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,๐)of](https://reader035.vdocuments.us/reader035/viewer/2022071610/6149dc5712c9616cbc69097f/html5/thumbnails/1.jpg)
University ofCambridge
Mathematics Tripos
Part IB
Groups, Rings and Modules
Lent, 2017
Lectures byO. Randal-Williams
Notes byQiangru Kuang
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,๐)of](https://reader035.vdocuments.us/reader035/viewer/2022071610/6149dc5712c9616cbc69097f/html5/thumbnails/2.jpg)
Contents
Contents
1 Groups 21.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Normal subgroups, Quotients and Homomorphisms . . . . . . . . 31.3 Actions & Permutations . . . . . . . . . . . . . . . . . . . . . . . 71.4 Conjugacy class, Centraliser & Normaliser . . . . . . . . . . . . . 101.5 Finite ๐-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Finite abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . 121.7 Sylowโs Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Rings 172.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Homomorphism, Ideals and Isomorphisms . . . . . . . . . . . . . 192.3 Integral domain, Field of fractions, Maximal and Prime ideals . . 242.4 Factorisation in integral domains . . . . . . . . . . . . . . . . . . 272.5 Factoriation in polynomial rings . . . . . . . . . . . . . . . . . . . 312.6 Gaussian integers . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.7 Algebraic integers . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.8 Hilbert Basis Theorem . . . . . . . . . . . . . . . . . . . . . . . . 38
3 Modules 403.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2 Direct Sums and Free Modules . . . . . . . . . . . . . . . . . . . 433.3 Matrices over Euclidean Domains . . . . . . . . . . . . . . . . . . 463.4 F[๐]-modules and Normal Form . . . . . . . . . . . . . . . . . . 533.5 Conjugacy* . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Index 60
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1 Groups
1 Groups
1.1 Definitions
Definition (Group). A group is a triple (๐บ, โ , ๐) of a set ๐บ, a functionโ โ โ โถ ๐บ ร ๐บ โ ๐บ and ๐ โ ๐บ such that
โข associativity: for all ๐, ๐, ๐ โ ๐บ, (๐ โ ๐) โ ๐ = ๐ โ (๐ โ ๐),
โข identity: for all ๐ โ ๐บ, ๐ โ ๐ = ๐ = ๐ โ ๐,
โข inverse: for all ๐ โ ๐บ, there exists ๐ผโ1 โ ๐บ such that ๐ โ ๐โ1 = ๐ =๐โ1 โ ๐.
Definition (Subgroup). If (๐บ, โ , ๐) is a group, ๐ป โ ๐บ is a subgroup if
โข ๐ โ ๐ป,
โข for all ๐, ๐ โ ๐ป, ๐ โ ๐ โ ๐ป.
This makes (๐ป, โ , ๐) into a group. Write ๐ป โค ๐บ.
Lemma 1.1. If ๐ป โ ๐บ is non-empty and for all ๐, ๐ โ ๐ป, ๐ โ ๐โ1 โ ๐ป then๐ป โค ๐บ.
Example.
1. Additive groups: (N, +, 0), (R, +, 0), (C, +, 0).
2. Groups of symmetries: ๐๐, ๐ท2๐, GL๐(R). They have subgroups ๐ด๐ โค ๐๐,๐ถ๐ โค ๐ท2๐, SL๐(R) โค GL๐(R).
3. An group ๐บ is abelian is a group such that ๐ โ ๐ = ๐ โ ๐ for all ๐, ๐ โ ๐บ.
If ๐ป โค ๐บ, ๐ โ ๐บ, we define the left ๐ป-coset of ๐บ to be
๐๐ป = {๐โ โถ โ โ ๐ป}.
As we have seen in IA Groups, the ๐ป-cosets form a partition of ๐บ and are inbijection with each other via
๐ป โ ๐๐ปโ โฆ ๐โ
๐โ1โ โค โ
We write ๐บ/๐ป for the set of left cosets.
Theorem 1.2 (Lagrange). If ๐บ is a finite group and ๐ป โค ๐บ then
|๐บ| = |๐ป| โ |๐บ/๐ป|.
We call |๐บ/๐ป| the index of ๐ป in ๐บ.
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1 Groups
Definition (Order). Given ๐ โ ๐บ, the order of ๐ is the smallest ๐ such that๐๐ = ๐. We write ๐ = ๐(๐) = |๐|. If no such ๐ exists then ๐ has infiniteorder.
Recall that if ๐๐ = ๐ then |๐| โฃ ๐.
Lemma 1.3. If ๐บ is finite and ๐ โ ๐บ then |๐| โฃ |๐บ|.
Proof. The setโจ๐โฉ = {๐, ๐, โฆ , ๐|๐|โ1}
is a subgroup of ๐บ. The result follows from Lagrange.
1.2 Normal subgroups, Quotients and HomomorphismsRecall that ๐๐ป = ๐โฒ๐ป if and only if ๐โ1๐โฒ โ ๐ป. In particular, if โ โ ๐ป then๐โ๐ป = ๐๐ป.
Given a subgroup ๐ป โค ๐บ, we want to define a group structure on its cosets.Argurably the most natural candidate for the group operation would be
โ โ โ โถ ๐บ/๐ป ร ๐บ/๐ป โ ๐บ/๐ป(๐๐ป, ๐โฒ๐ป) โฆ ๐๐โฒ๐ป
But is this well-defined? Suppose ๐โฒ๐ป = ๐โฒโ๐ป, then
(๐๐ป) โ (๐โฒโ๐ป) = ๐๐โฒโ๐ป = ๐๐โฒ๐ป
so it is well-defined in the second coordinate. Suppose ๐๐ป = ๐โ๐ป, then
(๐โ๐ป) โ (๐โฒ๐ป) = ๐โ๐โฒ๐ป ?= ๐๐โฒ๐ป
where the last step holds if and only if (๐โฒ)โ1โ๐โฒ โ ๐ป for all โ โ ๐ป, ๐โฒ โ ๐บ.Thus we need this true to define a group structure on the cosets. This motivatesus to define
Definition (Normal subgroup). A subgroup ๐ป โค ๐บ is normal if for allโ โ ๐ป, ๐ โ ๐บ, ๐โ1โ๐ โ ๐ป. Write ๐ป โด ๐บ.
Definition (Quotient group). If ๐ป โด ๐บ, then ๐บ/๐ป equipped with theproduct
๐บ/๐ป ร ๐บ/๐ป โ ๐บ/๐ป(๐๐ป, ๐โฒ๐ป) โฆ ๐๐โฒ๐ป
and identity ๐๐ป is a group. This is the quotient group of ๐บ by ๐ป.
Now we have defined and seen quite a few groups. We are interested notin the internal structure of groups but how they relate to each other. Thismotivates to define morphisms between groups:
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1 Groups
Definition (Homomorphism). If (๐บ, โ , ๐๐บ) and (๐ป, โ, ๐๐ป) are groups, afunction ๐ โถ ๐บ โ ๐ป is a homomorphism if for all ๐, โ โ ๐บ,
๐(๐ โ ๐โฒ) = ๐(๐) โ ๐(๐โฒ).
This implies that ๐(๐๐บ) = ๐๐ป and ๐(๐โ1) = ๐(๐)โ1. We define
ker๐ = {๐ โ ๐บ โถ ๐(๐) = ๐๐ป},Im๐ = {๐(๐) โถ ๐ โ ๐บ}.
Lemma 1.4.
โข ker๐ โด ๐บ,
โข Im๐ โค ๐ป.
Proof. Easy.
Definition (Isomorphism). A homomorphism ๐ is an isomorphism if it isa bijection. Say ๐บ and ๐ป are isomorphic if there exists some isomorphism๐ โถ ๐บ โ ๐ป. Write ๐บ โ ๐ป.
Exercise. If ๐ is an isomorphism then the inverse ๐โ1 โถ ๐ป โ ๐บ is also anisomorphism.
Theorem 1.5 (1st Isomorphism Theorem). Let ๐ โถ ๐บ โ ๐ป be a homomor-phism. Then ker โดโค ๐บ, Im๐ โค ๐บ and
๐บ/ ker๐ โ Im๐.
Proof. We have done the first part. For the second part, define
๐ โถ ๐บ/ ker๐ โ Im๐๐ ker๐ โฆ ๐(๐)
๐บ ๐ป
๐บ/ ker๐
๐
๐
Check this is well-defined: if ๐ ker๐ = ๐โฒ ker๐ then ๐โ1๐โฒ โ ker๐ so ๐๐ป =๐(๐โ1๐โฒ) = ๐(๐)โ1๐(๐โฒ), ๐(๐) = ๐(๐โฒ) and ๐(๐ ker๐) = ๐(๐โฒ ker๐).
๐ is a homomorphism:
๐(๐ ker๐ โ ๐โฒ ker๐) = ๐(๐๐โฒ ker๐) = ๐(๐๐โฒ) = ๐(๐)๐(๐โฒ) = ๐(๐ ker๐)๐(๐ ker๐).
๐ is surjective and finally to show it is injective, suppose ๐(๐ ker๐) = ๐๐ป. Then๐ โ ker๐ so ๐ ker๐ = ๐ ker๐.
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1 Groups
Example. Consider
๐ โถ C โ C \ {0}๐ง โฆ ๐๐ง
๐๐ง+๐ค = ๐๐ง โ ๐๐ค so ๐ โถ (C, +, 0) โ (C \ {0}, ร, 1) is a homomorphism. ๐ issurjective (as log is a left inverse).
ker๐ = {๐ง โ C โถ ๐๐ง = 1} = {2๐๐๐ โถ ๐ โ Z} = 2๐๐Z
so by 1st Isomorphism Theorem
C/2๐๐Z โ C \ {0}.
Theorem 1.6 (2nd Isomorphism Theorem). Let ๐ป โค ๐บ and ๐พ โด ๐บ. Then
๐ป๐พ โค ๐บ๐ป โฉ ๐พ โด ๐ป
๐ป๐พ/๐พ โ ๐ป/(๐ป โฉ ๐พ)
Proof. Let โ, โโฒ โ ๐ป, ๐, ๐โฒ โ ๐พ. Then
(โโฒ๐โฒ)(โ๐)โ1 = โโฒ๐โฒ๐โ1โโ1 = (โโฒโโ1)(โ๐โฒ๐โ1โโ1) โ ๐ป๐พ.
Consider
๐ โถ ๐ป โ ๐บ/๐พโ โฆ โ๐พ
This is the composition ๐ป๐
โช ๐บ๐โ ๐บ/๐พ so a homomorphism. Since
ker๐ = {โ๐พ โถ โ๐พ = ๐๐พ} = ๐ป โฉ ๐พ โด ๐ปIm๐ = {๐๐พ โถ ๐๐พ = โ๐พ for some โ โ ๐ป} = ๐ป๐พ/๐พ
so by 1st Isomorphism Theorem
๐ป/(๐ป โฉ ๐พ) โ ๐ป๐พ/๐พ.
As a corollary we have
Theorem 1.7 (Subgroup correspondence). Let ๐พ โด ๐บ. There is a bijectionbetween
{subgroups of ๐บ/๐พ} โ {subgroups of ๐บ containing ๐พ}๐ป โฆ {๐ โ ๐บ โถ ๐๐พ โ ๐ป}
๐ฟ/๐พ โค ๐พ โด ๐ฟ โค ๐บ
Moreover, the same map gives a bijection between
{normal subgroups of ๐บ/๐พ} โ {normal subgroups of ๐บ containing ๐พ}.
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1 Groups
Theorem 1.8 (3rd Isomorphism Theorem). Let ๐พ โค ๐ฟ โค ๐บ be normalsubgroups. Then
๐บ/๐พ๐ฟ/๐พ
โ ๐บ/๐ฟ.
Proof. Consider
๐ โถ ๐บ/๐พ โ ๐บ/๐ฟ๐๐พ โฆ ๐๐ฟ
Check it is well-defined: if ๐๐พ = ๐โฒ๐พ, ๐โ1๐โฒ โ ๐พ โ ๐ฟ so ๐๐ฟ = ๐โฒ๐ฟ. ๐ is clearlysurjective and has kernel
ker๐ = {๐๐พ โ ๐บ/๐พ โถ ๐๐ฟ = ๐๐ฟ} = ๐ฟ/๐พ
so by 1st Isomorphism Theorem
๐บ/๐พ๐ฟ/๐พ
โ ๐บ/๐ฟ.
Definition (Simple group). A group ๐บ is simple if its only normal subgroupsare {๐} and ๐บ.
Lemma 1.9. An abelian group is simple if and only if it is isomorphic to๐ถ๐ for some prime ๐.
Proof. In an abelian group every subgroup is normal. Let ๐ โ ๐บ be non-trivial.Then
โจ๐โฉ = {โฆ , ๐โ2, ๐โ1, ๐, ๐, ๐2, โฆ } โด ๐บ.If ๐บ is simple, this must be the whole group so ๐บ is cyclic. If ๐บ is infinite, it isisomorphic to Z which is not simple as 2Z โด ๐บ. Therefore ๐บ โ ๐ถ๐ for some ๐.If ๐ = ๐๐, ๐, ๐ โ N, ๐, ๐ โ 1 then โจ๐๐โฉ โด ๐บ. Absurd. Thus ๐ is a prime.
For the other directions, note that ๐ถ๐ is simple for prime ๐ by Lagrange.
Theorem 1.10. Let ๐บ be a finite group. Then there is a chain of subgroups
๐บ = ๐ป0 โฅ ๐ป1 โฅ ๐ป2 โฅ โฏ โฅ ๐ป๐ = {๐}
such that ๐ป๐+1 โด ๐ป๐ and ๐ป๐/๐ป๐+1 is simple for all ๐.
Proof. Let ๐ป1 be a normal subgroup of ๐ป0 = ๐บ of maximal order. If ๐ป0/๐ป1is not simple, there would be a proper normal subgroup ๐ โด ๐ป1/๐ป2. Thiscorresponds to a normal subgroup of ๐ป0, ๐, which strictly contains ๐ป1. Absurd.Thus ๐ป0/๐ป1 is simple.
Choose ๐ป2 to be the maximal normal subgroup of ๐ป1 and continue. As๐ป๐+1 is a proper subgroup of ๐ป๐, |๐ป๐+1| < |๐ป๐| so this process terminates afterfinitely many steps.
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1 Groups
1.3 Actions & PermutationsPart of the reason we study groups is that they have interesting internal structures.However, more importantly, groups are interesting because many transformationsof an object can be described by a group. This is formalised by the concept ofgroup action in this section.
The symmetric group ๐๐ is the set of permutations of {1, โฆ , ๐}. Everypermutation is a product of transpositions. A permutation is even if it is aproduct of evenly-many transpositions and odd otherwise.
The sign of a permutation is a homomorphism
sgn โถ ๐๐ โ {ยฑ1}
๐ โฆ {1 ๐ is evenโ1 ๐ is odd
The kernel of sgn is the alternating group ๐ด๐ โด ๐๐ of index 2 for ๐ โฅ 2.For any set ๐, we let Sym(๐) denote the set of all permutations of ๐, with
composition as the group operation.Here is a definition that is included in the syllabus but seems to be never
used anywhere:
Definition. A group ๐บ is a permutation group of degree ๐ if
๐บ โค Sym(๐)
with |๐| = ๐.
Example.
1. ๐๐ is a permutation group of order ๐, so is ๐ด๐.
2. ๐ท2๐ acts on the ๐ vertices of a regular ๐-gon, so
๐ท2๐ โค ๐({๐ vertices}).
Definition (Group action). An action of a group (๐บ, โ , ๐) on a set ๐ is afunction โ โ โ โถ ๐บ ร ๐ โ ๐ such that
1. For all ๐, โ โ ๐บ, ๐ฅ โ ๐,
๐ โ (โ โ ๐ฅ) = (๐โ) โ ๐ฅ.
2. For all ๐ฅ โ ๐,๐ โ ๐ฅ = ๐ฅ.
Lemma 1.11. Giving an action of ๐บ on ๐ is the same as giving a homo-morphism ๐ โถ ๐บ โ Sym(๐).
Proof.
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โข โ: Let โ โ โ be an action. For all ๐ โ ๐บ, let
๐ โถ ๐ โ ๐๐ฅ โฆ ๐ โ ๐ฅ
This satisfies
๐(๐โ)(๐ฅ) = (๐โ) โ ๐ฅ = ๐ โ (โ โ ๐ฅ) = ๐(๐)(๐(โ)(๐ฅ)) = (๐(๐) โ ๐(โ))(๐ฅ)
so ๐(๐โ) = ๐(๐) โ ๐(โ).In addition ๐(๐)(๐ฅ) = ๐ โ ๐ฅ = ๐ฅ = id๐(๐ฅ) so ๐(๐) = id๐. Now we notethat
id๐ = ๐(๐) = ๐(๐๐โ1) = ๐(๐) โ ๐(๐โ1)
so ๐(๐โ1) is inverse to ๐(๐). In particular ๐(๐) is a bijection.
โข โ: Let ๐ โถ ๐บ โ Sym(๐) be a homomorphism. Define
โ โ โ โถ ๐บ ร ๐ โ ๐(๐, ๐ฅ) โฆ ๐(๐)(๐ฅ)
Verify that
๐ โ (โ โ ๐ฅ) = ๐(๐)(๐(โ)(๐ฅ)) = (๐(๐) โ ๐(โ))(๐ฅ) = ๐(๐โ)(๐ฅ) = (๐โ) โ ๐ฅ๐ โ ๐ฅ = ๐(๐)(๐ฅ) = id๐(๐ฅ) = ๐ฅ
Given a homomorphism ๐ โถ ๐บ โ Sym(๐) induced by an action, define ๐บ๐ =Im๐, ๐บ๐ = ker๐. Then by 1st Isomorphism Theorem ๐บ๐ โด ๐บ, ๐บ/๐บ๐ โ ๐บ๐.
If ๐บ๐ = {๐}, i.e. ๐ is injective then we say ๐ is a permutation representationof ๐บ. It follows that ๐บ โ ๐บ๐ โค Sym(๐).
Example.
1. Let ๐บ be the symmetries of a cube. Then ๐บ acts on the set ๐ of diagonals.|๐| = 4 and ๐ โถ ๐บ โ Sym(๐) is surjective so ๐บ๐ = Sym(๐) โ ๐4.๐บ๐ = {id, antipodal map} so by Lagrange
|๐บ| = |๐บ๐| โ |๐บ๐| = 48.
2. For any group ๐บ, left multiplication is a homomorphism:
๐ โถ ๐บ โ Sym๐บ๐ โฆ ๐ โ โ
๐บ๐ = {๐ โ ๐บ โถ ๐โ = โ for all ๐บ} = {๐} so ๐ is a permutation representa-tion. This is
Theorem 1.12 (Cayley). Every group is isomorphic to a subgroup ofa symmetric group.
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3. If ๐บ is a group and ๐ป โค ๐บ, we have
๐ โถ ๐บ โ Sym(๐บ/๐ป)๐ โฆ ๐ โ โ
๐บ๐ = {๐ โ ๐บ โถ ๐๐๐ป = ๐๐ป for all ๐๐ป} = โ๐โ๐บ ๐๐ป๐โ1. This is the largestsubgroup of ๐ป which is normal in ๐บ.
Theorem 1.13. Let ๐บ be a finite group and ๐ป โค ๐บ with index ๐. Thenthere is a ๐พ โด ๐บ, ๐พ โค ๐ป such that ๐บ/๐พ is isomorphic to a subgroup of ๐๐.In particular
|๐บ/๐พ| โฃ ๐!๐ โฃ |๐บ/๐พ|
Proof. Let ๐พ = ๐บ๐ for the action of ๐บ on ๐ = ๐บ/๐ป. Then
๐บ/๐บ๐ โ ๐บ๐ โค Sym(๐) โ ๐๐.
Theorem 1.14. Let ๐บ be a non-abelian simple group and ๐ป โค ๐บ is asubgroup of index ๐ > 1. Then ๐บ is isomorphic to a subgroup of ๐ด๐ for some๐ โฅ 5.
Proof. Let ๐บ act on ๐บ/๐ป, giving ๐ โถ ๐บ โ Sym(๐บ/๐ป). Then ker๐ โด ๐บ. As ๐บis simple, ker๐ = {๐} or ๐บ. But ker๐ = โ๐โ๐บ ๐โ1๐ป๐ โค ๐ป, a proper subgroupof ๐บ so ker๐ = {๐}. By 1st Isomorphism Theorem
๐บ = ๐บ/{๐} โ Im๐ = ๐บ๐ โค Sym(๐บ/๐ป) โ ๐๐.
Applying 2nd Isomorphism Theorem to ๐ด๐ โด ๐๐, ๐บ๐ โค ๐๐, we get
๐บ๐ โฉ ๐ด๐ โด ๐บ๐, ๐บ๐/(๐บ๐ โฉ ๐ด๐) โ ๐บ๐๐ด๐/๐ด๐.
As ๐บ๐ โ ๐บ is simple, ๐บ๐ โฉ ๐ด๐ is either trivial or ๐บ๐, i.e. ๐บ๐ โค ๐ด๐. But if๐บ๐ โฉ ๐ด๐ = {๐},
๐บ๐ โ ๐บ๐๐ด๐/๐ด๐ โค ๐๐/๐ด๐ โ ๐ถ2
which contradicts ๐บ๐ โ ๐บ being non-abelian. Hence ๐บ โ ๐บ๐ โค ๐ด๐.
1 ๐บ Sym(๐บ/๐ป) ๐ถ2๐ sgn
To see that we must have ๐ โฅ 5, observe that ๐ด2, ๐ด3 and ๐ด4 have nonon-abelian simple subgroup.
Corollary 1.15. If ๐บ is non-abelian simple, ๐ป โค ๐บ of index ๐, then
|๐บ| โฃ ๐!2
.
Some futher definitions we have already met in IA Groups:
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Definition (Orbit & Stabiliser). If ๐บ acts on ๐, the orbit of ๐ฅ โ ๐ is
๐บ โ ๐ฅ = {๐ โ ๐ฅ โถ ๐ โ ๐บ}.
and the stabiliser of ๐ฅ is
๐บ๐ฅ = {๐ โ ๐บ โถ ๐ โ ๐ฅ = ๐ฅโ๐ฅ โ ๐}.
Theorem 1.16 (Orbit-stabiliser). If ๐บ acts on ๐, for all ๐ฅ โ ๐ there is abijection
๐บ โ ๐ฅ โ ๐บ/๐บ๐ฅ
๐ โ ๐ฅ โ ๐๐บ๐ฅ
1.4 Conjugacy class, Centraliser & NormaliserIn the previous section we use a group action of a group on itself, namely leftmultiplication, to study the structure of a group. In this section we studyconjugation, another group action that gives much richer results.
There is an action of ๐บ on ๐ = ๐บ via ๐ โ ๐ฅ = ๐๐ฅ๐โ1, giving ๐ โถ ๐บ โ Sym(๐บ).
Remark.
๐(๐)(๐ฅ๐ฆ) = ๐๐ฅ๐ฆ๐โ1 = (๐๐ฅ๐โ1)(๐๐ฆ๐โ1) = ๐(๐)(๐ฅ)๐(๐)(๐ฆ)
so ๐(๐) is a group homomorphism. In fact this is an automorphism and ๐(๐) โInn(๐บ), which is the group of all automorphisms arising from conjugation.
Denote
Aut(๐บ) = {๐ โถ ๐บ โ ๐บ โถ ๐ is an isomorphism} โค Sym(๐บ).
We have shown ๐ โถ ๐บ โ Sym(๐บ) has image in Aut(๐บ) โค Sym(๐บ), i.e. Inn(๐บ) โคAut(๐บ).
Definition (Conjugacy class). The conjugacy clss of ๐ฅ โ ๐บ is
๐บ โ ๐ฅ = Cl๐บ(๐ฅ) = {๐๐ฅ๐โ1 โถ ๐ โ ๐บ}.
Definition (Centraliser). The centraliser of ๐ฅ โ ๐บ is
๐ถ๐บ(๐ฅ) = {๐ โ ๐บ โถ ๐๐ฅ = ๐ฅ๐}.
Definition (Centre). The centre of ๐บ is
๐(๐บ) = ker๐ = {๐ โ ๐บ โถ ๐๐ฅ๐โ1 = ๐ฅโ๐ฅ โ ๐บ}.
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Definition (Normaliser). The normaliser of ๐ป โค ๐บ is
๐๐บ(๐ป) = {๐ โ ๐บ โถ ๐๐ป๐โ1 = ๐ป}.
By Orbit-stabiliser, there is a bijection between
Cl๐บ(๐ฅ) โ ๐บ/๐ถ๐บ(๐ฅ)
so if ๐บ is finite, |Cl๐บ(๐ฅ)| = |๐บ/๐ถ๐บ(๐ฅ)| divides |๐บ|.Recall from IA Groups that in the permutation group ๐๐
1. every element can be written as a product of disjoint cycles,
2. permutations are conjugations if and only if they have the same cycle type.
We will use these knowledge to make our first (and the only one in thiscourse) step towards classification of finite simple groups:
Theorem 1.17. ๐ด๐ is simple for ๐ โฅ 5.
Proof. First claim ๐ด๐ is generated by 3-cycles. Need to show that doubletranspositions are generated by 3-cycles. There are two cases:
โข (๐๐)(๐๐) = (๐๐๐),
โข (๐๐)(๐๐) = (๐๐๐)(๐๐๐).
Let ๐ป โด ๐ด๐. Suppose ๐ป contains a 3-cycle, say (๐๐๐). There exists ๐ โ ๐๐such that
(๐๐๐) = ๐โ1(123)๐.
If ๐ โ ๐ด๐ then (123) โ ๐ป. If ๐ โ ๐ด๐, let ๐โฒ = (45)๐ โ ๐ด๐. Here we use the factthat ๐ โฅ 5. Then
(๐๐๐) = ๐โฒโ1(45)(123)(45)๐โฒ = ๐โฒโ1(123)๐โฒ.
Hence ๐ป contains all 3-cycles and ๐ป = ๐ด๐. It then suffices to show any non-trivial๐ป โด ๐ด๐ contains a 3-cycle. Split into different cases:
โข Case I: ๐ป contains ๐ = (12 โฏ ๐)๐, written in disjoint cycle notation, forsome ๐ โฅ 4. Let ๐ = (123) and consider the commutator
[๐, ๐] = ๐โ1๐โ1๐๐ = ๐โ1(๐ โฏ 21)(132)(12 โฏ ๐)๐(123) = (23๐)
which is a 3-cycle in ๐ป.
โข Case II: ๐ป contains ๐ = (123)(456)๐. Let ๐ = (124) and consider
[๐, ๐] = ๐โ1(132)(465)(142)(123)(456)๐(124) = (12436)
which is a 5-cycle in ๐ป. This reduces to Cases I.
โข Case III: ๐ป contains ๐ = (123)๐ and ๐ is a product of 2-cycles. Then๐2 = (132) โ ๐ป.
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โข Case IV: ๐ป contains ๐ = (12)(34)๐ where ๐ is a product of 2-cycles. Let๐ = (123) and
๐ข = [๐, ๐] = (12)(34)(132)(12)(34)(123) = (14)(23).
Not let ๐ฃ = (125) where we used the fact ๐ โฅ 5 again. Then
[๐ข, ๐ฃ] = (14)(23)(152)(14)(23)(125) = (12345) โ ๐ป
which is a 5-cycle.
1.5 Finite ๐-groupsA finite group ๐บ is a ๐-group if |๐บ| = ๐๐ for some prime ๐.
Theorem 1.18. If ๐บ is a finite ๐-group then ๐(๐บ) โ {๐}.
Proof. The conjugacy classes partition ๐บ and |Cl(๐ฅ)| = |๐บ/๐ถ(๐ฅ)| which divides|๐บ|. Thus |Cl(๐ฅ)| is a power of ๐. Class equation reads
|๐บ| = |๐(๐บ)| + โother cclโs
|Cl(๐ฅ)|
Reduce mod ๐, we get |๐(๐บ)| = 0 mod ๐. But |๐(๐บ)| โฅ 1 so |๐(๐บ)| โฅ ๐.
Corollary 1.19. A group of order ๐๐, ๐ > 1 is never simple.
Lemma 1.20. For any group ๐บ, if ๐บ/๐(๐บ) is cyclic, ๐บ is abelian.
Proof. Let ๐บ/๐(๐บ) = โจ๐๐(๐บ)โฉ. Then every coset is of the form ๐๐๐(๐บ), ๐ โ Z.Thus every element of ๐บ is of the form ๐๐๐ง where ๐ง โ ๐(๐บ). Then
๐๐๐ง๐๐โฒ๐งโฒ = ๐๐๐๐โฒ๐ง๐งโฒ = ๐๐+๐โฒ๐งโฒ๐ง = ๐๐โฒ๐งโฒ๐๐๐ง
and hence ๐บ is abelian.
Corollary 1.21. If |๐บ| = ๐2, ๐บ is abelian.
Proof. ๐(๐บ) โ {๐} so |๐(๐บ)| = ๐ or ๐2. Suppose |๐(๐บ)| = ๐, |๐บ/๐(๐บ)| = ๐ so๐บ/๐(๐บ) โ ๐ถ๐ so by the lemma ๐บ is abelian. Absurd. Thus ๐(๐บ) = ๐บ and thus๐บ is abelian.
Theorem 1.22. If |๐บ| = ๐๐, ๐บ has a subgroup of order ๐๐ for all 0 โค ๐ โค ๐.
Proof. Induction on ๐. If ๐ = 1 then done. Suppose ๐ > 1. Then ๐(๐บ) โ {๐}.Let ๐ฅ โ ๐(๐บ) be non-identity. Then ๐ฅ has order a power of ๐, say ๐๐. Then๐ง = ๐ฅ๐๐โ1 has order precisely ๐. Let ๐ถ = โจ๐งโฉ โด ๐บ. Then ๐บ/๐ถ has order ๐๐โ1.By induction hypothesis we can find a subgroup ๐ป โค ๐บ/๐ถ of order ๐๐โ1. Then๐ป must be of the form ๐ฟ/๐ถ for some ๐ฟ โค ๐บ and |๐ฟ| = ๐๐.
1.6 Finite abelian groups
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Theorem 1.23. If ๐บ is a finite abelian group then
๐บ โ ๐ถ๐1ร ๐ถ๐2
ร โฏ ร ๐ถ๐๐
with ๐๐+1 โฃ ๐๐ for all ๐.
Proof. This will be a corollary of the main result on modules by consideringabelian groups as Z-modules.
Example. If |๐บ| = 8 and ๐บ is abelian, ๐บ is isomorphic to one of ๐ถ8, ๐ถ4 ร ๐ถ2and ๐ถ2 ร ๐ถ2 ร ๐ถ2.
Lemma 1.24 (Chinese Remainder Theorem). If ๐ and ๐ are coprime, then
๐ถ๐๐ โ ๐ถ๐ ร ๐ถ๐.
Proof. Let ๐ โ ๐ถ๐ has order ๐, โ โ ๐ถ๐ has order ๐. Consider
๐ฅ = (๐, โ) โ ๐ถ๐ ร ๐ถ๐.
If ๐ = ๐ฅ๐ = (๐๐, โ๐), then ๐ โฃ ๐, ๐ โฃ ๐ so ๐๐ โฃ ๐. Thus |๐ฅ| = ๐๐. The group iscyclic.
Corollary 1.25. If ๐บ is a finite abelian group then
๐บ โ ๐ถ๐1ร ๐ถ๐2
ร โฏ ร ๐ถ๐โ
with each ๐๐ a power of prime.
Proof. If ๐ = ๐๐11 ๐๐2
2 โฏ ๐๐๐๐ , a factorisation of distinct primes, the above lemmashows
๐ถ๐ โ ๐ถ๐๐11
ร ๐ถ๐๐22
ร โฏ ร ๐ถ๐๐๐๐ .
Apply this to the theorem above.
1.7 Sylowโs Theorem
Theorem 1.26 (Sylowโs Theorem). Let |๐บ| = ๐๐ โ ๐ with (๐, ๐) = 1 where๐ is a prime. Then
1. the setSyl๐(๐บ) = {๐ โค ๐บ โถ |๐ | = ๐๐}
of Sylow ๐-subgroups is not empty,
2. all elements of Syl๐(๐บ) are conjugates in ๐บ,
3. the number๐๐ = | Syl๐(๐บ)|
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satisfies๐๐ = 1 mod ๐, ๐๐ โฃ |๐บ|.
Lemma 1.27. If ๐๐ = 1 then the unqiue Sylow ๐-subgroup is normal in ๐บ.
Proof. Let ๐ โค ๐บ be the Sylow ๐-subgroup and ๐ โ ๐บ. As ๐๐๐โ1 โ Syl๐(๐บ),๐๐๐โ1 = ๐ so ๐ โด ๐บ.
Example. Let ๐บ be group of order 96 = 25 โ 3. Then
โข ๐2 = 1 mod 2 and ๐2 โฃ 3 so ๐2 = 1 or 3.
โข ๐3 = 1 mod 3 and ๐3 โฃ 32 so ๐3 = 1, 4 or 16.
๐บ acts on the set Syl๐(๐บ) by conjugation. The second part of Sylowโs Theoremsays that this action has precisely one orbit. The stabiliser of ๐ โ Syl๐(๐บ) isthe normaliser ๐๐บ(๐ ) โค ๐บ of index ๐๐ = | Syl๐(๐บ)|.
Corollary 1.28. If ๐บ is non-abelian simple, then |๐บ| โฃ (๐๐)!2 and ๐๐ โฅ 5.
Proof. ๐๐บ(๐ ) has index ๐๐ so Theorem 1.14 to get the result. Alternatively,consider the conjugation action of ๐บ on Syl๐(๐บ).
Example (Continued). |๐บ| โค 3!2 so ๐บ cannot be simple.
Example. Suppose ๐บ is a simple group of order 132 = 22 โ 3 โ 11. We have๐11 = 1 mod 11 and ๐11 โฃ 12. As ๐บ is simple we canโt have ๐11 = 1 so ๐11 = 12.Each Sylow 11-subgroup has order 11 so isomorphic to ๐ถ11, and thus contains10 elements of order 11. Such subgroups can only intersect in the identity so wehave 12 ร 10 = 120 elements of order 11.
In addition we know ๐3 = 1 mod 3 and ๐3 โฃ 44, so ๐3 = 4 or 22. If ๐3 = 4,we must have |๐บ| โฃ 4!
2 by the previous corollary. Absurd. Thus ๐3 = 22. Asabove, we get 22โ (3โ1) = 44 elements of order 3. This gives 164 > 132 elements.Absurd.
Thus there is no simple group of order 132.
Proof of Sylowโs Theorem. Let |๐บ| = ๐๐ โ ๐.
1. Letฮฉ = {๐ โ ๐บ โถ |๐| = ๐๐}
and ๐บ act on ฮฉ via
๐ โ {๐1, ๐2, โฆ , ๐๐๐} = {๐๐1, ๐๐2, โฆ , ๐๐๐๐}.
Let ฮฃ โ ฮฉ be an orbit of the action. If {๐1, โฆ , ๐๐๐} โ ฮฃ, then
(๐๐โ11 ) โ {๐1, โฆ , ๐๐๐} โ ฮฃ
so for all ๐ โ ๐บ there is an element of ฮฃ containing ๐. Thus |ฮฃ| โฅ |๐บ|๐๐ = ๐.
If there is some orbit ฮฃ with |ฮฃ| = ๐, its stabiliser ๐บฮฃ has order ๐๐ so wehave a Sylow ๐-subgroup.
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To show this happens, we must show it is not possible for every orbit tohave size strictly bigger than ๐. By Orbit-stabiliser, for any ฮฃ, |ฮฃ| โฃ ๐๐ โ ๐so if |ฮฃ| > ๐ then ๐ โฃ |ฮฃ|. If all orbits have size > ๐, ๐ divides all of themso ๐ โฃ |ฮฉ|.Let us calculate |ฮฉ|. We have
|ฮฉ| = (๐๐ โ ๐๐๐ ) =
๐๐โ1
โ๐=0
๐๐ โ ๐ โ ๐๐๐ โ ๐
.
The largest power of ๐ dividing ๐๐ โ ๐ โ ๐ is the same as the largest powerof ๐ dividing ๐, which is the same as the largest power of ๐ dividing ๐๐ โ ๐.Thus |ฮฉ| is not divisible by ๐.
2. Let us show something stronger: if ๐ โ Syl๐(๐บ) and ๐ is a ๐-subgroupthen there is a ๐ โ ๐บ such that ๐โ1๐๐ โค ๐.Let ๐ act on ๐บ/๐ by
๐ โ ๐๐ = ๐๐๐ .By Orbit-stabiliser, the size of an orbit divides |๐| = ๐๐ so it is either 1 ordivisible by ๐.On the other hand |๐บ/๐ | = |๐บ|
|๐ | = ๐ which is not divisible by ๐. Thusthere must be an orbit of size 1, say {๐๐}, i.e. for all ๐ โ ๐, ๐๐๐ = ๐๐ so๐โ1๐๐ โ ๐. ๐โ1๐๐ โค ๐.
3. By 2 ๐บ acts on Syl๐(๐บ) by conjugation with one orbit. By Orbit-stabiliser๐๐ = | Syl๐(๐บ)| divides |๐บ|, which is the second part of the statement.
Now we show ๐๐ = 1 mod ๐. Let ๐ โ Syl๐(๐บ) and let ๐ act on Syl๐(๐บ)by conjugation. By Orbit-stabiliser, the size of an orbit divides |๐ | = ๐๐
so each orbit either has size 1 or dividible by ๐. But {๐} is a singletonorbit. To show ๐๐ = 1 mod ๐ it suffices to show every other orbit has size> 1.Suppose that {๐} is another singleton orbit. Then for all ๐ โ ๐, ๐โ1๐๐ = ๐so ๐ โค ๐๐บ(๐). But we also have ๐ โด ๐๐บ(๐) (since the normaliser is thelargest subgroup of ๐บ in which ๐ is normal). Now ๐ and ๐ are Sylow๐-subgroups of ๐๐บ(๐) so are conjugates in ๐๐บ(๐). Thus there exists๐ โ ๐๐บ(๐) such that ๐ = ๐โ1๐๐ = ๐. Thus ๐ = ๐ which contradicts ๐being different from ๐.
Example. Let ๐บ = GL๐(F๐). It has order
|๐บ| = (๐๐ โ 1)(๐๐ โ ๐) โฏ (๐๐ โ ๐๐โ1) =๐โ1โ๐=0
(๐๐ โ ๐๐) = ๐๐(๐โ1)
2
๐โ1โ๐=0
(๐๐โ๐ โ 1).
Let ๐ be the set of upper triangular matrices with diagonal entries 1, whichforms a subgroup of ๐บ. |๐| = ๐
๐(๐โ1)2 so ๐ is a Sylow ๐-subgroup.
Consider GL2(F๐). It has order (๐2 โ 1)(๐2 โ ๐) = ๐(๐ + 1)(๐ โ 1)2. Let โbe an odd prime dividing ๐ โ 1. Then โ โค ๐, โ โค ๐ + 1 so โ2 is the largest powerof โ dividing |GL2(F๐)|.
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1 Groups
Define the unit group
(Z/๐Z)ร = {๐ฅ โ Z/๐Z โถ โ๐ฆ โ Z/๐Z, ๐ฅ๐ฆ = 1} = {๐ฅ โ Z/๐Z โถ ๐ฅ โ 0}
which is isomorphic to ๐ถ๐โ1. Thus it has a subgroup ๐ถโ โค ๐ถ๐โ1, i.e. we can find๐ฅ โ (Z/๐Z)ร such that ๐ฅโ = 1.
Let
๐ป = {( ๐ 00 ๐ ) โถ ๐, ๐ โ (Z/๐Z)ร, ๐โ = ๐โ = 1} โ ๐ถโ ร ๐ถโ โค GL2(F๐).
Then ๐ป is a Sylow โ-subgroup.Example. Let
SL2(F๐) = ker(det โถ GL2(F๐) โ (Z/๐Z)ร).
det is surjective as det( ๐ 00 1 ) = ๐ so SL2(F๐) โด GL2(F๐) has index ๐ โ 1. Thus
|SL2(F๐)| = (๐ โ 1)๐(๐ + 1).
Further definePSL2(F๐) = SL2(F๐)/{( ๐ 0
0 ๐ )}.If ( ๐ 0
0 ๐ ) โ SL2(F๐), then ๐2 = 1. As long as ๐ > 2, there are two such ๐โs, ยฑ1 so
|PSL2(F๐)| = (๐ โ 1)๐(๐ + 1)2
.
Let (Z/๐Z)โ = Z/๐Z โช {โ}. Then PSL2(F๐) acts on (Z/๐Z)โ by theMรถbius map
[๐ ๐๐ ๐] โ ๐ง = ๐๐ง + ๐
๐๐ง + ๐.
Take ๐ = 5 for example, this actions gives a homomorphism
๐ โถ PSL2(F5) โ ๐6.
|PSL2(F5)| = 60. Claim ๐ is injective.
Proof. Suppose ๐๐ง+๐๐๐ง+๐ = ๐ง for all ๐ง. Set ๐ง = 0, ๐ = 0. ๐ง = โ, ๐ = 0. ๐ง = 1, ๐ = ๐.
Thus[๐ ๐
๐ ๐] = [1 00 1] โ PSL2(F5).
Further claim Im๐ โค ๐ด6.
Proof. Consider
1 PSL2(F5) ๐6 ๐ถ2.๐ sgn
Need to show ๐ = sgn โ๐ is trivial. We already know elements of odd order inPSL2(F5) has be be sent to 1.
Note that ๐ป = {[ ๐ 00 ๐โ1 ], [ 0 ๐
โ๐โ1 0 ]} has order 4, so it is a Sylow 2-subgroupof PSL2(F5). Any elemnt of order 2 or 4 is conjugate to an element in the group.We will show ๐(๐ป) = {๐}.
๐ป is generated by [ โ2 00 2 ] and [ 0 1
โ1 0 ]. [ โ2 00 2 ] acts on (Z/5Z)โ via ๐ง โฆ โ๐ง.
It is thus an even permutation. [ 0 1โ1 0 ] acts via ๐ง โฆ โ 1
๐ง , which is also an evenpermutation.
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2 Rings
2 Rings
2.1 Definitions
Definition (Ring). A ring is a quintuple (๐ , +, โ , 0๐ , 1๐ ) such that
โข (๐ , +, 0๐ ) is an abelian group,
โข the operation โ โ โ โถ ๐ ร ๐ โ ๐ is associative and satisfies
1๐ โ ๐ = ๐ = ๐ โ 1๐
โข ๐ โ (๐1 + ๐2) = ๐ โ ๐1 + ๐ โ ๐2 and (๐1 + ๐2) โ ๐ = ๐1 โ ๐ + ๐2 โ ๐.
A ring is commutative if for all ๐, ๐ โ ๐ , ๐ โ ๐ = ๐ โ ๐. We will only considercommutative rings in this course.
Definition (Subring). If (๐ , +, โ , 0๐ , 1๐ ) is a ring and ๐ โ ๐ , then it is asubring if 0๐ , 1๐ โ ๐ and +, โ make ๐ into a ring. Write ๐ โค ๐ .
Example.
1. Z โค Q โค R โค C with usual 0, 1, + and โ .
2. Z[๐] = {๐ + ๐๐ โถ ๐, ๐ โ Z} is the subring of Gaussian integers.
3. Q[โ
2] = {๐ +โ
2๐ โถ ๐, ๐ โ Q} โค R.
Definition (Unit). An element ๐ โ ๐ is a unit if there exists ๐ โ ๐ suchthat ๐ โ ๐ = 1๐ .
Note that being a unit depends on the ambient ring: 2 โ Z is not a unit but2 โ Q is.
If every ๐ โ ๐ , ๐ โ 0๐ is a unit, then ๐ is a field.
Notation. If ๐ฅ โ ๐ , write โ๐ฅ โ ๐ for the inverse of ๐ฅ in (๐ , +, 0๐ ). Write๐ฆ โ ๐ฅ = ๐ฆ + (โ๐ฅ).
Example. 0๐ + 0๐ = 0๐ so
๐ โ 0๐ = ๐ โ (0๐ + 0๐ ) = ๐ โ 0๐ + ๐ โ 0๐
so ๐ โ 0๐ = 0๐ . Thus if ๐ โ {0}, 0๐ โ 1๐ since choosing ๐ โ 0๐ , we would get๐ = ๐ โ 1๐ = ๐ โ 0๐ = 0๐ . Absurd.
However, ({0}, +, โ , 0, 0) is indeed a ring.
Example. If ๐ and ๐ are rings, then ๐ ร ๐ is a ring via
(๐1, ๐ 1) + (๐2, ๐ 2) = (๐1 + ๐2, ๐ 1 + ๐ 2)(๐1, ๐ 1) โ (๐2, ๐ 2) = (๐1 โ ๐2, ๐ 1 โ ๐ 2)
1๐ ร๐ = (1๐ , 1๐)0๐ ร๐ = (0๐ , 0๐)
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2 Rings
Let ๐1 = (1๐ , 0), ๐2 = (0, 1๐), then1
๐21 = ๐1
๐22 = ๐2
๐1 + ๐2 = 1๐ ร๐
Example (Polynomial). Let ๐ be a ring. A polynomial ๐ over ๐ is an expression
๐ = ๐0 + ๐1๐ + โฏ ๐๐๐๐
with ๐๐ โ ๐ for all ๐. Note that ๐ is just a symbol and the sum is formal. Wewill consider ๐ and
๐0 + ๐1๐ + โฏ ๐๐๐๐ + 0๐ โ ๐๐+1
as equal.The degree of ๐ is the largest ๐ such that ๐๐ โ 0. If in addition ๐๐ = 1๐ , we
say ๐ is monic.Write ๐ [๐] for the set of all polynomials over ๐ . If
๐ = ๐0 + ๐1๐ + โฏ + ๐๐๐๐,
we define
๐ + ๐ =max (๐,๐)
โ๐=0
(๐๐ + ๐๐)๐๐
๐ โ ๐ = โ๐
๐โ๐=0
๐๐๐๐โ๐๐๐
which make ๐ [๐] a ring.We consider ๐ as a subring of ๐ [๐], given by the polynomials of degree 0.
In particular, 1๐ โ ๐ gives 1๐ [๐].
Example. Conisder Z/2Z[๐], ๐ = ๐ + ๐2 โ 0. We have
๐(0) = 0 + 0 = 0๐(1) = 1 + 1 = 0
This shows that a polynomial vanishing everywhere on a finite ring is notnecessarily zero (but necessarily so for an infinite ring).
Example. Write ๐ [[๐]] for the ring of formal power series with elements
๐ = ๐0 + ๐1๐ + ๐2๐2 + โฆ
with the same addition and multiplication as above.
Example. The Laurent polynomials ๐ [๐, ๐โ1] is the set of expressions
๐ = โ๐โZ
๐๐๐๐
such that only finitely many ๐๐โs are non-zero.1This is known as orthogonal idempotents.
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2 Rings
Example. The ring of Laurent series are elements of the form
๐ = โ๐โZ
๐๐๐๐
with only finitely many ๐ < 0 such that ๐๐ โ 0.
Example. If ๐ is a ring and ๐ is a set, the set ๐ ๐ of all functions ๐ โถ ๐ โ ๐ is a ring via
(๐ + ๐)(๐ฅ) = ๐(๐ฅ) + ๐(๐ฅ)(๐ โ ๐)(๐ฅ) = ๐(๐ฅ) โ ๐(๐ฅ)(1๐ ๐)(๐ฅ) = 1๐
(0๐ ๐)(๐ฅ) = 0๐
For example, we have the following chain
R[๐] = {๐ โถ R โ R polynomial} < {๐ โถ R โ R continuous} < RR.
2.2 Homomorphism, Ideals and Isomorphisms
Definition (Homomorphism). A function ๐ โถ ๐ โ ๐ between rings is ahomomorphism if
โข ๐(๐1 + ๐2) = ๐(๐1) + ๐(๐2), i.e. ๐ โถ (๐ , +, 0๐ ) โ (๐, +, 0๐) is a grouphomomorphism,
โข ๐(๐1๐2) = ๐(๐1)๐(๐2),
โข ๐(1๐ ) = 1๐.
If in addition ๐ is a bijection, it is an isomorphism.
The kernel of ๐ โถ ๐ โ ๐ is
ker๐ = {๐ โ ๐ โถ ๐(๐) = 0๐}.
Lemma 2.1. ๐ โถ ๐ โ ๐ is injective if and only if ker๐ = {0๐ }.
Proof. ๐ โถ (๐ , +, 0๐ ) โ (๐, +, 0๐) is a group homomorphism and its kernel asgroup homomorphism is also ker๐.
Definition (Ideal). A subset ๐ผ โ ๐ is an ideal if
โข ๐ผ is a subgroup of (๐ , +, 0๐ ),
โข strong (multiplicative) closure: for all ๐ฅ โ ๐ผ, ๐ โ ๐ , ๐ฅ โ ๐ โ ๐ผ.
Write ๐ผ โด ๐ .
We say ๐ผ โด ๐ is proper if ๐ผ โ ๐ .
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2 Rings
Lemma 2.2. If ๐ โถ ๐ โ ๐ is a homomorphism then ker๐ โด ๐ .
Proof. The first axiom holds since ๐ is a group homomorphism. Let ๐ฅ โ ker๐, ๐ โ๐ , then
๐(๐ โ ๐ฅ) = ๐(๐) โ ๐(๐ฅ) = ๐(๐) โ 0๐ = 0๐
so ๐ โ ๐ฅ โ ker๐.
Example.
1. If ๐ผ โด ๐ and 1๐ โ ๐ผ, then for all ๐ โ ๐ , ๐ = ๐ โ 1๐ โ ๐ผ so ๐ผ = ๐ .Equivalently, if ๐ผ is a proper ideal then 1๐ โ ๐ผ. Consequenctly, properideals are never subrings.
2. This can be generalaised to units: if ๐ข is a unit in ๐ with inverse ๐ฃ โ ๐ ,then if ๐ข โ ๐ผ, so is 1๐ = ๐ข โ ๐ฃ โ ๐ so ๐ผ = ๐ .Equivalently, if ๐ผ is a proper ideal then it contains no unit.
Example. If ๐ is a field then {0} and ๐ are the only ideals.
Example. In the ring Z, all ideals are of the form
๐Z = {โฆ , โ2๐, โ๐, 0, ๐, 2๐, โฆ }.
Proof. ๐Z is certainly an ideal.Let ๐ผ โด Z be an ideal. Let ๐ โ ๐ผ be the smallest positive element. Then
๐Z โ ๐ผ. If this is not an equality, choose ๐ โ ๐ผ \ ๐Z. By Euclidean algorithm,๐ = ๐๐ + ๐ with 0 โค ๐ < ๐. So ๐ = ๐ โ ๐๐ โ ๐ผ. But ๐ is the smallest positiveelement in ๐ผ, so ๐ = 0. Thus ๐ โ ๐Z.
Definition (Generated ideal). For an element ๐ โ ๐ , write
(๐) = {๐ โ ๐ โถ ๐ โ ๐ } โด ๐ ,
the ideal generated by ๐.More generally, for a set of elements {๐1, โฆ , ๐๐ }, write
(๐1, โฆ , ๐๐ ) = {๐1๐1 + โฏ + ๐๐ ๐๐ โถ ๐1, โฆ , ๐๐ โ ๐ } โด ๐ .
Definition (Principal ideal). If ๐ผ โด ๐ if of the form (๐), we say it is aprincipal ideal.
Example.
1. ๐Z = (๐) โด Z is ideal. In fact we have shown that all ideals of Z areprincipal.
2. (๐) = {polynomials with constant coefficient 0} โด C[๐].
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2 Rings
Proposition 2.3 (Quotient ring). Let ๐ผ โด ๐ be an ideal. The quotient ringis the set of cosets ๐ + ๐ผ (i.e. (๐ , +, 0)/๐ผ). Addition and multiplication aregiven by
(๐1 + ๐ผ) + (๐2 + ๐ผ) = ๐1 + ๐2 + ๐ผ(๐1 + ๐ผ) โ (๐2 + ๐ผ) = ๐1๐2 + ๐ผ
with 0๐ /๐ผ = 0๐ + ๐ผ, 1๐ /๐ผ = 1๐ + ๐ผ. This is a ring, and the quotient map
๐ โ ๐ /๐ผ๐ โฆ ๐ + ๐ผ
is a ring homomorphism.
Proof. We already knew (๐ /๐ผ, +, 0๐ /๐ผ) is an abelian group and addition asdescribed above is well-defined. Suppose
๐1 + ๐ผ = ๐โฒ1 + ๐ผ
๐2 + ๐ผ = ๐โฒ2 + ๐ผ
then ๐โฒ1 โ ๐1 = ๐1 โ ๐ผ, ๐โฒ
2 โ ๐2 = ๐2 โ ๐ผ. So
๐โฒ1๐โฒ
2 = (๐1 + ๐1)(๐2 + ๐2) = ๐1๐2 + ๐1๐2 + ๐2๐1 + ๐1๐2โโโโโโโโโโ๐ผ
.
Thus ๐โฒ1๐โฒ
2 + ๐ผ = ๐1๐2 + ๐ผ. This shows multiplication is well-defined. The ringaxioms for ๐ /๐ผ then follow from those of ๐ .
Example.
1. ๐Z โด Z so Z/๐Z is a ring. It has elements
0 + ๐Z, 1 + ๐Z, โฆ , (๐ โ 1) + ๐Z
and addition and multiplication are modular arithmetic mod ๐.
2. (๐) โด C[๐] so C[๐]/(๐) is a ring. We have
๐0 + ๐1๐ + ๐2๐2 + โฏ + ๐๐๐๐โโโโโโโโโโโโ(๐)
+(๐) = ๐0 + ๐.
If ๐0 + (๐) = ๐0 + (๐) then ๐0 โ ๐0 โ (๐) so ๐0 โ ๐0 is divisible by ๐,๐0 โ ๐0 = 0. Consider
๐ โถ C โ C[๐]/(๐)๐ โฆ ๐ + (๐)
which is a bijection. Observe that ๐ is a bijection and its inverse is givenby the map ๐ + (๐) โฆ ๐(0).
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2 Rings
Proposition 2.4 (Euclidean algorithm for polynomials). Let ๐น be a fieldand ๐, ๐ โ ๐น [๐]. Then we may write
๐ = ๐ โ ๐ + ๐
with deg ๐ < deg ๐.
Proof. Let deg ๐ = ๐, deg ๐ = ๐ so
๐ = ๐0 + ๐1๐ + โฏ + ๐๐๐๐
๐ = ๐0 + ๐1๐ + โฏ + ๐๐๐๐
with ๐๐, ๐๐ โ 0.If ๐ < ๐, let ๐ = 0, ๐ = ๐ so done. Suppose ๐ โฅ ๐ and proceed by induction
on ๐. Let๐1 = ๐ โ ๐๐๐โ๐๐๐๐โ1
๐
where ๐โ1๐ exists since ๐๐ โ ๐น and ๐๐ โ 0. This has degree < ๐. If ๐ = ๐ then
๐ = ๐(๐๐โ๐๐๐๐โ1๐ ) + ๐1
with deg ๐1 < ๐ = ๐ = deg ๐. If ๐ > ๐, by induction we have ๐1 = ๐๐1 + ๐ withdeg ๐ < deg ๐ so
๐ = ๐(๐๐โ๐๐๐๐โ1๐ ) + ๐๐1 + ๐ = ๐(๐๐โ๐๐๐๐โ1
๐ + ๐1) + ๐
as required.
Example. Consider (๐2 + 1) โด R[๐] and let ๐ = R[๐]/(๐2 + 1). It haselements of the form ๐ + (๐2 + 1). By Euclidean algorithm for polynomials๐ = (๐2 + 1)๐ + ๐ with deg ๐ โค 1 so ๐ + (๐2 + 1) = ๐ + (๐2 + 1). Any elementcan be represented by a polynomial of degree โค 1, say ๐ + ๐๐ + (๐2 + 1). If๐1 + ๐1๐ + (๐2 + 1) = ๐2 + ๐2๐ + (๐2 + 1) then (๐1 + ๐1๐) โ (๐2 + ๐2๐) isdivisible by ๐2 + 1. But degrees add in multiplication so ๐1 + ๐1๐ = ๐2 + ๐2๐.Consider the bijection
๐ โถ ๐ โ C๐ + ๐๐ + (๐2 + 1) โฆ ๐ + ๐๐
It obviously send addition to addition. For multiplication,
๐((๐ + ๐๐ + (๐2 + 1)) โ (๐ + ๐๐ + (๐2 + 1)))=๐(๐๐ + (๐๐ + ๐๐)๐ + ๐๐๐2 + (๐2 + 1))=๐(๐๐ + (๐๐ + ๐๐)๐ + ๐๐(๐2 + 1) โ ๐๐ + (๐2 + 1)=(๐๐ โ ๐๐) + (๐๐ + ๐๐)๐=(๐ + ๐๐) โ (๐ + ๐๐)=๐(๐ + ๐๐ + (๐2 + 1)) โ ๐(๐ + ๐๐ + (๐2 + 1))
Thus we have shown that C โ R[๐]/(๐2 + 1).
Remark. The key idea in the proof is to force ๐2 + 1 to vanish by quotient thepolynomial ring by the generated ideal so that โ๐ = ยฑ๐โ. Similarly Q[๐]/(๐2 โ2) โ Q[
โ2] โค R.
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2 Rings
This is a nice result. However, the proof is too cumbersome to be generalisedas we have to check well-definedness for each case. Instead, we have the followingtheorems stating the general results for abstract rings and ideals. The proofsare similar to those for groups and are omitted.
Theorem 2.5 (1st Isomorphism Theorem). Let ๐ โถ ๐ โ ๐ be a ringisomorphism. Then ker๐ โด ๐ , Im๐ โค ๐ and
๐ / ker๐ โ Im๐๐ + ker๐ โฆ ๐(๐)
is a ring isomorphism.
Theorem 2.6 (2nd Isomorphism Theorem). Let ๐ โค ๐ and ๐ฝ โด ๐. Then๐ โฉ ๐ฝ โด ๐ and
๐ + ๐ฝ๐ฝ
โ ๐ ๐ โฉ ๐ฝ
as rings.
Theorem 2.7 (Subring and ideal correspondence). Let ๐ผ โด ๐ . Then thereis a bijection between
{subrings of ๐ /๐ผ} โ {subrings of ๐ containing ๐ผ}๐ฟ โค ๐ /๐ผ โฆ {๐ โ ๐ โถ ๐ + ๐ผ โ ๐ฟ}
๐/๐ผ โค ๐ /๐ผ โค ๐ผ โด ๐ โค ๐
and
{ideals of ๐ /๐ผ} โ {ideals of ๐ containing ๐ผ}๐ฟ โด ๐ /๐ผ โฆ {๐ โ ๐ โถ ๐ + ๐ผ โ ๐ฟ}
๐ฝ/๐ผ โด ๐ /๐ผ โค ๐ผ โด ๐ฝ โด ๐
Theorem 2.8 (3rd Isomorphism Theorem). Let ๐ผ, ๐ฝ โด ๐ , ๐ผ โ ๐ฝ. Then๐ฝ/๐ผ โด ๐ /๐ผ and
๐ /๐ผ๐ฝ/๐ผ
โ ๐ /๐ฝ.
Example. Consider the homomorphism
๐ โถ R[๐] โ C
โ ๐๐๐๐ โฆ โ ๐๐๐๐
i.e. evaluation at ๐. It is surjective and
ker๐ = {๐ โ R[๐] โถ ๐(๐) = 0} = (๐2 + 1)
because real polynomials with ๐ as a root also have โ๐ as aroot, so are divisibleby (๐ โ ๐)(๐ + ๐) = ๐2 + 1. By 1st Isomorphism Theorem
R[๐]/(๐2 + 1) โ C.
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Example (Characteristic of a ring). For any ring ๐ there is a unique homomor-phism
๐ โถ Z โ ๐
๐ โฆโง{โจ{โฉ
1๐ + 1๐ + โฏ + 1๐ โโโโโโโโโ๐ times
๐ > 0
โ (1๐ + 1๐ + โฏ + 1๐ )โโโโโโโโโ๐ times
๐ < 0
ker ๐ โด Z so ker ๐ = ๐Z for some ๐ โฅ 0. This number ๐ is called the characteristicof ๐ , denoted ch๐ .
For example, Z โค Q โค R โค C all have characteristic 0. Z/๐Z have character-istic ๐.
2.3 Integral domain, Field of fractions, Maximal and Primeideals
Definition (Integral domain). A non-zero ring ๐ is an integral domain iffor all ๐, ๐ โ ๐ , ๐ โ ๐ = 0 implies that ๐ = 0 or ๐ = 0.
Definition (Zero divisor). ๐ฅ is a zero divisor in ๐ if ๐ฅ โ 0 and there exists๐ฆ โ 0 such that ๐ฅ โ ๐ฆ = 0.
Example.
1. All fields are integral domains: if ๐ฅ๐ฆ = 0 with ๐ฆ โ 0, then ๐ฆโ1 exists and
0 = 0 โ ๐ฆโ1 = (๐ฅ๐ฆ) โ ๐ฆโ1 = ๐ฅ.
2. A subring of an integral domain is an integral domain. Thus Z โค Q,Z[๐] โคC are integral domains.
Definition (Principal ideal domain). A ring ๐ is a principal ideal domain(PID) if it is an integral domain and every ideal is principal, i.e. for all ๐ผ โด ๐ ,there exists ๐ โ ๐ such that ๐ผ = (๐).
Example. Z is a PID.
Lemma 2.9. A finite integral domain is a field.
Proof. Let ๐ โ 0 โ ๐ and consider
๐ โ โ โถ ๐ โ ๐ ๐ โฆ ๐๐
This is a group homomorphism and its kernel is
ker(๐ โ โ) = {๐ โ ๐ โถ ๐๐ = 0} = {0}.
Thus ๐ โ โ is injective. As |๐ | < โ, ๐ โ โ must also be surjective. Thus thereexists ๐ โ ๐ such that ๐๐ = 1. ๐ = ๐โ1.
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Lemma 2.10. Let ๐ be an integral domain. Then ๐ [๐] is an integraldomain.
Proof. Let
๐ =๐
โ๐=0
๐๐๐๐
๐ =๐
โ๐=0
๐๐๐๐
with ๐๐, ๐๐ โ 0 be non-zero polynomials. Then the largest power of ๐ in ๐๐ is๐๐+๐ and its coefficient is ๐๐๐๐ โ ๐ . This is a product of non-zero elementson an integral domain so non-zero. Thus ๐๐ โ 0.
This gives us a way to produce a new integral domain from old ones. Moreover,iterating this, ๐ [๐1, โฆ , ๐๐] = ((๐ [๐1])[๐2] โฆ [๐๐]) is an integral domain.
Theorem 2.11 (Field of fractions). Let ๐ be an integral domain. There isa field of fractions ๐น of ๐ with the following properties:
1. ๐น is a field,
2. ๐ โค ๐น,
3. every element of ๐น is of the form ๐ โ ๐โ1 where ๐, ๐ โ ๐ โค ๐น.
Proof. Consider ๐ = {(๐, ๐) โ ๐ 2 โถ ๐ โ 0} with an equivalence relation
(๐, ๐) โผ (๐, ๐) โ ๐๐ = ๐๐ โ ๐ .
This is reflexive and symmetric. To show it is transitive, suppose (๐, ๐) โผ(๐, ๐), (๐, ๐) โผ (๐, ๐). Then
(๐๐)๐ = (๐๐)๐ = ๐(๐๐) = ๐(๐๐)
so ๐(๐๐ โ ๐๐) = 0. As ๐ โ 0 and ๐ is an integral domain, ๐๐ โ ๐๐ = 0, i.e.(๐, ๐) โผ (๐, ๐).
Let ๐น = ๐/ โผ and write [(๐, ๐)] = ๐๐ . Define
๐๐
+ ๐๐
= ๐๐ + ๐๐๐๐
๐๐
โ ๐๐
= ๐๐๐๐
0๐น = 01
1๐น = 11
These are well-defined. If ๐๐ โ 0๐น = 0
1 then ๐ โ 1 โ 0 โ ๐ = 0. Then ๐๐ โ ๐น and
๐๐ โ ๐
๐ = 11 = 1๐น so ๐
๐ โ ๐น has an inverse. ๐น is a field.๐ is a subring of ๐น via
๐ โช ๐น
๐ โฆ ๐1
which is injective as ๐ is an integral domain.
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Example.
1. The field of fractions of Z is Q.
2. The field of fractions of C[๐] is
C(๐) = {๐(๐)๐(๐)
โถ ๐(๐), ๐(๐) โ C[๐], ๐(๐) โ 0} ,
the field of rational functions.
As we have mentioned before, {0} is a bona fide ring although it is a (trivial)counterexample to many results. However, it is not a field as we require 0 โ 1.To emphasise this, we declare
Fiat. The ring {0} is not a field.
Lemma 2.12. A non-zero ring ๐ is a field if and only if its only ideals are{0} and ๐ .
Proof.
โข โ: Suppose ๐ผ โด ๐ is a non-zero ideal, then it contains ๐ โ 0. But an idealcontaining a unit must be the whole ring.
โข โ: Let ๐ฅ โ 0 โ ๐ . Then (๐ฅ) = ๐ as it is not the zero ideal. Thus thereexists ๐ฆ โ ๐ such that ๐ฅ๐ฆ = 1๐ so ๐ฅ is a unit.
Definition (Maximal ideal). An ideal ๐ผ โด ๐ is maximal if there is no properideal which properly contains ๐ผ.
Lemma 2.13. An ideal ๐ผ โด ๐ is maximal if and only if ๐ /๐ผ is a field.
Proof. ๐ /๐ผ is a field if and only if ๐ผ/๐ผ and ๐ /๐ผ are the only ideals in ๐ /๐ผ, if andonly if ๐ผ, ๐ โด ๐ are the only ideals containing ๐ผ.
Definition (Prime ideal). An ideal ๐ผ โด ๐ is prime if ๐ผ is proper and if๐, ๐ โ ๐ such that ๐๐ โ ๐ผ then ๐ โ ๐ผ or ๐ โ ๐ผ.
Example. The ideal ๐Z is prime if and only if ๐ is 0 or a prime number: if ๐ isprime and ๐, ๐ โ ๐Z then ๐ โฃ ๐๐ so ๐ โฃ ๐ or ๐ โฃ ๐, i.e. ๐ โ ๐Z or ๐ โ ๐Z. Conversely,if ๐ = ๐ข๐ฃ is composite, ๐ข < ๐ then ๐ข๐ฃ โ ๐Z but ๐ข โ ๐Z.
Lemma 2.14. ๐ผ โด ๐ is prime if and only if ๐ /๐ผ is an integral domain.
Proof.
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โข โ: Supose ๐ผ โด ๐ is prime. Let ๐ + ๐ผ, ๐ + ๐ผ โ ๐ /๐ผ be such that (๐ + ๐ผ)(๐ +๐ผ) = 0๐ /๐ผ. Since ๐๐ + ๐ผ = 0๐ /๐ผ, ๐๐ โ ๐ผ. As ๐ผ is prime, ๐ โ ๐ผ or ๐ โ ๐ผ, i.e.๐ + ๐ผ = 0๐ /๐ผ or ๐ + ๐ผ = 0๐ /๐ผ. Thus ๐ /๐ผ is an integral domain.
โข โ: Suppose ๐ /๐ผ is an integral domain. Let ๐, ๐ โ ๐ such that ๐๐+๐ผ = 0๐ /๐ผ.(๐ + ๐ผ)(๐ + ๐ผ) = 0๐ /๐ผ. As ๐ /๐ผ is an integral domain, ๐ + ๐ผ = 0๐ /๐ผ or๐ + ๐ผ = 0๐ /๐ผ, i.e. ๐ โ ๐ผ or ๐ โ ๐ผ.
Corollary 2.15. Maximal ideals are prime.
Proof. Fields are integral domains.
Lemma 2.16. If ๐ is an integral domain then its characteristic is 0 or aprime number.
Proof. Consider ker(๐ โถ Z โ ๐ ) = ๐Z. By 1st Isomorphism Theorem
Z/๐Z โ Im ๐ โค ๐ .
As a subring of an integral domain is an integral domain, Z/๐Z is an integraldomain so ๐Z โด Z is prime. Thus ๐ = 0 or a prime number.
2.4 Factorisation in integral domainsLet ๐ be an integral domain in this section.
We begin with several definitions. Note that for every statement about anelement of the ring there is an equivalent one in terms of ideals.
Definition (Unit, divisibility, associates, irreducible, prime).
โข An element ๐ โ ๐ is a unit if there is ๐ โ ๐ such that ๐๐ = 1๐ .Equivalently, (๐) = ๐ .
โข ๐ โ ๐ divides ๐ โ ๐ if there is a ๐ โ ๐ such that ๐ = ๐๐. Equivalently,(๐) โ (๐). Write ๐ โฃ ๐.
โข ๐, ๐ โ ๐ are associates if ๐ โฃ ๐ and ๐ โฃ ๐. Equivalently, (๐) = (๐).
โข ๐ โ ๐ is irreducible if it is not zero, not a unit and if ๐ = ๐๐ then ๐ or๐ is a unit.
โข ๐ โ ๐ is prime if it is not zero, not a unit and if ๐ โฃ ๐๐ then ๐ โฃ ๐ or๐ โฃ ๐. Equivalently, ๐๐ โ (๐) โ ๐ โ (๐) or ๐ โ (๐).
Remark. Being a unit/irreducible/prime depends not only on the element butalso on the ambient ring: 2๐ โ Z[๐] is not irreducible but 2๐ โ Q[๐] is.
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Lemma 2.17. (๐) โด ๐ is prime if and only if ๐ is zero or prime.
Proof.
โข โ: Let (๐) โด ๐ be a prime ideal and ๐ โฃ ๐๐. Then ๐๐ โ (๐) so ๐ โ (๐) or๐ โ (๐) as (๐) is prime. So ๐ โฃ ๐ or ๐ โฃ ๐. ๐ is 0 or a prime.
โข โ: If ๐ = 0 then (0) โด ๐ is a prime ideal since ๐ โ ๐ /(0) is an integraldomain. Let ๐ โ 0 be a prime and ๐๐ โ (๐). Then ๐ โฃ ๐๐ so ๐ โฃ ๐ or ๐ โฃ ๐.๐ โ (๐) or ๐ โ (๐) as required.
Lemma 2.18. If ๐ โ ๐ is prime then it is irreducible.
Proof. Let ๐ = ๐๐. Then ๐ โฃ ๐๐ so ๐ โฃ ๐ or ๐ โฃ ๐. Suppose ๐ โฃ ๐ wlog. Then ๐ = ๐๐.๐ = (๐๐)๐, ๐(๐๐ โ 1) = 0. As ๐ โ 0 and ๐ is an integral domain, ๐๐ โ 1 = 0 so ๐is a unit.
Example. Let ๐ = Z[โ
โ5] = {๐ + ๐โ
โ5 โถ ๐, ๐ โ Z} โค C. This is a subring ofa field so an integral domain. Define
๐ โถ ๐ โ Zโฅ0
๐ + ๐โ
โ5 โฆ ๐2 + 5๐2
so ๐(๐ง) = ๐ง๐ง. Note ๐(๐1๐2) = ๐(๐1)๐(๐2). If ๐ is a unit then there exists ๐ โ ๐ such that ๐๐ = 1, then ๐(๐)๐(๐ ) = ๐(1) = 1, so ๐(๐) = 1. So ๐ = ๐ + ๐
โโ5
such that ๐2 + 5๐2 = 1. The only possibility is ๐ = ยฑ1. Claim that 2 โ ๐ isirreducible:
Proof. Let 2 = ๐๐ so ๐(๐)๐(๐) = 4. ๐(๐) = 1, 2 or 4. But ๐(๐) โ 2 so๐(๐) = 1 or 4, ๐(๐) = 4 or 1 so ๐ or ๐ is a unit.
Similarly we can show that 3 and 1 ยฑโ
โ5 are irreducible.Note that
(1 +โ
โ5)(1 โโ
โ5) = 1 + 5 = 6 = 2 โ 3
so 2 โฃ (1 +โ
โ5)(1 โโ
โ5) but ๐(1 ยฑโ
โ5) = 6 is not divisible by ๐(2) = 4 so2 โค 1 ยฑ
โโ5. Thus 2 โ ๐ is not prime.
We also find that 6 = 2 โ 3 = (1 +โ
โ5)(1 โโ
โ5) has two different factori-sations into irreducibles.
Definition (Euclidean domain). An integral domain ๐ is a Euclidean domain(ED) if there is a function ๐ โถ ๐ \ {0} โ Zโฅ0, a Euclidean function such that
1. โ๐, ๐ โ ๐ \ {0}, ๐(๐๐) โฅ ๐(๐),
2. โ๐, ๐ โ ๐ , ๐ โ 0, we have ๐ = ๐๐ + ๐ with ๐ = 0 or ๐(๐) < ๐(๐).
Example.
1. Z is a Euclidean domain with ๐(๐) = |๐|.
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2. For a field F, F[๐] is a Euclidean domain with ๐(๐) = deg ๐.
3. Z[๐] is a Euclidean domain with ๐(๐ + ๐๐) = ๐2 + ๐2 = (๐ + ๐๐)(๐ โ ๐๐).
Proof. Let ๐ง1, ๐ง2 โ Z[๐], ๐ง2 โ 0. Consider ๐ง1๐ง2
โ C. By considering thelattice of Gaussian integers on the complex plane, we can find ๐ โ Z[๐]such that โฃ ๐ง1
๐ง2โ ๐โฃ < 1. Consider ๐ = ๐ง1 โ ๐๐ง2 โ Z[๐],
โฃ ๐๐ง2
โฃ = โฃ๐ง1๐ง2
โ ๐โฃ < 1
so |๐| < |๐ง2| so ๐(๐) = |๐|2 < |๐ง2|2 = ๐(๐ง2).
4. Similarly we can show Z[โ
โ2] is a Euclidean domain.
Proposition 2.19. If ๐ is a ED then it is a PID.
This proof is a generalisation of the proof that Z is a PID.
Proof. Let ๐ผ โด ๐ and choose 0 โ ๐ โ ๐ผ such that ๐(๐) is minimal. If ๐ โ ๐ผthen Euclidean property gives ๐ = ๐๐ + ๐ with ๐(๐) < ๐(๐) or ๐ = 0. Then๐ = ๐ โ ๐๐ โ ๐ผ but if ๐ โ 0 then minimality of ๐(๐) is contradicted. Thus ๐ = 0and ๐ โ (๐). ๐ผ = (๐).
Example. Z,F[๐] and Z[๐] are PIDs.
Example. Z[๐] is not a PID. Consider (2, ๐) โด Z[๐]. Suppose (2, ๐) = (๐)for some ๐ โ Z[๐], then ๐ โฃ 2. Degrees of polynomials on an integral domain addunder multiplication so if ๐ divides a constant polynomial it must be constant.Thus ๐ = ยฑ1, ยฑ2. If ๐ = ยฑ2, ยฑ2 โค ๐. Absurd. Thus ๐ = ยฑ1, (๐) = Z[๐]. But1 โ (2, ๐). Absurd.
Example. Let F be a field and ๐ด โ โณ๐(F). Consider
๐ผ = {๐ โ F[๐] โถ ๐(๐ด) = 0}.
If ๐, ๐ โ ๐ผ, (๐ + ๐)(๐ด) = ๐(๐ด) + ๐(๐ด) = 0. If โ โ F[๐], (๐โ)(๐ด) = ๐(๐ด)โ(๐ด) = 0so ๐ผ โด F[๐]. As F[๐] is a PID, ๐ผ = (๐๐ด) for some ๐๐ด โ F[๐]. This ๐๐ด is theminimal polynomial of ๐ด and it follows that it is unique up to a unit.
Definition (Unique factorisation domain). An integral domain is a uniquefactoriation domain (UFD) if
โข every non-zero, non-unit is a product of irreducibles,
โข if ๐1 โฏ ๐๐ = ๐1 โฏ ๐๐ are factorisations into irreducibles, then ๐ = ๐and ๐๐ is an associate of ๐๐ up to reordering.
We will show that PIDs are UFDs.
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Lemma 2.20. If ๐ is a PID then irreducibles are primes.
Proof. Let ๐ โ ๐ be irreducible and suppose ๐ โฃ ๐๐. Need to show that ๐ โฃ ๐ or๐ โฃ ๐. Suppose ๐ โค ๐. Consider (๐, ๐) โด ๐ . As ๐ is a PID, there exists ๐ โ ๐ such that (๐) = (๐, ๐), so ๐ = ๐1๐, ๐ = ๐2๐. As ๐ is irreducible, either ๐1 or ๐ isa unit. If ๐1 is a unit then ๐ = ๐2๐ = ๐2(๐โ1
1 ๐) so ๐ โฃ ๐. Thus ๐ must be a unitand (๐, ๐) = (๐) = ๐ . Thus 1๐ = ๐๐ + ๐ ๐ for some ๐ and ๐ . ๐ = ๐๐๐ + ๐๐๐ so๐ โฃ ๐.
Lemma 2.21. Let ๐ be a PID and ๐ผ1 โ ๐ผ2 โ ๐ผ2 โ โฏ be an increasingsequence of ideal. Then there exists ๐ โ N such that for all ๐ โฅ ๐, ๐ผ๐ = ๐ผ๐+1.
Definition (Noetherian). The above condition is the ascending chain con-dition. A chain satisfying the above condition is Noetherian.
Proof. Let ๐ผ = โโ๐=1 ๐ผ๐ which is again an ideal so ๐ผ = (๐) for some ๐ โ ๐ . Then
๐ โ ๐ผ so there exists ๐ โ N such that ๐ โ ๐ผ๐. Then
(๐) โ ๐ผ๐ โ ๐ผ๐+1 โฏ โ (๐)
so equality throughout.
Theorem 2.22. PID is UFD.
Proof. Let ๐ be a PID. The proof consists of two parts: first show the existenceof factorisation in ๐ (the proof thereof generalises to all Noetherian rings), andthen show its uniqueness.
1. Suppose for contradiction there exists ๐ โ ๐ which cannot be written as aproduct of irreducibles. then ๐ is not irreducible so ๐ = ๐1๐1 with ๐1, ๐1not units and one of then cannot be written as a product of irreducibles(otherwise ๐ would be), say it is ๐1. Hence ๐1 = ๐2๐2 where ๐2, ๐2 arenot units and wlog ๐2 could not be written as a product of irreducibles.Continue this way. Now
(๐) โ (๐1) โ (๐2) โ โฏ
is an ascending chain so by ACC we must have (๐๐) = (๐๐+1) for some ๐,i.e. ๐๐ = ๐๐+1๐๐+1 with ๐๐+1 a unit.
2. Let ๐1 โฏ ๐๐ = ๐1 โฏ ๐๐ be factorisations into irreducibles. Thus ๐1 โฃ ๐1 โฏ ๐๐.In a PID irreducibles are primes so ๐1 โฃ ๐๐ for some ๐. After reordering๐1 โฃ ๐1 so ๐1 = ๐1 โ ๐. As ๐1 is irreducible, ๐ is a unit so ๐1 and ๐1are associates. Now ๐1(๐2 โฏ ๐๐ โ ๐๐2 โฏ ๐๐) = 0. As ๐ is an integraldomain ๐2 โฏ ๐๐ = (๐๐2) โฏ ๐๐. Continue this way, we get ๐ โค ๐ and1 = (unit) โ ๐๐+1 โฏ ๐๐. Thus ๐๐+1, โฆ ๐๐ are units. Absrud. Thus ๐ = ๐and ๐๐โs and ๐๐โs are associates up to reordering.
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Definition (gcd, lcm).
โข ๐ is a greatest common divisor (gcd) of ๐1, โฆ , ๐๐, written gcd(๐1, โฆ , ๐๐),if ๐ โฃ ๐๐ for all ๐ and if ๐โฒ โฃ ๐๐ for all ๐ then ๐โฒ โฃ ๐.
โข ๐ is a lowest common multiple (lcm) of ๐1, โฆ , ๐๐, written lcm(๐1, โฆ , ๐๐),if ๐๐ โฃ ๐ for all ๐ and if ๐๐ โฃ ๐โฒ for all ๐ then ๐ โฃ ๐โฒ.
It is easy to see that if gcd or lcm exists then it is unique up to associates.
Proposition 2.23. If ๐ is a UFD then gcdโs and lcmโs exist.
Proof. Write each ๐๐ as a product
๐๐ = ๐ข๐ โ โ๐
๐๐๐๐๐
where ๐ข๐ is a unit and ๐๐โs are (the same) irreducibles which are not associatesof each other. Set
๐ = โ๐
๐๐๐๐
where ๐๐ = min๐ ๐๐๐. Certainly ๐ โฃ ๐๐ for all ๐. If ๐โฒ โฃ ๐๐ for all ๐ then write
๐โฒ = ๐ข โ โ๐
๐๐ก๐๐
for some ๐ก๐. As ๐โฒ โฃ ๐ we must have ๐ก๐ โค ๐๐๐ for all ๐ so ๐ก๐ โค min๐ ๐๐๐ = ๐๐ forall ๐. Thus ๐โฒ โฃ ๐.
The argument for lcm is similar.
2.5 Factoriation in polynomial ringsFor a field F we know F[๐] is a ED, so also a PID and UFD so
1. any ๐ผ โด F[๐] is principal, i.e. ๐ผ = (๐) for some ๐;
2. ๐ โ F[๐] is irreducible if and only if ๐ is prime;
3. let ๐ โ F[๐] be irreducible and (๐) โ ๐ฝ โด F[๐] be a larger ideal. Then๐ฝ = (๐) for some ๐ โ F[๐] so (๐) โ (๐), i.e. ๐ โฃ ๐. But ๐ is irreducible soeither ๐ is a unit, then (๐) = F[๐], or ๐ is an associate of ๐, so (๐) = (๐).Thus (๐) is maximal;
4.(๐) prime โน ๐ prime โน ๐ irreducible โน (๐) maximal
so prime ideals of F[๐] are precisely the maximal ideals;
5. ๐ โ F[๐] is irreducible if and only if (๐) is maximal, if and only if F[๐]/(๐)is a field.
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Definition (Content). Let ๐ be a UFD and
๐ = ๐0 + ๐1๐ + โฆ ๐๐๐๐ โ ๐ [๐]
with ๐๐ โ 0. The content is
๐(๐) = gcd(๐0, โฆ , ๐๐).
Definition (Primitive). ๐ above is primitive if ๐(๐) is a unit, i.e. ๐๐โs arecoprime.
Theorem 2.24 (Gaussโ Lemma). Let ๐ be a UFD and ๐น be its field offractions. Let ๐ โ ๐ [๐] be primitive. Then ๐ is irreducible in ๐ [๐] if andonly if ๐ is irreducible in ๐น[๐].
Example. Let ๐ = 1 + ๐ + ๐3 โ Z[๐]. ๐(๐) = 1 so ๐ is primitive. Suppose๐ = ๐โ, a product of irreducibles in Z[๐]. As ๐ is primitive, neither ๐ nor โcan be a constant polynomial so they have degree 1 and 2 respectively. Wlogsuppose ๐ = ๐0 +๐1๐, โ = ๐0 +๐1๐ +๐2๐2 โ Z[๐]. Expanding out and equatingthe coefficients, ๐0๐0 = 1, ๐1๐2 = 1 so ๐0๐1 = ยฑ1. Thus ๐ has one of ยฑ1 as aroot and so does ๐. But it doesnโt so such factorisation does not exist. ThusQ[๐]/(1 + ๐ + ๐3) is a field.
Lemma 2.25. Let ๐ be a UFD. If ๐, ๐ โ ๐น [๐] are primitives then so is ๐๐.
Proof. Let
๐ = ๐0 + ๐1๐ + โฏ + ๐๐๐๐
๐ = ๐0 + ๐1๐ + โฏ + ๐๐๐๐
with ๐๐, ๐๐ โ 0. If ๐๐ is not primitive, then ๐(๐๐) is not a unit so there is anirreducible ๐ โฃ ๐(๐๐). As ๐(๐) and ๐(๐) are units, we have
๐ โฃ ๐0, ๐ โฃ ๐1, โฆ , ๐ โฃ ๐๐โ1, ๐ โค ๐๐
๐ โฃ ๐0, ๐ โฃ ๐1, โฆ , ๐ โฃ ๐โโ1, ๐ โค ๐โ
The coefficients of ๐๐+โ in ๐๐ is
โ๐+๐=๐+โ
๐๐๐๐ = โฏ + ๐๐+1๐โโ1 + ๐๐๐โ + ๐๐โ1๐โ+1 + โฏ
where LHS is divisible by ๐ so ๐ โฃ ๐๐๐โ but ๐ is prime so ๐ โฃ ๐๐ or ๐ โฃ ๐โ. Absurd.Thus ๐(๐๐) is a unit and ๐๐ is a primitive.
Corollary 2.26. Let ๐ be a UFD. Then ๐(๐๐) is an associate of ๐(๐)๐(๐).
Proof. Let ๐ = ๐(๐) โ ๐1, ๐ = ๐(๐) โ ๐1 with ๐1, ๐1 primitive. Then
๐๐ = ๐(๐)๐(๐) โ (๐1๐1)
where ๐1๐1 is primitive by the lemma above. Thus ๐(๐)๐(๐) is a gcd of thecoefficients of ๐๐.
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Proof of Gaussโ Lemma. Let ๐ โ ๐ [๐] be primitive. If ๐ = ๐โ is reducible in๐ [๐] then ๐, โ cannot be constants as otherwise ๐ would not be primitive. Thus๐, โ โ ๐น [๐] are not units so ๐ โ ๐น [๐] is reducible.
Suppose instead ๐ is reducible in ๐น[๐], say ๐ = ๐โ. We can โclear thedenominatorsโ: find ๐, ๐ โ ๐ such that ๐๐, ๐โ โ ๐ [๐], then
๐๐๐ = (๐๐) โ (๐โ) โ ๐ [๐].
Take contents, ๐๐ = ๐(๐๐) โ ๐1, ๐โ = ๐(๐โ) โ โ1 with ๐1, โ1 primitive. Then
๐๐ โ ๐ = ๐(๐๐)๐(๐๐) ๐1โ1โprimitive
so ๐๐ is an associate of ๐(๐๐)๐(๐โ) so ๐(๐๐)๐(๐โ) = ๐ข๐๐ where ๐ข is a unit. Thus๐๐๐ = ๐ข๐๐๐1โ1 and cancel to get ๐ = (๐ข๐1)โ1 is reducible in ๐ [๐].
Proposition 2.27. Let ๐ be a UFD and ๐ โ ๐ [๐] primitive. Let ๐ผ = (๐) โด๐น[๐] where ๐น is the field of fraction of ๐ and ๐ฝ = (๐) โด ๐ [๐]. Then
๐ฝ = ๐ผ โฉ ๐ [๐].
Equivalently, if ๐ โ ๐ [๐] is divisible by a primitive ๐ โ ๐น [๐] then it isdivisible by ๐ in ๐ [๐].
Proof. The โ inclusion is clear. To show the other direction, let ๐ = ๐โ โ ๐น[๐].Clear denominators by find ๐ โ ๐ such that ๐โ โ ๐ [๐] so ๐๐ = (๐โ) โ ๐ โ ๐ [๐].Thus ๐๐ = ๐(๐โ)โ1๐ with โ1 primitive. Now it follows that ๐ โฃ ๐(๐โ), as๐๐(๐) = ๐(๐โ), so we get ๐ = ๐(๐) โ โ1๐ โ ๐ [๐]. ๐ divides ๐ in ๐ [๐].
Theorem 2.28. If ๐ is a UFD then so is ๐ [๐].
Proof. To show existence, let ๐ โ ๐ [๐] and write ๐ = ๐(๐) โ ๐1 with ๐1 primitive.As ๐ is a UFD we can write ๐(๐) = ๐1 โฏ ๐๐ โ ๐ with ๐๐ irreducible in ๐ , soalso irreducible in ๐ [๐]. If ๐1 is not irreducible, write ๐1 = ๐2 โ ๐3with ๐2, ๐3 notunits and are primitive. Thus ๐2, ๐3 are not constants so have degree smallerthan that of ๐1. If ๐2 or ๐3 is irreducible, factor again. The degree continues tostrictly decrease and this stops eventually. So
๐ = ๐1 โฏ ๐๐๐1 โฏ ๐๐,
a product of irreducibles.Now for the uniqueness part, note ๐1 โฏ ๐๐ = ๐(๐) โ ๐ , a UFD so the
๐๐โs are unique up to reordering and associates. Thus it suffices to show if๐1 โฏ ๐๐ = ๐1 โฏ ๐โ as products of primitive polynomials then ๐ = โ and the ๐๐โsand ๐๐โs are the same up to reordering and associates. Since ๐น[๐] is a PID andthus UFD, ๐1 โฏ ๐๐ = ๐1 โฏ ๐โ โ ๐ [๐] โ ๐น[๐] implise that ๐ = โ and ๐๐โs equalto ๐๐โs in ๐น[๐]. If ๐1 is an associate of ๐1 in ๐น[๐] then ๐1 = ๐ข๐1 for some unit๐ข โ ๐น[๐]. Then ๐ข โ ๐น is a unit, write ๐ข = ๐
๐ . Get ๐๐1 = ๐๐1 โ ๐ [๐]. Takingcontents, it follows that ๐ is an associate of ๐ in ๐ . Cancel to get ๐1 = ๐๐1 โ ๐ [๐].Repeat for ๐๐โs and ๐๐โs.
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Example.
1. Z[๐] is a UFD.
2. If ๐ is a UFD then so is ๐ [๐1, โฆ , ๐๐].
Proposition 2.29 (Eisensteinโs criterion). Let ๐ be a UFD and ๐ = ๐0 +๐1๐ + โฏ + ๐๐๐๐ โ ๐ [๐] with ๐๐ โ 0 be primitive. Suppose ๐ โ ๐ is anirreducible such that
โข ๐ โค ๐๐,
โข ๐ โฃ ๐๐ for ๐ = 0, 1, โฆ , ๐ โ 1,
โข ๐2 โค ๐0
then ๐ is irreducible in ๐ [๐], so also in ๐น[๐].
Proof. Let ๐ = ๐โ with
๐ = ๐0 + ๐1๐ + โฏ + ๐๐๐๐
โ = ๐ 0 + ๐ 1๐ + โฏ + ๐โ๐โ
with ๐๐, ๐ โ โ 0. Then ๐ + โ = ๐ and ๐๐ = ๐๐๐ โ. As ๐ โค ๐๐, ๐ โค ๐๐ and๐ โค ๐ โ. Since ๐ โฃ ๐0 and ๐2 โค ๐0, suppose wlog that ๐ โฃ ๐0, ๐ โค ๐ 0. Suppose๐ โฃ ๐0, ๐ โฃ ๐1, ๐ โฃ ๐๐โ1, ๐ โค ๐๐. Then
๐๐ = ๐ 0๐๐ + ๐ 1๐๐โ1 + ๐ 2๐๐โ2 + โฏ + ๐ ๐๐0
so ๐ โค ๐๐ and by 2 ๐ = ๐. Thus deg ๐ = ๐ and โ is a constant. As ๐ (and hence ๐and โ) is a primitive โ is a unit.
Example. For ๐ โ Z prime, ๐ = ๐๐ โ ๐ โ Z[๐] is irreducible in Z[๐] andQ[๐] so ๐ does not have a root in Q. In particular, this shows that ๐
โ๐ โ Q).This will be important in IID Galois Theory.
Example. For ๐ โ Z prime, let
๐ = ๐๐โ1 + ๐๐โ2 + โฏ + ๐ + 1 โ Z[๐].
Note that (๐ โ 1)๐ = ๐๐ โ 1. Consider the ring isomorphism
๐ โถ Z[๐] โ Z[๐]๐ โฆ ๐ + 1
Then
๐(๐) = ๐๐โ1โ๐โค
+ (๐1)
โ๐โฃ
๐๐โ1 + โฏ + ( ๐๐ โ 2
)โ
๐โฃ
๐ + ( ๐๐ โ 1
)โ
=๐
so Eisensteinโs criterion says that ๐(๐) is irreducible, so is ๐.
Remark. The hypothesis of Eisensteinโs criterion depends on the ambient ringwhile the conclusiohn does not. As a heuristics, we can apply ring isomorphismsto reduce the problem sometimes.
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2.6 Gaussian integersRecall
Z[๐] = {๐ + ๐๐ โถ ๐, ๐ โ Z} โค C.
It has a norm ๐(๐ + ๐๐) = ๐2 + ๐2, making it a ED, and thus a PID and UFD.In particular primes and irreducibles agree. The units in Z[๐] are ยฑ1, ยฑ๐ as theyare the only elements of norm 1. In addition, we have the following observations:
1. 2 = (1 + ๐)(1 โ ๐) is not a prime.
2. ๐(3) = 9. If 3 = ๐ฅ๐ฆ then 9 = ๐(๐ฅ)๐(๐ฆ). Either ๐ฅ or ๐ฆ is a unit or๐(๐ฅ) = ๐(๐ฆ) = 3. But the norm is never 3 so 3 is a prime.
3. 5 = (2 + ๐)(2 โ ๐) is not a prime.
4. 7 is a prime.
Proposition 2.30. A prime ๐ โ Z is a prime in Z[๐] if and only if ๐ โ ๐2+๐2
for ๐, ๐ โ Z.
Proof.
โข โ: If ๐ = ๐2 + ๐2 = (๐ + ๐๐)(๐ โ ๐๐), it is reducible and thus not a prime.
โข โ: Note ๐(๐) = ๐2. If ๐ factors as ๐ข๐ฃ with ๐ข, ๐ฃ not units then ๐(๐ข) =๐(๐ฃ) = ๐. Write ๐ข = ๐ + ๐๐, we have ๐ = ๐(๐ข) = ๐2 + ๐2.
Now we prove a lemma regarding the multiplicative group of a finite field:
Lemma 2.31. Let F๐ = Z/๐Z be a field with ๐ elements and ๐ prime. ThenFร
๐ = F๐ \ {0} is a group under multiplication and is isomorphic to ๐ถ๐โ1.
Proof. Certainly Fร๐ is an abelian group of order ๐ โ 1. By the classification
theorem of finite abelain groups, Fร๐ is either cyclic or contains ๐ถ๐ ร ๐ถ๐ as a
subgroup for some ๐ โฅ 2.Suppose ๐ถ๐ ร ๐ถ๐ โค Fร
๐ . Consider ๐ = ๐๐ โ 1 โ F๐[๐]. Each element of๐ถ๐ ร ๐ถ๐ โค Fร
๐ โ F๐ gives a root of ๐ so it has at least ๐2 distinct roots. Butas F๐[๐] is a ED and thus UFD, it can be factorised into at most ๐ uniqueirreducibles. Thus it has at most ๐ distinct roots in F๐. Thus there is nosubgroup ๐ถ๐ ร ๐ถ๐ in Fร
๐ and Fร๐ is cyclic.
Proposition 2.32. The primes in Z[๐] are, up to associates,
1. prime ๐ โ Z with ๐ = 3 mod 4,
2. ๐ง โ Z[๐] such that ๐(๐ง) = ๐ where ๐ is a prime and ๐ = 2, or ๐ = 1mod 4.
Proof. First show what we claimed are indeed primes, i.e. irreducibles:
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1. if ๐ = 3 mod 4, ๐ โ ๐2 + ๐2 so ๐ โ Z[๐] is a prime.
2. suppose ๐ง = ๐ข๐ฃ then ๐(๐ข)๐(๐ฃ) = ๐ so ๐(๐ข) or ๐(๐ฃ) = 1. ๐ข or ๐ฃ is a unitso ๐ง is irreducible.
Now let ๐ง โ Z[๐] be a prime. Then ๐ง is irreducible too so ๐(๐ง) = ๐ง๐ง is afactorisation of ๐(๐ง) into irreducibles in Z[๐]. Let ๐ โ Z be a prime dividing๐(๐ง).
โข Case 1: ๐ = 3 mod 4. Then ๐ is irreducible in Z[๐]. As ๐ โฃ ๐(๐ง), ๐ โฃ ๐ง or๐ โฃ ๐ง. Wlog ๐ โฃ ๐ง. As ๐ and ๐ง are both irreducibles, they are associates.
โข Case 2: ๐ = 2, or ๐ = 1 mod 4. If ๐ = 1 mod 4, consider Fร๐ โ ๐ถ๐โ1 =
๐ถ4๐. It has a unique element of order 2, namely [โ1]. As 4 โฃ ๐ โ 1, there isalso an element [๐] โ Fร
๐ order 4. Then [๐2] has order 2 and thus ๐2 = โ1mod ๐. Thus there exists ๐ โ Z such that ๐2 + 1 = ๐๐, ๐ โฃ (๐ + ๐)(๐ โ ๐).If ๐ = 2 then ๐ โฃ (1 + ๐)(1 โ ๐).But ๐ โค ๐ + ๐, ๐ โค ๐ โ ๐ so ๐ โ Z[๐] is not prime and thus not irreducible.Hence ๐ = ๐ง1๐ง2 with ๐ง1, ๐ง2 not units. ๐2 = ๐(๐) = ๐(๐ง1)๐(๐ง2), ๐(๐ง1) =๐(๐ง2) = ๐ so ๐ = ๐ง1๐ง1 = ๐ง2๐ง2. But also ๐ = ๐ง1๐ง2 so ๐ง2 = ๐ง1.We choose ๐ such that ๐ โฃ ๐(๐ง) so ๐ง1๐ง1 โฃ ๐ง๐ง and ๐ง is prime so ๐ง โฃ ๐ง1 or๐ง โฃ ๐ง1. ๐ง1 or ๐ง1 is an associate of ๐ง. ๐(๐ง) = ๐(๐ง1) or ๐(๐ง1) = ๐.
Corollary 2.33. An integer ๐ โ Z > 0 can be written as ๐2 + ๐2, ๐, ๐ โ Zif and only if when we write ๐ = ๐๐1
1 ๐๐22 โฏ ๐๐๐
๐ with ๐๐โs distinct, if ๐๐ = 3mod 4 then ๐๐โs are even.
Proof. Let ๐ = ๐2 + ๐2 = (๐ + ๐๐)(๐ โ ๐๐) = ๐(๐ + ๐๐). Let ๐ง = ๐ + ๐๐.Then ๐ง = ๐ผ1 โฏ ๐ผ๐ as a product of irreducibles (i.e. primes) in Z[๐]. Then๐ = ๐(๐ผ1) โฏ ๐(๐ผ๐ ). Each ๐ผ๐ is either a prime ๐ congruent to 3 mod 4 so๐(๐ผ๐) = ๐2, or has ๐(๐ผ๐) = ๐, a prime not congruent to 3 mod 4. Thus ๐ canbe written as a product of primes as claimed.
Conversely, suppose ๐ = ๐๐11 โฏ ๐๐๐
๐ with ๐๐ even if ๐ = 3 mod 4. For each ๐if ๐๐ = 3 mod 4 then ๐(๐๐) = ๐2
๐ , ๐๐๐๐ = ๐(๐๐๐/2
๐ ). As ๐ is a product of norms ofGaussian integers, it is the norm of a Gaussian integer so is a sum of squares.
Example. In how many ways can 65 be written as a sum of two squares?65 = 5 ร 13, 5 = 12 + 22 = (2 + ๐)(2 + ๐), 13 = 22 + 32 = (2 + 3๐)(2 + 3๐) so
65 = (2 + ๐)(2 + 3๐)(2 + ๐)(2 + 3๐)= ๐((2 + ๐)(2 + 3๐)) = ๐(1 + 8๐) = 12 + 82
= ๐((2 + ๐)(2 โ 3๐)) = ๐(7 โ 4๐) = 72 + 42
Exercise (Challenge). Find conditions such that ๐ = ๐2 + 2๐2 and ๐2 + 3๐2 inZ[
โโ2] and Z[
โโ3].
2.7 Algebraic integers
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Definition (Algebraic integer). A complex number ๐ผ โ C is an algebraicinteger if it is a root of a monic polynomial with integer coefficients.
If ๐ผ is an algebraic integer, let Z[๐ผ] โค C be the smallest subring containing๐ผ, i.e. it is the image of the ring homomorphism
๐ โถ Z[๐] โ C๐ โฆ ๐ผ
Thus by 1st Isomorphism Theorem Z[๐ผ] โ Z[๐]/๐ผ where ๐ผ = ker๐.
Proposition 2.34 (Minimal polynomial). If ๐ผ is an algebraic integer then๐ผ = ker๐ is principal and is generated by an irreducible ๐๐ผ โ Z[๐], theminimal polynomial of ๐ผ.
Proof. As ๐ผ is an algebraic integer, it is a root of some ๐ โ Z[๐] so ๐ โ ๐ผ. Let๐๐ผ โ ๐ผ be a polynomial of minimal degree, which we may assume is positive. Wewant to show that
1. ๐ผ = (๐๐ผ),
2. ๐๐ผ is irreducible.
1. Let โ โ ๐ผ. Now Q[๐] is a ED so we can write โ = ๐๐๐ผ + ๐ โ Q[๐] with๐ = 0 or deg ๐ < deg ๐๐ผ. Clearing denominators, there is an ๐ โ Z suchthat ๐๐, ๐๐ โ Z[๐], so ๐โ = (๐๐)๐๐ผ + ๐๐ โ Z[๐]. ๐ผ is a root of โ and of ๐๐ผso is also a root of ๐๐. As ๐๐ผ has minimal degree among polynomials with๐ผ as a root, we must have ๐๐ = 0. Thus ๐โ = (๐๐)๐๐ผ. Now ๐(๐โ) = ๐โ ๐(โ),๐((๐๐)๐๐ผ) = ๐(๐๐) so ๐ โฃ ๐(๐๐) so ๐๐ = ๐๐ with ๐ โ Z[๐]. Cancelling showsthat ๐ = ๐. Thus โ = ๐๐๐ผ so โ โ (๐๐ผ).
2. Z[๐]/(๐๐ผ) โ Z[๐ผ] โค C. As C is an integral domain, so is Z[๐ผ]. Thus (๐๐ผ)is prime. Thus ๐๐ผ โ Z[๐] is a prime and hence irreducible.
Example.
1. ๐ผ = ๐, ๐๐ผ = ๐2 + 1.
2. ๐ผ =โ
2, ๐๐ผ = ๐2 โ 2.
3. ๐ผ = 1+โ
โ32 , ๐๐ผ = ๐2 โ ๐ + 1.
4. Less trivially, for ๐ โ Z, ๐5 โ ๐ + ๐ has a unique real root ๐ผ. This ๐ผcannot be constructed using (Z, +, ร,
โ). c.f. IID Galois Theory.
Lemma 2.35. If ๐ผ is an algebraic integer and ๐ผ โ Q then ๐ผ โ Z.
Proof. ๐๐ผ โ Z[๐] is irreducible and primitive. By Gaussโ Lemma ๐๐ผ โ Q[๐] isalso irreducible. But if ๐ผ โ Q, ๐ โ๐ผ โฃ ๐๐ผ in Q[๐] so ๐๐ผ = ๐ โ๐. But ๐๐ผ โ Z[๐]so ๐ผ โ Z.
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2.8 Hilbert Basis TheoremRecall that a ring ๐ satisfies the ascending chain condition (ACC) if whenever
๐ผ1 โ ๐ผ2 โ โฏ
ais an increasing sequence of ideals then there exists ๐ โ N such that for all๐ โฅ ๐, ๐ผ๐ = ๐ผ๐+1.
A ring satisfying ACC is called Noetherian.We have shown that a PID is Noetherian.
Lemma 2.36. A ring ๐ is Noetherian if and only if every ideal of ๐ iffinitely generated.
Proof.
โข โ: Let ๐ผ1 โ ๐ผ2 โ โฏ be an ascending chain of ideal and ๐ผ = โ๐ ๐ผ๐. Then๐ผ = (๐1, โฆ , ๐๐) for some ๐๐ โ ๐ . For all ๐ there exists ๐๐ โ N such that๐๐ โ ๐ผ๐๐
so(๐1, โฆ , ๐๐) โ ๐ผmax๐ ๐๐
โ ๐ผ.Take ๐ = max๐ ๐๐ and the result follows.
โข Suppose ๐ is Noetherian and ๐ผ โด ๐ . Choose ๐1 โ ๐ผ. If ๐ผ = (๐1) then done,so suppose not. Then choose ๐2 โ ๐ผ \ (๐1). If ๐ผ = (๐1, ๐2) then done, sosuppose not. If we are never finished by this process then we get
(๐1) โ (๐1, ๐2) โ โฏ
which is impossible as ๐ is Noetherian. Thus ๐ผ = (๐1, โฆ , ๐๐) for some ๐.
Theorem 2.37 (Hilbert Basis Theorem). If ๐ is Noetherian then so is๐ [๐].
Proof. Let ๐ฝ โด ๐ [๐]. Let ๐1 โ ๐ฝ be of minimal minimal degree. If ๐ฝ = (๐1)then done, else choose ๐2 โ ๐ฝ \ (๐1) of minimal degree. Suppose we have
(๐1) โ (๐1, ๐2) โ โฏ
as an ascending chain of non-stabilising ideas. Let ๐๐ โ ๐ be the coefficient ofthe largest power of ๐ in ๐๐ and consider
(๐1) โ (๐1, ๐2) โ โฏ โด ๐ .
As ๐ is Noetherian this chain stabilises, i.e. there exists ๐ โ N such that all ๐๐โslie in (๐1, โฆ , ๐๐). In particular, ๐๐+1 = โ๐
๐=1 ๐๐๐๐ for some ๐๐ โ ๐ . Let
๐ =๐
โ๐=1
๐๐๐๐๐deg ๐๐+1โdeg ๐๐
which has leading term๐
โ๐=1
๐๐๐๐๐deg ๐๐+1 = ๐๐+1๐deg ๐๐+1 .
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Thus deg(๐๐+1 โ ๐) < deg ๐๐+1. But ๐ โ (๐1, โฆ , ๐๐) but ๐๐+1 โ (๐1, โฆ , ๐๐)so ๐๐+1 โ ๐ โ (๐1, โฆ , ๐๐). This contradicts the minimality of the degree of๐๐+1.
Example. Z[๐1, โฆ , ๐๐], F[๐1, โฆ , ๐๐] are Noetherian.
Lemma 2.38. A quotient of a Noetherian ring is Noetherian.
Corollary 2.39. Any ring which may be generated by finitely many elementsis Noetherian.
Example (Non-example). Z[๐1, ๐2, โฆ ] is not Noetherian since
(๐1) โ (๐1, ๐2) โ โฏ
is an non-stabilising ascending chain.
Remark. Suppose โฑ โ F[๐1, โฆ , ๐๐] is a set of polynomials. ๐ผ = (๐1, โฆ , ๐๐) โF๐ is a solution of โฑ if and only if โฑ is contained in the kernel of
๐๐ โถ F[๐1, โฆ , ๐๐] โ F๐๐ โฆ ๐๐
As F[๐1, โฆ , ๐๐] is Noetherian, (โฑ) = (๐1, โฆ , ๐๐) for finitely many ๐๐โs. ๐ผ isa simultaneous solution to โฑ if and only if ker๐๐ผ โ (โฑ) = (๐1, โฆ , ๐๐), if andonly if ๐ผ is a simultaneous solution to ๐1, โฆ , ๐๐. That is to say, we only have toconsider a finite family of polynomials of which ๐ผ is a root. This is important inalgebraic geometry.
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3 Modules
3.1 Definitions
Definition (Module). Let ๐ be a commutative ring. A quadruple (๐, +, 0๐, โ )is an ๐ -module if (๐, +, 0๐) is an abelian group and the operation โ โ โ โถ๐ ร ๐ โ ๐ satisfies
โข (๐1 + ๐2) โ ๐ = ๐1 โ ๐ + ๐2 โ ๐,
โข ๐ โ (๐1 + ๐2) = ๐ โ ๐1 + ๐ โ ๐2,
โข ๐2 โ (๐1 โ ๐) = (๐2๐2) โ ๐,
โข 1๐ โ ๐ = ๐.
Example.1. If ๐ = F is a field then an ๐ -module is precisely an F-vector space.
2. For any ring ๐ , ๐ ๐ = ๐ ร โฏ ร ๐ โโโโโ๐ times
is an R-module via
๐ โ (๐1, โฆ , ๐๐) = (๐๐1, โฆ , ๐๐๐).
In particular for ๐ = 1, ๐ is an ๐ -module.
3. If ๐ผ โด ๐ then ๐ผ is an ๐ -module via
๐ โ ๐ = ๐๐ โ ๐ผ.
Also ๐ /๐ผ is an R-module via
๐ โ (๐1 + ๐ผ) = ๐๐1 + ๐ผ โ ๐ /๐ผ.
4. For ๐ = Z, an R-module is precisely an abelian group. This is because theaxiom for โ says that
โ โ โ โถ Z ร ๐ โ ๐
(๐, ๐) โฆโง{โจ{โฉ
๐ + โฏ + ๐โโโโโ๐ times
๐ โฅ 0
โ(๐ + โฏ + ๐โโโโโ๐ times
) ๐ < 0
so โ is determined by the abelian structure on ๐.
5. Let F be a field and ๐ be a vector space over F. Let ๐ผ โถ ๐ โ ๐ be a linearmap. Then we can make ๐ into an F[๐]-module via
F[๐] ร ๐ โ ๐(๐, ๐ฃ) โฆ ๐(๐ผ)(๐ฃ)
Different ๐ผโs make ๐ into different F[๐ฅ]-modules.
6. Restriction of scalars: if ๐ โถ ๐ โ ๐ is a ring homomorphism and ๐ is an๐-module, then ๐ becomes an ๐ -modules via
๐ โ ๐ ๐ = ๐(๐) โ ๐ ๐.
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Definition (Submodule). If ๐ is an R-module, ๐ โ ๐ is a submodule if๐ is a subgroup of (๐, +, 0๐) and for any ๐ โ ๐, ๐ โ ๐ , ๐ โ ๐ โ ๐. Write๐ โค ๐.
Example. A subset of ๐ is a submodule if and only if it is an ideal.
Definition (Quotient module). If ๐ โค ๐ is a submodule, the quotientmodule ๐/๐ is the set of ๐-cosets in (๐, +, 0๐), i.e. the quotient abeliangroup with
๐ โ (๐ + ๐) = ๐ โ ๐ + ๐.
Definition (Homomorphism). A function ๐ โถ ๐ โ ๐ is an R-modulehomomorphism if it is a homomorphism of abelian groups and ๐(๐ โ ๐) =๐ โ ๐(๐).
Example. If ๐ = F is a field and ๐ and ๐ are F-modules (i.e. F-vector spaces),then a map is an F-module homomorphism if and only if it is an F-linear map.
Theorem 3.1 (1st Isomorphism Theorem). If ๐ โถ ๐ โ ๐ is an ๐ -modulehomomorphism then
ker ๐ = {๐ โ ๐ โถ ๐(๐) = 0} โค ๐Im ๐ = {๐ โ ๐ โถ ๐ = ๐(๐)} โค ๐
and๐/ ker ๐ โ Im ๐.
Theorem 3.2 (2nd Isomorphism Theorem). Let ๐ด, ๐ต โค ๐ be submodules.Then
๐ด + ๐ต = {๐ โ ๐ โถ ๐ = ๐ + ๐, ๐ โ ๐ด, ๐ โ ๐ต} โค ๐๐ด โฉ ๐ต โค ๐
and(๐ด + ๐ต)/๐ด โ ๐ต/(๐ด โฉ ๐ต).
Theorem 3.3 (3rd Isomorphism Theorem). Let ๐ โค ๐ฟ โค ๐ be a chain ofsubmodules. Then
๐/๐๐ฟ/๐
โ ๐/๐ฟ.
Definition (Annihilator). If ๐ is an ๐ -module and ๐ โ ๐, the annihilatorof ๐ is
Ann(๐) = {๐ โ ๐ โถ ๐ โ ๐ = 0๐} โด ๐ .
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The annihilator of ๐ is
Ann(๐) = โ๐โ๐
Ann(๐) โด ๐ .
Definition (Generated submodule). If ๐ is an ๐ -module and ๐ โ ๐, thesubmodule generated by ๐ is
๐ ๐ = {๐ โ ๐ โ ๐ โถ ๐ โ ๐ }.
Note. Intuitively, the annihilator of an element is the stabiliser of a ring actionand that of a module is the kernel. We also have
๐ ๐ โ ๐ /Ann(๐).
Definition (Finitely generated). ๐ is finitely generated if there are ๐1, โฆ , ๐๐ โ๐ such that
๐ = ๐ ๐1 + โฆ ๐ ๐๐ = {๐1๐1 + โฏ + ๐๐๐๐ โถ ๐๐ โ ๐ }.
Lemma 3.4. An ๐ -module ๐ is finitely generated if and only if there is asurjection ๐ โถ ๐ ๐ โ ๐ for some ๐.
Proof.
โข โ: Suppose ๐ = ๐ ๐1 + โฏ + ๐ ๐๐. Define
๐ โถ ๐ ๐ โ ๐(๐1, โฆ , ๐๐) โฆ ๐1๐1 + โฏ + ๐๐๐๐
This is an ๐ -module homomorphism and is surjective.
โข โ: Let ๐๐ = ๐((0, โฆ , 0, 1, 0, โฆ , 0)) with 1 in the ๐th position. Then
๐((๐1, โฆ , ๐๐)) = ๐((๐1, 0, โฆ , 0) + โฏ + (0, โฆ , 0, ๐๐))= ๐((๐1, 0, โฆ , 0)) + โฏ + ๐((0, โฆ , 0, ๐๐))= ๐1๐((1, 0, โฆ , 0)) + โฏ + ๐๐๐((0, โฆ , 0, 1))= ๐1๐1 + โฏ + ๐๐๐๐
As ๐ is surjective, ๐ = ๐ ๐1 + โฏ + ๐ ๐๐.
Corollary 3.5. Let ๐ be an ๐ -module and ๐ โค ๐. If ๐ is finitely generatedthe so is ๐/๐.
Proof.๐ ๐ ๐ ๐/๐.๐ ๐
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Note. A submodule of a finitely generated ๐ -module need not be finitelygenerated. For example,
(๐1, ๐2, โฆ ) โด Z[๐1, ๐2, โฆ ] = ๐
is an ๐ -module but not finitely generated, as otherwise it would be a finitelygenerated ideal.
Example. For ๐ผ โ C, ๐ผ is an algebraic integer if and only if Z[๐ผ] is a finitelygenerated Z-module.
3.2 Direct Sums and Free Modules
Definition (Direct sum). If ๐1, โฆ , ๐๐ are ๐ -modules, the direct sum๐1 โ โฏ โ ๐๐ is the set ๐1 ร โฏ ร ๐๐ with addition
(๐1, โฆ , ๐๐) + (๐โฒ1, โฆ , ๐โฒ
๐) = (๐1 + ๐โฒ1, โฆ , ๐๐ + ๐โฒ
๐)
and ๐ -module structure
๐ โ (๐1, โฆ , ๐๐) = (๐๐1, โฆ , ๐๐๐).
Example.๐ ๐ = ๐ โ โฏ โ ๐ โโโโโ
๐ times.
Definition (Independence). Let ๐1, โฆ ๐๐ โ ๐. They are independent if
โ๐
๐๐ โ ๐๐ = 0
implies that ๐๐ = 0 for all 1 โค ๐ โค ๐.
Definition (Free generation). A subset ๐ โ ๐ generates ๐ freely if
1. ๐ generates ๐.
2. Any function ๐ โถ ๐ โ ๐ to an ๐ -module ๐ extends to an ๐ -modulehomomorphism ๐ โถ ๐ โ ๐.
๐ ๐ ๐
๐๐
๐
Note. We can show this extension is unique: given ๐1, ๐2 โถ ๐ โ ๐ twoextensions of ๐, ๐1 โ ๐2 โถ ๐ โ ๐ is an ๐ -module homomorphism so ker(๐1 โ๐2) โค ๐. But ๐1, ๐2 both extend ๐ so ๐ โ ker(๐1 โ ๐2). As ๐ generates ๐,๐ โค ker(๐1 โ ๐2) so ๐1 = ๐2.
An ๐ -module which is freely generated by ๐ โ ๐ is said to be free and ๐ iscalled a basis.
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Proposition 3.6. For a finite subset ๐ = {๐1, โฆ , ๐๐} โ ๐, TFAE:
1. ๐ is freely generated by ๐.
2. ๐ is generated by ๐ and ๐ is independent.
3. Every ๐ โ ๐ can be written as ๐1๐1 + โฏ + ๐๐๐๐ for some unique๐๐ โ ๐ .
Proof.
โข 1 โ 2: Let ๐ generate ๐ freely. If ๐ is not independent, then there is anon-trivial relation
๐โ๐=1
๐๐๐๐ = 0
with ๐๐ โ 0. Let
๐ โถ ๐ โ ๐
๐๐ โฆ {0๐ ๐ โ ๐1๐ ๐ = ๐
This extends to an ๐ -module homomorphism ๐ โถ ๐ โ ๐ . Then
0 = ๐(0) = ๐ (โ ๐๐๐๐) = โ ๐๐๐(๐๐) = ๐๐.
Absurd. Thus ๐ is independent.
โข The other steps follow similarly from those in IB Linear Algebra.
Example. Unlike vector spaces, a minimal generating set need not be inde-pendent. For example {2, 3} โ Z generates Z but is not linear independent as(โ3) โ 2 + (2) โ 3 = 0.
However, like vector spaces, in case a module is freely generated, it isisomorphic to direct sums of copies of the ring:
Lemma 3.7. If ๐ = {๐1, โฆ , ๐๐} โ ๐ freely generates ๐ then
๐ โ ๐ ๐
as an ๐ -module.
Proof. This is entirely analogous to vector spaces. Let
๐ โถ ๐ ๐ โ ๐
(๐1, โฆ , ๐๐) โฆ โ๐
๐๐๐๐
It is surjective as ๐ generates ๐ and injective as ๐๐โs are independent.
If an ๐ -module is generated by ๐1, โฆ , ๐๐, we have seen before that there isa surjection ๐ โถ ๐ ๐ โ ๐. We define
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Definition (Relation module). The relation module for the generators is
ker ๐ โค ๐ ๐.
As ๐ โ ๐ ๐/ ker ๐, knowing ๐ is equivalent to knowing the relation module.
Definition (Finitely presented). ๐ is finitely presented if there is a finitegenerating set ๐1, โฆ , ๐๐ for which the associated relation module is finitelygenerated.
Let {๐1, โฆ , ๐๐} โ ker ๐ โค ๐ ๐ be a set of generators. Then
๐๐ = (๐๐1, ๐๐2, โฆ , ๐๐๐)
and ๐ is generated by ๐1, โฆ , ๐๐ subject to relations๐
โ๐=1
๐๐๐๐๐ = 0
for 1 โค ๐ โค ๐.
Proposition 3.8 (Invariance of Dimension). If ๐ ๐ โ ๐ ๐ then ๐ = ๐.
Note. This does not hold in general for modules over non-commutative rings.Proof. As a general strategy, let ๐ผ โด ๐ . Then
๐ผ๐ = {โ ๐๐๐๐ โถ ๐๐ โ ๐ผ, ๐๐ โ ๐} โค ๐
is a submodule as๐ โ โ ๐๐๐๐ = โ(๐๐๐)๐๐ โ ๐ผ๐.
Thus we have a quotient ๐ -module ๐/๐ผ๐. We can make this into an ๐ /๐ผ-modulevia
(๐ + ๐ผ) โ (๐ + ๐ผ๐) = ๐๐ + ๐ผ๐.Let ๐ผ โด ๐ be a maximal proper ideal (this requires Zornโs Lemma). Then
๐ /๐ผ is a field and therefore ๐ ๐ โ ๐ ๐ implies
๐ ๐/๐ผ๐ ๐ โ ๐ ๐/๐ผ๐ ๐
(๐ /๐ผ)๐ โ (๐ /๐ผ)๐
This is a vector space isomorphism so ๐ = ๐.
We have classified all finite abelian groups (well, at least we claimed so), i.e.Z-modules. What if we want to classify all ๐ -modules? That is going to be thefinal goal we will build towards.
Recall that ๐ is finitely generated by ๐1, โฆ , ๐๐ if and only if there is asurjection ๐ โถ ๐ ๐ โ ๐. ๐ is finitely presentely if and only ker ๐ is finitelygenerated, say ๐1, โฆ , ๐โ. Let
๐๐ = (๐๐1, ๐๐2, โฆ , ๐๐๐)
then such an ๐ -module ๐ is determined by the matrix
โโโโโ
๐11 ๐12 โฏ ๐1โ๐๐1โฎ โฑ
๐๐1 ๐๐โ
โโโโโ
โ โณ๐,โ(๐ ).
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3.3 Matrices over Euclidean DomainsFor this section assume ๐ to be a Euclidean domain and let ๐ โถ ๐ \ {0} โ Zโฅ0be the Euclidean function. For ๐, ๐ โ ๐ , we have shown that gcd(๐, ๐) existsand is unique up to associates. In addition, the Euclidean algorithm shows thatgcd(๐, ๐) = ๐๐ฅ + ๐๐ฆ for some ๐ฅ, ๐ฆ โ ๐ .
What follows would be very similar to what we have learned in IB LinearAlgebra โ in fact identical except a single modification:
Definition (Elementary row operation). Elementary row operation on an๐ ร ๐ matrix with entries in ๐ are
1. Add ๐ โ ๐ times the ๐th row to the ๐th row where ๐ โ ๐. This can berealised by left multiplication by ๐ผ + ๐ถ where ๐ถ is ๐ in (๐, ๐)th positionand 0 elsewhere.
2. Swapping the ๐th and ๐th row where ๐ โ ๐. Realised by left multiplica-tion by
โโโโโโโโโโโโโ
1 0 โฏ 0โฎ โฑ โฎ
0 1 00 โฏ 0 โฑ 0 0
1 0 0โฎ โฑ0 โฏ โฏ 0
โโโโโโโโโโโโโ
3. Multiply the ๐th row by a unit ๐ โ ๐ . Realised by left multiplicationby
โโโโโโโโ
1 0 โฏ 0โฑ
โฎ ๐ โฎโฑ
0 โฏ 0 1
โโโโโโโโ
Definition (Elementary column operation). Defined analogously by replac-ing โrowโ with โcolumnโ.
Similarly to IB Linear Algebra, we define an equivalence relation
Definition (Equivalence). ๐ด, ๐ต โ โณ๐,๐(๐ ) are equivalent if there is asequence of elementary row and column operations taking ๐ด to ๐ต.
If ๐ด and ๐ต are equivalent then there are invertible square matrices ๐ and ๐such that
๐ต = ๐๐ด๐ โ1.
Theorem 3.9 (Smith Normal Form). An ๐ ร ๐ matrix over a Euclidean
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domain ๐ is equivalent to
โโโโโโโโโโโโโ
๐1๐2
โฑ๐๐
0โฑ
0
โโโโโโโโโโโโโ
where the ๐๐โs are non-zero and
๐1 โฃ ๐2 โฃ โฏ โฃ ๐๐.Proof. This proof is going to be algorithmic. If the matrix ๐ด = 0 we are done.Otherwise there is a ๐ด๐๐ โ 0. By swapping 1st and ๐th row, and 1st and ๐thcolumn we may suppose ๐ด11 โ 0. We want to reduce ๐(๐ด11) as much as possible.Split into three cases:
โข Case 1: if there is a ๐ด1๐ not divisible by ๐ด11 then have
๐ด1๐ = ๐๐ด11 + ๐
with ๐(๐) < ๐(๐ด11). Add โ๐ times the 1st column to the ๐th. This makesthe (1, ๐)th entry ๐. Swap 1st and ๐th column to get ๐ด11 = ๐. Thus wehave strictly decreased the ๐ value of the (1, 1) entry.
โข Case 2: if ๐ด11 does not divide some ๐ด๐1, do the analogous to entries in thefirst column to strictly reduce ๐(๐ด11).As ๐(๐ด11) can only strictly decrease finitely many times, after some appli-cations of Case 1 and 2 we can assumes ๐ด11 divides all the entries in the1st row and 1st column. If ๐ด1๐ = ๐๐ด11 then we can add โ๐ times the 1stcolumn to the ๐th row to make the (๐, ๐)th entry 0. Thus we obtain
๐ด = (๐ 00 ๐ถ)
โข Case 3: if there is an entry ๐๐๐ of ๐ถ not divisible by ๐, write
๐๐๐ = ๐๐ + ๐
where ๐(๐) < ๐(๐). Conduct the following series of elementary operations
โโโโโโโโโโ
๐ 0 โฏ 0 โฏ 00โฎ0 ๐๐๐โฎ0
โโโโโโโโโโ
EC 1โ
โโโโโโโโโโ
๐ 0 โฏ ๐ โฏ 00โฎ0 ๐๐๐โฎ0
โโโโโโโโโโ
ER 1โ
โโโโโโโโโโ
๐ 0 โฏ ๐ โฏ 00โฎ
โ๐๐ ๐โฎ0
โโโโโโโโโโ
ER 2,EC 2โ
โโโโโ
๐ โ โฏ โโโฎ โโ
โโโโโ
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Repeat Case 1 and 2, we finally get
(๐โฒ
๐ถโฒ)
where ๐(๐โฒ) < ๐(๐).
Eventually we can suppose that ๐โฒ divides every entry of ๐ถโฒ. By induction๐ถโฒ is equivalent to
โโโโโโโโโโโโโ
๐2๐3
โฑ๐๐
0โฑ
0
โโโโโโโโโโโโโ
with๐2 โฃ ๐3 โฃ โฏ โฃ ๐๐
and we must have ๐โฒ โฃ ๐๐ for ๐ > 1.
Remark. The ๐๐โs in Smith Normal Form are unique up to associates.
Certainly Smith Normal Form is a nice form and the algorithm guaranteesits existence and uniqueness (up to associates). However, the computation istoo cumbersome to be useful. However, if we could prove it is invariant undermatrix conjugation, we may apply some clever tricks to extract the ๐๐โs in SmithNormal Form without explicitly computing them.
Definition (Minor). A ๐ ร ๐ minor of a matrix ๐ด is the determinant of amatrix formed by forgetting all but ๐ rows and ๐ columns of ๐ด.
Definition (Fitting ideal). The ๐th Fitting ideal of ๐ด Fit๐(๐ด) โด ๐ is theideal generated by all ๐ ร ๐ minors of ๐ด.
Given a matrix ๐ด in Smith Normal Form as above with ๐1 โฃ โฏ โฃ ๐๐, theonly ๐ ร ๐ submatrices which do not have a whole row or column 0 are thosewhich keep both ๐1th row and ๐1th column, both ๐2th row and ๐2th column, etc.Therefore
Fit๐(๐ด) =โโโโโโ
detโโโโโโ
๐๐1๐๐2
โฑ๐๐๐
โโโโโโ
โโโโโโ
= (๐๐1โฏ ๐๐๐
โถ sequences ๐1, โฆ , ๐๐)= (๐1๐2 โฏ ๐๐)
as ๐๐ โฃ ๐๐๐for all ๐.
Therefore from the above computation Fit๐(๐ด) and Fit๐โ1(๐ด) determine ๐๐up to associates.
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Lemma 3.10. If ๐ด and ๐ต are equivalent matrices then Fit๐(๐ด) = Fit๐(๐ต)for all ๐.
Proof. It amounts to show that elementary operations does not change Fit๐(๐ด) โด๐ . We do the first type of row operation. Fix a ๐ ร ๐ submatrix ๐ถ in ๐ด. Recallthat this row operation adds ๐ times the ๐th row to the ๐th row. Depending on๐ and ๐, split into three cases:
โข Case 1: if the ๐th row is not in ๐ถ then ๐ถ is unchanged, so is its determinant.
โข Case 2: if the ๐th and ๐th rows are both in ๐ถ, the operation changes ๐ถ bya row operation so its determinant is unchanged.
โข Cases 3: if the ๐th row is in ๐ถ but the ๐th is not, suppose wlog the ๐th rowof ๐ด corresponding to columns of ๐ถ has entries (๐1, ๐2, โฆ , ๐๐). After therow operation, ๐ถ is changed to ๐ถโฒ whose ๐th row is
(๐๐,1 + ๐๐1, ๐๐,2 + ๐๐2, โฆ , ๐๐,๐ + ๐๐๐).
By expansion along the ๐th row,
det๐ถโฒ = det๐ถ ยฑ ๐ det๐ท
where ๐ท is the matrix obtained by replacing the ๐th row of ๐ถ with(๐1, โฆ , ๐๐), which is a ๐ ร ๐ submatrix of ๐ด up to reordering (which isaccounted for by the ยฑ sign), by multilinearity of det. So det๐ถโฒ โ Fit๐(๐ด)as it is a linear combination of minors. Therefore Fit๐(๐ดโฒ) โ Fit๐(๐ด)where ๐ดโฒ is obtained from ๐ด by this operation. As row operations areinvertible, we must have equality.
The other two types of row operations are similar but easier. Columnoperations follow analogously.
Example. Let
๐ด = (2 โ11 2 ) โ โณ2,2(Z).
Algorithmically, we can carry out the following sequence of operations to obtainSmith Normal Form:
(2 โ11 2 )
ER 2โ (1 2
2 โ1)ER 1โ (1 2
0 โ5)ER 1โ (1 0
0 โ5)ER 3โ (1 0
0 5)
Alternatively, using what we have just proved,
Fit1(๐ด) = (2, โ1, 2, 1) = (1)Fit2(๐ด) = (det๐ด) = (5)
so ๐1 = 1, ๐1๐2 = 5 so ๐2 = 5.
Recall that we have remarked that a submodule of a finitely genereatedmodule may not be finitely generated. However the following lemma tells us thatsubmodules of finitely generated free modules over some particular rings are so:
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Lemma 3.11. Let ๐ be a PID. Any submodule of ๐ ๐ is generated by atmost ๐ elements.
Proof. Let ๐ โค ๐ ๐ and consider the ideal
๐ผ = {๐ โ ๐ โถ โ๐2, โฆ , ๐๐ such that (๐, ๐2, โฆ , ๐๐) โ ๐},
which is the image of ๐๐
โ ๐ ๐๐1โ ๐ , a submodule of ๐ .
As ๐ is a PID, ๐ผ = (๐) โด ๐ for some ๐ โ ๐ . Thus there is some
๐1 = (๐, ๐2, ๐2, โฆ , ๐๐) โ ๐.
Suppose (๐1, ๐2, โฆ , ๐๐) โ ๐. Then there exists some ๐ฅ โ ๐ such that ๐1 = ๐๐ฅ.Then
(๐1, โฆ , ๐๐) โ ๐ฅ โ ๐1 = (0, ๐2 โ ๐ฅ๐2, โฆ , ๐๐ โ ๐ฅ๐๐) โ ๐ โฉ (0 โ ๐ ๐โ1).
By induction ๐ โฉ (0 โ ๐ ๐โ1) โ ๐ โฒ โค ๐ ๐โ1 is generated by ๐2, โฆ , ๐๐ so๐1, โฆ , ๐๐ generate ๐.
Theorem 3.12. Let ๐ be a Euclidean domain and ๐ โค ๐ ๐. Then there isa basis ๐ฃ1, โฆ , ๐ฃ๐ of ๐ ๐ such that ๐ is generated by ๐1๐ฃ1, โฆ , ๐๐๐ฃ๐ for some0 โค ๐ โค ๐ and some ๐1 โฃ โฆ โฃ ๐๐.
Proof. By the previous lemma there are ๐ฅ1, โฆ , ๐ฅ๐ โ ๐ which generate ๐ and0 โค ๐ โค ๐. Each ๐ฅ๐ is an element of ๐ ๐ so we can form an ๐ ร ๐ matrix whosefirst ๐ columns are ๐ฅ๐, i.e.
๐ด = โโโ
โ โ โ๐ฅ1 ๐ฅ2 โฏ ๐ฅ๐โ โ โ
โโโ
โ โณ๐,๐(๐ )
We can put ๐ด into Smith Normal Form with diagonal entries ๐1 โฃ โฏ โฃ ๐๐by elementary operations. Each row operation is given by a change of basisof ๐ ๐ and each column operation is given by rechoosing the generating set๐ฅ1, โฆ , ๐ฅ๐. Thus after a change of basis of ๐ ๐ to ๐ฃ1, โฆ , ๐ฃ๐, ๐ is generated by๐1๐ฃ1, โฆ , ๐๐๐ฃ๐.
Corollary 3.13. A submodule ๐ โค ๐ ๐ is isomorphic to ๐ ๐ for some๐ โค ๐.
Proof. By the theorem above, we can find a basis ๐ฃ1, โฆ , ๐ฃ๐ for ๐ ๐ such that ๐is generated by ๐1๐ฃ1, โฆ , ๐๐๐ฃ๐. These are linearly independent as a dependencebetween them would give a dependence between ๐ฃ1, โฆ , ๐ฃ๐.
Now we are ready for the big theorem in this course:
Theorem 3.14 (Classification Theorem for Finitely Generated Modulesover Euclidean Domain). Let ๐ be a Euclidean domain and ๐ a finitelygenerated ๐ -modules. Then
๐ โ ๐ (๐1)
โ ๐ (๐2)
โ โฏ โ ๐ (๐๐)
โ ๐ โ โฏ โ ๐
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for some ๐๐ โ 0 with ๐1 โฃ ๐2 โฃ โฏ โฃ ๐๐.
Proof. Let ๐ be generated by ๐1, โฆ , ๐๐ โ ๐, giving a surjection ๐ โถ ๐ ๐ โ ๐so ๐ โ ๐ ๐/ ker๐. By the previous theorem there is a basis ๐ฃ1, โฆ , ๐ฃ๐ of ๐ ๐
such that ker๐ is generated by ๐1๐ฃ1, โฆ , ๐๐๐ฃ๐ with ๐1 โฃ โฏ โฃ ๐๐. Thus by changingthe basis of ๐ ๐ to ๐ฃ๐โs, ker๐ is generated by columns of
โโโโโโโโโโโโโ
๐1๐2
โฑ๐๐
0โฑ
0
โโโโโโโโโโโโโ
so๐ โ ๐ ๐
ker๐โ (
๐โจ๐=1
๐ (๐๐)
) โ ๐ โ โฏ โ ๐
as required.
Example. Let ๐ = Z, a Euclidean domain, and ๐ด be the abelian group (i.e.Z-module) generated by ๐, ๐, ๐, subject to
โง{โจ{โฉ
2๐ + 3๐ + ๐ = 0๐ + 2๐ = 05๐ + 6๐ + 7๐ = 0
Thus ๐ด = Z3/๐ where ๐ โค Z3 is generated by (2, 3, 1)๐, (1, 2, 0)๐, (5, 6, 7)๐.The matrix ๐ด whose columns are these vectors
๐ด = โโโ
2 1 53 2 61 0 7
โโโ
has Smith Normal Form with diagonal entries 1, 1, 3:
Proof.
Fit1(๐ด) = (1)
Fit2(๐ด) โ (det(2 13 2)) = (1)
Fit3(๐ด) = (det๐ด) = 3
so ๐1 = 1, ๐1๐2 = 1, ๐1๐2๐3 = 3.
After change of basis, ๐ is generated by (1, 0, 0)๐, (0, 1, 0)๐, (0, 0, 3)๐ so
๐ด โ Z/1Z โ Z/1Z โ Z/3Z โ Z/3Z.
We can derive, as a corollary actually, what we stated earlier without proof
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Theorem 3.15 (Structure Theorem for Finitely Generated Abelian Groups).Any finitely generated abelian group is isomorphic to
๐ถ๐1ร ๐ถ๐2
ร โฏ ร ๐ถ๐๐ร ๐ถโ ร โฏ ร ๐ถโ
with ๐1 โฃ โฏ โฃ ๐๐.
Proof. โTrivialโ should suffice here but let us spell it out: apply ClassificationTheorem for Finitely Generated Modules over Euclidean Domain to Z, and notethat
Z/(๐) = ๐ถ๐, Z = ๐ถโ.
The above classification theorem decompose into modules whose relationmodulesโ principal ideals form a descending chain by divisibility. It turns outit is also possible to decompose by the coprime factors of the relation modules.Before that let us prove something we have known for a (very) long time, but ata higher level:
Lemma 3.16 (Chinese Remainder Theorem). Let ๐ be a Euclidean domainand ๐, ๐ โ ๐ with gcd(๐, ๐) = 1. Then
๐ /(๐๐) โ ๐ (๐) โ ๐ /(๐).
Proof. Consider the ๐ -module homomorphism
๐ โถ ๐ /(๐) โ ๐ /(๐) โ ๐ /(๐๐)(๐1 + (๐), ๐2 + (๐)) โฆ ๐๐1 + ๐๐2 + (๐๐)
As gcd(๐, ๐) = 1, (๐, ๐) = (1) so 1 = ๐ฅ๐ + ๐ฆ๐ for some ๐ฅ, ๐ฆ โ ๐ . Therefore for๐ โ ๐ , ๐ = ๐๐ฅ๐ + ๐๐ฆ๐ so
๐ + (๐๐) = ๐๐ฅ๐ + ๐๐ฆ๐ + (๐๐) = ๐((๐๐ฆ + (๐), ๐๐ฅ + (๐)))
and so ๐ is surjective.If ๐((๐1 + (๐), ๐2 + (๐))) = 0 then ๐๐1 + ๐๐2 โ (๐๐). Thus ๐ โฃ ๐๐1 + ๐๐2, ๐ โฃ ๐๐1.
As gcd(๐, ๐) = 1, ๐ โฃ ๐1 so ๐1 + (๐) = 0 + (๐). Similarly ๐2 + (๐) = 0 + (๐) so ๐ isinjective.
We thus have
Theorem 3.17 (Primary Decomposition Theorem). Let ๐ be a Euclideandomain and ๐ be a finitely generated ๐ -module. Then
๐ โ ๐
โจ๐=1
๐๐
with each ๐๐ either equal to ๐ or ๐ /(๐๐) for some prime ๐ โ ๐ and ๐ โฅ 1.
Proof. Note that if ๐ = ๐๐11 โฏ ๐๐๐
๐ with ๐๐ โ ๐ distinct primes, by the previouslemma
๐ (๐)
โ ๐ (๐๐1
1 )โ โฏ โ ๐
(๐๐๐๐ )
.
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Plug this into Classification Theorem for Finitely Generated Modules overEuclidean Domain to get the required result.
3.4 F[๐]-modules and Normal FormFor any field F, F[๐] is a Euclidean domain and so results of the last sectionapply. If ๐ is an F-vector space and ๐ผ โถ ๐ โ ๐ is an endomorphism, then wehave
F[๐] ร ๐ โ ๐(๐, ๐ฃ) โฆ ๐(๐ผ)(๐ฃ)
which makes ๐ into an F[๐]-module, call it ๐๐ผ. It turns out that F[๐]-module isthe correct tool to study endomorphisms and many results in IB Linear Algebra,as well as many further results in algebra, can be obtained by looking into theF[๐]-module structure.
Lemma 3.18. If ๐ is finite-dimensional then ๐๐ผ is finitely generated as anF[๐]-module.
Proof. ๐ is a finitely generated F-module and F โค F[๐] so ๐ is also a finitelygenerated F[๐]-module.
Example.
1. Suppose ๐๐ผ โ F[๐]/(๐๐) as an F[๐]-module. This has F-basis {๐๐}๐โ1๐=0
and the action of ๐ผ corresponds to multiplication by ๐. Thus in this basis๐ผ has matrix representation
โโโโโโโโ
01 0
1 0โฑ1 0
โโโโโโโโ
2. Suppose ๐๐ผ โ F[๐]/((๐ โ ๐)๐). Consider ๐ฝ = ๐ผ โ ๐ โ id. Then ๐๐ฝ โ F[๐ ]/(๐ ๐) as an F[๐ ]-module. By the previous example ๐ has a basis sothat ๐ฝ is given by the matrix above and ๐ผ is given by
โโโโโโโโ
๐1 ๐
1 ๐โฑ1 ๐
โโโโโโโโ
3. Suppose ๐๐ผ โ F[๐]/(๐) where
๐ = ๐๐ + ๐๐โ1๐๐โ1 + โฏ + ๐1๐ + ๐0.
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Then {๐๐}๐โ1๐=0 is an F-basis and in this basis ๐ผ is given by
โโโโโโโโโโ
0 โ๐01 0 โ๐1
1 0 โ๐21 0 โ๐3
โฑ โฎ1 โ๐๐โ1
โโโโโโโโโโ
This matrix is called the companion matrix for ๐, written ๐ถ(๐).
Theorem 3.19 (Rational Canonical Form). Let ๐ be a finite-dimensionalF-vector space and ๐ผ โถ ๐ โ ๐ be linear. Regard ๐ as an F[๐]-module ๐๐ผ, wehave
๐๐ผ โ F[๐](๐1)
โ โฏ โ F[๐](๐๐)
with ๐1 โฃ ๐2 โฃ โฏ โฃ ๐๐. There is a basis of ๐ with respect to which ๐ผ is given by
โโโโโ
๐ถ(๐1)๐ถ(๐2)
โฑ๐ถ(๐๐)
โโโโโ
Proof. Apply Classification Theorem for Finitely Generated Modules over Eu-clidean Domain to F[๐], a Euclidean domain. Note that no copies of F[๐] appearas it has infinite dimension over ๐.
Some observations:
1. If ๐ผ is represented by a matrix ๐ด in some basis, then ๐ด is conjugate to theabove matrix.
2. The minimal polynomial of ๐ผ is ๐๐ โ F[๐].
3. The characteristic polynomial of ๐ผ is ๐1๐2 โฏ ๐๐.
Recall that we have two classification theorems for modules over Euclideandomain. The above theorem corresponds to invariant decomposition. One mightnaturally ask what result follows from primary decomposition. Before that letโsconvince ourselves that primes in C[๐] are as simple as they can be:
Lemma 3.20. The primes in C[๐] are ๐ โ ๐ for ๐ โ C up to associates.
Proof. If ๐ โ C[๐] is irreducible then Fundamental Theorem of Algebra saysthat ๐ has a root ๐, or ๐ is a constant. If it is constant then it is 0 or a unit,absurd. Thus (๐ โ ๐) โฃ ๐, write ๐ = (๐ โ ๐)๐. But ๐ is irreducible so ๐ is a unit.Thus ๐ is an associate of ๐ โ ๐.
Remark. The lemma is equivalent to the statement that C is algebraicallyclosed, which says that every polynomial with coefficients in C factorises intolinear factors over C. In fact, every field can be extended to an algebraicallyclosed one. This will be discussed in detail in IID Galois Theory.
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Theorem 3.21 (Jordan Normal Form). Let ๐ be a finite-dimensional C-vector space and ๐ผ โถ ๐ โ ๐ linear. Consider ๐๐ผ as an C[๐]-module, then
๐๐ผ โ C[๐]((๐ โ ๐1)๐1)
โ C[๐]((๐ โ ๐2)๐2)
โ โฏ โ C[๐]((๐ โ ๐๐)๐๐)
where the ๐๐โs are not necessarily distinct. There is a basis of ๐ with respectto which ๐ผ is given by
โโโโโโ
๐ฝ๐1(๐1)
๐ฝ๐2(๐2)
โฑ๐ฝ๐๐
(๐๐)
โโโโโโ
where
๐ฝ๐(๐) =โโโโโโโโ
๐1 ๐
1 ๐โฑ1 ๐
โโโโโโโโ
has size ๐.Proof. Immediate from Primary Decomposition Theorem and knowing all theprimes in C[๐].
Remark.1. The ๐ฝ๐(๐) are called Jordan ๐-blocks.
2. The minimal polynomial of ๐ผ is
๐๐ผ(๐ก) = โ๐
(๐ โ ๐)๐๐
where ๐๐ is the largest ๐-block.
3. The characteristic polynomial of ๐ผ is
๐๐ผ(๐ก) = โ๐
(๐ โ ๐)๐๐
where ๐๐ is the sum of the sizes of the ๐-blocks.
4. Consider ker(๐ โ โ โถ ๐๐ผ โ ๐๐ผ). What is its dimension?On C[๐]/(๐ โ ๐)๐, if ๐ โ 0 then the map ๐ โ โ is an isomorphism since
ker(๐ โ โ) = {๐ + ((๐ โ ๐)๐) โถ ๐๐ โ ((๐ โ ๐)๐)}
so if ๐๐ = (๐ โ ๐)๐ โ ๐, as ๐ and ๐ โ ๐ are coprime, ๐ โฃ ๐, (๐ โ ๐)๐ โฃ ๐so ker(๐ โ โ) = 0.If ๐ = 0, ๐ โ โ โถ C[๐]/(๐๐) โ C[๐]/(๐๐) has matrix
โโโโโโโโ
01 0
1 0โฑ1 0
โโโโโโโโ
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so 1-dimensional kernel. Thus
dimker(๐ โ โ โถ ๐๐ผ โ ๐๐ผ) = #Jordan 0-blocks.
5. Similarly, ๐2 โ โ โถ C[๐]/((๐ โ ๐)๐) โ C[๐]/((๐ โ ๐)๐) is an isomorphismfor ๐ โ 0 and for ๐ = 0 is given by the matrix
โโโโโโโโโโ
00 01 0 0
1 0โฑ
0
โโโโโโโโโโ
which has 2-dimensional kernel if ๐ > 1 and 1-dimensional kernel if ๐ = 1.Therefore
dimker(๐2 โ โ โถ ๐๐ผ โ ๐๐ผ) = #Jordan 0-blocks+ #Jordan 0-blocks of size > 1.
so
#Jordan 0-blocks of size 1 = 2 dimker(๐ โ โ) โ dimker(๐2 โ โ).
Using the same method we can find Jordan 0-blocks of other sizes.
3.5 Conjugacy*
Lemma 3.22. If ๐ผ โถ ๐ โ ๐ and ๐ฝ โถ ๐ โ ๐ are endomorphism of F-vectorspaces, then ๐๐ผ โ ๐๐ฝ as F[๐]-modules if and only if there is an isomorphism๐พ โถ ๐ โ ๐ such that
๐พโ1๐ฝ๐พ = ๐ผ,
i.e. ๐ผ and ๐ฝ are conjugates.
Proof. Let ฬ๐พ โถ ๐๐ผ โ ๐๐ฝ be an F[๐]-module isomorphism. In particular ฬ๐พ givesan F-vector space isomorphism ๐พ โถ ๐ โ ๐. Then
๐ฝ โ ๐พ โถ ๐๐ฝ โ ๐๐ฝ
๐ฃ โฆ ๐ โ ๐พ(๐ฃ)
Now
๐ โ ๐พ(๐ฃ) = ๐ โ ฬ๐พ(๐ฃ) ฬ๐พ as an F[๐]-module map= ฬ๐พ(๐ โ ๐ฃ) in F[๐]-module ๐๐ผ
= ฬ๐พ(๐ผ(๐ฃ))= ๐พ(๐ผ(๐ฃ))
so ๐ฝ โ ๐พ = ๐พ โ ๐ผ, ๐พโ1 โ ๐ฝ โ ๐พ = ๐ผ. Therefore if ๐ = ๐ then ๐๐ผ โ ๐๐ฝ if and only if๐ผ and ๐ฝ are conjugates.
๐ ๐ ๐๐ผ ๐๐ผ
๐ ๐ ๐๐ฝ ๐๐ฝ
๐ผ
๐พ ๐พ
๐ผ=๐โ โ
๏ฟฝฬ๏ฟฝ ๏ฟฝฬ๏ฟฝ๐ฝ ๐ฝ=๐โ โ
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Applying Classification Theorem for Finitely Generated Modules over Eu-clidean Domain, we get
Corollary 3.23. There is a bijection
{conjugacy class of โณ๐(F)} โ { sequence of monic polynomials ๐1, โฆ , ๐๐where ๐1 โฃ โฏ โฃ ๐๐ and deg(๐1 โฏ ๐๐) = ๐ }
Example. Consider GL2(F). The conjugacy classes are described by ๐1 โฃ โฏ โฃ ๐๐where deg(๐1 โฏ ๐๐) = 2. Therefore we have one of the followings:
1. deg ๐1 = 2,
2. deg ๐1 = deg ๐2 = 1. As ๐1 โฃ ๐2, ๐1 = ๐2.
These give us respecively
1. F[๐]/(๐2 + ๐1๐ + ๐0),
2. F[๐]/(๐ โ ๐) โ F[๐]/(๐ โ ๐).
Therefore any ๐ด โ GL2(F) is conjugate to one of
(0 โ๐01 โ๐1
) , (๐ 00 ๐)
They are not conjugates.The first case be futher split into two cases. If ๐2 + ๐1๐ + ๐0 is reducible
they it factorises as either (๐ โ ๐)2 or (๐ โ ๐)(๐ โ ๐) where ๐ โ ๐. Thus weget one of
(๐ 01 ๐) , (๐ 0
0 ๐)
Example. Let F = Z/3Z. For what ๐1, ๐0 is ๐2 + ๐1๐ + ๐0 โ F[๐] irreducible?There are 3 ร 3 = 9 polynomials in total, of which (3
1) + (32) = 6 are reducible.
Guess (any verify!) that the irreducibles are ๐2 + 1, ๐2 + 2๐ + 2, ๐2 + 2๐ + 2.Therefore the conjugacy classes in GL2(Z/3Z) are
( 0 โ11 0 ) ( 0 โ2
1 โ1 ) ( 0 โ21 โ2 )
( ๐ 01 ๐ ) ๐ โ 0
( ๐ 00 ๐ ) ๐, ๐ โ 0
so there are in total 8 conjugacy classes. They have order
( 0 21 0 ) ( 0 1
1 2 ) ( 0 11 1 ) ( 1 0
1 1 ) ( 2 01 2 ) ( ๐ 0
0 ๐ )4 8 8 3 6 2
Just for fun, letโs use what we deduced above and knowledge about Sylow๐-subgroups way back in the beginning of the course to determine the groupstructure of GL2(Z/3/๐).
Recall that|๐บ๐ฟ2(Z/3Z)| = (32 โ 1)(32 โ 3) = 24 โ 3
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so the Sylow 2-subgroup has order 24 = 16. There are no elements of order 16so it cannot be cyclic. Let
๐ด = (0 21 0) , ๐ต = (0 1
1 2)
so๐ดโ1๐ต๐ด = (2 2
2 0) = ๐ต3.
Therefore โจ๐ตโฉ โด โจ๐ด, ๐ตโฉ โค GL2(Z/3Z). The 2nd Isomorphism Theorem saysthat
โจ๐ด, ๐ตโฉ/โจ๐ตโฉ โ โจ๐ดโฉ/(โจ๐ดโฉ โฉ โจ๐ตโฉ).
Now โจ๐ดโฉ โฉ โจ๐ตโฉ = โจ( 2 00 2 )โฉ, a group of order 2. Therefore
|โจ๐ด, ๐ตโฉ| = |โจ๐ดโฉ| โ |โจ๐ตโฉ||โจ๐ดโฉ โฉ โจ๐ตโฉ|
= 8 โ 42
= 16
which is a Sylow 2-subgroup of GL2(Z/3Z). It has presentation
โจ๐ด, ๐ต|๐ด4 = ๐ต8 = 1, ๐ดโ1๐ต๐ด = ๐ต3โฉ,
the semidihedral group of order 16.Since we still have time left, we can do one more fun example.
Example. Let ๐ = Z[๐]/(๐2 + 5) โ Z[โ
โ5] โค C. Then
(1 + ๐)(1 โ ๐) = 1 โ ๐2 = 1 + 5 = 6 = 2 โ 3.
As 1 ยฑ ๐, 2 and 3 are irreducibles ๐ is not a UFD. Let
๐ผ1 = (3, 1 + ๐), ๐ผ2 = (3, 1 โ ๐ฅ)
be submodules of ๐ . Consider
๐ โถ ๐ผ1 โ ๐ผ2 โ ๐ (๐, ๐) โ ๐ + ๐
Then Im๐ = (3, 1 + ๐, 1 โ ๐). Since 3 โ (1 + ๐) โ (1 โ ๐) = 1, Im๐ = ๐ . Also
ker๐ = {(๐, ๐) โ ๐ผ1 โ ๐ผ2 โถ ๐ + ๐ = 0} โ ๐ผ1 โฉ ๐ผ2
where the last isomorphism can be deduced from the map (๐ฅ, โ๐ฅ) โค ๐ฅ. Notethat (3) โ ๐ผ1 โฉ ๐ผ2. Let
๐ โ 3 + ๐ก โ (1 + ๐) โ (3, 1 โ ๐) โ ๐ = Z[๐]/(๐2 โ 5).
Reduce mod 3, we get
๐ก โ (1 + ๐) = (1 โ ๐)๐ mod (3, ๐2 + 5) = (3, ๐2 โ 1) = (2, (๐ + 1)(๐ โ 1))
so 1 โ ๐ โฃ ๐ก, (1 + ๐)(1 โ ๐) โฃ ๐ก(1 + ๐) so
๐ก(1 + ๐) = ๐(๐2 โ 1) = ๐(๐2 + 5 โ 6) = 3(โ2๐).
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Then ๐ โ 3 + ๐ก โ (1 + ๐) is divisible by 3 so ๐ผ1 โฉ ๐ผ2 โ (3). Equality follows.From example sheet 4 we know that if ๐ โค ๐ and ๐/๐ โ R๐ then
๐ โ ๐ โ ๐ ๐. Here๐ผ1 โ ๐ผ2/ ker๐ โ Im๐ = ๐
so๐ผ1 โ ๐ผ2 โ ๐ โ ker๐ = ๐ โ (3).
Consider
๐ โถ ๐ โ (3)๐ฅ โฆ 3๐ฅ
a surjective ๐ -module map. ker๐ = 0 as ๐ is an integral domain so ๐ is anisomorphism. Thus
๐ผ1 โ ๐ผ2 โ ๐ โ ๐ = ๐ 2.
In particular this shows that sums of non-free modules can be free.Next we claim that ๐ผ1 is not principal. If ๐ผ1 = (๐ + ๐๐) then ๐ผ2 = (๐ + ๐๐).
This is because ๐ผ1 = (3, 1 + ๐) and ๐ผ2 = (3, 1 โ ๐) and ๐ has automorphism๐ โฆ โ๐ which interchanging ๐ผ1 and ๐ผ2.1 But then
(3) = ๐ผ1 โฉ ๐ผ2 = ((๐ + ๐๐)(๐ โ ๐๐)) = (๐2 โ ๐2๐2) = (๐2 + 5๐2)
so ๐2 + 5๐2 โฃ 3, absurd.In summary, we have shown that
1. ๐ผ1 needs 2 elements to generate (as it is not principal), but it is not thefree module ๐ 2.
2. ๐ผ1 is a direct summand of ๐ 2.
1This technique will play a central role in IID Galois Theory.
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Index
algebraic integer, 37annihilator, 41ascending chain condition, 30associate, 27
basis, 43
Cayleyโs theorem, 8centraliser, 10centre, 10Chinese Remainder Theorem, 13conjugacy class, 10content, 32
direct sum, 43
Eisensteinโs criterion, 34equivalence, 46Euclidean domain, 28
field of fractions, 25finitely generated, 42finitely presented, 45Fitting ideal, 48free generation, 43free module, 43
Gaussโ Lemma, 32group, 2
abelian, 2homomorphism, 4index, 2isomorphism, 4quotient, 3subgroup, 2normal, 3
group action, 7
Hilbert Basis Theorem, 38
ideal, 19generated, 20maximal, 26prime, 26principal, 20
independence, 43
integral domain, 24irreducible, 27isomorphism theorem, 4, 23, 41
Jordan normal form, 55
Lagrangeโs Theorem, 2
minimal polynomial, 37minor, 48module, 40
homomorphism, 41quotient, 41submodule, 41
generated, 42
Noetherian, 30normaliser, 11
Orbit-stabiliser theorem, 10order, 3
PID, 24primary decomposition, 52prime, 27primitive, 32
Rational Canonical Form, 54relation module, 45ring, 17
homomorphism, 19quotient, 21subring, 17
sign, 7Simith Normal Form, 46simple group, 6Structural Theorem for modules, 50Sylow subgroup, 13Sylowโs Theorem, 13symmetric group, 7
unique factorisation domain, 29unit, 17, 27
zero divisor, 24
60