fields and galois theory fall 2004 professor yu-ru liu

Fields and Galois TheoryFall 2004

Professor Yu-Ru Liu

CHRIS ALMOST

Contents

1 Introduction 31.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Brief Review of Ring Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Field extensions 42.1 Degree of a Field Extention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Algebraic and Transcendental Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.4 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Splitting Fields 73.1 Existence of splitting fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Uniqueness of the splitting field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Separable Polynomials 94.1 Prime Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.2 Formal Derivative and Repeated Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94.3 Separable Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.4 Perfect Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Automorphism Groups 125.1 Automorphism Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125.2 Automorphism Groups of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125.3 Fixed Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

6 Galois Extensions 136.1 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136.2 Normal extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146.3 Conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.4 Galois Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.5 Artin’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1

2 CONTENTS

7 The Galois Correspondence 197.1 The Fundemental Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217.3 Brief Review of Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217.4 The Primitive Element Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

8 Ruler and Compass Constructions 248.1 Constructible Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248.2 Constructible Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

9 Cyclotomic Extensions 279.1 Cyclotomic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279.2 Cyclotomic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289.3 Abelian Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289.4 Constructible n-gons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

10 Galois Groups of Polynomials 3010.1 Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3010.2 Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3110.3 Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

11 Solvability by Radicals 3311.1 Cardano’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3311.2 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3511.3 Cyclic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3711.4 Radical Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3811.5 Solving polynomials by Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3911.6 Probabilistic Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

INTRODUCTION 3

1 Introduction

Galois Theory is the interplay between fields and groups.

1.1 Motivation

Consider the following historical problems.

• Construct an arbitrary regular n-gon using only a ruler and a compass. We know how to construct a triangleand square, but what about 5-gon, etc.?

• Square the circle using only a ruler and compass (i.e. construct a square of area π).• Solve an arbirary polynomial using only algebraic means (i.e. plus, minus, times, divides, and nthroot).

The quadratic formula gives a solution for quadratic equations. Cubic and quartic equations can be solvedsimilarily. e.g. if x3 + px = q then

x =3

È

q

2+

r

p3

27+

q2

4+

3

È

q

2−

r

p3

27+

q2

4

• For which quintic equations do we have radical solutions? If we know there is such a solution, what doesthe solution look like?

How can we solve these problems? The main steps in applying the theory that we develope in this course areas follows:

1. Associate the solution of interest, say α=pπ or α= the root of some quintic, with the field Q(α).

2. Associate Q(α) with the group of isomorphisms of Q(α) that fix Q, AutQ(Q(α)). If α is algebraic thenAutQ(Q(α)) is finite. If α is constructable then the order of AutQ(Q(α)) is in certain forms.

Hard Question: How many intermediate fields between Q and Q(α)? There is a 1-1 correspondence between theintermediate fields and the subgroups of AutQ(Q(α)) (this is the Fundemental Theorem of Galois theory.)

1.2 Brief Review of Ring Theory

For this course we will be dealing with commutative rings with identity.

1.1 Example. Let R be a ring. We denote by R[x] the polynomial ring over R in indeterminant x . The degreeof a polynomial is the exponent on the leading term. By convention, deg0 = −∞. If a polynomial has leadingcoefficient 1 then it is called “monic”.

A ring R is called a domain if it has no zero divisors. An element u ∈ R is called a unit if it is invertible. A fieldis a commutative ring in which each non-zero element is a unit and 0 6= 1.

1.2 Example. If F is a field, then F[x] is a domain (it is sufficient that F be a domain) and for f , g ∈ F[x], deg( f g) =deg( f ) + deg(g). This degree function actually makes F[x] into a Euclidean domain.

The rational (function) field over a field F is denoted F(x) and consists of all quotients of polynomials (withnon-zero denominator) from F[x]. It is the smallest field that contains F[x].

An ideal I of a ring R is a (not necessarily unital) subring of R that is absorbing with respect to multiplicationby elements of R. We can now construct R/I , the quotient ring modulo I .

I is said to be maximal if I 6= R and for any ideal J we have I ⊆ J ⊆ R⇒ I = J ∨ J = R. I is said to be primeif I 6= R and ab ∈ I ⇒ a ∈ I ∨ b ∈ I . Notice that every maximal ideal is prime, and in PIDs every prime ideal ismaximal. Fields have only trivial ideals.

4 FIELDS AND GALOIS

1.3 Theorem. Let I be a proper ideal of R. Then

1. R/I is a field if and only if I is maximal

2. R/I is a domain if and only if I is prime

1.4 Theorem. (First Isomorphism Theorem) If ϕ : R→ S is a ring homomorphism and kerϕ = I then there is anisomorphism

α : R/I → Imϕ : r + I 7→ ϕ(r)

2 Field extensions

2.1 Definition. If E is a field containing another field F then E is said to be a field extension of F , denoted byE/F

2.1 Degree of a Field Extention

If E/F is a field extension then we can view E as a vector space over F .

• Addition is given to agree with the field addition

• Scalar multiplication is given to agree with the field multiplication

2.2 Definition. The dimension of E viewed as a vector space over F is called the degree of E over F and isdenoted [E : F]. If this quantity happens to be finite, then E/F is said to be a finite extension, otherwise it is aninfinite extension.

2.3 Example. 1. C∼= R⊕ iR, so [C : R] = 2

2. [R :Q] =∞3. Let F be a field. The rational field is an infinite extension. An infinite linearly independent set is {. . . , x−1, 1, x , x2, . . .}

2.4 Theorem. If E/K and K/F are finite field extensions, then E/F is finite and

[E : F] = [E : K][K : F]

PROOF: Let {a1, . . . , am} be a basis for E over K and {b1, . . . , bn} be a basis for K over F . It suffices to proveα := {ai b j | 1≤ i ≤ m, 1≤ j ≤ n} is a basis for E over F . Every element of E is a linear combination of elementsof α since each element of E is a linear combination of elements of {a1, . . . , am}, and each of the ai ’s (beingelements of K) can be written as a linear combination of elements from {b1, . . . , bn}. α is linearly independentover F , for otherwise if

∑mi=1

∑nj=1 ci, j b jai = 0, then {a1, . . . , am} a basis implies that

∑nj=1 ci, j b j = 0 for all i.

Since {b1, . . . , bn} is also a basis, we get that ci, j = 0 for all i and j. �

2.5 Definition. Let E/F be a field extension. If K is a subfield of E that contains F then we say that K is anintermediate field of E/F .

2.6 Corollary. If E/F is a finite extension and K is an intermediate field then [E : K] and [K : F] are divisors of[E : F].

FIELD EXTENSIONS 5

2.2 Algebraic and Transcendental Numbers

2.7 Definition. Let E/F be a field extension and α ∈ E. We say that α is algebraic over F if there is f (x) ∈ F[x]such that f 6= 0 and f (α) = 0. Otherwise α is said to be transcendental over F .

In particular, for α ∈ C and α algebraic (transcendental) overQ, we say that α is an algebraic (transcendental)number. For example, all rational numbers are algebraic, as are

p2, 3p

2+ i, etc. The real numbers e (Hermite1873) and π (Lindemann 1882) are transcendental numbers.

2.8 Theorem. (Liouville 1884) Let α ∈ R \Q be a root of a polynomial f (x) ∈ Q[x] of degree n. Then thereexists a constant c > 0 such that for any rational number p

qwith q > 0

�

�

�

�

α−p

q

�

�

�

�

>c

qn

PROOF: Without loss of generality, we can assume |α− pq| < 1 and that f (x) ∈ Z[x] and f is irreducible. Then

f (α) = 0 and f ( pq) 6= 0. By the Mean Value theorem, | f ( p

q)| = | f (α)− f ( p

q)| ≤ M |α− p

q|, where M = sup | f ′(x)|

for |x−α|< 1. Since α is irrational, deg( f )≥ 2 and M 6= 0. Furthermore, | f ( pq)| ≥ 1/qn, and thus |α− p

q| ≥ 1

M1qn ,

so take c = 1M

. �

Remark. Liouville’s Theorem says that algebraic numbers are “harder” to approximate by rational numbers thantranscendental numbers. Thue (1909) and Siegel (1921) improved the above theorem by replacing n with n

2+1

and 2p

n, respectively. In 1955, Roth improved the above theorem to |α− pq| > c′

q2+ε . This won him the Fieldsmedal in 1958.

2.9 Example. z =∑

n≥11

10n! is trancendental.Suppose that z is algebraic and is a root of a polynomial of degree n. Then there is a constant c > 0 such that

for any rational number pq

with q > 0�

�

�

�

z−p

q

�

�

�

�

>c

qn

Consider∑s

n=11

10n! =p

10s! , q = 10s! We have

c

qn <

�

�

�

�

z−p

q

�

�

�

�

=∞∑

n=s+1

1

10n! <1

10(s+1)!−1

It follows that

0< c <10n·s!

10(s+1)!−1−→ 0

as s→∞. This implies that c = 0, a contradiction.

2.3 Simple Extensions

Let E/F be a field extension and α ∈ E. Let F[α] denote the smallest subring of E containing F and α and F(α)denote the smallest sufield of E containing F and α.

2.10 Definition. If E = F(α) then we say that E is a simple extension of F .

[E : F] can be either∞ or finite depending on whether α is transcendental or algebraic over F .

6 FIELDS AND GALOIS

2.11 Definition. If R and R′ are two rings containing a field F , then a ring homomorphism ψ : R→ R′ such thatψ(c) = c ∀ c ∈ F is said to be an F -homomorphism.

2.12 Theorem. Let E/F be a field extension and α ∈ E. If α is transcendental over F then F[α] ∼= F[x] andF(α)∼= F(x). In particular, F[α] 6∼= F(α).

PROOF: The F -homomorphism α 7→ x is clearly the desired isomorphism in each case. �

2.13 Theorem. Let E/F be a field extension and α ∈ E. If α is algebraic over F then there is a unique monicirreducible polynomial p(x) ∈ F[x] such that there is an F -isomorphism

ψ : F[x]/⟨p(x)⟩ → F[α]

with ψ(x) = α. From this we conclude that F[α] = F(α).

PROOF: Let ψ : F[x] → F(α) be the unique F -homomorphism with ψ(x) = α. Thus, Imψ = F[α] and letI = kerψ. Since α is algebraic, I 6= 0. We have F[x]/I ∼= Imψ, a subring of a field, so it is a (principal ideal)domain. Therefore I is a prime ideal, so it must be generated by some irreducible polynomial p(x). We mayassume that p(x) is monic without loss of generality. It follows that F[x]/⟨p(x)⟩ ∼= F[α] is a field. F(α) is also afield, and since it is the smallest field that contains F[α], we must have F[α] = F(α). �

2.14 Definition. The monic irreducible in the last theorem is called the minimal polynomial of α over F .

2.15 Theorem. Let E/F be a field extension and α ∈ E.

1. α is transcendental over F if and only if [F(α) : F] =∞2. α is algebraic over F if and only if [F(α) : F]<∞

If p(x) is the minimal polynomial of α over F then we have [F(α) : F] = deg p and {1,α, . . . ,αdeg p−1} is a basisof F(α)/F .

2.16 Example. Let p be a prime and ζp be the primitive pth root of unity. It is a root of the cyclotomic polynomialΦp(x). From the assignment, this polynomial is irreducible over Q and it is monic, so it is the minimal polynomialof ζp. Thus [Q(ζp) :Q] = p− 1. Q(ζp) is called the pth cyclotomic extension of Q.

2.4 Algebraic Extensions

2.17 Theorem. Let E/F be a field extension. If [E : F]<∞ there exists {α1, . . . ,αn} ⊆ E such that F $ F(α1)$F(α1,α2)$ · · · F(α1 . . . ,αn) = E

PROOF: By induction on [E : F]. If [E : F] = 1, E = F and we are done. Suppose that [E : F] > 1. Then there isα1 ∈ E \ F such that [E : F] = [E : F(α1)][F(α1) : F]. Since [F(α1) : F] > 1, we get that [E : F(α1)] < [E : F].Applying the induction hypothesis to [E : F(α1)], there is {α2, . . . ,αn} ⊆ E such that F(α1) = F1 $ F1(α2)$ · · ·$F1(α2 . . . ,αn) = E. It follows that E = F(α1)(α2 . . . ,αn) = F(α1 . . . ,αn). �

2.18 Definition. A field extension E/F is algebraic if every α ∈ E is algebraic over F . Otherwise the extension istranscendental.

2.19 Theorem. Let E/F be a field extension. If [E : F]<∞ then E/F is algebraic.

SPLITTING FIELDS 7

PROOF: Suppose that [E : F] = n. For α ∈ E the elements {1,α, . . . ,αn} are not linearly independent over F .Thus there exist ci ∈ F , not all zero, such that

n∑

i=0

ciαi = 0

Hence α is a root of the polynomial∑n

i=0 ci xi ∈ F[x]. �

2.20 Theorem. Let E/F be a field extension. Define the set of algebraic elements to be

L := {α ∈ E | [F(α) : F]<∞}

Then L is an intermediate field.

PROOF: If a, b ∈ L, then [F(a) : F] <∞ and [F(b) : F] <∞. Consider the field F(a, b). By assignment 1, wehave [F(a, b) : F(a)]≤ [F(b) : F]. It follows that

[F(a, b) : F] = [F(a, b) : F(a)][F(a) : F]≤ [F(b) : F][F(a) : F]<∞

Thus F(a, b)/F is algebraic, so a± b, ab, and a/b (b 6= 0) are all in L, so L is a field. �

2.21 Definition. Let E/F be a field extension. The set

F = {α ∈ E | [F(α) : F]<∞}

is called the algebraic closure of F in E.

2.22 Example. Let Q be the algebraic closure of Q over C. Then [Q :Q] =∞ (See assignment 2). In particular,the converse of Theorem 2.19 is false.

2.23 Definition. A field F is said to be algebraically closed if for any algebraic extension E/F , then E = F .

Bonus Question: Let F be a field with characteristic p, and assume that F ⊆ E, where E is algebraicallyclosed. Is there such a field E/F such that [E : F]<∞?

3 Splitting Fields

3.1 Definition. For a field F , we consider the polynomial ring F[x]. For f (x) ∈ F[x] and a field extension E/F ,we say that f (x) splits over E if it is a product of linear factors in E[x]. In other words, E contains all roots off (x). If furthermore there is no proper subfield of E that f (x) splits over, then we say that E is a splitting fieldof f (x) in E.

3.1 Existence of splitting fields

3.2 Theorem. Let p(x) ∈ F[x] be irreducible. The quotient ring F[x]/⟨p(x)⟩ is a field containing F and a rootof p(x).

PROOF: Since p(x) is irreducible, the ideal I = ⟨p(x)⟩ is maximal. Hence E := F[x]/I is a field. Consider themap

ψ : F → E : a 7→ a+ I

This map is injective since kerψ is an ideal of the field F (and hence trivial). By identifying F with ψ(F), F is asubfield of E. Moreover, let α= x + I ∈ E.

8 FIELDS AND GALOIS

Claim. α is a root of p(x)

Write p(x) = a0+a1 x+ · · ·+an xn ∈ F[x], so p(x) = (a0+ I)+(a1+ I)x+ · · ·+(an+ I)xn ∈ E[x]. Thus we have

p(α) = (a0 + I) + (a1 + I)(x + I) + · · ·+ (an + I)(x + I)n = p(x) + I = 0

in E. Thus α= x + I ∈ E is a root of p(x). �

3.3 Theorem. (Kronecker) Let f (x) ∈ F[x]. There exists a field E/F such that f (x) splits over E

PROOF: By induction on deg f . If deg f = 1, then E = F . If deg f > 1 then write f (x) = p(x)g(x) wherep(x) is irreducible. By the previous theorem there is a field K/F containing a root α of p(x). Hence f (x) =(x − α)h(x)g(x) ∈ K[x], for some h(x) ∈ K[x]. Since deg(hg) < deg f , by induction there is a field E/K overwhich gh is a product of linear factors. It follows that f (x) splits over E/F . �

3.4 Theorem. Every f (x) ∈ F[x] has a splitting field that is a finite extension of F .

PROOF: For f (x) ∈ F[x], there exists a field E/F such that f (x) splits over E. Say a1, . . . , an are the roots. Con-sider the algebraic extension F(a1, . . . , an). This extension is finite, and f (x) splits over F(a1, . . . , an). Moreover,f (x) does not split over any proper subfield of F(a1, . . . , an), since any such subfield will omit at least one of theai ’s. Therefore F(a1, . . . , an) is a splitting field of f (x) in E. �

3.2 Uniqueness of the splitting field

3.5 Lemma. Let ϕ : R→ R1 be a ring homomorphism. Then there is a unique ring homomorphism Φ : R[x]→R1[y] such that Φ|R = ϕ and Φ(x) = y . We say that Φ extends the map ϕ.

PROOF: Trivial. �

3.6 Theorem. Let ϕ : F → F1 be an isomorphism of fields, and f (x) ∈ F[x]. Let Φ : F[x]→ F1[x] be the uniquering isomorphism which extends ϕ and maps x to x . Let f1(x) = Φ( f (x)) and E/F and E1/F1 be splitting fieldsof f and f1, respectively. Then there exists an isomorphism ψ : E→ E1 which extends ϕ.

PROOF: By induction on [E : F]. If [E : F] = 1, f is a product of linear factors in F[x]. Thus E = F andE1 = F1. Take ψ = ϕ and we are done. If [E : F] > 1 then let p(x) be an irreducible factor of f (x) withdeg p ≥ 2. Write p1(x) = Φ(p(x)). Let α ∈ E and α1 ∈ E1 be roots of p and p1, respectively. Then we have anF -isomorphism F(α)∼= F[x]/⟨p(x)⟩ and an F1-isomorphism F1(α1)∼= F1[x]/⟨p1(x)⟩. Consider the isomorphismΦ. Since p1(x) = Φ(p1(x)) there must exist a field isomorphism

Φ1 : F[x]/⟨p(x)⟩ → F1[x]/⟨p1(x)⟩ ∼= F1(α1)

which extends ϕ. It follows that there exists a field isomorphism ϕ1 : F(α)→ F1(α1) which extends ϕ and sendsα to α1.

F ∼=

ϕ //� _

��

F1� _

��F(α)

ϕ1 //� _

��

F1(α1)� _

��E

ψ // E1

By induction, since [E : F(α)]< [E : F], there exists ψ : E→ E1 which extends ϕ1, and thus extends ϕ. �

SEPARABLE POLYNOMIALS 9

3.7 Corollary. Any two splitting fields of a non-zero polynomial f (x) ∈ F[x] over F are F -isomorphic.

3.8 Corollary. (E.H. Moore) Any two finite fields of order pn for some prime p are isomorphic.

PROOF: Any finite field F of order pn is a splitting field of x pn− x over Fp �

3.9 Theorem. Let F be a field and f (x) ∈ F[x] have degree n ≥ 1. Let E/F be a splitting field of f (x). Then[E : F] divides n!.

PROOF: By induction on deg f . If deg f = 1 then [E : F] = 1 and it’s trivial. Suppose deg f > 1. If f isirreducible and α ∈ E is a root of f , then there exists a simple extension F(α)/F such that F(α) ∼= F[x]/⟨ f (x)⟩and [F(α) : F] = deg f = n. Write f (x) = (x − α)g(x) ∈ F(α)[x] and deg g = n− 1. By induction, [E : F(α)]is a divisor of (n− 1)!. It follows that [E : F] = [E : F(α)][F(α) : F] divides n!. If f (x) is not irreducible, writef = g · h, where deg g = m and deg h= k. Let K be a splitting field of g over F . By induction, [K : F] divides m!.Also, [E : K] divides k! (E is a splitting field of h over K). Thus [E : F] divides m!k!, which is a factor of n!. �

4 Separable Polynomials

4.1 Prime Fields

4.1 Definition. The prime field of a field F is the intersection of all of the subfields of F .

4.2 Theorem. If F is a field, then its prime field is isomorphic to Q or to Fp for some prime p.

PROOF: Consider the ring mapχ : Z→ F : n 7→ 1+ 1+ · · ·+ 1

︸︷︷︸

n times

Let I = kerχ. Then Z/I is a domain (since it is isomorphic to the image of χ(Z), a subring of F). Hence I is aprime ideal of Z, and so either is ⟨0⟩ or ⟨p⟩ for some prime p. If I = ⟨0⟩ then Z ⊆ F . It follows that all subfieldsof F contain Frac(F) =Q, and so the prime field of F is Q. If I = ⟨p⟩ then by the first isomorphism theorem,

Fp∼= Z/⟨p⟩ ∼= Imχ ⊆ F

and so the prime field of F is Fp. �

4.3 Definition. Given a field F , if the prime field is isomorphic to Q then we say that F has characteristic 0,denoted ch F = 0. On the other hand, if the prime field is isomorphic to Fp then we say ch F = p. Notice that ifch F = p then (a+ b)p = ap + bp.

4.2 Formal Derivative and Repeated Roots

4.4 Definition. If F is a field, the monomials {1, x , x2, . . . } form an F -basis for F[x]. Define the linear operatorD : F[x] → F[x] by D1 = 0 and Dxn = nxn−1. D is called the formal derivative, and is also denoted with aprime.

The formal derivative has all the usual algebraic properties of the differential operator from calculus, inparticular

1. D( f + g) = D f +Dg2. D( f g) = (D f )g + f (Dg)

10 FIELDS AND GALOIS

4.5 Theorem. Let F be field and f (x) ∈ F[x].

1. If ch F = 0 and D f = 0 then f (x) = c for some c ∈ F2. If ch F = p and D f = 0 then f (x) = g(x p) for some g(x) ∈ F[x]

PROOF: Trivial. �

4.6 Definition. Let E/F be a field extension and f (x) ∈ F[x]. We say that α ∈ E is a repeated root of f (x) iff (x) = (x −α)2 g(x) for some g(x) ∈ E[x].

4.7 Lemma. If E[x], α is a repeated root of f (x) if and only if x −α divides both f and D f .

PROOF: If f (x) = (x −α)2 g(x) then D f (x) = 2(x −α)g(x) + (x −α)2Dg(x), so x −α is a common factor of fand D f . Suppose conversely that x −α divides both f and D f . Write f (x) = (x −α)h(x), for some h(x) ∈ E[x].Then D f (x) = h(x) + (x −α)Dh(x). D f (α) = 0 implies that h(α) = 0, and so we are done. �

4.8 Theorem. Let f (x) ∈ F[x]. Then f has no repeated roots in any extension of F if and only if gcd( f , D f ) = 1in F[x]

Notice that the condition of repeated roots depends on the extension of F , while the gcd condition involvesonly F .

PROOF: Let g = gcd( f ,D f ). Write g = s f + tD f for some polynomials s(x), t(x) ∈ F[x] (F[x] is a Euclideandomain). Suppose f (x) has a repeated root α in some extension E/F . Then clearly x −α is a common factor off and D f , and so g 6= 1. Suppose now that g 6= 1. Then there is an extension E/F such that E contains a root αof g. Then x −α divides both f and D f , and so α is a repeated root of f . �

4.3 Separable Polynomials

4.9 Definition. Let F be a field and f (x) ∈ F[x] not zero. If f (x) is irreducible, then we say f (x) is separableover F if it has no repeated roots in any extension of F . If f (x) is not irreducible, then we say it is separable ifall of it’s irreducible factors are separable.

4.10 Example. Consider the polynomial f (x) = x t − a ∈ F[x], with t ≥ 2. If a = 0, then f is clearly separable,as the only irreducible factor of f is x . A linear polynomial is always separable. Now we assume that a 6= 0. Notethat D f (x) = t x t−1.

1. If ch F = 0 then gcd( f ,D f ) = 1, so f is separable.2. If ch F = p and gcd(p, t) = 1 then gcd( f , D f ) = 1, so f is separable.3. If ch F = p and t = p then D f = 0, so gcd( f ,D f ) 6= 1. However, it is still possible that all of the irreducible

factors p(x) have the property that gcd(p, Dp) = 1. To decide, we need to find the irreducible factors off . Define F p = {ap | a ∈ F}, a subfield of F . If a ∈ F p then there is some b ∈ F such that a = bp, and sof (x) = (x − b)p, and f is separable. There is another case, although it only comes up if F is an infinitefield of characteristic p. If a 6∈ F p then we claim that f (x) = x p − a is irreducible. Assume that we maywrite x p − a = g(x)h(x), where g, h ∈ F[x] are monic. Let E/F be a extension such that x p − a has a rootβ ∈ E. Then β p = a, and so β 6∈ F . We have

x p − a = x p − β p = (x − β)p

Thus g(x) = (x − β)r and h(x) = (x − β)s for some r + s = p. Write g(x) = x r + rβ x r−1 + · · · . Thensince rβ ∈ F , r = 0 in F . Thus r = kp for some k. This shows that either r = 0 or s = 0, and so x p − a isirreducible over F . Therefore x p − a is not separable in this case. We say that f is purely inseparable sinceall of the roots of f are the same.

SEPARABLE POLYNOMIALS 11

4.4 Perfect Fields

4.11 Definition. A field F is called perfect if every irreducible polynomial f (x) ∈ F[x] is separable.

4.12 Theorem. Let F be a field.

1. If ch F = 0 then F is perfect.2. If ch F = p and F p = F then F is perfect.

PROOF: Let r(x) ∈ F[x] be irreducible. Then either gcd(r, Dr) = 1 or gcd(r, Dr) = r.

1. Let ch F = 0. Suppose that r is not separable, that is, gcd(r,Dr) = r. Then Dr = 0, and so deg r = 0, acontradiction. Therefore r is separable and F is perfect.

2. Let ch F = p. Suppose that r is not separable, that is, gcd(r,Dr) = r. Then Dr = 0 in F[x]. Write

r(x) = a0 + a1 x p + · · ·+ am xmp, ai ∈ F

Since F p = F , we can write ai = bpi for some bi ∈ F . Thus

r(x) = bp0 + bp

1 x p + · · ·+ bpm xmp = (b0 + b1 x + · · ·+ bm xm)p

which is a contradiction since r is irreducible. Thus r is separable and F is perfect. �

4.13 Corollary. Every finite field is perfect. (Assignment 3)

Recall that if E/F is a finite extension then there exist α1, . . . ,αn ∈ E such that

F $ F(α1)$ · · ·$ F(α1, . . . ,αn) = E

4.14 Theorem. If ch F = 0 and E/F is a finite extension then E/F is a simple extension.

PROOF: Since E = F(α1, . . . ,αn) for some α1, . . . ,αn ∈ E, it suffices to consider the case when E = F(α,β). Thegeneral case follows by induction. Let E = F(α,β). Our goal is to find γ ∈ E such that E = F(γ). It suffices to findλ ∈ F such that γ = α+λβ and β ∈ F(γ) because then we will have F(α,β) ⊆ F(γ) (the reverse containment isclear).

Let a(x) and b(x) be the minimal polynomials of α and β over F , respectively. Choose λ ∈ F such that

λ 6=α−αβ − β

where α runs over all the roots of a in E, and β runs over all of the roots of b in E that are not β . We can dothis because there are infinitely many elements in F , but only finitely many excluded choices. Let γ = α+ λβ .Consider h(x) = a(γ−λx) ∈ F(γ)[x]. Then β is a root of h. However, for all β 6= β , since

γ−λβ = α+λ(β − β) 6= α

by the choice of λ, we have that h(β) 6= 0. Thus h and b have β as a common root, but no others in any extensionof F(γ). The minimal polynomial of β in F(γ), call it b1(x), must divide h and b. Since ch F = 0 and b1 isirreducible, b1 has distinct roots. The roots of b1 are also roots of b and h. Since β is the only common root,b1(x) = x − β , and so β ∈ F(γ). �

Remark. This a special case of a more general result called the Primative Element Theorem that we will see later.


5 Automorphism Groups

5.1 Automorphism Groups

5.1 Definition. If E is a field, we say that a map ψ : E → E is an automorphism if it is an isomorphism of E. IfE/F is a field extension and ψ : E → E is an automorphism which fixes F , we say that ψ is an F -automorphismof E. By map composition, the set

AutF (E) = {ψ : E→ E |ψ is an F -automorphism}

is called the automorphism group of E/F . It may also be denoted Aut(E/F).

5.2 Lemma. Let f (x) ∈ F[x] and α ∈ E a root of f (x). For ψ ∈ AutF (E), ψ(α) is also a root of f (x). Noticethat E does not have to be the splitting field of f (x).

PROOF: If f (x) = a0 + a1 x + · · ·+ an xn then we have

f (ψ(α)) = a0 + a1ψ(α) + · · ·+ anψ(α)n

=ψ(a0) +ψ(a1α) + · · ·+ψ(anαn)

=ψ(a0 + a1α+ · · ·+ anαn)

=ψ(0) = 0

Thus ψ(α) is a root of f (x). �

5.3 Lemma. Let E = F(α1, . . . ,αn) be a field extension. For ψ1,ψ2 ∈ AutF (E), if ψ1(αi) = ψ2(αi) for alli = 1, . . . , n then ψ1 =ψ2.

PROOF: Trivial. �

5.4 Corollary. If E/F is a finite extension then AutF (E) is a finite group.

5.2 Automorphism Groups of Polynomials

5.5 Definition. Let F be a field and f (x) ∈ F[x]. The automorphism group of f (x) over F is defined to be thegroup AutF (E), where E is a splitting field of f (x). Notice that this definition does not depend on the choice ofE. By a previous theorem all splitting fields of f (x) are isomorphic, and hence their automorphism groups areisomorphic.

5.6 Theorem. Let E/F be a splitting field of a non-zero polynomial f (x) ∈ F[x]. Then |AutF (E)| ≤ [E : F], andequality holds if and only if f (x) is separable over F .

PROOF: Assignment 3. �

5.7 Example. 1. Let F be a field with ch F = p. Let a ∈ F \ F p and E/F a splitting field of the polynomialf (x) = x p − a. We have seen before that x p − a = (x −β)p, for some β ∈ E \ F . Thus E = F(β), and sinceβ can only map to β , AutF (E) is the trivial group. Notice that |AutF (E)|= 1 while [E : F] = p.

2. Consider F =Q(p

2,p

3), which is the splitting field of f (x) = (x2− 2)(x2− 3) ∈Q[x]. f (x) is separable,so |AutF (E)| = [E : F] = 4. It follows that AutF (E) is isomorphic to Z2 ⊕Z2, as AutF (E) has not elementsof order 4.

GALOIS EXTENSIONS 13

3. Consider the irreducible polynomial x3−2 ∈Q[x]. Let ζ3 = e2πi/3. The roots of x3−2 are { 3p

2, 3p

2ζ3, 3p

2ζ23},

and thus the splitting field of x3 − 2 is

E =Q( 3p2, 3p2ζ3, 3p2ζ23) =Q(

3p2,ζ3)

Let L = Q( 3p

2) be a subfield of E containing Q. We consider AutQ(L) and AutQ(E). L contains only oneroot of x3−2 since it is a real field, and so AutQ(L) is the trivial group. E is the splitting field of a separablepolynomial, so |AutQ(E)| = [E : Q] = 6. By the next theorem, we see that it is a subgroup of S3, soAutQ(E)∼=S3. We notice from this example that the automorphism group is not always Abelian.

Open Problem: Does every finite group occur as the automorphism group over Q of the splitting field ofsome polynomial? It is known that every finite Abelian group does occur.

5.8 Theorem. If f (x) ∈ F[x] has n distinct roots in its splitting field E then AutF (E) is isomorphic to a subgroupof the symmetric group Sn. In particular, |AutF (E)| divides n!.

PROOF: Let X = {α1, . . . ,αn} be the distinct roots of f (x) in E. If ψ ∈ AutF (E), then ψ(X ) = X . From thisobservation and the fact that ψ is uniquely determined by its action on X , it is clear that AutF (E) is isomorphicto a subgroup of the symmetric group on X , which itself is isomorphic to Sn, with an injective homomorphismgiven by ψ 7→ψ|X . �

5.3 Fixed Fields

5.9 Definition. Let E/F be a field extension and ϕ ∈ AutF (E). Define

Eϕ = {a ∈ E | ϕ(a) = a}

which is necessarily a subfield of E that contains F . We usually call Eϕ the fixed field of ϕ. Let G be a subgroupof AutF (E). The fixed field of G is defined to be

EG =⋂

ψ∈G

Eψ = {a ∈ E |ψ(a) = a ∀ψ ∈ G}

5.10 Theorem. Let f (x) ∈ F[x] be a separable polynomial and E/F its splitting field. Then EAutF (E) = F .

PROOF: Let G = AutF (E) and L = EG . Clearly F ⊆ L, and thus AutL(E) ⊆ AutF (E). If ψ ∈ AutF (E) = Gthen for all a ∈ L, ψ(a) = a. That is, ψ ∈ AutL(E), and thus AutL(E) = AutF (E). Because f (x) is separableover F and splits over E, f (x) is also separable over L and has E as its splitting field over L. It follows that[E : L] = |AutL(E)| = |AutF (E)| = [E : F] Since [E : F] = [E : L][L : F], it follows that [L : F] = 1 and soL = F . �

6 Galois Extensions

6.1 Separable Extensions

6.1 Definition. Let E/F be an algebraic field extension. For α ∈ E, let p(x) ∈ F[x] be the minimal polynomialof α. We say that α is separable over F if p(x) is separable. If α is separable for all α ∈ E then we say that theextension E/F is separable.

6.2 Theorem. Let E/F be a splitting field of f (x) ∈ F[x]. If f (x) is separable then E/F is a separable extension.


PROOF: If ch F = 0 then F is perfect and every extension is separable. If ch F = p then consider α ∈ E. Letp(x) ∈ F[x] be the minimal polynomial of α. Let α = α1, . . . ,αn be the distinct roots of p(x) that are containedin E. We claim that p(x) = (x −α1) · · · (x −αn). It suffices to show that

p(x) := (x −α1) · · · (x −αn)

is in F[x], since p(x) is the minimal polynomial of α and p(x) has α as a root. Let ψ ∈ AutF (E). ψ permutesα1, . . . ,αn and the coefficients of p are symmetric with respect to α1, . . . ,αn, so each coefficient of p(x) is fixedwith respect to ψ. Therefore p(x) ∈ Eψ[x]. Since ψ was arbitrary, p(x) ∈ EAutF (E)[x] = F[x]. �

6.3 Corollary. Let E/F be a finite extension and E = F(α1, . . . ,αn). If each αi is separable over F then E/F isseparable.

PROOF: For 1 ≤ i ≤ n, let pi(x) ∈ F[x] be the minimal polynomial of αi . Let f (x) =∏n

i=1 pi(x). Then f (x) isseparable. Let L be the splitting field of f , so that L/F is separable. Since E = F(α1, . . . ,αn) is a subfield of L, Eis also separable. �

6.4 Corollary. Let E/F be an algebraic extension and L be the set of all α ∈ E that are separable over F . Then Lis an intermediate field.

6.2 Normal extensions

6.5 Definition. Let E/F be an algebraic extension. We say that E/F is a normal extension if given any irreduciblepolynomial p(x) ∈ F[x], either p(x) has no root in E or E contains all of the roots of p(x). In other words, ifp(x) has a root in E then p(x) splits over E.

6.6 Example. Let α ∈ R such that α4 = 5 and let β = (1+ i)α. Consider the field extension Q(β)/Q. Noticethat β2 = 2iα2, and so β4 = −20. Hence the minimal polynomial of β over Q is x4 + 20 and [Q(β) : Q] = 4.The roots of x4 + 20 are ±β ,±iβ . It is sufficient to show that α 6∈ Q(β) to show that iβ 6∈ Q(β). The minimalpolynomial of α is x4 − 5, and so we have that [Q(α) : Q] = 4. Notice that if α ∈ Q(β) then Q(α) = Q(β), andthis is impossible since Q(α) is a real field while Q(β) is not. It follows that the prime factorization of x4 + 20over Q(β) is (x − β)(x + β)(x2 + β2), and hence it does not split over Q(β), so Q(β) is not a normal extensionof Q.

6.7 Theorem. A finite extension E/F is normal if and only if it is the splitting field of some polynomial f (x) ∈F[x].

PROOF: Suppose that E/F is a finite extension and is normal. Let E = F(α1, . . . ,αn). For each i, let pi(x) bethe minimal polynomial of αi . Define f (x) =

∏ni=1 pi(x). Since E/F is normal, each pi(x) splits over E, say

αi,1, . . . ,αi,riare the roots of pi(x) over E. Thus

E = F(α1, . . . ,αn) = F(α1,1, . . . ,α1,ri,α2,1, . . . ,αn,ri

)

Therefore E is a splitting field of f (x) over F .Now suppose that E/F is the splitting field of f (x) ∈ F[x]. Let p(x) ∈ F[x] be an irreducible polynomial

with a root α ∈ E. Let K/E be a splitting field of p(x) over E. Write

p(x) = c(x −α1) . . . (x −αn)

where 0 6= c ∈ F and α= α1, . . . ,αn ∈ K = E(α1, . . . ,αn). Define an F -isomorphism

θ : F(α)→ F(α2) : α 7→ α2


Note that p(x) ∈ F(α)[x], F(α2)[x]. Hence we can view K as a splitting field of p(x) f (x) over F(α) and F(α2)respectively. Thus there exists an isomorphism ψ : K → K which extends θ .

Kψ // K

E

F(α) θ // F(α2)

Fid // F

Since ψ ∈ AutF (K), ψ permutes the roots of f (x). Since E is generated over F by the roots of f (x), wehave ψ(E) = E. It follows that for α ∈ E, α2 = ψ(α) ∈ E. Since the choice of α2 was arbitrary, αi ∈ E for all i.Therefore K = E and p(x) splits over E and E is normal. �

6.8 Example. Every quadratic extension is normal. Let E/F be a quadratic extension. For α ∈ E \ F , E = F(α).Let p(x) = x2 + ax + b be the minimal polynomial of α over F . Then −a−α ∈ F(α) is the other root of p, andso E is the splitting field of p. Therefore E/F is normal.Q( 4p

2)/Q is not normal since the irreducible polynomial x4 − 2 does not split over Q( 4p

2) despite having aroot in Q( 4

p2). Note that the extension Q( 4

p2)/Q is made up of two quadratic extensions

Q( 4p2)/Q(p

2) and Q(p

2)/Q

Q( 4p

2)

not normal

normal

uuuuuuuuu

Q(p

2)

normal IIIIIIIIII

Q

6.9 Proposition. If E/F is a normal extension and K is an intermediate field then E/K is normal.

PROOF: Let p(x) ∈ K[x] be irreducible and have a root α ∈ E. Let f (x) ∈ F[x] be the minimal polynomial of αover F . Then f (x) splits over E since E/F is normal, and p(x)| f (x). It follows that p(x) splits over E as well, soE/K is a normal extension. �

Remark. K/F is not always normal. Take F = Q, K = Q( 3p

2), E = Q( 3p

2,ζ3). Then E/F is normal but K/F isnot.

Q( 3p

2,ζ3)

normal

normal

ssssssssss

Q( 3p

2)

not normalLLLLLLLLLLL

Q


6.3 Conjugates

6.10 Definition. Let E/F be a field extension and α,β ∈ E. If α and β have the same minimal polynomial thenthey are said to be conjugate over F .

It is clear that a field extension E/F is normal if and only if for every α ∈ E, E contains all of the conjugatesof α over F .

6.11 Proposition. Let E/F be a finite normal extension and α,β ∈ E. Then the following are equivalent

1. α and β are conjugate over F2. there exists ψ ∈ AutF (E) such that ψ(α) = β

PROOF: Suppose that p(x) ∈ F[x] is the minimal polynomial of both α and β . Then

F(α)∼= F[x]/⟨p(x)⟩ ∼= F(β)

and so there is an F -isomorphism θ : F(α) → F(β) : α 7→ β . Now E/F is a finite normal extension, so by anabove theorem, E is the splitting field of some polynomial f (x) ∈ F[x]. We can also view E as a splitting field off (x) over F(α) and F(β) respectively. Thus, there exists an isomorphism ψ : E→ E which extends θ . It followsthat ψ ∈ AutF (E) and ψ(α) = β .

Now suppose that there is ψ ∈ AutF (E) with ψ(α) = β . Let p(x) ∈ F[x] be the minimal polynomial of α overF . Then

p(β) = p(ψ(α)) =ψ(p(α)) =ψ(0) = 0

so β is a root of p(x). Therefore p(x) must be the minimal polynomial of β as well. �

6.12 Definition. A normal closure of a finite extension E/F is a finite normal extension N/F which has thefollowing properties

1. E is a subfield of N2. If L is any intermediate field of N/E and L is normal over F then L = N .

6.13 Theorem. Every finite extension E/F has a normal closure N/F . Moreover, N is unique up to E-isomorphism.

PROOF: (Existence) Write E = F(α1, . . . ,αn). Let pi(x) ∈ F[x] be the minimal polynomial of αi , and let f (x) =∏n

i=1 pi(x). Let N/E be the splitting field of f (x) over E. Then N is a normal extension of F (since is it alsothe splitting field of f (x) over F) that contains E. If N ⊃ L ⊃ E is normal then f (x) splits over L since eachirreducible factor of f (x) has a root in L. Thus L = N , so N is a normal closure of E/F .

(Uniqueness) Let N1 be another normal closure of E/F . Since N1 is normal over F and contains α1, . . . ,αn,N1 must contain a splitting field N2 of f (x) over F with E ⊆ N2. Since N2 is normal over F , we must haveN1 = N2. Therefore N1 are N are splitting fields of f (x) over F , and hence over E, so they are E-isomorphic byTheorem 3.6. �

6.4 Galois Extensions

6.14 Definition. An algebraic extension E/F is Galois if it is normal and separable. If E/F is a Galois extensionthen the Galois group of E over F is defined to be AutF (E), denoted GalF (E).

Remark. 1. Notice that by the last two sections, the finite Galois extensions of F are exactly the splitting fieldsof separable polynomials in F[x].

2. If E/F is a finite Galois extension then |GalF (E)|= [E : F]


3. If E/F the splitting field of a separable polynomial f (x) of degree n then GalF (E) is a subgroup of Sn.

6.15 Example. Let E be the splitting field of x5 − 7 over Q. Then E = Q( 5p

7,ζ5). The minimal polynomials of5p

7 and ζ5 over Q are x5− 7 and x4+ x3+ x2+ x + 1, respectively. Since [Q( 5p

7) :Q] = 5 and [Q(ζ5) :Q] = 4are divisors of [E : Q], [E : Q] is divisible by 20. Since [E : Q] = [E : Q(ζ5)][Q(ζ5) : Q] and Q(ζ5) : Q] = 4, wemay conclude that [E : Q(ζ5)] ≥ 5. Also, E = Q( 5

p7,ζ5) = Q(ζ5)(

5p

7) and the minimal polynomial of 5p

7 overQ(ζ5) is a factor of x5 − 7. Thus [E :Q(ζ5)]≤ 5, and so [E :Q(ζ5)] = 5.

E =Q( 5p

7,ζ5)5

MMMMMMMMMMM4

ppppppppppp

Q( 5p

7)

5NNNNNNNNNNNNN

Q(ζ5)

4pppppppppppp

Q

Then for ψ ∈ GalQ(E), ψ is determined by its action on the roots of x5 − 7, so denote ψ = ψk,s with1≤ s, k ≤ 5 if ψ( 5

p7) = 5

p7ζk

5 and ψ(ζ5) = ζs5. We have the following identity (Check this)

ψk1,s1◦ψk2,s2

=ψk1+s1k2,s1s2

There are two ways to view GalQ(E)

1. GalQ(E) can be viewed as a group of permutations of the roots of x5 − 7. Identity the roots of x5 − 7 withthe elements of {1,2, 3,4, 5} as `↔ 5

p7ζ`5. Then, for example, we may view ψ2,3 as (5 2 3 1).

2. We can also understand GalQ(E) in terms of matrix groups. notice that

�

s1 k10 1

�

·�

s2 k20 1

�

=�

s1s2 k1 + s1k20 1

�

Thus we can associate ψk,s ∈ GalQ(E) with the matrix

�

s k0 1

�

∈ GL2(F5)

and the map composition law in GalQ(E) is preserved by the matrix mulitplication. Thus we have that

GalQ(E)∼=

¨

�

s k0 1

��

�

�

�

s ∈ F∗5, k ∈ F5

«

6.5 Artin’s Theorem

6.16 Theorem. (E. Artin) Let E be a field and G a finite subgroup of Aut(E). Then E/EG is a finite Galoisextension with G = GalEG (E). In particular, [E : EG] = |G|.

PROOF: Let n= |G| and F = EG . For any α ∈ E, consider the G-orbit of α, that is, the set

{ψ(α) |ψ ∈ G}= {α= α1, . . . ,αm}


where the αi are distinct and m ≤ n. Let f (x) = (x − α1) . . . (x − αm). For any ψ ∈ G, ψ permutes the roots{α1, . . . ,αm}. Thus f (x) ∈ EG[x] = F[x]. Let g(x) be a factor of f (x). Without loss of generality, we may writeg(x) = (x−α1) . . . (x−α`) for some `≤ m. If ` 6= m, chooseψ ∈ G such that {α1, . . . ,αm} 6= {ψ(α1), . . . ,ψ(αm)}.It follows that ψ(g(x)) = (x −ψ(α1)) . . . (x −ψ(α`)) 6= g(x). Thus, if ` 6= m then g(x) /∈ F[x]. Thus f (x) isirreducible over F , and so is the minimal polynomial of α over F . Since f (x) is separable and splits over E, thisshows that E/F is Galois.

Now consider [E : F]. We show first that [E : F]≤ n. If [E : F]> n= |G| then we can choose α1, . . . ,αn+1 ∈ Ewhich are linearly independent over F . Consider the system

ψ(α1)v1 + · · ·+ψ(αn+1)vn+1 = 0 as ψ ranges over G

of linear equations in n + 1 variables v1, . . . , vn+1. It has a non-trivial solution in (β1, . . . ,βn+1) in E. Assumethat (β1, . . . ,βn+1) has the minimal number of non-zero coordinates, say r. Clearly, r > 1 and without loss ofgenerality we may assume that β1, . . . ,βr 6= 0 and βr+1, . . . ,βn+1 = 0. Furthermore, we may assume that βr = 1.Thus

ψ(α1)β1 + · · ·+ψ(αr)βr = 0 for all ψ ∈ G (∗)

and taking ψ = idE we get that α1β1 + · · ·+ αrβr = 0, so we may assume that β1 6∈ F since α1, . . . ,αn+1 arelinearly independent in F . Choose φ ∈ G such that φ(β1) 6= β1. Applying φ to (∗) yeilds

(φ ◦ψ)(α1)φ(β1) + · · ·+ (φ ◦ψ)(αr)φ(βr) = 0 for all ψ ∈ G

But βr = 1, so φ(βr) = βr , and subtracting this equation from (1) gives us a solution with strictly fewer non-zerocoordinates. This contradiction shows that [E : F] ≤ n. We have seen that E/F is a finite Galois extension, thusE is a splitting field of some separable polynomial g(x) ∈ F[x]. Also, since F = EG , G is a subgroup of GalF (E).But then n= |G| ≤ |GalF (E)|= [E : F]≤ n. Therefore [E : F] = n and G = GalF (E). �

Remark. Let E/F be a Galois extension with Galois group G. For α ∈ E let {α = α1, . . . ,αn} be the G-orbit of α.This is the set of all conjugate roots of α. Then the minimal polynomial of α over F is (x −α1) . . . (x −αn).

6.17 Example. Let E = F(t1, . . . , tn) be the function field in n variables over F . Consider the symmetric groupSn as a subgroup of AutF (E) which permutes the variables t1, . . . , tn. We would like to find ESn . The Sn-orbit oft1 is {t1, . . . , tn}. It follows that the minimal polynomial of t1 over ESn is

f (x) = (x − t1) . . . (x − tn)

Recall the the elementary symmetric functions in t1, . . . , tn are

s0 = 1

s1 = t1 + · · ·+ tn

s1 =∑

1≤i< j≤n

t i t j

...

sn = t1 . . . tn

Thus f (x) =∑n

i=0(−1)n−isn−i xi . Define L = F(s1, . . . , sn) ⊆ ESn . We have f (x) ∈ L[x] and E is a splitting field

of f (x) over L. Since deg f ≤ n, [E : L]≤ n!. On the other hand, [E : ESn] = |Sn|= n! by Artin’s theorem. SinceL ⊆ ESn , we have n!= [E : ESn]≤ [E : L]≤ n!, and so ESn = L.

THE GALOIS CORRESPONDENCE 19

6.18 Example. Let E = F(t) be the function field in one variable over F . Let G be the subgroup of AutF (E)generated by involutions σ and τ defined by

σ : g(t) 7→ g�

1

t

�

and τ : g(t) 7→ g(1− t)

Let ρ = στ. Then ρ(g(t)) = g( 11−t), ρ2(g(t)) = g( t−1

t), and ρ3(g(t)) = g(t). Hence ρ3 = 1 in G. We have

G = ⟨σ,τ⟩= ⟨ρ,σ⟩ ∼=S3. To consider EG , notice that the G-orbit of t is

tρ //

σ

��

11−t

ρ //

σ

��

t−1t

σ

��1t 1− t t

t−1

Hence the minimal polynomial of t in EG[x] is

f (x) = (x − t)�

x −1

1− t

��

x −t − 1

t

��

x −1

t

�

�

x −t

t − 1

�

(x − (1− t))

= x6 − 3x5 + (6− h)(x4 + x2) + (2h− 7)x3 − 3x + 1

where h= (t2−t+1)3

t2(t−1)2. Now h ∈ EG (check this) and we have that F ⊆ F(h)⊆ EG ⊆ E. Since

(t2 − t + 1)3 − ht2(t − 1)2 = 0

t ∈ E is a root of g(x) = (x2− x +1)3−hx2(x −1)2 ∈ F(h)[x]. Since deg g = 6 and E = F(h)(t), [E : F(h)]≤ 6.Also, [E : EG] = |G| = 6 by Artin’s theorem. Since 6 = [E : EG] ≤ [E : F(h)] ≤ 6, we have that EG = F(h) andg(x) is the minimal polynomial of t over F(h).

7 The Galois Correspondence

7.1 The Fundemental Theorem

7.1 Theorem. (Fundemental Theorem of Galois Theory) Let E/F be a finite Galois extension and G = GalF (E).Then there is an order reversing bijection between the intermediate fields of E/F and the subgroups of G. Moreprecisely, let Int(E/F) denote the set of intermediate fields of E/F and Sub(G) the set of subgroups of G. Thenthe maps

• Int(E/F)→ Sub(G) : L 7→ L∗ := GalL(E)

• Sub(G)→ Int(E/F) : H 7→ H∗ := EH

are inverses of each other and reverse the inclusion relation. In particular, for L1 ⊇ L2 ∈ Int(E/F) and H1 ⊆ H2 ∈Sub(G) then we have

[L1 : L2] = [L∗2 : L∗1] and [H1 : H2] = [H

∗2 : H∗1]


E {1}= GalE(E)

L1 L∗1 = GalL1(E)

L2 L∗2 = GalL2(E)

F G = GalF (E)

PROOF: Recall the following theorems:

1. If f (x) ∈ F[x] is separable and E/F is its splitting field then EAutF (E) = F .

2. If E is a field and G is finite subgroup of Aut(E) then E/EG is a finite Galois extension and GalEG (E) = G.

3. If E/F is Galois and L is an intermediate field then E/L is also Galois.

Let L ∈ Int(E/F) and let H ∈ Sub(G). Then

EGalL(E) = L so (L∗)∗ = (GalL(E))∗ = L

Also,

GalEH (E) = H so (H∗)∗ = (EH)∗ = H

Hence we have

H 7→ H∗ 7→ (H∗)∗ = H and L 7→ L∗ 7→ (L∗)∗ = L

so the maps L 7→ L∗ and H 7→ H∗ are inverses of each other. For L1, L2 ∈ Int(E/F), E/L1 and E/L2 are also Galois.If L2 ⊆ L1 then we have GalL1

(E)⊆ GalL2(E). Thus L2 ⊆ L1 =⇒ L∗1 ⊆ L∗2. Also,

[L1 : L2] =[E : L2][E : L1]

=|GalL2

(E)||GalL1

(E)|=|L∗2||L∗1|= [L∗2 : L∗1]

For H1, H2 ∈ Sub(G), if H2 ⊆ H1 then we have EH1 ⊆ EH2 . Thus H2 ⊆ H1 =⇒ H∗1 ⊆ H∗2. Also,

[H1 : H2] =|H1||H2|

=|GalEH1 (E)||GalEH2 (E)|

=[E : EH1][E : EH2]

= [EH2 : EH1] = [H∗2 : H∗1] �

Remark. Given a finite Galois extension E/F , we can ask how many intermediate fields are between E and F .Without the Fundemental Theorem of Galois Theory, this would be a hard question to answer. In particular, sinceGalF (E) is finite for finite Galois extensions, there are only finitely many intermediate fields. This is exactly thespirit of Galois theory: transform a question of infiniteness (fields), which is hard to answer, to a question offiniteness (groups), which is easier to understand.


7.2 Applications

7.2 Lemma. Let E/F be a finite Galois extension with Galois group G. Let L be an intermediate field. For ψ ∈ G,we have

Galψ(L)(E) =ψGalL(E)ψ−1

PROOF: For any α ∈ψ(L),ψ−1(α) ∈ L. Ifφ ∈ GalL(E), we haveφ◦ψ−1(α) =ψ−1(α). That is to say,ψ◦φ◦ψ−1 ∈Galψ(L)(E) for any φ ∈ GalL(E). Thus ψGalL(E)ψ−1 ⊆ Galψ(L)(E). Since the groups have the same order weconclude that they are the same. �

7.3 Theorem. Let E/F , L, G be defined as in the last theorem. Then L/F is Galois if and only if L∗ is a normalsubgroup of G. In this case

GalF (L)∼= G/L∗

PROOF:

L/F is normal ⇐⇒ ψ(L) = L ∀ψ ∈ GalF (E)⇐⇒ Galψ(L)(E) = GalL(E) ∀ψ ∈ GalF (E)

⇐⇒ ψGalL(E)ψ−1 = GalL(E) ∀ψ ∈ GalF (E)

⇐⇒ L∗ = GalL(E) is a normal subgroup of G

If L/F is a Galois extension, the restriction map ψ 7→ ψ|L from G to GalF (L) is well-defined. Moreover, it issurjective and has kernel L∗. We are done by the first isomorphism theorem. �

7.4 Example. For a prime p, let q = pn. Consider Fq, which is an extension of Fp of degree n. The FrobeniusAutomorphism of Fq is defined by

σp : Fq → Fq : α 7→ αp

Notice that the above map is really an automorphism (see assignment 3). For all α ∈ Fq, we have that σnp(α) =

αpn= α. Thus σn

p = 1. For 1≤ m< n, σmp (α) = α implies that α is a root of x pm

− x , which has at most pm roots.Therefore σm

p 6= 1. Hence σnp has order n. It follows that

n= |⟨σp⟩| ≤ |GalFp(Fq)|= [Fq : Fp] = n

Thus GalFp(Fq) = ⟨σp⟩.

Consider a subgroup H of GalFp(Fq) of order d. Then d|n and [G : H] = n

d. By the Fundemental Theorem,

we haven

d= [G : H] = [H∗ : G∗] = [FH

q : Fp]

and thus H∗ = Fp

nd.

7.3 Brief Review of Group Theory

7.5 Theorem. (Cauchy) Let p be prime and G a finite group. If p divides |G| then G contains an element of orderp.

7.6 Definition. Let p be prime. A group in which every element has order a power of p is called a p-group. Itfollows by Cauchy’s theorem that a finite group G is a p-group if and only if |G| is a power of p.


7.7 Theorem. (First Sylow Theorem) Let G be a group with order pnm where p is prime, n> 0, and gcd(p, m) =1. Then G contains a subgroup of order pi for each 1 ≤ i ≤ n and every subgroup of G of order pi for i < n isnormal in some subgroup of order pi+1.

7.8 Definition. A subgroup P of a group G is a Sylow p-subgroup if P is a maximal p-subgroup of G. By the firstSylow theorem, if |G|= pnm (as in the theorem) then |P|= pn.

7.9 Theorem. (Second Sylow Theorem) If H is a p-subgroup of a finite group G and P is any Sylow p-subgroupof G, then there exists g ∈ G such that H ⊆ gP g−1. In particular, any two Sylow p-subgroups of G are conjugate.

7.10 Theorem. (Third Sylow Theorem) Let G be a finite group and p be a prime. Then the number of Sylowp-subgroups of G divides |G| and is of the form 1+ kp for some k ≥ 0.

7.11 Example. Determine the lattice of subfields of the splitting field of x5 − 7.We have seen in the previous section that the splitting field of x5 − 7 over Q is Q(α,ζ5) where α = 5

p7. We

already know that [Q(ζ5) : Q] = 4 and [E : Q(ζ5)] = 5. It follows that [E : Q] = 20 and GalQ(E) is a subgroupof S5 of order 20. Also, for each ψ ∈ GalQ(E), we write ψ=ψk,s if ψ(α) = αζk

5 and ψ(ζ5) = ζs5. Define

σ : α 7→ αζ5 : ζ5 7→ ζ5 and τ : α 7→ α : ζ5 7→ ζ25

So σ =ψ1,1 and τ=ψ0,2. It can be checked that τσ = στ2. We have

G := GalQ(E) = ⟨σ,τ | σ5 = τ4 = 1,τσ = στ2⟩

Since |G| = 20, the possible subgroups of G are of orders 1, 2,4, 5,10, 20. Since 20 = 4 · 5, by the first Sylowtheorem, G has Sylow 2-subgroups and Sylow 5-subgroups. By the third Sylow theorem, there must be onlyone Sylow 5-subgroup, and it is normal by the second Sylow theorem. Using the same argument, the number ofSylow 2-subgroups of G is either 1 or 5. But if there is only one Sylow 2-subgroup then it would be normal andhence we would have that G ∼= Z5 ⊕ Z4, a contradiction since G is not Abelian. Hence there must be 5 Sylow2-subgroups, and they must all be cyclic (since ⟨τ⟩ is cyclic and all Sylow 2-subgroups are conjugate). Noticethat all the elements of G are of the form σaτb. Conjugating τ gives σaτσa, and using the relation τσ = στ2

we get ⟨στσ−1⟩= ⟨σ4τ⟩= ⟨ψ4,2⟩

{1}

uuuuuuuuuu

EEEE

EEEE

SSSSSSSSSSSSSSSSSS

WWWWWWWWWWWWWWWWWWWWWWWWWWWW

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

⟨ψ1,1⟩ ⟨ψ20,2⟩ ⟨ψ2

4,2⟩ ⟨ψ23,2⟩ ⟨ψ2

2,2⟩ ⟨ψ21,2⟩

⟨ψ1,1,ψ0,2⟩

kkkkkkkkkkkkkkkk

gggggggggggggggggggggggggg

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

dddddddddddddddddddddddddddddddddddddddddddddddd

ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc ⟨ψ0,2⟩ ⟨ψ4,2⟩ ⟨ψ3,2⟩ ⟨ψ2,2⟩ ⟨ψ1,2⟩

G

KKKKKKKKKK

xxxxxxxxx

kkkkkkkkkkkkkkkkkk

ggggggggggggggggggggggggggggg

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

ddddddddddddddddddddddddddddddddddddddddddddddddddd

The corresponding diagram of subfields is


Q(α,ζ5)

uuuuuuuuu

KKKKKKKKKK

UUUUUUUUUUUUUUUUUUU

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[

Q(ζ5) Q(α,β) Q(αζ5,β) Q(αζ25,β) Q(αζ3

5,β) Q(αζ45,β)

Q(β)

jjjjjjjjjjjjjjjjjjjj

ffffffffffffffffffffffffffffffff

dddddddddddddddddddddddddddddddddddddddddddddd

cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb Q(α) Q(αζ5) Q(αζ25) Q(αζ3

5) Q(αζ45)

Q

JJJJJJJJJJ

sssssssssss

iiiiiiiiiiiiiiiiiiiiiii

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

ddddddddddddddddddddddddddddddddddddddddddddddddddd

ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

where β = ζ5 + ζ−15 (notice that β2 + β − 1= 0).

7.4 The Primitive Element Theorem

Given a field extension E/F , we may ask

1. Is it simple? That is, is E = F(α) for some α ∈ E? If this is the case, we say that α is a primitive element ofE.

2. Are there infinitely many intermediate fields?

We have see that in characteristic zero every finite extension is simple. However, in characteristic p there arefinite extensions which are not simple.

7.12 Example. Let F be a field with ch (F) = p and let F(s, t) be the rational function field in two variables. Wehave F(sp, t p) ⊆ F(s, t p) ⊆ F(s, t) Since t is a root of the irreducible polynomial x p − t p ∈ F(s, t p)[x] (note thatt p 6∈ F(s, t p)p) we have that [F(s, t) : F(s, t p)] = p, and similarily [F(s, t p) : F(sp, t p)] = p. Thus F(s, t) is a finiteextension of F(sp, t p) of degree p2. Let u ∈ F(s, t). Notice that up ∈ F(sp, sp). Thus [F(sp, t p)(u) : F(sp, t p)] ≤ psince u is a root of x p − up ∈ F(sp, t p)[x]. Hence the extension cannot be simple.

7.13 Theorem. A finite extension E/F is simple if and only if it has finitely many intermediate fields.

PROOF: Suppose that E = F(α) is a simple extension. Let K be any intermediate field. We denote by f (x) andg(x) the minimal polynomials of α over F and K respectively. Thus g(x) is a monic factor of f (x) in E[x]. Writeg(x) = xm + cm−1 xm−1 + · · ·+ c0, where ci ∈ K . Let L = F(c0, . . . , cm−1), a subfield of K . Then g(x) ∈ L[x].Notice that E = F(α) = L(α) = K(α). We have

m= [E : K]≤ [E : L] = [L(α) : L]≤ m

Hence K = L = F(c0, . . . , cm−1), so K is completely determined by g(x), a factor of f (x). There are only finitelymany choices for g(x), so there can only be finitely many different intermediate fields.

Suppose conversly that E/F has only finitely many intermediate fields. Since E/F is a finite extension,E = F(α1, . . . ,αn). Without loss of generality, we may assume that E = F(α,β) (the general case follows byinduction).

Claim. There exists λ ∈ F such that F(α+λβ) = F(α,β)


Since we undertand completely a finite extension of a finite field, we may assume that F is an infinite field. Byassumption there are only finitely many intermediate fields, so we can find some λ,λ′ ∈ F such that λ 6= λ′ andF(α+ λβ) = F(α+ λ′β). Hence α+ λβ ,α+ λ′β ∈ F(α+ λβ), so β ∈ F(α+ λβ) (since λ− λ′ 6= 0). ThusE = F(α,β)⊆ F(α+λβ). The other inclusion is clear, so E = F(α+λβ). �

7.14 Theorem. (Primitive Element Theorem) Every finite separable extension is simple.

PROOF: Exercise. �

8 Ruler and Compass Constructions

8.1 Constructible Points

Consider the Euclidean plane R2. Let O, I ∈ R2 be two distinct points. We take the distance OI as the unit oflength. Introduce an orthogonal coordinate system in R2 with the origin O and I on the x-axis with coordinates(1, 0)

8.1 Definition. Let S be any set of points in R2. We call a line L an S-line if |S ∩ L| ≥ 2. We call a circle C anS-circle if the centre of C is in S and the radius of C is equal to the distance between two points in S.

Notation. We denote by S′ the set of points which are either in S or lie in the intersection of two distinct S-lines,two distinct S-circles, or an S-line and an S-circle.

8.2 Definition. A point P ∈ R2 is constructible if there exists a finite sequence of points {P1, . . . , Pn} such thatPn = P and Pi ∈ {O, I , P1, . . . , Pi−1}′ ∀1≤ i ≤ n.

8.3 Lemma. All rational numbers (i.e. points in Q× {0}) are constructible.


8.4 Theorem. For a point P = (α,β) ∈ R2, the following are equivalent

1. P is constructible2. there exists a tower of fields Q = F0 ⊆ F1 ⊆ · · · ⊆ Fn ⊂ R such that α,β ∈ Fn and [Fi : Fi−1] ≤ 2 for all

1≤ i ≤ n

PROOF: Suppose that P is constructible. Then there exists a finite sequence of points {P1, . . . , Pn} such that

Pn = P and Pi ∈ {O, I , P1, . . . , Pi−1}′ ∀1≤ i ≤ n

Write Pi = (αi ,βi) and define F0 = Q and Fi = Fi−1(αi ,βi). Let S = {O, I , P1, . . . , Pi−1}, so that P ∈ S′. There aretwo cases

Case 1: If Pi ∈ S then Fi = Fi−1

Case 2: Suppose Pi ∈ S′ \ S. Then Pi is the intersection point of two S-lines, two S-circles, or an S-line and anS-circle. Notice that given two points (a, b), (c, d), the equation of the line that contains them is

(b− d)x + (c− a)y(ad − bc) = 0

Similarily, given the center of a circle (a, b) and a radius r then the equation of the circle is

(x − a)2 + (y − b)2 = r2

There are three subcases

RULER AND COMPASS CONSTRUCTIONS 25

(a) If Pi is on the intersection of two S-lines then we may clearly use the equations of these lines to solvefor the coordinates Pi , and see that Fi = Fi−1.

(b) If Pi is on the intersection of an S-line and an S-circle then αi and βi are solutions to a equation ofdegree at most two. Hence [Fi : Fi−1]≤ 2.

(c) Suppose Pi is on the intersection of two S-circles. By subtracting the equations of the circles we get alinear equation that is satisfied by αi and βi , so we may use the last case to see that [Fi : Fi−1]≤ 2.

Now suppose that (2) holds. We prove that P is constructible by induction on n. If n = 0 then α,β ∈ Q, so Pis constructible by the last lemma. Suppose that for all P = (α,β) with α,β ∈ Fn−1 are constructible. ConsiderFn.

1. Fn = Fn−1 trivially implies that P is constructible.

2. [Fn : Fn−1] = 2 implies that Fn = Fn−1(pγ) for some γ ∈ Fn−1, γ > 0.

pγ is constructible (see diagram). In

general, for α ∈ Fn, α = a+ bpγ with a, b ∈ Fn−1. Since all of these are constructible so is α. Therefore P

is constructible. �

8.2 Constructible Numbers

8.5 Definition. For α ∈ R, α is constructible if the point P = (α, 0) is constructible. For γ = α+ iβ ∈ C, γ isconstructible if the point P = (α,β) is constructible.

8.6 Corollary. If α ∈ R is constructible then α is algebraic and the degree of the minimal polynomial polynomialof α over Q is a power of 2.

Remark. The converse of this corollary is false, as we shall see later.

8.7 Lemma. Let γ= α+ iβ . Suppose there is a real field L ⊆Q(γ) such that [Q(γ) : L] = 2. If all elements of Lare constructible then γ is constructible.

PROOF: Since [Q(γ) : L] = 2, γ is a root of a polynomial ax+bx + c ∈ L[x] where a 6= 0. Then

γ=−b±

p

b2 − 4ac

2a

so that

α=

(

−b±p

b2−4ac2a

if b2 − 4ac ≥ 0−b2a

otherwiseand β =

(

0 if b2 − 4ac ≥ 0±p

4ac−b2

2aif b2 − 4ac < 0

Recall that if δ ∈ R is constructible, then so ispδ. �

8.3 Applications

8.8 Example. 1. The regular pentagon is constructible. It is enough to show that ζ5 is constructible. Theminimal polynomial of ζ5 is Φ5(x) = x4 + x3 + x2 + x + 1. Let β = ζ5 + ζ−1

5 =p

5−12

, a real number. Theminimal polynomial of β is x2 + x − 1, so Q ⊆ Q(β) ⊆ Q(ζ) is a tower of fields such that the increase ofdegree at each step is 2.


2. The regular 9-gon is not constructible. Consider ζ9 and λ = ζ9 + ζ−19 . Then ζ9 is a root of the polynomial

x2−λx+1 ∈Q(λ)[x]. Therefore [Q(ζ9) :Q(λ)] = 2, so ζ9 is constructible if and only if λ is constructible.Since x9 − 1= (x3 − 1)(x6 + x3 + 1) the minimal polynomial of ζ9 is x6 + x3 + 1. Notice that

λ3 = (ζ9 + ζ−19 )

3

= ζ39 + ζ

−39 + 3(ζ9 + ζ

−19 )

= ζ39 + ζ

69 + 3λ

=−1+ 3λ

Therefore λ is a root of the irreducible polynomial x3 − 3x + 1, so λ cannot be constructible since 3 is nota power of 2.

Consequently, the angle of 2π3

can not be trisected by ruler and compass.

3. The cicle cannot be squared. Specifically,pπ is not constructible. It is sufficient to show that π is not

constructible. But π is not algebraic, so it is not constructible.4. The unit cube cannot be doubled. Specifically, 3

p2 is not constructible. The minimal polynomial of 3

p2 is

x3 − 2, which is of degree 3, not a power of two.

8.9 Theorem. Let α ∈ R be an algebraic number and p(x) its minimal polynomial over Q. Let E/Q be thesplitting field of p(x). Then α is constructible if and only if GalQ(E) is a 2-group.

PROOF: Assume that α is constructible. Let

Q= F0 ⊆ F1 ⊆ · · · ⊆ Fn ⊆ R

be a tower of real quadratic extensions and α ∈ Fn. Since we are in characteristic zero, there is β ∈ Fn such thatFn = Q(β). Let pβ(x) ∈ Q[x] be the minimal polynomial of β . Let β = β1, . . . ,βm be the roots of pβ(x). LetEβ = Q(β1, . . . ,βm), which is a Galois extension. For each i = 1, . . . , m, define ψi : Q(β)→ Q(βi) : β 7→ βi suchthat ψi fixes Q. This is a field isomorphism. We have

Q= F0 ⊆ F1 ⊆ · · · ⊆ Fn =Q(β)=Q(β)(ψ2(F0))⊆Q(β)(ψ2(F1))⊆ · · · ⊆Q(β)(ψ2(Fn)) =Q(β1,β2)=Q(β1,β2)(ψ3(F0))⊆ · · · ⊆Q(β1,β2,β3)...

⊆Q(β1, . . . ,βm) = Eβ

which is a chain of quadratic extensions. Therefore [Eβ : Q] is a power of 2. Since α ∈ Q(β) ⊆ Eβ and Eβ isGalois, all of the conjugates of α are in Eβ . It follows that E is a subfield of Eβ , and so the degree of E over Q isa power of 2. Hence |GalQ(E)| is a power of 2.

Conversely, let G = GalQ(E). If |G| = 2n for some n, by the first Sylow theorem there exists a subgroupHn−1 ⊆ G of order 2n−1. Applying the Sylow theorem repeatedly, we get a chain of subgroups of G

{1}= H0 ⊆ H1 ⊆ · · · ⊆ Hn−1 ⊆ Hn = G

Let H∗i = EHi . By the Fundemental Theorem of Galois Theory,

E = H∗0 ⊇ H∗1 ⊇ · · · ⊇ H∗n−1 ⊇ H∗n = G∗ =Q

where [H∗i−1 : H∗i ] = 2 for i = 1, . . . , m. Since α ∈ E, α is constructible. �

CYCLOTOMIC EXTENSIONS 27

9 Cyclotomic Extensions

9.1 Cyclotomic Polynomials

For a prime p, the pth cyclotomic polynomial

Φp(x) =x p − 1

x − 1= x p−1 + x p−2 + · · ·+ x + 1

is irreducible. However, for general n the polynomial xn−1x−1

is not irreducible if n is not prime. To generalize thedefinition of cyclotomic polynomial to general n, we notice that

Φp(x) = (x − ζp)(x − ζ2p) . . . (x − ζp−1

p )

For each k = 1, . . . , p− 1 we have that gcd(k, p) = 1. Hence

Φp(x) =∏

1≤k≤p(k,p)=1

(x − ζkp)

Thus, a natural way to define Φn(x) isΦn(x) =

∏

1≤k≤n(k,n)=1

(x − ζkn)

9.1 Definition. Let n ∈ N and ζn = e2πin . For any k ∈ N with (k, n) = 1, we call ζk

n a primitive nth root of unity inC.

9.2 Proposition. xn − 1=∏

d|nΦd(x), where d runs through all positive divisors of n.

9.3 Example. x6 − 1 = (x − 1)(x + 1)(x2 + x + 1)(x2 − x + 1), so the sixth cyclotomic polynomial is Φ6(x) =x2 − x + 1.

Notice that if ψ ∈ GalQ(Q(ζn)) then ψ(ζn) = ζkn, where (k, n) = 1. It follows that Φn(x) ∈Q[x].

9.4 Theorem. The polynomial Φn(x) has integer coefficients and is irreducible over Q.

PROOF: The following statement is an application of Gauß’s Lemma.

Claim. Let h(x) ∈ Z[x] be monic and h(x) = f (x)g(x), where f (x), g(x) ∈ Q[x]. If f (x), g(x) are both monicthen f (x), g(x) ∈ Z[x].

Now let ζn be a primitive nth root of unity and f (x) be the minimal polynomial of ζn over Q. Then xn − 1 =f (x)g(x) for some g(x) ∈ Q[x]. Since f (x) is monic, g(x) is monic, so f (x), g(x) ∈ Z[x]. Let p be a primewith (n, p) = 1. Reduce the above equation modulo p to get xn − 1 = f (x)g(x) in Fp. Since (n, p) = 1, xn − 1

has no multiple roots in any extension of Fp. In particular, f (x) and g(x) are relatively prime.Notice that f (ζp

n)g(ζpn) = (ζ

pn)

n − 1 = 0. Suppose that g(ζpn) = 0. Since f (x) is the minimal polynomial of

ζn and g(ζpn) = 0, we have g(x p) = f (x)h(x) for some h(x) ∈ Z[x]. Then g(x)p = g(x p) = f (x)h(x), and this

is a contradiction because if r(x) is an irreducible factor of f (x) then r(x) divides g(x), contradicting that f (x)and g(x) are relatively prime. Therefore f (ζp

n) = 0. Now for 1 ≤ k ≤ n with (k, n) = 1, let k = p1 . . . ps it’sprime factorization (where the pi ’s are not necessarily distinct). Notice that if ζn is a primitive root, then ζp

n with(p, n) = 1 is also a primitive root. Hence we have

0= f (ζn) = f (ζp1n ) = · · ·= f (ζps

n ) = f (ζp1 p2n ) = · · ·= f (ζk

n)

Thus all primitive nth roots ζkn are roots of f (x), so Φn(x)| f (x). The other direction is obvious, so Φn(x) = f (x)

is the minimal polynomial of ζn over Q. �


9.2 Cyclotomic Fields

9.5 Definition. The nth cyclotomic field is Q(ζn), a splitting field of xn − 1.

9.6 Theorem. The Galois group of xn − 1 over Q is isomorphic to Z∗n, the group of invertible elements of Zn. Itfollows that [Q(ζn) :Q] = ϕ(n), where ϕ is the Euler function.


9.7 Theorem. Every quadratic extension of Q in C is contained in some cyclotomic extension Q(ζn).

PROOF: Every quadratic extension is of the form Q(p

D), where D 6= 1 square-free integer. Notice that for distinctprimes p1 and p2, if Q(pp1) ⊆ Q(ζn1

) and Q(pp2) ⊆ Q(ζn2) then Q(pp1p2) ⊆ Q(ζn1

,ζn2) ⊆ Q(ζn1n2

). Hence it

is enough to consider Q(p

±p) for prime p.

If p = 2, since (1+ i)2 = 2i and 1+ i ∈ Q(ζ4) = Q(i), we havep

2i ∈ Q(ζ4). Also, i ∈ Q(ζ4), sop

i ∈ Q(ζ8).It follows that

p2,p−2 ∈Q(ζ8), and so Q(

p±2)⊆Q(ζ8).

Let p be an odd prime. Consider Q(ζp). The minimal polynomial of ζp over Q is

Φp(x) =∏

1≤k<p

(x − ζkp)

The discriminant of Φp(x) is

D(Φp) =∏

1≤i< j<p

(ζip − ζ

jp)

2

It can be shown that D(Φp) = (−1)p−1

2 pp−2. Thus we have

∏

1≤i< j<p

(ζip − ζ

jp) =±p

p−32

Æ

(−1)p−1

2 p

Since p−32∈ Z and

∏

1≤i< j<p(ζip−ζ

jp) ∈Q(ζp), if p ≡ 1 (mod 4) then

pp ∈Q(ζp) and

p

−p ∈Q(ζ4p). Otherwise,

if p ≡ 3 (mod 4) thenp

−p ∈Q(ζp) andp

p ∈Q(ζ4p).

Hence in all cases, Q(p

±p)⊆Q(ζ4p). �

Remark. Notice that GalQ(Q(p

D)) ∼= {1} or Z2, which are Abelian groups. We call these type of extensionsAbelian extensions. It turns out that all Abelian extensions of Q in C are contained in some cyclotomic extension(Kronecker-Weber). The proof of this theorem is beyond the scope of this course. The proof of the converse isnot too difficult.

9.3 Abelian Extensions

9.8 Lemma. Let p be prime and m ≥ 1 with p - m. Let Φm(x) ∈ Z[x] be the mth cyclotomic polynomial anda ∈ Z. Then p|Φm(a) if and only if a is not divisible by p and a has order m in F∗p.

PROOF: Assume p|Φm(a). Then since m and p are coprime, xm−1 ∈ Fp[x] has no multiple roots in any extensionof Fp. Write

xm − 1=∏

d|m

Φd(x) = Φm(x)∏

d|md<m

Φd(x) ∈ Fp[x]

CYCLOTOMIC EXTENSIONS 29

We have p|Φm(a), so Φm(a) = 0, and hence (a)m = 1. It follows that p - a. Since p - m, xm − 1 ∈ Fp[x] has nomultiple roots in any extension. We have already seen that the order of a divides m. Assume d < m is the orderof a. Then ad − 1 = 0, so a is a root of Φd ′ for some d ′|d. But then d ′|m, and so a is a double root of xm − 1, acontradiction. Therefore the order of a is m in F∗p.

Suppose conversely. If d|m and d < m then ad − 1 6= 0 so Φd(a) 6= 0 either. Since am − 1 = 0, we must haveΦm(a) = 0, so p|Φm(a). �

We have all seen Euclid’s theorem that there are infinitely many primes. We may generalize this slightly andsay that there are infinitely many primes congruent to 1 modulo 2. Can we generalize this further?

9.9 Lemma. If f (x) ∈ Z[x] is a monic polynomial and deg f ≥ 1, the set of prime divisors of the non-zerointegers in the sequence f (1), f (2), f (3), . . . is infinite.

PROOF: Suppose p1, . . . , pk are the prime divisors of the non-zero integers in the sequence f (1), f (2), f (3), . . . .Choose s ∈ Z such that m = f (s) 6= 0. Define g(x) = 1

mf (s+mp1 . . . pk x). Notice that g(0) = 1

mf (s) = 1. Also,

since all terms involving x in f (s+mp1 . . . pk x) have m in the coefficients, g(x) ∈ Z[x]. Moreover, for any n ∈ Z,g(n) ≡ 1 (mod p1 . . . pk). Choose n ∈ Z such that |g(n)| > 1. Since pi |g(n)− 1 and |g(n)| > 1 it follows thatpi - g(n) for all i = 1, . . . , k. Hence g(n) has a prime divisor p /∈ {p1, . . . , pk}, and so p| f (s + mp1 . . . pkn), acontradiction. Therefore there are infinitely many divisors of this sequence. �

9.10 Theorem. (Dirichlet’s Theorem, weak version) Let m be a positive integer. Then there are infinitely manyprimes p such that p ≡ 1 (mod m).

PROOF: Consider Φm(x) ∈ Z[x], which has degree at least 1. By the above lemma there are infinitely many primedivisors p of Φm(1),Φm(2), . . . . If p|Φm(a) for some a > 1 then a has order m in F∗p. Since F∗p has order p − 1,m|p− 1, so p ≡ 1 (mod m). �

Remark. The actual statement of Dirichlet’s Theorem is much stronger. Considering modulo m, for almost allprimes p, p ≡ k (mod m) where (k, m) = 1. There are ϕ(m) equivalence classes for each m. Let π(x) denote thenumber of primes less than or equal to x . Consider π(x , k, m), the number of primes less than or equal to m andcongruent to k modulo m. Dirichlet’s Theorem says that π(x , k, m) = 1

ϕ(m)π(x)+error.

9.11 Theorem. Given a finite Abelian group A, there is a subfield E of a cyclotomic field with GalQ(E)∼= A.

PROOF: We have A ∼= Ck1× · · · × Cks

where Ck is the cyclic group of order k. Choose odd primes p1 < · · · < pssuch that p1 ≡ 1 (mod k1),. . . ,ps ≡ 1 (mod ks). Such primes exist by Dirichlet’s Theorem. Let n = p1 . . . ps andconsider the nth cyclotomic field L =Q(ζn). Then

G = GalQ(L)∼= Z∗n∼= (Zp1

× · · · ×Zps)∗

∼= Cp1−1 × · · · × Cps−1

Write p1 − 1 = k1d1,. . . ,ps − 1 = ksds. Since Cpi−1 is cyclic, there exists a subgroup Ddiof Cpi−1 which is of

order di . Moreover, Cpi−1/Ddi∼= Cki

. Define H ∼= Dd1× · · · × Dds

, which is a normal subgroup of G. Also,G/H ∼= Ck1

× · · · × Cks∼= A.

L =Q(ζn) oo // {1}

LH = H∗ oo // H

Q oo // G


Let E = H∗ = LH . Since H is normal, by Theorem 7.3, E/Q is Galois. Also, GalQ(E)∼= G/H ∼= A. �

9.4 Constructible n-gons

9.12 Definition. A Fermat prime is a Fermat number Fn = 22n+ 1 which is prime.

Remark. 1. Fermat conjectured in 1650 that every Fermat number is prime. The conjecture is false sinceF5 = 225

+ 1= 641 · 6700417.2. Are there infinitely many Fermat primes? This question is still open. The only Fermat primes known to

date are F0 = 3, F1 = 5, F2 = 17, F3 = 257, and F4 = 65537.

9.13 Theorem. (Gauss) The regular n-gon is constructible if and only if n = 2k p1 . . . pm where k ≥ 0 and the piare distinct Fermat primes.

PROOF: Let ζn be a primative nth root of unity. We have seen that the minimal polynomial of ζn has degree ϕ(n).By Corollary 8.6, the regular n-gon is constructible if and only if ϕ(n) is a power of 2. Write n = 2k pd1

1 . . . pdrr

where k ≥ 0, di ≥ 1, and pi are distinct odd primes. Then ϕ(n) = ϕ(2k)ϕ(pd11 ) . . .ϕ(pdr

r ). Now ϕ(2k) is always a

power of 2. ϕ(pdii ) = pdi−1

i (pi − 1), and so is a power of 2 if and only if di = 1 and pi − 1 is a power of 2. Write

pi = 2ri + 1. Notice that if q is an odd prime dividing r then 2r + 1 = (2rq + 1)(2

rq(q−1) − 2

rq(q−2) + · · · ± 1). Thus

since pi is prime, it must be the case that ri is a power of 2 as well. �

10 Galois Groups of Polynomials

10.1 Discriminant

10.1 Definition. Let F be a field and f (x) ∈ F[x] a separable polynomial. Let E be the splitting field of f (x)over F . The Galois group of f (x) is GalF (E). We denote it by GalF ( f ).

10.2 Definition. Let F be a field and let f (x) ∈ F[x] be a square-free separable polynomial of degree n. Letα1, . . . ,αn be the n distinct roots of f (x) in some splitting field E of F . The discriminant D( f ) of f (x) is

D( f ) =∏

i< j

(αi −α j)2

Remark. We do not lose generality by assuming that f (x) is square-free. If p(x)2| f (x), the splitting field of f (x)is the same as the splitting field of f (x)

p(x).

10.3 Proposition. Let F be a field of characteristic not 2. Let f (x) ∈ F[x] be a square-free separable polynomialof degree n. Let D( f ) be the discriminant of f (x), d2 = D( f ), and G = GalF ( f ). Then

1. D( f ) ∈ F2. For each ψ ∈ G ⊆Sn, ψ(d) =±d, and moreover ψ is even if and only if ψ(d) = d.3. In the Galois correspondence of subgroups of G with intermediate fields of E/F (E is a splitting field of

f (x) over F) we haveF(d)∗ = G ∩ An

In particular, G consists of even permutations if and only if d ∈ F (which is to say that D( f ) is a square inF).

PROOF: Assignment 6. �

GALOIS GROUPS OF POLYNOMIALS 31

10.2 Cubic Polynomials

Let F be a field of characteristic not 2. A general cubic polynomial in F[x] is of the form

p(x) = x3 + ax2 + bx + c ∈ F[x]

If ch (F) 6= 3, by replacing x with (x − a3) it suffices to consider

p(x) = x3 + bx + c

If p(x) is separable and square-free, say α1,α2,α3 are the distinct roots of p(x). Then

D(p) = (α1 −α2)2(α1 −α3)

2(α2 −α3)2 =−4b3 − 27c2

Since deg p = 3, GalF (p)⊆S3. By Propostion 10.3 we get

10.4 Theorem. Let F be a field with ch (F) 6= 2,3. Let p(x) = x3 + bx + c ∈ F[x] be an irreducible polynomialand D(p) its discriminant. Then

GalF (p) =

¨

A3∼= C3 if D(p) is a square in F

S3 otherwise

10.5 Definition. A subgroup G of the symmetric group Sn is transitive if for any 1 ≤ i 6= j ≤ n, there is ψ ∈ Gsuch that ψ(i) = j.

10.6 Lemma. Let F be a field and f (x) ∈ F[x]. Let G = GalF ( f ). If f (x) is an irreducible separable polynomialof degree n then G is isomorphic to a transitive subgroup of Sn and n divides the order of G.

PROOF: Let α = α1, . . . ,αn be distinct roots of f (x) and E = F(α1, . . . ,αn) be the splitting field. Since F(α) ⊆ E,[F(α) : F] is a divisor of [E : F]. Hence n= [F(α) : F] divides |G|= [E : F].

For any i 6= j there is a field isomorphism σ : F(αi) → F(α j) : αi 7→ α j such that σ|F = idF . Since E is asplitting field of f (x) over F(αi) and F(α j) there is ψ : E → E which extends σ. Clearly ψ is an automorphismof E that maps αi to α j . Hence GalF ( f ) is a transitive subgroup of Sn. �

10.3 Quartic Polynomials

Now we consider a quartic polynomial. Let F be a field of characteristic not 2. A general quartic polynomial inF[x] is of the form

p(x) = x4 + ax3 + bx2 + c x + d ∈ F[x]

By replacing x with (x − a4) it suffices to consider

p(x) = x4 + bx2 + cx + d

If p(x) is irreducible and separable, by the above theorem G = GalF ( f ) is a transitive subgroup of S4, the orderof which is divisible by 4. The possibilities are S4, A4, D4, V , and C4. Let α1,α2,α3,α4 be the roots of p(x). Set

u= α1α2 +α3α4

v = α1α3 +α2α4

w = α1α4 +α2α3


Notice that u, v, w are all distinct. Every ψ ∈ GalF (p) permutes the roots of p(x), and so permutes {u, v, w}.Hence we have

gp(x) := (x − u)(x − v)(x −w) ∈ F[x]

It can be computed thatgp(x) = x3 − bx2 − 4d x + 4bd − c2

Notice that

u− v = (α1 −α4)(α2 −α3)v−w = (α1 −α2)(α3 −α4)w− u= (α1 −α3)(α4 −α2)

and hence D(gp) = D(p). We call gp the resolvent cubic of p(x).

10.7 Lemma. Let F be a field of characteristic not 2. Let p(x) = x4 + bx2 + cx + d ∈ F[x] be irreducible andseparable and gp be its resolvent cubic (as above). Let

E = F(α1,α2,α3,α4) and L = F(u, v, w)

be the splitting fields of p and gp respectively. Under the Galois correspondence for G = GalF (p) = GalF (E), Lcorresponds to the subgroup G ∩ V . It follows that

GalF (gp) = GalF (L)∼= G/G ∩ V

PROOF: (Sketch) Since all elements of V fix u, v, w, we have G∩V ⊆ L∗ = GalF (L). Hence to show that G∩V = L∗

it suffices to show that all elements of G \ V move at least one of u, v, w. Just check all 20 possibilities (or check5 representatives from the cosets of S4/V ). Notice that V is a normal subgroup of S4 and so is G, so G ∩ V isnormal. By Theorem 7.3 L is a Galois extension of F and GalF (L)∼= G/G ∩ V . �

Let m= |GalL(E)|= |G/G ∩ V |. We have the following table

G S4 A4 D4 V C4G ∩ V V V V V C2

G/G ∩ V S3 C3 C2 C1 C2m 6 3 2 1 2

In the case m= 2, gp(x) has exactly one root in F , say u ∈ F and v, w 6∈ F . Since either G ∼= D4 or C4 and both D4and C4 contain a 4-cycle, there is an element in G of order 4. Since u = α1α2 +α3α4 we have σ = (1 2 3 4) ∈ Gand σ2 = (1 2)(3 4) ∈ G. Consider

x2 − ux + d = (x −α1α2)(x −α3α4)

Notice that(α1 +α2)(α3 +α4) + (α1α2 +α3α4) = b

Hence we havex2 + (b− u) = (x − (α1 +α2))(x − (α3 +α4))

since the roots sum to zero. Assume that G ∼= C4 = ⟨σ⟩. Then GalL(E) = G ∩ V = ⟨σ2⟩. Also, σ2 fixesα1α2,α3α4,α1 +α2,α3 +α4. Hence x2 − ux + d, x2 + b− u ∈ F[x] and they split over L.

Conversely, if x2 − ux + d, x2 + b − u split over L then α1 + α2,α1α2 ∈ L. Since α1 is a root of x2 − (α1 +α2)x + α1α2, we have [L(α1) : L] = 2. Consider L(α1). Since α1 + α2 ∈ L, we have α2 ∈ L. Also, v, w ∈ Lgive a system of linear equations for α3,α4 which can be solved in L. Hence L(α1) = E. Hence [E : L] = 2 and[L : F] = m= 2 we have [E : F] = 4. Thus G ∼= C4. We have proven

SOLVABILITY BY RADICALS 33

10.8 Theorem. Let F be a field of characteristic not 2. Let p(x) = x4 + bx2 + cx + d ∈ F[x] be irreducible andseparable and gp = x3 − bx2 − 4d x + 4bd − c2 be its resolvent cubic. Let m= |GalF (gp)|. Then

GalF (p)∼=

S4 if m= 6

A4 if m= 3

D4 or C4 if m= 2

V if m= 1

In the case of m= 2, let u be the root of gp that belongs to F . We have GalF (p)∼= C4 if and only if the polynomialsx2 − ux + d and x2 + (b− u) split over L, the splitting field of gp.

10.9 Example. The polynomial p(x) = x4 − 2x − 2 ∈ Q[x] is irreducible by Eisenstein’s criterion. Its resolventcubic is gp(x) = x3+8x −4 and is irreducible over Q. We have D(gp) =−4(83)−27(−4)2 =−155 ·44, which isnot a square in Q. Hence by Theorem 10.4 we have GalQ(gp)∼=S3, i.e. m= 6. Hence by Theorem 10.8 we haveGalQ(p)∼=S4.

Remark. We have seen that α ∈ R is constructible only if the minimal polynomial of α has degree a power of 2.The converse of this is false. For example, let α be a real root of p(x) = x4 − 2x − 2. If E is the splitting field ofp(x) then GalQ(E) ∼= S4. By Theorem 8.9, α is constructible if and only if GalQ(E) is a 2-group. Hence α is notconstructible even though it’s minimal polynomial has degree 4, a power of 2.

10.10 Example. 1. Consider the irreducible polynomial p(x) = x4 − 10x2 + 1 ∈ Q[x]. Its resolvent cubic isgp(x) = x3 + 10x2 − 4x − 40= (x + 10)(x − 2)(x + 2). Hence GalQ(gp) is trivial and so GalQ(p)∼= V .

2. Consider the irreducible polynomial p(x) = x4 + 5x + 5 ∈Q[x]. Its resolvent cubic is gp(x) = x3 − 20x −25 = (x − 5)(x2 + 5x + 5). Hence m = 2. Let L be the splitting field of gp. Since the roots of gp are

5, −5±p

52

, we have L =Q(p

5). Hence Galp(p)∼= C4.

11 Solvability by Radicals

11.1 Cardano’s Formula

For simplicity, we will assume that F is a field of characteristic not 2 or 3. We all know the quadratic formula:

the roots of x2 + bx + c ∈ F[x] are −b±p

b2−4c2

. An expression of this type, involving only +,−,×,÷, and np·

is called a radical. We consider the cubic equation x3 + bx + c = 0 ∈ F[x]. Set x = u+ v, where u and v areindeterminates. We obtain

0= x3 + bx + c

= (u+ v)3 + b(u+ v) + c

= u3 + v3 + (3uv+ b)(u+ v) + c

= u3 + v3 + c

by imposing the condition that uv = −b3

. Letting α= u3 and β = v3 we have α+β =−c and αβ =�

−b3

�3. Hence

α and β are roots of the quadratic

y2 + c y −�

b

3

�3

= 0


Thus by the above formula we have

α,β =−c±

p

c2 + 4(b/3)3

2=−c

2±

r

c2

4+

b3

27

There seems to be 3 choices for each of u and v, but the imposed conditions narrow them down to just 3. Wehave proven

11.1 Theorem. (Tartaglia, del Ferro, Fontana) The solutions fo the cubic equation x3 + bx2 + c = 0 are of theform

α1 =3

È

−c

2+

r

c2

4+

b3

27+

3

È

−c

2−

r

c2

4+

b3

27

α2 = ζ3

3

È

−c

2+

r

c2

4+

b3

27+ ζ2

3

3

È

−c

2−

r

c2

4+

b3

27

α3 = ζ23

3

È

−c

2+

r

c2

4+

b3

27+ ζ3

3

È

−c

2−

r

c2

4+

b3

27

Where the cubic roots are chosen such that

3

È

−c

2+

r

c2

4+

b3

27·

3

È

−c

2−

r

c2

4+

b3

27=−b

3

Consider x4 + bx2 + cx + d ∈ F[x]. Let α1,α2,α3,α4 be the roots. We have seen before the that resolventcubic is defined to be g(x) = x3 − bx2 − 4d x + 4bd − c2 where the roots of g are

u= α1α2 +α3α4

v = α1α3 +α2α4

w = α1α4 +α2α3

Applying the Cardano formula for cubics, we can obtain u, v, w. Notice that

u+ v =−(α1 +α4)2 ←→ α1 +α4 =±

pu+ v

v +w =−(α1 +α2)2 ←→ α1 +α2 =±

pv+w

w+ u=−(α1 +α3)2 ←→ α1 +α3 =±

pw+ u

It appears as though there are 8 choices for the signs. However, we know that

(α1 +α4)(α1 +α2)(α1 +α3) =−c

and this cuts down the choices. Now

(α1 +α4) + (α1 +α2) + (α1 +α3) = 2α1

and we can get similar expressions for the other roots. We have almost proven


11.2 Theorem. (Ferrari) The solutions of the quartic equation x4 + bx2 + cx + d = 0 are of the form

α1 =1

2

�p−u− v+

p−v−w+

p−w− u

�

α2 =1

2

�

−p−u− v−

p−v−w+

p−w− u

�

α3 =1

2

�

−p−u− v+

p−v−w−

p−w− u

�

α4 =1

2

�p−u− v−

p−v−w−

p−w− u

�

where the square roots are chosen such that

(p−u− v)(

p−v−w)(

p−w− u) =−c

11.2 Solvable groups

11.3 Definition. If G is a group and N is a subgroup of G then N is normal if gN g−1 = N for all g ∈ G. We writeN Ã G. A group G is solvable if there is a tower

G = G0 ⊇ G1 ⊇ · · · ⊇ Gm = {1}

where Gi+1 Ã Gi and Gi/Gi+1 is Abelian for i = 0, . . . , m− 1.

11.4 Example. The symmetric group S4 is solvable. Notice that A4 and V are normal subgroups of S4.

S4 ⊇ A4 ⊆ V ⊇ {1}

and S4/A4∼= C2 and A4/V ∼= C3. These quotients are Abelian, so S4 is solvable.

11.5 Theorem. (Second Isomorphism Theorem) If H, N are subgroups of G with N Ã G then

H/H ∩ N ∼= NH/N

11.6 Theorem. (Third Isomorphism Theorem) If G a group and H, N Ã G such that N ⊆ H then H/N Ã G/N and

(G/N)/(H/N)∼= G/H

11.7 Theorem. If G is a solvable group, then every subgroup and every quotient group of G is solvable. Con-versely, if N Ã G and both N and G/N are solvable then G is solvable.

PROOF: Suppose that G is a solvable group with tower

G = G0 ⊇ G1 ⊇ · · · ⊇ Gm = {1}

where Gi+1 Ã Gi and Gi/Gi+1 is Abelian for i = 0, . . . , m− 1.Let H be a subgroup of G. Define Hi = h∩ Gi . Since Gi+1 Ã Gi we have Hi+1 Ã Hi for i = 0, . . . , m− 1 and

H = H0 ⊇ H1 ⊇ · · · ⊇ Hm = {1}

Notice that Hi and Gi+1 are subgroups of Gi and Hi+1 = H ∩Gi+1 = Hi ∩Gi+1. Applying the second isomorphismtheorem to Gi , we have

Hi/Hi+1 = Hi/Hi ∩ Gi+1∼= HiGi+1/Gi+1 ⊆ Gi/Gi+1


Since Gi/Gi+1 is Abelian, so is Hi/Hi+1. It follows that H is solvable.Let N be a normal subgroup of N . We want that G/N is normal. Mulitplying by N , we have a tower

G = G0N ⊇ G1N ⊇ · · · ⊇ GmN = N

taking the quotient givesG/N = G0N/N ⊇ G1N/N ⊇ · · · ⊇ GmN/N = {1}

Since Gi+1 Ã Gi and N Ã G, we have Gi+1N Ã GiN , which implies that Gi+1N/N Ã GiN/N . By the thirdisomorphism theorem, we have

(Gi+1N/N)/(GiN/N)∼= Gi+1N/GiN

Apply the second isomorphism theorem to get

Gi+1N/GiN ∼= Gi/Gi ∩ Gi+1N

Since Gi+1 ⊆ Gi ∩ Gi+1N , there is a natural injection

Gi/Gi ∩ Gi+1N −→ Gi/Gi+1 : g + (Gi ∩ Gi+1N) 7−→ g + Gi+1

Gi/Gi+1 is Abelian, so as is Gi/Gi ∩ Gi+1N . Thus (Gi+1N/N)/(GiN/N) is Abelian and hence G/N is solvable.Let N be a normal subgroup of G and suppose that N and G/N are solvable. Since N is solvable there is a

towerN = N0 ⊇ N1 ⊇ · · · ⊇ Nm = {1}

where Ni+1 Ã Ni and Ni/Ni+1 is Abelian for i = 0, . . . , m− 1. For a subgroup H ⊆ G with N ⊆ H, we denoteH = H/N . Since G/N is solvable, we have a tower

G/N = G0 ⊇ G1 ⊇ · · · ⊇ G r = {1}

where G i+1 Ã G i and G i/G i+1 is Abelian for i = 0, . . . , r − 1. Let σ : G → G/N , H → H/N . For all i = 0, . . . , r,define Gi = σ−1(G i). Since N Ã G and G i+1 Ã G i , we have Gi+1 Ã Gi . Moreover, by the third isomorphismthreorem, Gi/Gi+1

∼= G i/G i+1 is Abelian. It follows that we have the tower

G = G0 ⊇ G1 ⊇ · · · ⊇ Gr = N = N0 ⊇ N1 ⊇ · · · ⊇ Nm = {1}

which shows that G is solvable. �

11.8 Example. Since S2 ⊆S3 ⊆S4, we have that S2 and S3 are solvable.

11.9 Corollary. If G is a finite solvable group then there is a tower

G = G0 ⊇ G1 ⊇ · · · ⊇ Gm = {1}

Gi+1 Ã Gi and Gi/Gi+1 is cyclic of prime order for i = 0, . . . , m− 1.

11.10 Definition. A group G is simple if it is not the trivial group and it has no normal subgroups other than Gand {1}.

The alternating group A5 is simple, hence is not solvable. By Theorem 11.7, we conclude that S5 is notsolvable. Hence for all n≥ 5, since Sn contains a subgroup isomorphic to S5, so Sn is not solvable.

Given a polynomial f (x) ∈ F[x] of degree n, its Galois group Gal( f ) is a subgroup of Sn. We will prove laterthat f (x) has radical solutions if and only if Gal( f ) is solvable. It follows (as had already been proven) that anypolynomial of degree 2, 3, or 4 has radical solutions. Since Sn is not solvable for n ≥ 5, there are no radicalsolutions for a general polynomial of degree n.


11.3 Cyclic Extensions

11.11 Definition. A Galois extension E/F is Abelian/cyclic/solvable if GalF (E) has the corresponding property.

11.12 Lemma. (Dedekind’s Lemma) Let E and F be fields and ψi : F → E be distinct homomorphisms for1≤ i ≤ n. If ci ∈ E and

c1ψ1(α) + · · ·+ cnψn(α) = 0 ∀α ∈ F

then c1 = · · ·= cn = 0.

PROOF: Suppose conversely. Let m≥ 2 be the smallest positive integer such that

c1ψ1(α) + · · ·+ cmψm(α) = 0 ∀α ∈ F

for some c1, . . . , cm ∈ E non-zero. Choose β ∈ F such that ψ1(β) 6=ψ2(β) and ψ1(β) 6= 0. We have

c1ψ1(βα) + · · ·+ cmψm(βα) = 0 ∀α ∈ F

Dividing by ψ1(β) gives

c1ψ1(α) +c2

ψ1(β)ψ2(βα) + · · ·+

cm

ψ1(β)ψm(βα) = 0 ∀α ∈ F

Subtracting this equation from the original equation gives us

c2

�

1−ψ2(β)ψ1(β)

�

ψ2(βα) + · · ·+ cm

�

1−ψm(β)ψ1(β)

�

ψm(βα) = 0 ∀α ∈ F

a contradiction (since not all of these coefficients are zero). �

11.13 Theorem. Let F be a field and n be a positive integer. Suppose that ch (F) = 0 or p, where p - n. Assumethat xn − 1 splits over F .

1. If the Galois extension E/F is cyclic of degree n then E = F(α) for some α ∈ E and αn ∈ F . It follows thatxn −αn is the minimal polynomial of α over F .

2. If E = F(α) and αn ∈ F then E/F is a cyclic extension of degree d, where d|n and αd ∈ F . It follows thatxd −αd is the minimal polynomial of α over F .

PROOF: Let ζn ∈ F be a primitive nth root of unity.

1. Let G = GalF (E) = ⟨ψ⟩ ∼= Cn. Apply Dedekind’s lemma to domain and codomain E, ψi = ψi−1, 1 ≤ i ≤ n,and ci = ζ1−i

n . There exists u ∈ E such that

α := u+ ζ−1n ψ(u) + · · ·+ ζ

−(n−1)n ψn−1(u) 6= 0

We haveψ(α) =ψ(u) + ζ−1

n ψ2(u) + · · ·+ ζ−(n−1)

n ψn(u) = αζn

Since ζn ∈ F it follows that ψi(α) = αζin. Also, ψ(αn) = αn, so αn ∈ EG = F (since ψ generates G).

Therefore α,αζn, . . . ,αζn−1n are roots of xn − αn ∈ F[x]. If p(x) ∈ F[x] is the minimal polynomial of α,

then all of the conjugates of α are also roots of p(x), so we must have p(x) = xn − αn. Moreover, sinceF(α)⊆ E and [F(α) : F] = deg p = n= [E : F] we must have E = F(α).


2. Let p(x) ∈ F[x] be the minimal polynomial of α over F . Since αn ∈ F , α is a root of xn −αn ∈ F[x]. Thusp(x)|xn − αn, and the roots of p(x) are of the form αζi

n for some i and ζn a primitive nth root of unity inF . We have p(0) = ±αdζk

n for some k and d = deg p. Since p(0),ζkn ∈ F , it follows that αd ∈ F , and so α

is a root of xd − αd ∈ F[x]. This polynomial has the same degree as p and is monic, so p(x) = xd − αd .d|n because if n= qd + r for r < d then we have αr = αn−qd = αn(α−d)q ∈ F , a contradiction unless r = 0(since otherwise α would be a root of x r − αr ∈ F[x], contradicting that α has degree d over F). Writen = md, and the roots of p are α,αζm

n , . . . ,αζ(d−1)mn . If ψ ∈ G satisfies ψ(α) = αζm

n , then G = ⟨ψ⟩ is cyclicof order d. �

11.14 Theorem. Let F be a field of characteristic p.

1. If x p − x − a ∈ F[x] is irreducible, then its splitting field E/F is cyclic of degree p.2. Theo converse of (1) is also true, that is, every cyclic extension of F of degree p is the splitting field of some

irreducible polynomial x p − x − a ∈ F[x].

PROOF: Assignment. �

11.4 Radical Extensions

For simplicity, we assume in this section that F is a field of characteristic 0.

11.15 Definition. A finite extension E/F is called a radical extension if there exists a tower of subfields

F = F0 ⊆ F1 ⊆ · · · ⊆ Fk = E

and αi ∈ Fi , i = 1, . . . , k, such that Fi = Fi−1(αi) and αdii ∈ Fi−1 for some integer di ≥ 1.

Notice in particular that every constructible extension is a radical extension. In this case, di = 1 or 2 for eachi.

11.16 Lemma. If E/F is a radical extension, then its normal closure N/F is also a radical extension.

PROOF: Since ch (F) = 0 and E/F is a finite extension, by Theorem 4.14, E/F is a simple extension. WriteE = F(α). Since E/F is a radical extension, there is a tower of subfields

F = F0 ⊆ F1 ⊆ · · · ⊆ Fk = E

and αi ∈ Fi , i = 1, . . . , k, such that Fi = Fi−1(αi) and αdii ∈ Fi−1 for some integer di ≥ 1. Let p(x) ∈ F[x] be

the minimal polynomial of α and N/E a splitting field of p(x) over E. Then N/F is a splitting field of p(x)over F and is a normal closure of E/F . Let α = α1, . . . ,αn be the roots of p in N . There is a field isomorphismσi : F(α)→ F(αi) such that σi |F = id and α 7→ αi for i = 2, . . . , n. Since N can be viewed as a splitting field ofp over F(α) and F(αi) respectively, there is ψi : N → N which extends σi . Hence ψi ∈ GalF (N) and ψi(α) = αi .We have

F = F0 ⊆ F1 ⊆ · · · ⊆ Fk = E = F(α) = F(α1)ψ2(F0)⊆F(α1)ψ2(F1)⊆ · · · ⊆ F(α1)ψ2(Fk) = F(α1,α2)⊆ · · · ⊆ F(α1, . . . ,αn) = N

Notice that since Fi = Fi−1(βi) and β dii ∈ Fi−1 for some β ∈ Fi \ Fi−1, we have

F(α1, . . . ,α j−1)ψ j(Fi) = F(α1, . . . ,α j−1)ψ j(Fi−1(βi)) = F(α1, . . . ,α j−1)ψ j(Fi−1)ψ j(βi)

and (ψ j(βi))di =ψ j(βdii ) ∈ψ j(Fi−1). This shows that N/F is a radical extension. �


11.5 Solving polynomials by Radicals

11.17 Definition. Let f (x) ∈ F . We say that f is solvable by radicals if there is a radical extension E/F such thatf splits over E. It follows that the equation f (x) = 0 has radical solutions.

11.18 Lemma. If K , L are intermediate fields of E/F with K/F a finite Galois extension, then K L is a finite Galoisextension over L and GalL(K L) is isomorphic to a subgroup of GalF (K).

PROOF: Suppose that K is the splitting field of f (x) ∈ F[x] over F . Then K L is a splitting field of f (x) over L.Hence K L/L is a finite Galois extension. Consider

Γ : GalL(K L)→ GalF (K) :ψ 7→ψ|K

This map is well defined since K is normal. Moreover, if ψ|K = idK then ψ is trivial on K and L, so must be equalto idK L . Thus Γ is an injection. Therefore GalL(K L) is isomorphic to a subgroup of GalF (K). �

11.19 Theorem. Let F be a field of characteristic zero and let f (x) ∈ F[x] with f 6= 0. Then f (x) is solvable byradicals if and only if its Galois group Gal( f ) is a solvable group.

PROOF: Assume that G = Gal( f ) is solvable. Let E/F be a splitting field of f over F . Let n = |G| and L/E be asplitting field of xn − 1 over E (so that L = E(ζn) for some primative nth root of unity). Let K = F(ζn) be thesplitting field of xn − 1 over F . We have L = KE. Since E/F is a finite Galois extension, by the previous lemmaL/K is a finite Galois extension and H = GalK(L) is isomorphic to a subgroup of G. Hence H is solvable since Gis solvable. Write

H = H0 ⊇ H1 ⊇ · · · ⊇ Hm = {1}

where Hi+1 Ã Hi and Hi/Hi+1∼= Cdi

(cyclic of order di). Let Ki = H∗i = LHi for i = 0, . . . , m. Then GalKi(L) ∼= Hi ,

so we have a tower of fields

F ⊆ F(ζn) = K = K0 ⊆ K1 ⊆ · · · ⊆ Km = L = E(ζn)

Since Hi+1 Ã Hi , Ki+1/Ki is Galois and the Galois group is isomorphic to Hi/Hi+1∼= Cdi

. By Theorem 11.13 there

is αi+1 ∈ Ki+1 such that Ki+1 = Ki(αi+1) and αdi+1i+1 ∈ Ki . It follows that L/F is a radical extension. Since all the

roots of f are in E and hence in L, we conclude that f is solvable by radicals.Suppose f (x) is solvable by radical, so that f splits over some extension E/F with

F = F0 ⊆ F1 ⊆ · · · ⊆ Fm = E

where Fi = Fi−1(αi) and αdii ∈ Fi−1. By lemma 11.16 we may assume that E/F is Galois. Let n =

∏mi=1 di and let

L/E be the splitting field of xn−1 over E. Set K = F(ζn) and we have L = E(ζn) = KE. Define Ki = Fi(ζn) = KFi ,so that Ki = Ki−1(αi) and αdi

i ∈ Fi−1 ⊆ Ki−1. Since αdii ∈ Ki−1, Ki is a splitting field of xdi − αdi over Ki−1. Then

Ki/Ki−1 is cyclic, and so we have

F ⊆ F(ζn) = K ⊆ K1 ⊆ · · · ⊆ Km = Fm(ζn) = L

Notice that L is a splitting field of f (x)(xn − 1) over F , hence L/F is Galois. Each Ki is an intermediate field ofL/F , so Ki is Galois. Applying the Galois correspondence we have

G = GalF (L)⊇ GalK(L)⊇ GalK1(L)⊇ · · · ⊇ GalKm

(L) = {1}

For each σ ∈ GalKi(L), ψ ∈ GalKi+1

(L), we have

σψσ−1�

�

�

Ki+1

= idKi+1


Hence GalKi+1(L) Ã GalKi

(L), and moreover we have GalKi(L)/GalKi+1

(L) ∼= GalKi(Ki+1), which is cyclic (and

hence Abelian). Also, GalF (L)GalK0(L) ∼= GalF (F(ζn)), which is also Abelian. Therefore GalF (L) is solvable.

Since GalF (E)∼= GalF (L)/GalE(L), Gal( f ) = GalF (E) is solvable as well. �

11.20 Proposition. Let f (x) ∈ Q[x] be irreducible of prime degree p. If f (x) contains precisely two non-realroots in C then Gal( f )∼=Sp.

PROOF: Recall that the symmetric group Sn is generated by (1 2) and (1 2 . . . n). Hence to show that Gal( f ) isisomorphic to Sp it suffices to find a 2-cycle and a p-cycle. Since f is irreducible with degree p, p divides theorder of Gal( f ). By Cauchy’s Theorem there is an element of Gal( f ) of order p – a p-cycle. Complex conjugationwill juxtapose the non-real roots of f and leave all other (real) roots fixed. Hence complex conjugation is a2-cycle in Gal( f ). �

Consider f (x) = x5+2x3−24x−2 ∈Q[x], which is irreducible by Eisensteins’s criterion. Since f (−1) = 19,f (1) = −23, limx→∞ f (x) =∞, and limx→−∞ f (x) = −∞, f has at least three real roots. Let a1, . . . , a5 be theroots of f (x). We have a1 + · · ·+ a5 = 0 and

∑

i< j aia j = 2. From the first sum,

0=

5∑

i=1

ai

!2

=5∑

i=1

a2i + 2

∑

i< j

aia j

so∑5

i=1 a2i = −4, and not all of the roots of f can be real. Therefore f has exactly three real roots and two

non-real roots. By the above proposition, Gal( f )∼=S5. Since S5 is not solvable, the equation

x5 + 2x3 − 24x − 2= 0

does not have radical solutions.

11.21 Theorem. (Abel) The general polynomial equation f (x) = 0 with deg f ≥ 5 is not solvable by radicalsolutions. In other words, we have radical solutions for f (x) = 0 if and only if f (x)≤ 4.

11.6 Probabilistic Galois Theory

(Extra Section)Indeed, for almost all f (x) ∈ Z[x] with degree n, Gal( f ) ∼= Sn. Since Sn is not solvable for n ≥ 5, by

Theorem 11.19, f is not solvable by radicals for almost all f (x) ∈ Z[x] of degree n ≥ 5. The study of “density”of polynomials f (x) of degree n with Gal( f ) isomorphic to certain subgroups of Sn is called probabilistic Galoistheory.

Notation. Let f (x) and g(x) be two functions. If there exists a constant C such that | f (x)| ≤ C g(x) when x issufficiently large, we write f (x)� g(x) or f (x) = O(g(x)).

For example, since limx→∞xn−1(log x)r

xn = 0 we have xn−1(log x)r � xn for any r.Consider En(N) = #{ f (x) = xn + an−1 xn−1 + · · ·+ a0 ∈ Z[x] | |ai | ≤ N , Gal( f ) $ Sn}. Notice that if a0 = 0

then f (x) = x(xn−1 + an−1 xn−2 + · · · + a1). Since x = 0 ∈ Q, Gal( f ) = Gal( f /x) ⊆ Sn−1 $ Sn. For eachan−1, . . . , a1 with |ai | ≤ N there are 2N + 1 choices for each of them, so there are (2N + 1)n−1 polnomials witha0 = 0 and Galois group a proper subgroup of Sn. If follows that

En(N)≥ (2N + 1)n−1 = 2n−1N n−1 +O(N n−2)>> N n−1

11.22 Conjecture. (van der Waerden) En(N)� N n−1.


This question remains open today. The best result known for this problem is due to Gallagher, who provesthat En(N) � N n− 1

2 (log N) by the large sieve method. In any case (i.e. whether the conjecture is true or not),since there are (2N + 1)n many polynomials of the form f (x) = xn + an−1 xn−1 + · · ·+ a0 ∈ Z[x] with |ai | ≤ N ,we have

#{ f (x) = xn + an−1 xn−1 + · · ·+ a0 ∈ Z[x] | |ai | ≤ N , Gal( f )∼=Sn}= (2N + 1)n +O(N n− 12 (log N))

Since

limN→∞

(2N + 1)n +O(N n− 12 (log N))

(2N + 1)n= 1

we conclude that for almost all (i.e. with probability 1) f (x) ∈ Z[x] of degree n, Gal( f )∼=Sn.Consider the special case of the Galois group of cubics. Define

E3(N) = { f (x) = x3 + bx2 + cx + d ∈ Z[x] | H( f )≤ N , Gal( f )$S3}

where H( f ) = height of f =max{|b|, |c|, |d|}. Our goal is prove that E3(N)� N2+ε.

11.23 Theorem. (van der Waerden)

#{ f (x) = x3 + bx2 + cx + d ∈ Z[x] | H( f )≤ N , f is reducible} � N2

Hence, to prove E3(N)� N2+ε it suffices to consider irreducible polynomials. Let f (x) = x3 + bx2 + cx + dbe irreducible. If Gal( f ) $ S3, then Gal( f ) ∼= A3. We recall that the discriminant D( f ) is b2c2 − 4c3 − 4b3d −27d2 + 18bcd. By Theorem 10.4, Gal( f ) ∼= A3 ⇐⇒ D( f ) = z2 for some z ∈ Z. Hence, to compute E3(N) ifsuffices to compute the number of z ∈ Z such that b2c2 − 4c3 − 4b3d − 27d2 + 18bcd = z2. That is,

27(d)2 + (4b3 − 18bc)d + z2 + (4c3 − b2c2) = 0 (1)

11.24 Theorem. Suppose that Q(x , y) = ax2+ bx y+c y2+d x+e y+ f is a quadratic polynomial with coefficientsin Z. Assume that the absolute values of all coefficients of Q(x , y) are bounded by N . Then

#{(x , y) ∈ Z2 |Q(x , y) = 0, |x |, |y| ≤ M} � (MN)ε

Consider equation (1). Since |d| ≤ N and |z| ≤ N2, for fixed b, c, the number of choices of d and z is� (NN2)ε � N ε. It follows that E3(N)� N2+ε.

fields and galois theory fall 2004 professor yu-ru liu

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