(K) = p = 0 and p ∤ [L : K].
79.4 Corollary. If K is a field, χ(K) = 0, f(x) ∈ K[x] and L is a splitting fieldof f(x) then f(x) can be solved by radicals iff L/K is a solvable extension.
Proof. Follows from Proposition 78.5 and Theorem 79.2.
303
Proof of Theorem 79.2.
(i) ⇒ (ii) It is enough to show that for any radical Galois extension N/K thegroup Gal(N/K) is solvable.
Indeed, if L ⊆ M where M/K is a radical extension then let N be a normalclosure of M/K. By Proposition 78.7 N/K is a radical Galois extension. Wehave
K ⊆ L ⊆ N
Since L/K is a Galois extension the group Gal(N/L) is a normal subgroup ofGal(N/K), and
Gal(L/K) ∼= Gal(N/K)/Gal(N/L)
Therefore if Gal(N/K) is solvable then so is Gal(L/K).
Assume then that N/K is a radical Galois extension. Then we have a1, . . . , an ∈N such that
N = K(a1, . . . , an)
and amii ∈ K(a1, . . . , ai−1) for some mi ≥ 1. Denote Li := K(a1, . . . , ai). We
haveK = L0 ⊆ L1 ⊆ . . . ⊆ Ln = N
This gives
Gal(N/K) = Gal(N/L0) ⊇ Gal(N/L1) ⊇ . . . ⊇ Gal(N/Ln) = {idN} (∗)
Let ζ ∈ K be a primitive root of unity of order m1m2 · · ·mn. Assume first thatζ ∈ K.
Claim. For every i = 1, . . . , n the group Gal(N/Li) is a normal subgroup ofGal(N/Li−1) and Gal(N/Li−1)/Gal(N/Li) is a cyclic group.
Indeed, Li = Li−1(ai) and amii ∈ Li−1, and Li−1 contains a primitive root of
unity of degree mi, so by Theorem 77.9 Li/Li−1 is a cyclic Galois extension. Wehave
Li−1 ⊆ Li ⊆ N
304
Since Li/Li−1 is a Galois extension thus Gal(N/Li) is a normal subgroup ofGal(N/Li−1) and
Gal(N/Li−1)/Gal(N/Li) ∼= Gal(Li/Li−1)
As a consequence we obtain that the sequence (∗) is a normal series ofGal(N/K)with cyclic quotients, and so Gal(N/K) is a solvable group.
Next, assume that K does not contain the primitive root ζ. In such case considerthe extensions
K ⊆ K(ζ) ⊆ N(ζ)
The extensions N(ζ)/K, K(ζ)/K are Galois extensions (check!), so we have
Gal(N(ζ)/K(ζ))◁Gal(N(ζ)/K)
andGal(K(ζ)/K) ∼= Gal(N(ζ)/K)/Gal(N(ζ)/K(ζ))
By Theorem 77.5 the group Gal(K(ζ)/K) is abelian. Also, since N(ζ)/K(ζ)is a radical extension thus by the Claim above the group Gal(N(ζ)/K(ζ)) issolvable. As a consequence Gal(N(ζ)/K) is also a solvable group. Finally wehave
K ⊆ N ⊆ N(ζ)
and since N/K is a Galois extension we have
Gal(N/K) ∼= Gal(N(ζ)/K)/Gal(N(ζ)/N)
so Gal(N/K) is also a solvable group.
(ii) ⇒ (i) Let L/K be a solvable extension. We want to show that L ⊆ Mwhere M/K is a radical extension.
Let [L : K] = m and let ζ ∈ L be a primitive root of unity of degree m. Assumefirst that ζ ∈ K.
Since Gal(L/K) is a solvable group we have a normal series
Gal(L/K) = G0 ⊇ G1 ⊇ . . . ⊇ Gn = {idL}
305
such that Gi/Gi+1 is a cyclic group for all i. Take Li := LGi . This gives asequence of field extensions:
K = L0 ⊆ L1 ⊆ . . . ⊆ Ln = L
such that Gi = Gal(L/Li).
Consider the extensions Li ⊆ Li+1 ⊆ L. Since Gi+1 = Gal(L/Li+1) is anormal subgroup of Gi = Gal(L/Li) thus Li+1/Li is a Galois extension, andGal(Li+1/Li) ∼= Gi/Gi+1 is a cyclic group. If [Li+1 : Li] = mi+1 then mi+1 | m,and so Li contains a primitive root of unity of degree mi+1. By Theorem 77.9there exists bi+1 ∈ Li+1 such that Li+1 = Li(bi+1) and b
mi+1
i+1 ∈ Li. As aconsequence L/K is a radical extension.
If ζ ∈ K then consider the extension L(ζ)/K(ζ). This extension is solvable(exercise) and [L(ζ) : K(ζ)] | [L : K]. By the argument above we obtain thatL(ζ)/K(ζ) is a radical extension. Since, K(ζ)/K is also radical extension weobtain that L(ζ)/K is a radical extension and L ⊆ L(ζ). Therefore we can takeM = L(ζ).
79.5 Note. Theorem 79.2 can be modified to hold when χ(K) = p = 0 asfollows. Let L/K be a Galois extension such that [L : K] <∞.
1) If L/K is a radical extension then it is a solvable extension.
2) If L/K is a solvable extension then there exist elements a1, . . . , an ∈ L suchthat L = K(a1, . . . , an) and for each i = 1, . . . , n we have either
amii ∈ K(a1, . . . , ai−1)
for some mi ≥ 1, orapi − ai ∈ K(a1, . . . , ai−1)
(exercise).
306
80 Solvability of polynomials by radicals
80.1 Theorem. If K is a field, χ(K) = 0 and f(x) ∈ K[x] is a polynomialsuch that deg f(x) ≤ 4 then f(x) is solvable by radicals.
80.2 Lemma. If G is a group such that |G| < 60 then G is a solvable group.
Proof. By the Jordan – Holder Theorem (19.10) we have a composition series
{e} ⊆ G1 ⊆ . . . ⊆ Gk = G
The groups Gi/Gi−1 are simple and |Gi/Gi−1| < 60. By Theorem 20.4 it followsthat the groups Gi/Gi−1 are abelian and so G is a solvable group.
Proof of Theorem 80.1. Let L be a splitting field of f(x). Since deg f(x) = 4we have
[L : K] ≤ 4! = 24
so by Lemma 80.2 Gal(L/K) is a solvable group. Therefore by Corollary 79.4f(x) is solvable by radicals.
Note. See e.g. Section 4.1 of
J. J. Rotman Advanced modern algebra, Prentice Hall, 2002
for explicit formulas for roots of polynomials f(x) ∈ Q[x], deg f(x) ≤ 4.
Next goal. There are polynomials f(x) ∈ Q[x], deg f(x) > 4 that are notsolvable by radicals.
307
80.3 Lemma. Let p be a prime number and letG be a subgroup of the symmetricgroup Sp. If p | |G| and G contains any transposition then G = Sp.
Proof. Exercise.
80.4 Proposition. Let p be a prime number and let K be a subfield of R. Letf(x) ∈ K[x] be an irreducible polynomial such that deg f(x) = p, and let L ⊆ Cbe a splitting field of f(x). If f(x) has exactly p− 2 roots in R then
Gal(L/K) ∼= Sp
Proof. Let a1, . . . , ap−2 ∈ L be the real roots of f(x), and let ap−1, ap ∈ Lbe the complex roots. Notice that ap is the complex conjugate of ap−1, i.e.ap = ap−1. We have
L = K(a1, . . . , ap)
and Gal(L/K) can be identified with a subgroup of the group of permutationsof the set {a1, . . . , ap}. Since f(x) is an irreducible polynomial of degree p weget that p divides [L : K] = |Gal(L/K)| (check!). By Lemma 80.3 it will sufficeto show that Gal(L/K) contains a transposition of the set {a1, . . . , ap}.
Take the mapφ : C→ C, φ(z) = z
Then φ is a field homomorphism. We have φ(ai) = ai for 1 ≤ i ≤ p − 2,φ(ap−1) = ap, and φ(ap) = ap−1. If follows that φ|L is an automorphism of L.Since K ⊆ R we also have φ|K = idK . Therefore φ|L ∈ Gal(L/K) and φ|Lcorresponds to the transposition (ap−1, ap) of the set {a1, . . . , ap}.
80.5 Lemma. Let f(x) ∈ R[x] be a polynomial of the form
f(x) = xn + an−1xn−1 + · · ·+ a1x+ a0
If an−1 = an−2 = 0 and all roots of f(x) are real numbers then f(x) = xn.
308
Proof. We havef(x) = (x− b1)(x− b2) · . . .(x− bn)
for some bi ∈ R. Then∑i
bi = an−1 = 0,∑i<j
bibj = an−2 = 0
This gives
0 =
(∑i
bi
)2
=∑i
b2i + 2∑i<j
bibj =∑i
b2i
Therefore bi = 0 for i = 1, . . . , n and so f(x) = xn.
80.6 Proposition. Let q be a prime number and let L be a splitting field of thepolynomial
f(x) = x5 − 2qx− q ∈ Q[x]
Then Gal(L/Q) = S5. As a consequence f(x) cannot be solved by radicals overQ.
Proof. By the Eisenstein Irreducibility Criterion (38.12) f(x) is irreducible inQ[x]. By Proposition 80.4 it is enough to show that f(x) has exactly 3 roots inR.
By Lemma 80.5 f(x) has at least 2 roots in C − R, and so at most 3 roots inR. Moreover we have
f(−q) = −q(q4 − 2q + 1) < 0
f(−1) = q − 1 > 0
f(0) = −1 < 0
f(q) = q(q4 − 2q + 1) > 0
Thus f(x) has a real root in each of the intervals (−q,−1), (−1, 0), and (0, q).It follows that f(x) has exactly 3 real roots.
Propostion 80.6 can be generalized as follows:
309
80.7 Proposition. Let p, q be prime numbers such that p ≥ 5 and let L be asplitting field of the polynomial
f(x) = xp − 2qx− q ∈ Q[x]
Then Gal(L/Q) is not a solvable group and as a consequence f(x) cannot besolved by radicals over Q.
80.8 Lemma. Let G be a solvable subgroup of Sp that acts transitively on theset {1, . . . , p}. If 1 = σ ∈ G then σ(i) = i for at most one i ∈ {1, . . . , p}.
Proof. Exercise.
80.9 Lemma. Let K be a field, let f(x) ∈ K[x] be an irreducible, separablepolynomial, and let L be a splitting field of f(x). If deg f(x) is a prime numberand Gal(L/K) is a solvable group then L = K(a1, a2) where a1, a2 ∈ L are anytwo distinct roots of f(x).
Proof. Let deg f(x) = p and let {a1, . . . , ap} ⊆ L be the set of all roots of f(x).Then Gal(L/K) can be identified with a subgroup of the group of permutationsof {a1, . . . , ap}. Also, since f(x) is an irreducible polynomial Gal(L/K) actstransitively on the set {a1, . . . , ap} (check!).
Take φ ∈ Gal(L/K(a1, a2)). We have φ(ai) = ai for i = 1, 2. By Lemma 80.8we get φ = idL. Therefore Gal(L/K(a1, a2)) = {idL}. Since L/K(a1, a2) is aGalois extension we obtain
K(a1, a2) = LGal(L/K(a1,a2)) = L
310
80.10 Lemma. Let K be a field, let f(x) ∈ K[x] be an irreducible, separablepolynomial, and let L be a splitting field of f(x). If p = deg f(x) is a primenumber and Gal(L/K) is a solvable group then any intermediate field
K ⊆M ⊆ L
contains either 0, 1, or p roots of f(x).
Proof. By Lemma 80.9 if f(x) contains two roots of f(x) then M = L, and soM contains all roots of f(x).
Proof of Proposition 80.7.
The polynomial f(x) is irreducible inQ[x] by the Eisenstein Irreducibility Criterion38.12). Let L ⊆ C be a splitting field of f(x) and let M = L ∩ R.
If Gal(L/K) were a solvable group then by Lemma 80.10 f(x) would have either0, 1, or p roots in M . By Lemma 80.5 f(x) can’t have all real roots thus thenumber of roots in M would have to be either 0 or 1. On the other hand wehave
f(−q) = −qp + 2q2 − q < 0
f(−1) = q − 1 > 0
f(0) = −1 < 0
so f(x) has a real root in each of the intervals (−q,−1) and (−1, 0). It followsthat Gal(L/K) cannot be a solvable group.
311
81 Straightedge and compass constructions
81.1. Motivation. Three great problems of antiquity.
1) Squaring of a circle. Using a straightedge and a compass construct a squarewhose area is equal to the area of a given circle.
2) Doubling of a cube. Using a straightedge and a compass construct a cubewhose volume is double the volume of a given cube.
3) Trisection of an angle Using a straightedge and a compass construct anangle whose measure is one third of the measure of a given angle.
Cartesian reformulation.
1) Squaring of a circle. We can assume that the given circle has its centerat the point (0, 0) and that it passes through the point (1, 0). Squaring of thiscircle amounts to constructing the point with coordinates (
√π, 0)
(1, 0)(0, 0) (√π, 0)
2) Doubling of a cube. We can assume that two vertices of the front face ofthe cube have coordinates (0, 0) and (1, 0). Doubling of this cube amounts toconstructing the point with coordinates (3
√2, 0)
312
(0, 0) (3√
2, 0)(1, 0)
3) Trisection of an angle. We can assume that one of the arms of the givenangle θ coincides with the x-axis. Being given such an angle is equivalent to beinggiven the point with coordinates (cos θ, 0). Trisection of this angle is equivalentto constructing the point with coordinates (cos 1
3θ, 0)
(1, 0)(0, 0)
θ
13θ
(cos θ, 0)
(cos 13θ, 0)
313
81.2. General constructibility problem. Assume that on the coordinate planeR2 we are given points a set of points
S0 = {P1, . . . , Pn}
where Pi = (ai, bi). Assume also that (0, 0), (1, 0) ∈ S0. We can perform twooperations:
• draw a line through any two points in S0
• draw a circle with the center in one point of S0 and passing through anotherpoint of S0.
Let S1 be the set of all intersection points of these lines and circles. Recursively,for each k ≥ 1 let Sk be the set of all points obtained in this way from the setSk−1. Denote
S =∞⋃k=0
Sk
81.3 Definition. A point Q ∈ R2 is constructible from the set S0 if Q ∈ S. Apoint Q ∈ R2 is constructible if it is constructible from the set {(0, 0), (1, 0)}.
Problem. Given a set S0 ∈ R2 determine which points of R2 are constructiblefrom S0.
81.4 Note.
1) Squaring of the circle can be performed if the point (√π, 0) is constructible.
2) Doubling of a cube can be performed if the point (3√2, 0) is constructible.
3) Trisection of the angle θ can be performed if the point (cos 13θ, 0) is con-
structible from the set {(0, 0), (1, 0), (cos θ, 0)}.
314
Constructible points and radical extensions.
81.5 Definition. A field extension L/K is a degree 2 radical extension if thereexist elements a1, . . . , an ∈ L such that L = K(a1, . . . , an) and for every i =1, . . . , n we have
a2i ∈ K(a1, . . . , ai−1)
81.6 Theorem. Let S0 = {P1, . . . , Pn} be set of points in R2 where Pi =(ai, bi). Let
K = Q(a1, b1, . . . , an, bn)
A point Q = (c, d) is constructible from the set S0 iff there exists a degree 2radical extension L/K, L ⊆ R such that c, d ∈ L.
In particular a point Q = (c, d) is constructible iff there exists a degree 2 radicalextension L/Q, L ⊆ C such that c, d ∈ L.
Proof. Exercise.
81.7 Corollary. Let S0 = {P1, . . . , Pn} be a set of points in R2 where Pi =(ai, bi). and let
K = Q(a1, b1, . . . , an, bn)
If a point Q = (c, d) is constructible then [K(c, d) : K] = 2k for some k ≥ 1.
Proof. By Theorem 81.6 we have K(c, d) ⊆ L where L is some degree 2 radicalextension of K. We have
[L : K] = 2m
for some m ≥ 1, so [K(c, d) : K] = 2k for some 1 ≥ k ≥ m.
315
81.8 Example.
Since√π is a transcendental number thus
√π ∈ L for any degree 2 radical
extension of Q. It follows that the point (√π, 0) is not constructible and it is
impossible to square a circle using straightedge and compass.
81.9 Example.
We have [Q(3√2) : Q] = 3. By Corollary 81.7 we obtain that the point (3
√2, 0)
is not constructible. As a consequence it is impossible to double a cube usingstraightedge and compass.
81.10 Example.
Recall that we can perform a trisection of an angle θ iff the point (0, cos 13θ) can
be constructed from the set S0 = {(0, 0), (1, 0), (cos θ, 0)}. By Theorem 81.6this is possible iff cos 1
3θ ∈ L for some degree 2 radical extension L of the field
K := Q(cos θ).
Claim. cos 13θ ∈ L where L is some degree 2 radical extension of K iff the
polynomial
f(x) = x3 − 3
4x− 1
4cos θ
is not irreducible in K[x].
Indeed, by de Moivre formula for any angle α we have:
cos 3α = 4 cos3 α− 3 cosα
It follows from here that cos 13θ is a root of f(x).
If f(x) is irreducible over K then [K(cos 13θ) : K] = 3, and so by Corollary 81.7
the point (cos 13θ, 0) is not constructible from S0.
Conversely, if f(x) is not irreducible then we have
f(x) = (x− a)g(x)
316
where a ∈ K, g(x) ∈ K[x], deg g(x) = 2. Since cos 13θ is a root f(x) we obtain
that either cos 13θ = a or cos 1
3is a root of g(x), and so it can be expressed by
elements of K and square roots of such elements. In both cases cos 13θ belongs
to some degree 2 radical extension of K.
Some special cases:
• θ =π
4
We have cos θ =√22, and so Q(cos θ) = Q(
√2). Moreover
f(x) = x3 − 3
4x−√2
8
Check: f(x) has a root −√22∈ Q(
√2) so it is not irreducible. Thus the angle
θ = π4can be trisected using straightedge and compass.
• θ =π
3
We have cos θ = 12, so Q(cos θ) = Q and
f(x) = x3 − 3
4x− 1
8
Check: f(x) has no roots in Q, and so it is irreducible in Q[x]. As a consequencethe angle θ = π
3cannot be trisected using straightedge and compass.
317
82 Construction of regular polygons
Problem. For which n it is possible to construct a regular polygon with n sidesusing a straightedge and a compass?
Note. Construction of a regular polygon with n sides is equivalent to the con-struction of the point with coordinates (cos 2π
n, sin 2π
n).
(1, 0)(0, 0)
2π
n
(cos 2πn
, sin 2πn
)
82.1 Lemma. The numbers cos 2πn, sin 2π
nbelong to some degree 2 radical
extension L of Q iff ζn = cos 2πn+ i sin 2π
nome degree 2 radical extension of Q.
Proof.
(⇒) If cos 2πn, sin 2π
n∈ L where L is a degree 2 radical extension of Q then
ζn ∈ L(i), and L(i) is a degree 2 radical extension of Q.
(⇐) Assume that ζn ∈ L for some degree 2 radical extension L of Q. We canassume that i ∈ L. We have ζ−1 = cos 2π
n− i sin 2π
n, which gives
cos2π
n=
1
2
(ζn +
1
ζn
), sin
2π
n=
1
2i
(ζn −
1
ζn
)318
Therefore cos 2πn, sin 2π
n∈ L.
82.2 Note. The element ζn is a primitive root of unity of degree n, so Q(ζn)/Qis the cyclotomic Galois extension of degree n (77.3). By (77.6) we have
Gal(Q(ζn)/Q) ∼= (Z/nZ)∗
82.3 Lemma. If L/Q is a Galois extension and [L : Q] < ∞ then L ⊆ M forsome degree 2 radical extension M iff Gal(L/Q) = 2m for some m ≥ 0
Proof. Exercise.
82.4 Corollary. A regular polygon with n sides can be constructed using astraightedge and a compass iff |(Z/nZ)∗| = 2m for some m.
82.5 Note.
1) We have |(Z/nZ)∗| = φ(n) where φ(n) is the Euler function:
φ(n) = |{m | 1 ≥ m ≥ n, gcd(m,n) = 1}|
2) If n = pk11 · . . . · pkrr where p1, . . . , pr are distinct primes then
φ(n) = φ(pk11 ) · . . . · φ(pkrr )
and φ(pkii ) = (pi − 1) · pki−1i .
319
Upshot. The number φ(n) is a power of two iff
n = 2kp1 · . . . · pr
where p1, . . . , pr are distinct primes such that pi = 2mi + 1 for some mi ≥ 1.
Note. If 2m + 1 is a prime then m must be of the form 2s for some s ≥ 0.Prime numbers of the form
Fs = 22s
+ 1
are called Fermat primes.
We obtain:
82.6 Theorem. A regular polygon with n sides can be constructed using astraightedge and a compass iff
n = 2kp1 · . . . · pr
where p1, . . . , pr are distinct Fermat primes.
82.7 Note. The only known Fermat primes are
F0 = 3, F1 = 5, F2 = 17, F3 = 257, F4 = 65537
It is not known whether there are any other prime numbers of this form.
320
83 Transcendental extensions
83.1 Definition. Let L/K be a field extension. A set S ⊆ L is algebraicallyindependent over K if for every finite sequence (s1, . . . , sn) of distinct elementsof S and for every non-zero polynomial f(x1, . . . , xn) ∈ K[x1, . . . , xn] we have
f(s1, . . . , sn) = 0
Otherwise we say that the set S is algebraically dependent over K.
83.2 Note. If L/K is a field extension then an element s ∈ L is transcendentalover K iff the set S = {s} is algebraically independent over K.
83.3 Proposition. Let L/K be a field extension. A set S ⊆ L is algebraicallyindependent over K iff every element s0 ∈ S is transcendental over the fieldK(S − {s0}).
Proof.
(⇒) Assume that there exists an element s0 ∈ S which is algebraic over thefield K(S −{s0}). Then there exists a finite subset {s1, . . . , sn} such that s0 isalgebraic over K(s1, . . . , sn). Let 0 = f(x) ∈ K(s1, . . . , sn)[x] be a polynomialsuch that f(s0) = 0. Notice thatK(s1, . . . , sn) is the field of fractions of the ringK[s1, . . . , sn]. As a consequence we can assume that f(x) ∈ K[s1, . . . , sn][x]:
f(x) = fr(s1, . . . , sn)xr + . . .+ f1(s1, . . . , sn)x+ f0(s1, . . . , sn)
for some fr, . . . , f0 ∈ K[s1, . . . , sn]. Take F ∈ K[x1, . . . , xn, xn+1] given by
F (x1, . . . , xn+1) =
fr(x1, . . . , xn)xrn+1 + . . .+ f1(x1, . . . , xn)xn+1 + f0(x1, . . . , xn)
Then F = 0 and F (s1, . . . , sn, s0) = 0. This shows that the set S is notalgebraically independent over K.
321
(⇐) Assume that the set is not algebraically independent over K, and let n bethe minimal number such that there exists a polynomial 0 = F ∈ K[x1, . . . , xn]satisfying F (s1, . . . , sn) = 0 for some distinct elements s1, . . . , sn ∈ S. We have
F (x1, . . . , xn) =
fr(x1, . . . , xn−1)xrn + . . .+ f1(x1, . . . , xn−1)xn + f0(x1, . . . , xn−1)
for some fr, . . . , f0 ∈ K[x1, . . . , xn−1]. Since F = 0 we must have fi = 0 forsome 0 ≤ i ≤ r. Then by the minimality of n we have fi(s1, . . . , sn−1) = 0.Take G(x) ∈ K(s1, . . . , sn−1)[x] defined by
G(x) = fr(s1, . . . , sn−1)xr + . . .+ f1(s1, . . . , sn−1)x+ f0(s1, . . . , sn−1)
We have G(x) = 0 and G(sn) = F (s1, . . . , sn) = 0. Therefore sn is an algebraicelement over K(s1, . . . , sn−1), and so it is also algebraic over K(S − {sn}).
83.4 Note. If L = K(x1, . . . , xn) is the field of rational functions in variablesx1, . . . , xn with coefficients in K then the set S = {x1, . . . , xn} is algebraicallyindependent over K.
83.5 Proposition. If L/K is a field extension and S = {s1, . . . , sn} is analgebraically independent set over K then we have an isomorphism
K(x1, . . . , xn) ∼= K(s1, . . . , sn)
Proof. Define φ : K(x1, . . . , xn)→ K(s1, . . . , sn) by
φ
(f(x1, . . . , xn)
g(x1, . . . , xn)
)=f(s1, . . . , sn)
g(s1, . . . , sn)
Check that this an isomorphism of fields.
322
83.6 Definition. Let L/K be a field extension. A transcendence basis of theextension L/K is a set S ⊆ L such that
1) S is algebraically independent over K
2) if S ′ ⊆ L is an algebraically independent set such that S ⊆ S ′ then S = S ′.
83.7 Proposition. For any field extension L/K there exists a transcendencebasis of L/K.
Proof. Exercise.
83.8 Theorem. Let L/K be a field extension. A subset S ⊆ L is a transcen-dence basis of L/K iff S a set algebraically independent over K and L/K(S) isan algebraic extension.
83.9 Lemma. Let L/K be a field extension. If S, T ⊆ L are sets such that S isalgebraically independent over K and T is algebraically independent over K(S)then S ∪ T is algebraically independent over K.
Proof. Let s1, . . . , sm ∈ S, t1, . . . , tn ∈ T be distinct elements, and let
0 = F ∈ K[x1, . . . , xm, y1, . . . , yn]
We need to show that F (s1, . . . , sm, t1, . . . , tn) = 0 We have
F =∑i1,...,in
fi1,...,inyi11 · . . . · yinn
for some fi1,...,in ∈ K[x1, . . . , xm]. Since F = 0 we have fi1,...,in = 0 for somei1, . . . in, and so by the algebraic independence of the set S over K we get thatfi1,...,in(s1, . . . , sm) = 0. As a consequence the polynomial
G(y1, . . . , ym) =∑i1,...,in
fi1,...,in(s1, . . . , sm)yi11 · . . . · yinn
323
is a non-zero polynomial in K(S)[y1, . . . , yn]. By the algebraic independence ofthe set T over K(S) we obtain
F (s1, . . . , sm, t1, . . . , tn) = G(t1, . . . , tn) = 0
Proof of Theorem 83.8.
(⇒) Let S be a transcendence basis of L/K. It is enough to show that anyelement a ∈ L−K(S) is algebraic over K(S).
Assume by contradiction that a is transcendental over K(S). Then the set {a}is algebraically independent over K(S), and so, by Lemma 83.9 the set S ∪ {a}is algebraically independent over K. This is impossible since S is a proper subsetof S ∪ {a}.
(⇐) Assume that S is a set algebraically independent over K and that L/K(S)is an algebraic extension. It is enough to show that for any a ∈ L − S the setS ∪ {a} is not algebraically independent over K.
Assume, by contradiction, that for some a ∈ L − S the set S ∪ {a} is alge-braically independent over K. By Proposition 83.3 we would have then that a isa transcendental element over K(S). This contradicts the assumption that allelements of L are algebraic over K(S).
83.10 Definition. A field extension L/K is purely transcendental if L = K(S)where S ⊆ L is a set algebraically independent over K.
83.11 Corollary. For any field extension L/K there exists an intermediate fieldK ⊆ M ⊆ L such that M/K is a purely transcendental extension and L/M isan algebraic extension.
324
Proof. By Theorem 83.8 is it enough to take M = K(S) where S ⊆ L is atranscendence basis of L/K.
83.12 Proposition. If L/K is a field extension and L = K(A) for some setA ⊆ L then there exists S ⊆ A such that S is a transcendence basis of L/K.
Proof. By Zorn’s Lemma 29.10 we can find a maximal subset S ⊆ A that isalgebraically independent over K. We will show S is a transcendence basis ofL/K.
By Theorem 83.8 it suffices to show that L/K(S) is an algebraic extension.Since L = K(A) = K(S)(A − S) this amounts to showing that every elementa ∈ A− S is algebraic over K(S).
Assume, by contradiction, that there exists an element a ∈ A − S that is tran-scendental over K(S). Then the set {a} is algebraically independent over K(S),and so by Lemma 83.9 the set S ∪{a} ⊆ A is algebraically independent over K.This however is impossible by the maximality of S.
83.13 Proposition. If If K ⊆ L ⊆ M are field extensions, S ⊆ L be atranscendence basis of L/K and T ⊆M be a transcendence basis of M/L thenS ∪ T is a transcendence basis of M/K.
Proof. The set T is algebraically independent over L, so it is also algebraicallyindependent over K(S) ⊆ L. By Lemma 83.9 it follows that the set S ∪ T isalgebraically independent over K.
By Theorem 83.8 the field L is an algebraic extension of K(S). It follows thatL(T ) is an algebraic extension of K(S)(T ) = K(S ∪ T ) (check!). Moreover, byTheorem 83.8 again, M is an algebraic extension of L(T ). As a consequence Mis an algebraic extension of K(S ∪ T ). Applying Theorem 83.8 one more timewe get from here that S ∪ T is a transcendence basis of M over K.
325
83.14 Corollary. If L/K is a field extension and S ⊆ L is a set algebraicallyindependent over K then there exists a transcendence basis S ′ of L/K such thatS ⊆ S ′.
Proof. The set S is a transcendence basis of the extension K(S)/K. Let T bea transcendence basis of the extension L/K(S). By Proposition 83.13 S ∪ T isa transcendence basis of L/K. Therefore we can take S ′ = S ∪ T .
83.15 Theorem. If L/K is a field extension then any two transcendence basesof L over K have the same cardinality.
Proof. See Hungerford Theorem 1.8 and Theorem 1.9.
83.16 Definition. The transcedence degree of a field extension L/K is thecardinality of a transcendence basis of L over K. It is denoted by tr(L/K).
83.17 Proposition. If K ⊆ L ⊆M are field extensions then
tr(M/K) = tr(L/K) + tr(M/L)
Proof. Let S ⊆ L be a transcendence basis of L/K and let T ⊆ M be atranscendence basis of M/L over L. By Proposition 83.13 the set S ∪ T is atranscendence basis of M/K. Since S ∩ T = ∅ we obtain
tr(M/K) = |S ∪ T | = |S|+ |T | = tr(L/K) + tr(M/L)
326
84 Algebraic sets
84.1 Definition. Let K be an algebraically closed field, and let
Kn = K × . . .×K︸ ︷︷ ︸n times
A subset A ⊆ Kn is an algebraic set if A is the set of solution of some systemof polynomial equations
f1(x1, . . . , xn) = 0...
fr(x1, . . . , xn) = 0
for some f1, . . . , fr ∈ K[x1, . . . , xn]. In such case we write:
A = V (f1, . . . , fr)
84.2 Note. Every finite subset A = {a1, . . . , am} ⊆ K is an algebraic set sincethis is the set of solutions of the equation
(x− a1) · . . . · (x− am) = 0
84.3 Note. Different sets of equations may give the same algebraic set. Takee.g. f1, f2, g1, g2 ∈ K[x1, x2] given by:
f1(x1, x2) = x1, f2(x1, x2) = x2, g1(x1, x2) = x1+x2, g2(x1, x2) = x1−x2Then
V (f1, f2) = V (g1, g2) = {(0, 0)} ⊆ K2
84.4 Definition. LetK be a field and let I be an ideal ofK[x1, . . . , xn]. Denote:
V (I) := {(a1, . . . , an) | f(a1, . . . , an) = 0 for all f ∈ I}
327
84.5 Proposition. Let K be a field, let f1, . . . , fr ∈ K[x1, . . . , xn], and letI = ⟨f1, . . . , fr⟩ be the ideal generated by f1, . . . fr. Then
V (I) = V (f1, . . . , fr)
Proof. Since {f1, . . . , fr} ⊆ I we have V (I) ⊆ V (f1, . . . , fr).
Conversely, assume that (a1, . . . , an) ∈ V (f1, . . . , fr). If g ∈ I then we have
g =r∑i=1
hifi
for some gi ∈ K[x1, . . . , xn]. Therefore
g(a1, . . . , an) =r∑i=1
gi(a1, . . . , an)fi(a1, . . . , an) = 0
Therefore (a1, . . . , an) ∈ V (I), and as a consequence V (f1, . . . , fr) ⊆ V (I).
84.6 Corollary. If K is a field, f1, . . . , fr, g1, . . . , gs ∈ K[x1, . . . , xn] and
⟨f1, . . . , fr⟩ = ⟨g1, . . . , gs⟩
then V (f1, . . . , fr) = V (g1, . . . , gs).
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85 Hilbert basis theorem
Goal:
85.1 Theorem. If K is a field and I ◁K[x1, . . . , xn] then the ideal I is finitelygenerated:
I = ⟨f1, . . . , fr⟩for some f1, . . . , fr ∈ K[x1, . . . , xn]
85.2 Note. Let K be an algebraically closed set. If I ◁ K[x1, . . . , xn] andI = ⟨f1, . . . , fr⟩ then by Proposition 84.5 we have
V (I) = V (f1, . . . , fr)
i.e. V (I) is an algebraic set. Thus by Theorem 85.1 we obtain an epimorphism:
V :
(ideals
I ◁K[x1, . . . , xn]
)−→
(algebraic sets
A ⊆ Kn
)
which sends an ideal I to the algebraic set V (I).
85.3 Definition. Let R be a commutative ring with identity. The ring R is aNoetherian ring if every ideal I ◁R is finitely generated:
I = ⟨r1, . . . , rn⟩
for some r1, . . . , rn ∈ R.
85.4 Example. If R is a PID then R is a Noetherian ring. In particular
1) Z is Noetherian
2) If K is a field then K[x] is Noetherian.
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85.5 Theorem. Let R be a commutative ring with identity. The followingconditions are equivalent.
1) R is a Noetherian ring.
2) IfI1 ⊆ I2 ⊆ I3 ⊆ . . .
is any increasing sequence of ideals of R then there exists n ≥ 1 such thatIn = Im for all m ≥ n.
3) Every family of ideals of R contains a maximal element.
85.6 Note. The condition 2) is called the ascending chain condition on idealsof R.
Proof of Theorem 85.5.
1) ⇒ 2) Let I1 ⊆ I2 ⊆ I3 ⊆ . . . be a chain of ideals in R. Take
I :=∞⋃k=1
Ik
Then I is an ideal of R, and since R is a Noetherian ring this I = ⟨r1, . . . , rn⟩for some r1, . . . , rn ∈ R. For j = 1, . . . , n there exists kj such that rj ∈ Ikj .Take
n = max{k1, . . . , kn}For any m ≥ n we have r1, . . . , rn ∈ In and so
I ⊆ Im ⊆ I
As a consequence for any m ≥ n we get In = I = Im.
2) ⇒ 3) Let J be a family of ideals in R. Assume that J does not contain amaximal element. Take any I1 ∈ J. Since I1 is not a maximal element of J
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therefore there exists I2 ∈ J such that I1 ⊂ I2 and I1 = I2. By induction wecan find in J an infinite sequence of ideals
I1 ⊂ I2 ⊂ I3 ⊂ . . .
such that Ik = Ik+1 for all k ≥ 1. This however contradicts the assumption thatideals of R satisfy the ascending chain condition.
3) ⇒ 2) Let I be an ideal of R. We need to show that I is finitely generated.
Let J be the family of all finitely generated ideals J such that J ⊆ I. Byassumption J contains a maximal element J0 = ⟨r1, . . . , rn⟩. Assume the J0 = Iand let s ∈ I − J0. Then J1 := ⟨r1, . . . , rn, s⟩ is an ideal such that J1 ∈ J,J0 ⊆ J1, and J0 = J1. This is impossible since J0 is a maximal element of J. Asa consequence we get that I = J0, and so I is fintely generated.
85.7 Hilbert Basis Theorem. Let R be a commutative ring with identity. IfR is a Noetherian ring then R[x] is also Noetherian.
Proof. Let I ◁R[x]. We will show that I is generated by a finite set.
Let In ⊆ R be the set such that a ∈ In if either a = 0 or if there existsf(x) = anx
n + · · ·+ a1x+ a0 such that f(x) ∈ I and a = an. Check: In ◁R.Moreover we have
I0 ⊆ I1 ⊆ . . .
Since R is a Noetherian ring we obtain:
1) for any n ≥ 0 there exists an,1, . . . , an,mn ∈ R such that
In = ⟨an,1, . . . , an,mn⟩
2) there exists N ≥ 0 such that IN = IN+1 = . . . .
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By the definition of In for any 0 ≤ n ≤ N and 1 ≤ i ≤ mn there exists apolynomial fn,i(x) ∈ I such that fn,i(x) = an,ix
n + . . . . Define
S := {fn,i(x) | 0 ≤ n ≤ N, 1 ≤ i ≤ mn}
Notice that S is a finite set and S ⊆ I. We will show that I = ⟨S⟩.
It suffices to show that if g(x) ∈ I then g(x) ∈ ⟨S⟩. We will argue by inductionwith respect to the degree of g(x).
If deg g(x) = 0 then g(x) = b0 for some b0 ∈ I0. Then
b0 =
m0∑i=1
ria0,i
for some ri ∈ R. Since f0,i(x) = a0,i this gives
g(x) =
m0∑i=1
rif0,i(x)
and so g(x) ∈ ⟨S⟩.
Next, assume that for some n > 0 if h(x) ∈ I and deg h(x) ≤ n − 1 thenh(x) ∈ ⟨S⟩. Let g(x) ∈ I be a polynomial of degree n:
g(x) = bnxn + . . . b1x+ b0
Assume first that n ≤ N By the definition of In we have bn ∈ In, and so
bn =mn∑i=1
rian,i
for some ri ∈ R. Definef(x) :=
mn∑i=1
rifn,i(x)
Notice that f(x) = bnxn + . . . , and so deg(g(x) − f(x)) < n. Also, since
g(x), f(x) ∈ I we have g(x) − f(x) ∈ I and thus by the inductive assumptiong(x)− f(x) ∈ ⟨S⟩. Therefore
g(x) = f(x)︸︷︷︸∈⟨S⟩
+(g(x)− f(x))︸ ︷︷ ︸∈⟨S⟩
332
is an element of ⟨S⟩.
If n > N then In = IN so we have
bn =
mN∑i=1
riaN,i
Then define
f(x) := xn−N
(mN∑i=1
rifN,i(x)
)= bnx
n + . . .
Then f(x) ∈ ⟨S⟩ and deg(g(x)− f(x)) < n. Similarly as before we obtain fromhere that g(x) ∈ ⟨S⟩.
85.8 Corollary. If R is a Noetherian ring then the ring R[x1, . . . , xn] is alsoNoetherian for any n ≥ 0.
Proof. This follows from Theorem 85.7 by induction with respect to n.
Proof of Theorem 85.1.
Since K is a field it is a Noetherian ring. Thus we can apply Corollary 85.8
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86 Radical ideals
86.1 Note. Let K be an algebraically closed field. For any subset X ⊆ Kn
define
J(X) := {f ∈ K[x1, . . . , xn] | f(a1, . . . , an) = 0 for all (a1, . . . , an) ∈ X}
Check: J(X) is an ideal of K[x1, . . . , xn].
We get maps of sets:
V :
(ideals
I ◁K[x1, . . . , xn]
)⇄
(algebraic sets
A ⊆ Kn
): J
Notice that
1) if I, I ′ ◁K[x1, . . . , xn] and I ⊆ I ′ then V (I) ⊇ V (I ′)
2) if A,A′ ⊆ Kn are algebraic sets and A ⊆ A′ then J(A) ⊇ J(A′).
86.2 Proposition. If A ⊆ Kn is an algebraic set then V (J(A)) = A.
Proof. If (a1, . . . , an) ∈ A then f(a1, . . . , an) = 0 for all f ∈ J(A). Therefore(a1, . . . , an) ∈ V (J(A)). This show that A ⊆ V (J(A)).
On the other hand, A is an algebraic set, so A = V (I) for some I◁K[x1, . . . , xn].We have I ⊆ J(A) which gives
A = V (I) ⊇ V (J(A))
86.3 Note. It is not true that for any ideal I ∈ K[x1, . . . , xn] we have J(V (I)) =I. Take e.g. I = ⟨x2⟩◁K[x]. We have
V (I) = {a ∈ K | a2 = 0} = {0}
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ThenJ(V (I)) = {g(x) ∈ K[x] | g(0) = 0} = ⟨x⟩
Therefore J(V (I)) = I.
86.4 Definition. Let R be a commutative ring with identity. An ideal I ◁ Ris a radical ideal if for every a ∈ R such that an ∈ I for some n ≥ 1 we havea ∈ I.
86.5 Proposition. If K is an algebraically closed field and A ⊆ Kn is analgebraic set then J(A) is a radical ideal of K[x1, . . . , xn].
Proof. Exercise.
Goal. The following maps are inverse bijections of sets:
V :
(radical ideals
I ◁K[x1, . . . , xn]
)⇄
(algebraic sets
A ⊆ Kn
): J
335
Note. In Sections 87-88 all rings are commutative rings with identity 1 = 0.
87 Integral extensions of rings
87.1 Definition. Let R, S be rings. We say that S is a ring extension of R ifR is a subring of S.
87.2 Notation. If R ⊆ S is a ring extension and A ⊆ S is a subset of S thenR[A] is the smallest subring of S such that R ∪ A ⊆ R[A]. If A is a finite set,A = {a1, . . . , an} then we write
R[A] = R[a1, . . . , an]
87.3 Note. We have
R[a] = {b ∈ S | b = r0 + r1a+ . . .rkak for some r0, . . . , rk ∈ R}
and R[a1, . . . , an] = R[a1, . . . , an−1][an].
87.4 Definition. Let R ⊆ S be a ring extension. An element a ∈ S is anintegral element over R if a is a root of a monic polynomial f(x) ∈ R[x].
87.5 Example.
1) If K, L are fields and K ⊆ L then an element a ∈ L is an integral elementover L iff a is an algebraic element over K.
2)√2 is integral over Z since it is a root of the monic polynomial f(x) =
x2 − 2 ∈ Z[x].
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3) 12is not integral over Z. Indeed, if f(x) ∈ Z[x] is a monic polynomial
f(x) = xn + an−1xn−1 + . . .+ a0
then2nf(1
2) = 1 + 2an−1 + · · ·+ 2na0︸ ︷︷ ︸
odd
= 0
so f(12) = 0.
87.6 Definition. Let R be a ring. An R-module M is faithful if for every0 = r ∈ R there exists m ∈M such that rm = 0.
87.7 Theorem. Let R ⊆ S be a ring extension and let a ∈ S. The followingconditions are equivalent.
1) The element a is integral over R.
2) R[a] is a finitely generated R-module.
3) There exists M ⊆ S such that M is a a faithful R[a]-module and it isfinitely generated as an R-module.
87.8 Lemma. Let S be a ring and let A = (aij) be an n × n matrix withcoefficients in S. If s1, . . . , sn ∈ S are elements such that
A
s1...sn
= 0
then (detA) · si = 0 for i = 1, . . . , n.
Proof. See Hungerford p.354 Exercise 8.
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Proof of Theorem 87.7.
1) ⇒ 2) Since the element a is integral over R there exists a monic polynomialf(x) ∈ R[x] such that f(0). Assume that deg f(x) = n.
If b ∈ R[a] then b = g(a) for some g(x) ∈ R[x]. By (38.1) there existsq(x), r(x) ∈ R[x] such that deg r(x) ≤ n− 1 and
g(x) = q(x)f(x) + r(x)
This gives g(a) = r(a). As a consequence we obtain
R[a] = {r(a) | r(x) ∈ R[x], deg r(x) ≤ n− 1}= {b0 + b1a+ . . .+ bn−1a
n−1 | b1, . . . , bn−1 ∈ R}
Therefore the elements 1, a, . . . , an−1 generate R[a] as an R-module.
2)⇒ 3) R[a] is a faithful R[a]-module and by assumption it is finitely generatedas an R-module. Thus we can take M = R[a].
3) ⇒ 1) Assume that the set {s1, . . . sn} generates M as an R-module. Forevery i = 1, . . . n there exists ai,j ∈ R such that
asi = ai1s1 + . . .+ ainsn
This gives a system of equations:
(a11 − a)s1 + a12s2 + . . . a1nsn = 0
a21s1 + (a22 − a)s2 + . . . a2nsn = 0
......
an1s1 + an2s2 + . . . (ann − a)sn = 0
In the matrix notation this gives:
((aij)− aI)
s1...sn
= 0
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By Lemma 87.8 we getdet((aij)− aI)si = 0
for i = 1, . . . , n. Since the elements s1, . . . , sn generate M this shows thatdet((aij)− aI)m = 0 for all m ∈M . Furthermore, since det((aij)− aI) ∈ R[a]and M is a faithful R[a]-module we obtain that det((aij)− aI) = 0.
Take f(x) = det((aij)− xI) ∈ R[x]. This is a monic polynomial and f(a) = 0.This shows that a is an integral element over R.
87.9 Definition. A ring extension R ⊆ S is an integral extension if every elementof S is integral over R.
87.10 Lemma. Let R ⊆ S ⊆ T be ring extensions. If T is finitely generatedas an S-module and S is finitely generated as an R-module then T is finitelygenerated as an R-module.
Proof. Exercise.
87.11 Proposition. Let R ⊆ S be a ring extension and let a1, . . . , an ∈ S beelements integral over R. Then R[a1, . . . , an] is finitely generated R-module andit is an integral extension of R.
Proof. We argue by induction with respect to n.
If n = 1 then R[a1] is finitely generated R-module by Theorem 87.7. Also, ifb ∈ R[a1] then R[a1] is a faithful R[b]-module. Using Theorem 87.7 again weobtain that b is integral over R.
Assume the statement of the proposition holds for some n, and let a1, . . . , an+1 ∈S be elements integral over R. Then an+1 is integral over R[a1, . . . , an], so
339
R[a1, . . . , an+1] is a finitely generated R[a1, . . . , an]-module. By Lemma 87.10we obtain that R[a1, . . . , an+1] is then a finitely generated R-module.
If b ∈ R[a1, . . . , an+1] then R[a1, . . . , an+1] is a faithful R[b] module, so byTheorem 87.7 b is integral over R.
87.12 Corollary. Let R ⊆ S be a ring extension and let
Rint := {a ∈ S | a is integral over R}then Rint is a subring of S.
Proof. Let a, b ∈ Rint. It is enough to show that a± b, ab are integral elementsover R. This holds since a± b, ab ∈ R[a, b], and by Proposition 87.11 R[a, b] isan integral extension of R.
87.13 Proposition. Let R ⊆ S ⊆ T be ring extensions. If T is an integralextension of S and S is an integral extension of R then T is an integral extensionof R.
Proof. It is enough to show that if b ∈ T then b is integral over R.
Since b is integral over S there exists a monic polynomial f(x) ∈ S[x]f(x) = xn + an−1x
n−1 + . . .+ a0
such that f(b) = 0. Since f(x) ∈ R[an−1, . . . , a0] it follows that b is inte-gral over R[an−1, . . . , a0]. By Theorem 87.7 R[an−1, . . . , a0, b] is then a finitelygenerated R[an−1, . . . , a0]-module. Also, since an−1, . . . , a0 are integral over Rby Proposition 87.11 R[a1, . . . , an−1] is a finitely generated R-module. There-fore, by Lemma 87.10 R[an−1, . . . , a0, b] is a finitely generated R-module. SinceR[an−1, . . . , a0, b] is a faithful R[b] module applying Theorem 87.7 we get thatb is integral over R.
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88 Noether Normalization Lemma
88.1 Noether Normalization Lemma. Let K ⊆ S be a ring extension suchthat K is a field and S is finitely generated over K:
S = K[a1, . . . , an]
for some a1, . . . , an ∈ S. There exists a ring K ⊆ R ⊆ S such that
R ∼= K[x1, . . . , xm]
for some m ≤ n and that R ⊆ S is an integral extension
88.2 Note. Compare this with the decomposition of field extensions into a purelytranscendental extension and an algebraic extension (Corollary 83.11).
Proof of Theorem 88.1.
We argue by induction with respect to n.
For n = 1 we have S = K[a1]. Consider the homomorphism
φ : K[x]→ K[a1]
given by φ(f(x)) = f(a). This is an epimorphism, so if Ker(φ) = {0} we getthat K[x] ∼= K[a1], and then we can take R = S.
If Ker(φ) = {0} then f(a) = 0 for some 0 = f(x) ∈ K[x]. Since K is a fieldwe can assume that f(x) is a monic polynomial. It follows that a is an integralelement over K, and so by (87.11) S is an integral extension of K. In this casetake R = K.
Next, assume that the statement of the theorem holds for some n − 1 and letS = K[a1, . . . , an]. We have a homomorphism
φ : K[x1, . . . , xn]→ K[a1, . . . , an]
341
given by φ(f(x1, . . . , xn)) = f(a1, . . . , an). If Ker(φ) = {0} then as before weobtain K[x1, . . . , xn] ∼= K[a1, . . . , an], so we can take R = S.
Assume then that there exists 0 = f ∈ Ker(φ). We have
f(x1, . . . , xn) =∑
(i1,...,in)∈I
bi1,...,inxi11 . . . x
inn
for some 0 = bi1,...,in ∈ K. Let (j1, . . . , jn) be the element of I that is thebiggest with respect to the lexicographic order. Since K is a field we can assumethat aj1,...,jn = 1.
Taked = max { ik | 1 ≤ k ≤ n, (i1, . . . , in) ∈ I }
and let h ∈ K[x1, . . . , xn] be a polynomial given by
h(x1, . . . , xn) = f(x1 + xdn
n , x2 + xdn−1
n , . . . , xn−1 + xdn, xn)
Claim. Consider h as a polynomial of the variable xn with coefficients in the ringK[x1, . . . , xn−1]. Then h is a monic polynomial.
Indeed, notice that
h =∑
(i1,...,in)∈I
bi1,...,in(x1 + xdn
n )i1(x2 + xdn−1
n )i2 . . . xinn
=∑
(i1,...,in)∈I
bi1,...,in(xi1dn+i2dn−1+...inn + lower powers of xn)
By the choice of (j1, . . . , jn) the highest degree monomial of h is then
bj1,...,jn(xj1dn+j2dn−1+...jnn )
and by assumption bj1,...,jn = 1
Next, let g(x) be the polynomial given by
g(x) = h(a1 − adn
n , a2 − adn−1
n , . . . , an−1 − adn, x)
Notice that
342
(i) g(x) ∈ K[a1 − adnn , . . . , an−1 − adn][x](ii) g(x) is a monic polynomial (since h is monic in xn)
(iii) g(an) = f(a1, . . . , an) = 0 (since f ∈ Ker(φ))
Denote S ′ := K[a1 − adnn , . . . , an−1 − adn]. By (i)-(iii) above we obtain thatan is an integral element over S ′, and so S ′[an] = K[a1, . . . , an] is an integralextension of S ′. On the other hand S ′ is generated over K by n − 1 elements,so by the inductive assumption there is a ring
K ⊆ R ⊆ S ′
such that R ∼= K[x1, . . . xm] for some m ≤ n − 1 and that S ′ is an integralextension of R. Consider the extensions
K ⊆ R ⊆ S
To finish the proof it is enough to notice that since R ⊆ S ′ and S ′ ⊆ S areintegral extensions thus, by Proposition 87.13, the extension R ⊆ S is alsointegral.
88.3 Corollary. Let K be a field and let L be a finitely generated ring extensionof K:
L = K[a1, . . . , an]
If L is a field then it is an algebraic extension of K.
Proof. By the Noether Normalization Lemma 88.1 there exists a ring R suchthat K ⊆ R ⊆ L, R ∼= K[x1, . . . , xm] for some 0 ≤ m ≤ n, and that R ⊆ L isan integral extension.
It is enough to show that R is a field. Indeed, this will imply that m = 0, (i.e.K = R), and that as a consequence K ⊆ L is an integral (and thus algebraic)extension.
Take b ∈ R. Since L is a field we have b−1 ∈ L. We need to show that b−1 ∈ R.
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Since L is an integral extension of R there exists a monic polynomial f(x) =xk + rk−1x
k−1 + . . .+ r1x+ r0 ∈ R[x] such that f(b−1) = 0. This gives
b−k = −rk−1b−(k−1) − rk−2b
−(k−2). . .− r1b−1 − r0
Multiplying both sides by bk−1 we obtain
b−1 = −rk−1 − rk−2b− . . .− r1bk−2 − r0bk−1
Since ri, b ∈ R therefore b−1 ∈ R.
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89 Hilbert Nullstellensatz
Recall.
1) Let K be an algebraically closed field. We have maps
V :
(ideals
I ◁K[x1, . . . , xn]
)⇄
(algebraic sets
A ⊆ Kn
): J
where
V (I) = {(a1, . . . , an) ∈ Kn | f(a1, . . . , an) = 0 ∀f ∈ I}J(A) = {f ∈ K[x1, . . . , xn] | f(a1, . . . , an) = 0 ∀(a1, . . . an) ∈ A }
2) For any algebraic set A ⊆ Kn we have V (J(A)) = A (86.2).
3) If A ⊆ Kn is an algebraic set then J(A) is a radical ideal (86.5).
Goal. If I◁K[x1, . . . , xn] is a radical ideal then J(V (I)) = I. As a consequencewe have inverse bijections of sets
V :
(radical ideals
I ◁K[x1, . . . , xn]
)⇄
(algebraic sets
A ⊆ Kn
): J
89.1 Theorem (Weak Nullstellensatz). Let K be an algebraically closed field.If I ∈ K[x1, . . . , xn] then V (I) = ∅ iff I = K[x1, . . . , xn]
89.2 Note. If I ◁K[x] then I = ⟨f(x)⟩ for some f(x) ∈ K[x] since K[x] is aPID. This gives
V (I) = {a ∈ K | f(a) = 0}
345
Therefore in the case of a single variable Theorem 89.1 says that a polynomialf(x) ∈ K[x] has no roots iff ⟨f(x)⟩ = K[x], i.e. iff f(x) is a non-zero con-stant polynomial. This property defines algebraically closed fields, so in this caseTheorem 89.1 is tautologically true.
Proof of Theorem 89.1.
If I = K[x1, . . . , xn] then 1 ∈ I, and so V (I) = ∅.
Assume then that I = K[x1, . . . , xn]. By (29.1) there exists a maximal idealP such that I ⊆ P . We have V (P ) ⊆ V (I), so it will suffice to show thatV (P ) = ∅.
Letφ : K[x1, . . . , xn]→ K
be a ring homomorphism and let ai = φ(xi) for i = 1, . . . , n. Notice that ifφ|K = idK then φ is given by
φ(f(x1, . . . , xn)) = f(a1, . . . , an)
Moreover, (a1, . . . , an) ∈ V (P ) iff P ⊆ Kerφ. This shows that we have abijection
Φ:
ring homomorphisms
φ : K[x1, . . . , xn]→ K
φ|K = idK , P ⊆ Ker(φ)
−→ points
(a1, . . . , an) ∈ V (P )
where Φ(φ) = (φ(x1), . . . , φ(xn)). In order to show that V (P ) = ∅ is suf-fices then to prove that there exists a homomorphism φ : K[x1, . . . , xn] → Ksatisfying the above conditions.
Take the quotient ring K[x1, . . . , xn]/P and let
π : K[x1, . . . , xn]→ K[x1, . . . , xn]/P
346
be the canonical epimorphism. Identifying a ∈ K with π(a) = a + P we getthat π|K = idK and we can consider K[x1, . . . , xn]/P as a ring extension of K.This extension is finitely generated:
K[x1, . . . , xn]/P = K[π(x1), . . . , π(xn)]
Since P is a maximal ideal K[x1, . . . , xn]/P is a field. By Corollary 88.3 weobtain that K[x1, . . . , xn]/P is an algebraic extension of K. Since K is analgebraically closed field this implies that K[x1, . . . , xn]/P = K. Therefore wecan take φ := π.
89.3 Corollary. If K is an algebraically closed field and P ◁K[x1, . . . , xn] is amaximal ideal then
P = ⟨x1 − a1, . . . , xn − an⟩for some a1, . . . , an ∈ K.
Proof. Since P = K[x1, . . . , xn] by Theorem 89.1 we get V (P ) = ∅. Let(a1, . . . , an) ∈ V (P ). Denote
I := ⟨x1 − a1, . . . , xn − an⟩
Notice that the set V (I) = {(a1, . . . , an)}. Since V (I) = ∅ the ideal I is aproper ideal of K[x1, . . . , xn]. Also, since V (I) ⊆ V (P ) we have P ⊆ I. Bymaximality of P we obtain P = I.
89.4 Definition. Let R be a commutative ring and let I ◁R. The radical of Iis the ideal √
I = {a ∈ R | an ∈ I for some n ≥ 1}
89.5 Note. If I ◁R then√I = I iff I is a radical ideal.
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89.6 Theorem (Hilbert Nullstellensatz). Let K be an algebraically closed field.If I ◁K[x1, . . . , xn] then
J(V (I)) =√I
In particular if I is a radical ideal then J(V (I)) = I.
Proof. Since J(V (I)) is a radical ideal and I ⊆ J(V (I)) we have√I ⊆
J(V (I)). It remains to show then that J(V (I)) ⊆√I.
Let 0 = F ∈ J(V (I)). It will be enough to show that FN ∈ I for some N ≥ 1.
Consider the inclusion homomorphism
K[x1, . . . , xn] ↪→ K[x1, . . . , xn, y]
Let g := (1− yF ) ∈ K[x1, . . . , xn, y] and let J ◁K[x1, . . . , xn, y] be the idealgiven by the set I ∪ {g}.
Claim. V (J) = ∅.
Indeed, assume that (a1, . . . , an, b) ∈ V (J). Since F ∈ J(V (I)) we haveF (a1, . . . , an) = 0. Therefore
g(a1, . . . , an, b) = 1− bF (a1, . . . , an) = 1
which is a contradiction since g ∈ J .
By the Weak Nullstellensatz (89.1) we obtain that J = K[x1, . . . , xn, y]. Thismeans that there exists f1, . . . , fk ∈ I and h0, h1, . . . hk ∈ K[x1, . . . , xn, y] suchthat
gh0 +k∑i=1
fihi = 1 (∗)
Consider the ring homomorphism
φ : K[x1, . . . , xn, y]→ K(x1, . . . , xn)
such that φ|K = idK , φ(xi) = xi, φ(y) =1F. Notice that φ(g) = 1− 1
FF = 0.
Therefore from the equation (∗) we obtain
1 = φ(1) =k∑i=1
fiφ(hi)
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For i = 1, . . . k we have hi =∑ni
j=0 hijyj for some hij ∈ K[x1, . . . , xn]. This
gives:
1 =k∑i=1
fiφ(hi) =k∑i=1
fi
(ni∑j=1
hij1
F j
)Take N = max{n1, . . . , nk}. We obtain
FN = FN
(k∑i=1
fi
(ni∑j=1
hij1
F j
))=
k∑i=1
fi
(ni∑j=1
hijFN−j
)
where N − j ≥ 0. Since fi ∈ I and hijFN−j ∈ K[x1, . . . , xn] for all i, j this
shows that FN ∈ I.
89.7 Corollary. Let K be an algebraically closed field. We inverse bijections ofsets:
V :
(radical ideals
I ◁K[x1, . . . , xn]
)⇄
(algebraic sets
A ⊆ Kn
): J
Proof. This follows directly from Proposition 86.2 and Theorem 89.6.
89.8 Corollary. If I1, I2 ◁K[x1, . . . , xn] then V (I1) = V (I2) iff√I1 =
√I2.
Proof. If I ◁K[x1, . . . xn] then by (86.2) and (89.6) we have
V (I) = V (J(V (I))) = V (√I)
Therefore if√I1 =
√I2 then V (I1) = V (I2).
Conversely, if V (I1) = V (I2) then√I1 = J(V (I1)) = J(V (I2)) =
√I2
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90 Zariski topology
Recall. If R is a commutative ring and I, I ◁R then
I + J := {a+ b | r ∈ I, s ∈ J}
In general, if {Ii}i∈S then
∑i∈S
Ii :=
{∑i∈S
ai
∣∣∣∣ ai ∈ Ii, ai = 0 for finitely many i only
}∑
i∈S Ii is the smallest ideal of R containing Ii for all i ∈ S.
90.1 Proposition. Let K be an algebraically closed set.
1) If {I1, . . . , Ik} is a finite family of ideals of K[x1, . . . , xn] then
V (I1) ∪ · · · ∪ V (Ik) = V (I1 ∩ · · · ∩ Ik)
2) If {Ii}i∈S is an arbitrary family of ideals of K[x1, . . . , xn] then⋃i∈S
V (Ii) = V (∑
i∈S Ii)
Proof.
1) Denote J := I1 ∩ · · · ∩ Ik. Since J ⊆ Ii for i = 1, . . . , k thus V (Ii) ⊆ V (J)and so
n⋃i=1
V (Ii) ⊆ V (J)
Conversely, assume that a ∈ ⋃ni=1 V (Ii). Then for i = 1, . . . , k there exists
fi ∈ Ii such that fi(a) = 0. Take f = f1 · . . . · fn. We have f ∈ J , andf(a) = 0. Thus a ∈ V (J). Therefore we obtain
V (J) ⊆n⋃i=1
V (Ii)
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2) Denote J :=∑
i∈S Ii. Since Ii ⊆ J for all i ∈ S thus V (J) ⊆ V (Ii) and so
V (J) ⊆⋂i∈S
V (Ii)
Conversely, if a ∈ ⋂i∈S V (Ii) then f(a) = 0 for all f ∈ Ii Since the set⋃i∈S Ii
generates J we get that f(a) = 0 for all f ∈ J , and so a ∈ V (J).
90.2 Corollary. Let K be an algebraically closed field.
1) ∅ and Kn are algebraic sets in Kn.
2) If {A1, . . . An} is a finite family of algebraic sets in Kn then⋃ni=1Ai is
also an algebraic set.
3) If {Ai}i∈S is an arbitrary family of algebraic sets in Kn then⋂i∈S Ai is
also an algebraic set.
Proof.
1) We have ∅ = V (⟨1⟩) and Kn = V ({0}).
2) – 3) This follows directly from Proposition 90.1.
90.3 Definition/Proposition. Let K be an algebraically closed field. Thereexists a topology on Kn such that closed sets in this topology are algebraic setsA ⊆ Kn. This topology is called the Zariski topology on Kn.
Proof. This follows directly from Corollary 90.2.
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90.4 Note. A set A ⊆ K1 is algebraic iff A is the set of zeros of some polynomialf(x) ∈ K[x]. If f(x) = 0 then A = K1, if f(x) = 1, then A = ∅, and ifdeg f(x) > 0 then A is a finite set. Thus the only closed sets in the Zariskitopology on K1 are K1, ∅ and all finite subsets A ⊆ K1.
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91 Algebraic varieties
91.1 Definition. Let K be an algebraically closed field. An algebraic set A ⊆Kn is irreducible if for any algebraic sets A1, A2 ⊆ Kn satisfying A = A1 ∪ A2
we have either A = A1 or A = A2.
An irreducible algebraic set is called an algebraic variety.
91.2 Theorem. Let K be an algebraically closed field. Every algebraic set inKn is a union of a finite number of algebraic varieties.
Proof. We argue by contradiction. Let J be the family of all radical idealsI ◁K[x1, . . . , xn] such that V (I) is not a union of a finite number of algebraicvarieties. Assume that J = ∅.
Since the ring K[x1, . . . , xn] is Noetherian, by Theorem 85.5 there exists an idealJ that is a maximal element in J. The set V (J) is not an algebraic variety (sinceby the definition of J it is not a union of a finite number of algebraic varieties),so
V (J) = V (I1) ∪ V (I1)
for some radical ideals I1, I2◁K[x1, . . . , xn]. such that V (I1) = V (J) = V (I2).Since V (Ii) ⊆ V (J) for i = 1, 2 we have J ⊆ Ii. Also, since V (J) = V (Ii)and J , Ii are radical ideals, thus by Corollary 89.7 we have J = Ii for i = 1, 2.Since J is a maximal element in J this implies that I1, I2 ∈ J. As a consequencethe sets V (I1) and V (I2) are finite unions of algebraic varieties. Since V (J) =V (I1)∪V (I2) this implies that V (J) is also a union of a finite number of algebraicvarieties which is a contradiction.
91.3 Definition. Let K be an algebraically closed field, let A ⊆ Kn be analgebraic set and let
A = A1 ∪ · · · ∪ Ar (∗)
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where A1, . . . , Ar are algebraic varieties. The decomposition (∗) is irredundantif
A = A1 ∪ · · · ∪ Ai−1 ∪ Ai+1 ∪ · · · ∪ Arfor all i = 1, . . . , r.
91.4 Theorem. Let K be an algebraically closed field and let A ⊆ Kn bean algebraic set. Assume we have two irredundant decompositions of A intoalgebraic varieties:
B1 ∪ · · · ∪Br = A = C1 ∪ · · · ∪ Cs
Then s = r and there exists a permutation σ : {1, . . . , r} → {1, . . . , r} suchthat Bi = Cσ(i) for i = 1, . . . , r.
Proof. Let 1 ≤ i ≤ r. We have
Bi = (Bi ∩ C1) ∪ · · · ∪ (Bi ∩ Cs)
and since Bi is an algebraic variety thus Bi = Bi ∩Cσ(i) for some 1 ≤ σ(i) ≤ s.Therefore Bi ⊆ Cσ(i). By the same argument we obtain that Cσ(i) ⊆ Bk forsome 1 ≤ k ≤ r. This gives
Bi ⊆ Cσ(i) ⊆ Bk
Since the decomposition A = B1 ∪ · · · ∪ Br is irredundant we have i = k, andso Bi = Cσ(i). Since the decomposition A = C1 ∪ · · · ∪ Cs is irredundant themap
σ : {1, . . . , r} → {1, . . . , s}is onto, and in particular s ≤ r. On the other hand, since the decompositionA = B1∪· · ·∪Br is irredundant σ must be 1-1, and so r ≤ s. As a consequencer = s and σ is a bijection.
91.5 Theorem. LetK be an algebraically closed field. An algebraic set V ⊆ Kn
is an algebraic variety iff J(V )◁K[x1, . . . , xn] is a prime ideal.
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91.6 Lemma. Let K be an algebraically closed field. If I1, I2 ◁K[x1, . . . , xn]then √
I1I2 =√I1 ∩ I2
As a consequence we have
V (I1I2) = V (I1 ∩ I2) = V (I1) ∪ V (I2)
Proof. Exercise.
Proof of Theorem 91.5.
(⇒) Assume that V is an algebraic variety, and let I1, I2 ◁ K[x1, . . . , xn] beideals such that I1I2 ⊆ J(V ). By Proposition 28.7 it suffices to show that eitherI1 ⊆ J(V ) or I2 ⊆ J(V ).
By Lemma 91.6 we have
V ⊆ V (I1I2) = V (I1 ∩ I2) = V (I1) ∪ V (I2)
As a consequenceV = (V ∩ V (I1)) ∪ (V ∩ V (I1))
Since V is a variety we obtain that either V = V ∩ V (I1) or V = V ∩ V (I2).We can assume that V = V ∩V (I1). This gives V ⊆ V (I1), and so I1 ⊆ J(V ).
(⇐) Assume that J(V ) is a prime ideal and let
V = V1 ∪ V2where V1, V2 are algebraic sets. For i = 1, 2 let Ji = J(Vi). Since J1, J2 areradical ideals the ideal J1 ∩ J2 is also radical (check). By Proposition 90.1 wehave
V (J1 ∩ J2) = V (J1) ∪ V (J2) = V1 ∪ V2 = V
so the bijective correspondence (89.7) implies that J(V ) = J1∩J2. This in turngives
J1J2 ⊆ J(V )
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Since J(V ) is a prime ideal by (28.7) we have that either J1 ⊆ J(V ) or J2 ⊆J(V ). This gives that either V ⊆ V (J1) = V1 or V ⊆ V (J2) = V2. Thereforeeither V = V1 or V = V2.
91.7 Corollary. Let K be an algebraically closed set. We have inverse bijectionsof sets:
V :
(prime ideals
I ◁K[x1, . . . , xn]
)⇄
(algebraic varieties
A ⊆ Kn
): J
Proof. This follows from Corollary 89.7 and Theorem 91.5.
91.8 Corollary. Let K be an algebraically closed field and let I◁K[x1, . . . , xn]be a radical ideal. Then there exist prime ideals P1, . . . , Pm◁K[x1, . . . , xn] suchthat
I = P1 ∩ · · · ∩ Pn
Proof. By Theorem 91.2 we have
V (I) = V1 ∪ · · · ∪ Vk
where V1, . . . , Vk are algebraic varieties. Let Pi = J(Vi). By Theorem 91.5P1, . . . , Pn are prime ideals. Since Vi = V (Pi) we obtain
V (I) = V (P1) ∪ · · · ∪ V (Pk) = V (P1 ∩ · · · ∩ Pk)
Both I and P1 ∩ · · · ∩ Pk are radicals ideals so by (89.7) we have
I = P1 ∩ · · · ∩ Pk
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91.9 Corollary. Let K be an algebraically closed field and let I◁K[x1, . . . , xn].Then √
I =⋂i∈S
Pi
where {Pi}i∈S is the family of all prime ideals of K[x1, . . . , xn] such that I ⊆ Pi.
Proof. Since prime ideals are radical ideals, for any prime ideal P such thatI ⊆ P we have
√I ⊆ P . Therefore
√I ⊆
⋂i∈S
Pi
Conversely, by Corollary 91.8 there exist ideals Pi1 , . . . , Pik ∈ {Pi}i∈S such that√I = Pi1 ∩ · · · ∩ Pik . This gives⋂
i∈S
Pi ⊆ Pi1 ∩ · · · ∩ Pik =√I
As a consequence√I =
⋂i∈S Pi.
91.10 Note.
1) The statement of Corollary 91.9 holds if we replace K[x1, . . . , xn] by anarbitrary commutative ring with identity (exercise).
2) As a consequence for the zero ideal {0}◁ R we obtain that√{0} is the
intersection of all prime ideals of R. Notice that we have√{0} = {a ∈ R | ak = 0 for some k ≥ 0}
The ideal√{0} is called the nilradical of R. We denote:
nil(R) :=√{0}
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92 Regular functions
92.1 Definition. Let K be an algebraically closed field and let V ⊆ Kn be analgebraic set. A regular function is a function V → K given by
(a1, . . . , an) 7→ f(a1, . . . , an)
where f is some polynomial in K[x1, . . . , xn].
92.2 Note.
1) Different polynomials may define the same regular function. E.g. if V ={0} ⊆ K1 then f(x) = x and g(x) = x2 + x define the same regularfunction since f(0) = g(0) = 0.
2) Polynomials f, g ∈ K[x1, . . . , xn] define the same regular function on Viff for every a ∈ V we have
f(a)− g(a) = 0
Equivalently, f, g define the same regular function on V iff f − g ∈ J(V ).This gives a bijection of sets:(
regular functions
V → K
)⇄
(elements of the rings
K[x1, . . . , xn]/J(V )
)
92.3 Definition. Let K be an algebraically closed field and let V ⊆ be analgebraic set. The coordinate ring of V is the ring
K[V ] := K[x1, . . . , xn]/J(V )
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92.4 Note.
1) The ring K[V ] is a finitely generated K-algebra.
2) Check: since J(V ) is a radical ideal we have nil(K[V ]) = {0} (see 91.10).3) If V is an algebraic variety then J(V ) is a prime ideal and so K[V ] is an
integral domain.
92.5 Proposition. Let K be an algebraically closed field. If A is a finitelygenerated K-algebra such that nil(A) = {0} then there exists an algebraic setV such that A = K[V ]. Moreover, if A is in integral domain then V is analgebraic variety.
Proof. Since A is finitely generated we have A = K[a1, . . . , an] for some ele-ments a1, . . . , an ∈ A. We have a ring epimorphism
φ : K[x1, . . . , xn]→ A
given by φ(xi) = ai for i = 1, . . . , n and φ|K = idK . Take V := V (Kerφ).Since nil(A) = {0} thus Ker(φ) is a radical ideal (check!) and so by Theorem89.6 we have J(V ) = Ker(φ). Therefore we have
K[V ] = Ker[x1, . . . , xn]/Kerφ ∼= A
If A is an integral domain then Ker(φ) is a prime ideal and so V is an algebraicvariety.
92.6 Definition. Let K be an algebraically closed field and let V ⊆ Kn, W ⊆Km be algebraic sets. Let pi : W → K be the map given by
pi(a1, . . . , am) = ai
A mapφ : V → W
is a morphism of algebraic sets if piφ : V → K is a regular function for i =1, . . . ,m.
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92.7 Note. If V is an algebraic set over K then morphisms V → K coincidewith regular functions on V (check).
92.8 Proposition. IfK is an algebraically closed field and φ : V → W , ψ : W →Z are morphism of algebraic sets over K then ψφ : V → Z is also a morphismof algebraic sets.
Proof. Exercise.
92.9 Definition. For an algebraically closed field K denote by AlgSet(K) thecategory of algebraic sets over K and morphisms of algebraic sets. Let Var(K)denote the subcategory of algebraic varieties over K.
92.10 Note. Let K be an algebraically closed field and let φ : V → W be amorphism of algebraic sets over K. For any regular function f : W → K themap fφ : V → K is a regular function on V . Using the bijective correspondencebetween regular functions on W and elements of the coefficient ring K[W ] weobtain a map
φ♯ : K[W ]→ K[V ]
Check: φ♯ is a homomorphism of K-algebras.
92.11 Proposition. Let K be an algebraically closed field and let FDK be thecategory of finitely generated K-algebras that are integral domains. The functor
F : Var(K)→ FDK
given by F (V ) = K[V ] and F (φ) = φ♯ is an equivalence of categories.
92.12 Corollary. If K is an algebraically closed field and V,W are algebraicvarieties over K then V ∼= W in Var(K) iff K[V ] ∼= K[W ] as K-algebras.
360
93 Suggested further reading
1) M. Aschbacher, Finite group theory, 2000.
2) J.P. Serre, Linear representations of finite groups, GTM 42, 1977
3) J.P. Serre, Trees, 2003
4) K. S. Brown, Cohomology of groups, GTM 87, 1982.
5) Ch. Weibel, Homological algebra, 1995.
6) M.I. Atiyah, I.G. MacDonald, Introduction to commutative algebra, 1969.
7) D. Eisenbud, Commutative algebra with a view toward algebraic geometry,GMT 150, 1995.
8) I.N. Herstein, Noncommutative rings, 1968.
9) J.W. Milnor, Introduction to algebraic K-theory, 1972.
10) S. Mac Lane, Categories for the working mathematician, GTM 5, 1998.
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