II Galois Theory (Full)

Part II — Galois Theory

Based on lectures by C. Birkar

Notes taken by Dexter Chua

Michaelmas 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Groups, Rings and Modules is essential

Field extensions, tower law, algebraic extensions; irreducible polynomials and relation

with simple algebraic extensions. Finite multiplicative subgroups of a field are cyclic.

Existence and uniqueness of splitting fields. [6]

Existence and uniqueness of algebraic closure. [1]

Separability. Theorem of primitive element. Trace and norm. [3]

Normal and Galois extensions, automorphic groups. Fundamental theorem of Galois

theory. [3]

Galois theory of finite fields. Reduction mod p. [2]

Cyclotomic polynomials, Kummer theory, cyclic extensions. Symmetric functions.

Galois theory of cubics and quartics. [4]

Solubility by radicals. Insolubility of general quintic equations and other classical

problems. [3]

Artin’s theorem on the subfield fixed by a finite group of automorphisms. Polynomial

invariants of a finite group; examples. [2]

Contents

0 Introduction

1 Solving equations

2 Field extensions

2.1 Field extensions

2.2 Ruler and compass constructions

2.3 K-homomorphisms and the Galois Group

2.4 Splitting fields

2.5 Algebraic closures

2.6 Separable extensions

2.7 Normal extensions

2.8 The fundamental theorem of Galois theory

2.9 Finite fields

3 Solutions to polynomial equations

3.1 Cyclotomic extensions

3.2 Kummer extensions

3.3 Radical extensions

3.4 Solubility of groups, extensions and polynomials

3.5 Insolubility of general equations of degree 5 or more

4 Computational techniques

4.1 Reduction mod p

4.2 Trace, norm and discriminant

0 Introduction

The most famous result of Galois theory is that there is no general solution

to polynomial equations of degree 5 or above in terms of radicals. However,

this result was, in fact, proven before Galois theory existed, and goes under the

name of the Abel–Ruffini theorem. What Galois theory does provides is a way

to decide whether a given polynomial has a solution in terms of radicals, as well

as a nice way to prove this result.

However, Galois theory is more than equation solving. In fact, the funda-

mental theorem of Galois theory, which is obviously an important theorem in

Galois theory, has completely nothing to do with equation solving. Instead, it is

about group theory.

In modern days, Galois theory is often said to be the study of field extensions.

The idea is that we have a field

, and then add more elements to get a field

When we want to study solutions to polynomial equations, what we add is the

roots of the polynomials. We then study the properties of this field extension,

and in some cases, show that this field extension cannot be obtained by just

adding radicals.

For certain “nice” field extensions

K ⊆ L

, we can assign to it the Galois group

Gal

(

L/K

). In general, given any group

, we can find subgroups of

. On the

other hand, given a field extension

K ⊆ L

, we can try to find some intermediate

field

that can be fitted into

K ⊆ F ⊆ L

. The key idea of Galois theory is

that these two processes are closely related — we can establish a one-to-one

correspondence between the subgroups of G and the intermediate fields F .

Moreover, many properties of (intermediate) field extensions correspond to

analogous ideas in group theory. For example, we have the notion of normal

subgroups, and hence there is an analogous notion of normal extensions. Similarly,

we have soluble extensions (i.e. extensions that can be obtained by adding

radicals), and these correspond to “soluble groups”. In Galois theory, we will

study how group-theoretic notions and field-theoretic notions interact.

Nowadays, Galois theory is an important field in mathematics, and finds its

applications in number theory, algebraic geometry and even cryptography.

1 Solving equations

Galois theory grew of the desire to solve equations. In particular, to solve

polynomial equations. To begin with, we will come up with general solutions to

polynomial equations of up to degree 4. However, this is the best we can do, as

we will later show in the course — there is no general solution to polynomial

equations of degree 5 or above.

Before we start, we will define some notations that we will frequently use.

is a ring, then

[

] is the polynomial ring over

in the variable

Usually, we take

and consider polynomials

(

)

∈ Q

[

]. The objective

is then to find roots to the equation

(

) = 0. Often, we want to restrict our

search domain. For example, we might ask if there is a root in

. We will thus

use Root

(X) to denote the set of all roots of f in X.

Linear equations

Suppose that

a ∈ Q

[

] (with

a ∈ Q

). This is easy to solve — we have

Root

(Q) = {−a}.

Quadratic equations

Consider a simple quadratic

+ 1

∈ Q

[

]. Then

Root

(

) =

∅

since the

square of all rationals are positive. However, in the complex plane, we have

Root

√

−1, −

√

−1}.

In general, let

b ∈ Q

[

]. Then as we all know, the roots are

given by

Root

(

−a ±

√

− 4b

)

Cubic equations

Let f = t

+ c ∈ Q[t]. The roots are then

Root

√

−c, µ

√

−c, µ

√

−c},

where

−1+

√

−3

is the 3rd root of unity. Note that

is defined by the

equation µ

− 1 = 0, and satisfies µ

+ µ + 1 = 0.

In general, let

c ∈ Q

[

], and let

Root

(

) =

{α

, α

}

not necessarily distinct.

Our objective is to solve

= 0. Before doing so, we have to make it explicit

what we mean by “solving” the equation. As in solving the quadratic, we want

to express the roots α

, α

and α

in terms of “radicals” involving a, b and c.

Unlike the quadratic case, there is no straightforward means of coming up

with a general formula. The result we currently have is the result of many

many years of hard work, and the substitutions we make seemingly come out of

nowhere. However, after a lot of magic, we will indeed come up with a general

formula for it.

We first simplify our polynomial by assuming

= 0. Given any polynomial

, we know

is the negative of the sum of the roots. So we

can increase each root by

so that the coefficient of

vanishes. So we perform

the change of variables

t 7→ t −

, and get rid of the coefficient of

. So we can

assume a = 0.

Let µ be as above. Define

β = α

+ µα

+ µ

γ = α

+ µ

+ µα

These are the Lagrange resolvers. We obtain

βγ = α

+ α

+ (µ + µ

)(α

+ α

)

Since µ

+ µ + 1 = 0, we have µ

+ µ = −1. So we can simplify to obtain

= (α

+ α

)

− 3(α

+ α

)

We have α

+ α

= −a = 0, while b = α

+ α

. So

= −3b

Cubing, we obtain

= −27b

On the other hand, recalling again that α

+ α

= 0, we have

+ γ

= (α

+ µα

+ µ

)

+ (α

+ µ

+ µα

)

+ (α

+ α

)

= 3(α

+ α

) + 18α

We have

−c

, and since

bα

= 0 for all

, summing gives

+ α

+ 3c = 0. So

= −27c

Hence, we obtain

(t − β

)(t − γ

) = t

+ 27ct − 27b

We already know how to solve this equation using the quadratic formula. We

obtain

{β

, γ

} =

(

−27c ±

(27c)

+ 4 × 27b

)

We now have

and

in terms of radicals. So we can find

and

in terms of

radicals. Finally, we can solve for α

using

0 = α

+ α

β = α

+ µα

+ µ

γ = α

+ µ

+ µα

In particular, we obtain

(β + γ)

(µ

β + µγ)

(µβ + µ

γ)

So we can solve a cubic in terms of radicals.

This was a lot of magic involved, and indeed this was discovered through a

lot of hard work throughout many many years. This is also not a very helpful

result since we have no idea where these substitutions came from and why they

intuitively work.

Quartic equations

Assume

d ∈ Q

[

]. Let

Root

(

) =

{α

, α

}

Can we express all these in terms of radicals? Again the answer is yes, but the

procedure is much more complicated.

We can perform a similar change of variable to assume a = 0. So α

+ α

= 0.

This time, define

β = α

+ α

γ = α

+ α

λ = α

+ α

Doing some calculations, we see that

= −(α

+ α

)(α

+ α

)

= −(α

+ α

)(α

+ α

)

= −(α

+ α

)(α

+ α

)

Now consider

g = (t − β

)(t − γ

)(t − λ

)

= t

+ 2bt

+ (b

− 4d)t − c

This we know how to solve, and so we are done.

Quintics and above

So far so good. But how about polynomials of higher degrees? In general, let

f ∈ Q

[

]. Can we write down all the roots of

in terms of radicals? We know

that the answer is yes if deg f ≤ 4.

Unfortunately, for

deg f ≥

5, the answer is no. Of course, this “no” means

no in general. For example,

= (

t −

1)(

t −

···

(

t −

∈ Q

[

] has the obvious

roots in terms of radicals.

There isn’t an easy proof of this result. The general idea is to first associate

a field extension

F ⊇ Q

for our polynomial

. This field

will be obtained

by adding all roots of

. Then we associate a Galois group

to this field

extension. We will then prove a theorem that says

has a solution in terms

of radicals if and only if the Galois group is “soluble”, where “soluble” has a

specific algebraic definition in group theory we will explore later. Finally, we

find specific polynomials whose Galois group is not soluble.

2 Field extensions

After all that (hopefully) fun introduction and motivation, we will now start

Galois theory in a more abstract way. The modern approach is to describe these

in terms of field extensions.

2.1 Field extensions

Definition (Field extension). A field extension is an inclusion of a field

K ⊆ L

where

inherits the algebraic operations from

. We also write this as

L/K

Alternatively, we can define this by a injective homomorphism

K → L

. We say

L is an extension of K, and K is a subfield of L.

Example.

(i) R/Q is a field extension.

(ii) C/Q is a field extension.

(iii) Q(

√

2) = {a + b

√

2 : a, b ∈ Q} ⊆ R is a field extension over Q.

Given a field extension

L/K

, we want to quantify how much “bigger”

is compared to

. For example, to get from

, we need to add a lot of

elements (since

is countable and

is uncountable). On the other hand, to get

from R to C, we just need to add a single element

√

−1.

To do so, we can consider

as a vector space over

. We know that

already comes with an additive abelian group structure, and we can define scalar

multiplication by simply multiplying: if

a ∈ K, α ∈ L

, then

a · α

is defined as

multiplication in L.

Definition (Degree of field extension). The degree of

over

is [

] is the

dimension of

as a vector space over

. The extension is finite if the degree is

finite.

In this course, we are mostly concerned with finite extensions.

Example.

(i)

Consider

C/R

. This is a finite extension with degree [

] = 2 since we

have a basis of {1, i}.

(ii) The extension Q(

√

2)/Q has degree 2 since we have a basis of {1,

√

2}.

(iii) The extension R/Q is not finite.

We are going to use the following result a lot:

Theorem (Tower Law). Let F/L/K be field extensions. Then

[F : K] = [F : L][L : K]

Proof.

Assume [

] and [

] are finite. Let

{α

, ··· , α

}

be a basis for

over

, and

{β

, ··· , β

}

be a basis for

over

. Pick

γ ∈ F

. Then we can

write

γ =

, b

∈ L.

For each b

, we can write as

, a

∈ K.

So we can write

γ =









i,j

{α

}

i,j

spans

over

. To show that this is a basis, we have to show

that they are linearly independent. Consider the case where

= 0. Then we

must have

= 0 since

{β

}

is a basis of

over

. Hence each

= 0 since

{α

} is a basis of L over K.

This implies that T is a basis of F over K. So

[F : K] = |T | = nm = [F : L][L : K].

Finally, if [

] =

∞

or [

] =

∞

, then clearly [

] =

∞

as well. So

equality holds as well.

Recall that in IA Numbers and Sets, we defined a real number

to be

algebraic if it is a root of some polynomial in integer (or rational) coefficients.

We can do this for general field (extensions) as well.

Definition (Algebraic number). Let

L/K

be a field extension,

α ∈ L

. We

define

= {f ∈ K[t] : f (α) = 0} ⊆ K[t]

This is the set of polynomials for which

is a root. It is easy to show that

an ideal, since it is the kernel of the ring homomorphism

[

]

→ L

g 7→ g

(

We say

is algebraic over



= 0. Otherwise,

is transcendental over

We say L is algebraic over K if every element of L is algebraic.

Example.

(i)

√

is algebraic over

because

(

√

) = 0, where

−

7. In general,

any number written with radicals is algebraic over Q.

(ii) π is not algebraic over Q.

These are rather simple examples, and the following lemma will provide us a

way of generating much more examples.

Lemma. Let L/K be a finite extension. Then L is algebraic over K.

Proof.

Let

= [

], and let

α ∈ L

. Then 1

, α, α

, ··· , α

are linearly

dependent over

(since there are

+ 1 elements). So there exists some

∈ K

(not all zero) such that

+ a

n−1

+ ··· + a

α + a

= 0.

So we have a non-trivial polynomial that vanishes at

. So

is algebraic over

Since α was arbitrary, L itself is algebraic.

L/K

is a field extension and

α ∈ L

is algebraic, then by definition, there is

some polynomial

such that

(

) = 0. It is a natural question to ask if there is

a “smallest” polynomial that does this job. Obviously we can find a polynomial

of smallest degree (by the well-ordering principle of the natural numbers), but

we can get something even stronger.

Since

is a field,

[

] is a PID (principal ideal domain). This, by definition,

implies we can find some (monic)

∈ K

[

] such that

⟨P

⟩

. In other words,

every element of I

is just a multiple of P

Definition (Minimal polynomial). Let

L/K

be a field extension,

α ∈ L

. The

minimal polynomial of

over

is a monic polynomial

such that

⟨P

⟩

Example.

(i) Consider R/Q, α =

√

2. Then the minimal polynomial is P

= t

− 2.

(ii) Consider C/R, α =

√

2. Then the minimal polynomial is P

= t −

√

It should be intuitively obvious that by virtue of being “minimal”, the

minimal polynomial is irreducible.

Proposition. Let

L/K

be a field extension,

α ∈ L

algebraic over

, and

the minimal polynomial. Then P

is irreducible in K[t].

Proof.

Assume that

[

]. So 0 =

(

) =

(

)

(

). So

(

) = 0

(

) = 0. Say

(

) = 0. So

Q ∈ I

. So

is a multiple of

. However, we

also know that

is a multiple of

. This is possible only if

is a unit in

K[t], i.e. R ∈ K. So P

is irreducible.

It should also be clear that if

is irreducible and

(

) = 0, then

is the

minimal polynomial. Often, it is the irreducibility of P

that is important.

Apart from the minimal polynomial, we can also ask for the minimal field

containing α.

Definition (Field generated by

). Let

L/K

be a field extension,

α ∈ L

. We

define

(

) to be the smallest subfield of

containing

and

. We call

(

)

the field generated by α over K.

This definition by itself is rather abstract and not very helpful. Intuitively,

(

) is what we get when we add

, plus all the extra elements needed to

make

(

) a field (i.e. closed under addition, multiplication and inverse). We

can express this idea more formally by the following result:

Theorem. Let L/K a field extension, α ∈ L algebraic. Then

(i) K

(

) is the image of the (ring) homomorphism

[

]

→ L

defined by

f 7→ f (α).

(ii) [K(α) : K] = deg P

, where P

is the minimal polynomial of α over K.

Note that the kernel of the homomorphism

is (almost) by definition the

ideal ⟨P

⟩. So this theorem tells us

K[t]

⟨P

⟩

∼

K(α).

Proof.

(i)

Let

be the image of

. The first step is to show that

is indeed a field.

Since

is the image of a ring homomorphism, we know

is a subring of

L. Given β ∈ F non-zero, we have to find an inverse.

By definition,

(

) for some

f ∈ K

[

]. The idea is to use B´ezout’s

identity. Since

β 

= 0,

(

)



= 0. So

f ∈ I

⟨P

⟩

. So

∤ f

[

]. Since

is irreducible,

and

are coprime. Then there exists some

g, h ∈ K

[

]

such that

= 1. So

(

)

(

) =

(

)

(

) +

(

)

(

) = 1. So

βg(α) = 1. So β has an inverse. So F is a field.

From the definition of

, we have

K ⊆ F

and

α ∈ F

, using the constant

polynomials f = c ∈ K and the identity f = t.

Now, if

K ⊆ G ⊆ L

and

α ∈ G

, then

contains all the polynomial

expressions of α. Hence F ⊆ G. So K(α) = F .

(ii)

Let

deg P

. We show that

{

, α, α

, ··· , α

n−1

}

is a basis for

(

)

over K.

First note that since deg P

= n, we can write

n−1

i=0

So any other higher powers are also linear combinations of the

s (by

induction). This means that

(

) is spanned by 1

, ··· , α

n−1

as a

vector

space.

It remains to show that

{

, ··· , α

n−1

}

is linearly independent. Assume

not. Then for some b

, we have

n−1

i=0

= 0.

Let

. Then

(

) = 0. So

f ∈ I

⟨P

⟩

. However,

deg f <

deg P

. So we must have

= 0. So all

= 0. So

{

, ··· , α

n−1

}

is a basis

for K(α) over K. So [K(α) : K] = n.

Corollary. Let

L/K

be a field extension,

α ∈ L

. Then

is algebraic over

and only if K(α)/K is a finite extension.

Proof.

is algebraic, then [

(

) :

] =

deg P

< ∞

by above. So the

extension is finite.

K ⊆ K

(

) is a finite extension, then by previous lemma, the entire

(

)

is algebraic over K. So α is algebraic over K.

We can extend this definition to allow more elements in the generating set.

Definition (Field generated by elements). Let

L/K

be a field extension,

, ··· , α

⊆ L

. We define

(

, ··· , α

) to be the smallest subfield of

containing K and α

, ··· , α

We call K(α

, ··· , α

) the field generated by α

, ··· , α

over K.

And we can prove some similar results.

Theorem. Suppose that L/K is a field extension.

(i)

, ··· , α

∈ L

are algebraic over

, then

(

, ··· , α

)

is a finite

extension.

(ii)

If we have field extensions

L/F/K

and

F/K

is a finite extension, then

F = K(α

, ··· , α

) for some α

, ··· , α

∈ L.

Proof.

(i)

We prove this by induction. Since

is algebraic over

K ⊆ K

(

) is a

finite extension.

For 1

≤ i < n

i+1

is algebraic over

. So

i+1

is also algebraic

over

(

, ··· , α

). So

(

, ··· , α

)

⊆ K

(

, ··· , α

)(

i+1

) is a finite

extension. But

(

, ··· , α

)(

i+1

) =

(

, ··· , α

i+1

). By the tower law,

K ⊆ K(α

, ··· , α

i+1

) is a finite extension.

(ii)

Since

is a finite dimensional vector space over

, we can take a basis

{α

, ··· , α

}

over

. Then it should be clear that

(

, ··· , α

When studying polynomials, the following result from IB Groups, Rings and

Modules is often helpful:

Proposition (Eisenstein’s criterion). Let

···

∈ Z

[

Assume that there is some prime number p such that

(i) p | a

for all i < n.

(ii) p ∤ a

(iii) p

∤ a

Then f is irreducible in Q[t].

Example. Consider the field extensions

Q ⊆ Q(

√

2) ⊆ Q(

√

2) ⊆ R,

Q ⊆ Q(

√

2) ⊆ Q(

√

2) ⊆ R.

We have [Q(

√

2) : Q] = 2 since {1,

√

2} is a basis of Q(

√

2) over Q.

How about [

(

√

) :

]? By the Eisenstein criterion, we know that

−

2 is

irreducible in

[

]. So the minimal polynomial of

√

over

−

2 which has

degree 3. So [Q(

√

2) : Q] = 3.

These results immediately tells that

√

2 ∈ Q

(

√

). Otherwise, this entails

that Q(

√

2) ⊆ Q(

√

2). Then the tower law says that

[Q(

√

2) : Q] = [Q(

√

2) : Q(

√

2)][Q(

√

2) : Q].

In particular, plugging the numbers in entails that that 3 is a factor of 2, which

is clearly nonsense. Similarly,

√

2 ∈ Q(

√

2).

How about the inclusion

(

√

)

⊆ Q

(

√

)? We now show that the

minimal polynomial P

√

2 over Q(

√

2) is t

− 2.

Suppose not. Then

−

2 is reducible, with the real

√

as one of its factors.

Let t

− 2 = P

√

· R for some non-unit polynomial R.

We know that

√

does not have degree 3 (or else it would be

−

2), and

not degree 1, since a degree 1 polynomial has a root. So it has degree 2. So

has degree 1. Then

has a root, i.e.

(

) = 0 for some

β ∈ Q

(

√

). So

− 2 = 0. Hence [Q(β) : Q] = 3. Again, by the tower law, we have

[Q(

√

2) : Q] = [Q(

√

2) : Q(β)][Q(β) : Q].

Again, this is nonsense since it entails that 3 is a factor of 2. So the minimal

polynomial is indeed t

− 2. So [Q(

√

2) : Q] = 6 by the tower law.

Alternatively, we can obtain this result by noting that the tower law on

Q ⊆ Q

(

√

)

⊆ Q

(

√

) and

Q ⊆ Q

(

√

)

⊆ Q

(

√

) entails that 2 and 3 are

both factors of [

(

√

) :

]. So it is at least 6. Then since

−

∈ Q

(

√

)[

]

has

√

2 as a root, the degree is at most 6. So it is indeed 6.

2.2 Ruler and compass constructions

Before we develop our theory further, we first look into a rather unexpected

application of field extensions. We are going to look at some classic problems in

geometry, and solve them using what we’ve learnt so far. In particular, we want

to show that certain things cannot be constructed using a compass and a ruler

(as usual, we assume the ruler does not have markings on it).

It is often easy to prove that certain things are constructible — just exhibit

an explicit construction of it. However, it is much more difficult to show that

things are not constructible. Two classical examples are

(i)

Doubling the cube: Given a cube, can we construct the side of another

cube whose volume is double the volume of the original cube?

(ii)

Trisecting an angle: Given an angle, can we divide the angle into three

equal angles?

The idea here is to associate with each possible construction a field extension,

and then prove certain results about how these field extensions should behave.

We then show that if we could, say, double the cube, then this construction

would inevitable break some of the properties it should have.

Firstly, we want to formulate our problem in a more convenient way. In

particular, we will view the plane as

, and describe lines and circles by

equations. We also want to describe “compass and ruler” constructions in a

more solid way.

Definition (Constructible points). Let

S ⊆ R

be a set of (usually finite) points

in the plane.

A “ruler” allows us to do the following: if

P, Q ∈ S

, then we can draw the

line passing through P and Q.

A “compass” allows us to do the following: if

P, Q, Q

′

∈ S

, then we can draw

the circle with center at P and radius of length QQ

′

Any point

R ∈ R

is 1-step constructible from

belongs to the in-

tersection of two distinct lines or circles constructed from

using rulers and

compasses.

A point

R ∈ R

is constructible from

if there is some

, ··· , R

R ∈ R

such that R

i+1

is 1-step constructible from S ∪ {R

, ··· , R

} for each i.

Example. Let

{

}

. What can we construct? It should be easy

to see that (

0) for all

n ∈ Z

are all constructible from

. In fact, we can show

that all points of the form (m, n) ∈ Z are constructible from S.

(0, 0) (1, 0)

Definition (Field of S). Let S ⊆ R

be finite. Define the field of S by

Q(S) = Q({coordinates of points in S}) ⊆ R,

where we put in the

coordinate and

coordinate separately into the generating

set.

For example, if S = {(

√

3)}, then Q(S) = Q(

√

3).

The key theorem we will use to prove our results is

Theorem. Let S ⊆ R

be finite. Then

(i) If R is 1-step constructible from S, then [Q(S ∪ {R}) : Q(S)] = 1 or 2.

(ii)

T ⊆ R

is finite,

S ⊆ T

, and the points in

are constructible from

Then [Q(S ∪ T ) : Q(S)] = 2

for some k (where k can be 0).

Proof.

By assumption, there are distinct lines or circles

C, C

′

constructed from

using ruler and compass, such that

R ∈ C ∩ C

′

. By elementary geometry,

and C

′

can be given by the equations

C : a(x

+ y

) + bx + cy + d = 0,

′

: a

′

+ y

) + b

′

x + c

′

y + d

′

= 0.

where

a, b, c, d, a

′

, b

′

, c

′

, d

′

∈ Q

(

). In particular, if we have a line, then we can

take a = 0.

Let

= (

, r

). If

′

= 0 (i.e.

and

′

are lines), then solving the two

linear equations gives r

, r

∈ Q(S). So [Q(S ∪ {R}) : Q(S)] = 1.

So we can now assume wlog that a = 0. We let

p = a

′

b − ab

′

, q = a

′

c − ac

′

,  = a

′

d − ad

′

which are the coefficients when we perform

′

×C −a ×C

′

. Then by assumption,

p 

= 0 or

q 

= 0. Otherwise,

and

′

would be the same curve. wlog

p 

= 0. Then

since (r

, r

) satisfy both equations of C and C

′

, they satisfy

px + qy +  = 0.

In other words, pr

+ qr

+  = 0. This tells us that

= −

+ 

. (∗)

If we put

, r

into the equations of

and

′

and use (

∗

), we get an equation

of the form

αr

+ βr

+ γ = 0,

where

α, β, γ ∈ Q

(

). So we can find

(and hence

using linear relations)

using only a single radical of degree 2. So

[Q(S ∪ {R}) : Q(S)] = [Q(S)(r

) : Q(S)] = 1 or 2,

since the minimal polynomial of r

over Q(S) has degree 1 or 2.

Then (ii) follows directly from induction, using the tower law.

Corollary. It is impossible to “double the cube”.

Proof.

Consider the cube with unit side length, i.e. we are given the set

{

}

. Then doubling the cube would correspond to constructing a side

of length



such that



= 2, i.e.



√

. Thus we need to construct a point

R = (

√

2, 0) from S.

If we can indeed construct this R, then we need

[Q(S ∪ {R}) : Q(S)] = 2

for some k. But we know that Q(S) = Q and Q(S ∪ {R}) = Q(

√

2), and that

[Q(

√

2) : Q] = 3.

This is a contradiction since 3 is not a power of 2.

2.3 K-homomorphisms and the Galois Group

Usually in mathematics, we not only want to study objects, but maps between

objects. Suppose we have two field extensions

K ⊆ L

and

K ⊆ L

′

. What should

a map between these two objects look like? Obviously, we would like this map

to be a field homomorphisms between

and

′

. Moreover, since this is a map

between the two field extensions, and not just the fields themselves, we would

like this map to preserve things in

, and is just a map between the “extended

parts” of L and L

′

Definition (

-homomorphism). Let

L/K

and

′

be field extensions. A

-homomorphism

L → L

′

is a ring homomorphism such that

φ|

, i.e. it

fixes everything in

. We write

Hom

(

L, L

′

) for the set of all

-homomorphisms

L → L

′

-isomorphism is a

-homomorphism which is an isomorphism of rings.

-automorphism is a

-isomorphism

L → L

. We write

Aut

(

) for the set

of all K-automorphism L → L.

There are a couple of things to take note of

(i) Given any φ ∈ Hom

(L, L

′

), we know that

(a)

Since

φ|

, we know that

ker φ 

. Since we know that

ker φ

an ideal, and a field only has two ideals, we must have

ker φ

= 0. So

is injective. It is, in fact, true that any homomorphisms of fields is

injective.

(b) φ

gives an isomorphism

L → φ

(

). So

(

) is a field and we get the

field extensions K ⊆ φ(L) ⊆ L

′

(ii)

If [

] = [

′

]

< ∞

, then any homomorphism in

Hom

(

L, L

′

) is in

fact an isomorphism. So

{K-homomorphisms : L → L

′

} = {K-isomorphisms : L → L

′

This is since any

-homomorphism

L → L

′

is an injection. So

[

] = [

(

) :

]. Hence we know that [

′

] = [

(

) :

]. But we

know that

(

) is a subfield of

′

. This is possible only if

′

(

). So

φ is a surjection, and hence an isomorphism.

In particular, Aut

(L) = Hom

(L, L).

Example. We want to determine Aut

(C). If we pick any ψ ∈ Aut

(C), then

(ψ(

√

−1))

+ 1 = ψ(

√

−1

+ 1) = ψ(0) = 0.

So under any automorphism

, the image of

√

−1

is a root of

+ 1. Therefore

(

√

−1

) =

√

−1

−

√

−1

. In the first case,

is the identity. In the second

case, the automorphism is

√

−1 7→ a −b

√

−1

, i.e. the complex conjugate.

So Aut

Similarly, we can show that

Aut

(

√

)) =

{id, φ}

, where

swaps

√

with

−

√

Example. Let

= 1 but

µ 

= 1 (i.e.

is a third root of unity). We want to

determine A = Hom

(Q(

√

2), C).

First define φ, ψ by

φ(

√

2) =

√

2µ

ψ(

√

2) =

√

2µ

We have φ, ψ ∈ A. Are there more?

Let λ ∈ A. Then we must have

(λ(

√

2))

− 2 = 0.

(

√

) is a root of

−

2. So it is either

√

2µ

√

2µ

. So

is either

φ or ψ. So A = {id, φ, ψ}.

Note that in general, if

is algebraic over

, then

(

)

∼

[

]

/⟨P

⟩

. Hence

to specify a

-homomorphism from

(

), it suffices to specify the image of

, or

just the image of α.

We will later see that the number of automorphisms

|Aut

(

)

is bounded

above by the degree of the extension [

]. However, we need not always have

[

] many automorphisms. When we do have enough automorphisms, we call

it a Galois extension.

Definition (Galois extension). Let

L/K

be a finite field extension. This is a

Galois extension if |Aut

(L)| = [L : K].

Definition (Galois group). The Galois group of a Galois extension

L/K

defined as

Gal

(

L/K

) =

Aut

(

). The group operation is defined by function

composition. It is easy to see that this is indeed a group.

Example. The extension

(

√

)

is Galois. The degree [

(

√

) :

] = 2, and

the automorphism group is

Aut

(

√

)) =

{id, φ}

, where

swaps

√

with

−

√

Example. The extension

(

√

)

is not Galois. The degree is [

(

√

) :

] = 3,

but the automorphism group is Aut

(Q(

√

2)) = {id}.

To show that there is no other automorphism, note that the automorphism

group can be viewed as a subset of

Hom

(

√

)

, C

). We have just seen that

Hom

(

√

)

, C

) has three elements, but only the identity maps

(

√

) to itself,

while the others map

√

2 to

√

2µ

∈ Q(

√

2). So this is the only automorphism.

The way we should think about this is that there is something missing in

(

√

), namely

. Without the

, we cannot get the other automorphisms we

need. In fact, in the next example, we will show that Q ⊆ Q(

√

2, µ) is Galois.

Example.

(

√

2, µ

)

is a Galois extension. Firstly, we know that [

(

√

2, µ

) :

(

√

)] = 2 because

−

1 = 0 implies

+ 1 = 0. So the minimal

polynomial has degree 2. This also means that

µ ∈ Q

(

√

). We also know that

[Q(

√

2) : Q] = 3. So we have

[Q(

√

2, µ) : Q] = 6

by the Tower law.

Now denote

√

2µ

and

√

2µ

. Then

(

√

2, µ

) =

(

α, β, γ

Now let

φ ∈ Aut

(

√

2, µ

)), then

(

) and

(

) are roots of

−

2. These

roots are exactly α, β, γ. So

{φ(α), φ(β), φ(γ)} = {α, β, γ}.

Hence

is completely determined by a permutation of the roots of

−

2. So

Aut

(

√

2, µ)

∼

and |Aut

(

√

2, µ)| = 6.

Most of the time, we will only be interested in Galois extensions. The main

reason is that Galois extensions satisfy the fundamental theorem of Galois theory,

which roughly says: if

L/K

is a finite Galois extension, then there is a one-to-one

correspondence of the set of subgroups

H ≤ Gal

(

L/K

) and the intermediate

fields

K ⊆ F ⊆ L

. In particular, the normal subgroups corresponds to the

“normal extensions”, which is something we will define later.

However, just as we have seen, it is not straightforward to check if an extension

is Galois, even in specific cases like the examples above. Fortunately, by the

time we reach the proper statement of the fundamental theorem, we would have

developed enough machinery to decide easily whether certain extensions are

Galois.

2.4 Splitting fields

As mentioned in the introduction, one major motivation for Galois theory is to

study the roots of polynomials. So far, we have just been talking about field

extensions. The idea here is given a field

and a polynomial

f ∈ K

[

], we

would like to study the field extension obtained by adding all roots of

. This is

known as the splitting field of f (over K).

Notation. Let

L/K

be a field extension,

f ∈ K

[

]. We write

Root

(

) for the

roots of f in L.

First, we establish a correspondence between the roots of a polynomial and

K-homomorphisms.

Lemma. Let

L/K

be a field extension,

f ∈ K

[

] irreducible,

deg f >

0. Then

there is a 1-to-1 correspondence

Root

(L) ←→ Hom

(K[t]/⟨f⟩, L).

Proof.

Since

is irreducible,

⟨f⟩

is a maximal ideal. So

[

]

/⟨f⟩

is a field. Also,

there is a natural inclusion

K → K

[

]

/⟨f⟩

. So it makes sense to talk about

Hom

(K[t]/⟨f⟩, L).

To any

β ∈ Root

(

), we assign

[

]

/⟨f⟩ → L

where we map

t 7→ β

(

the equivalence class of

). This is well defined since if

¯g

, then

for

some h ∈ K[t]. So φ(¯g) = φ(t + hf) = β + h(β)f(β) = β.

Conversely, given any

-homomorphism

[

]

/⟨f⟩ → L

, we assign

φ(

t). This is a root since f(β) = f (φ(

t)) = φ(f(

t)) = φ(0) = 0.

This assignments are inverses to each other. So we get a one-to-one corre-

spondence.

Recall that if

K ⊆ F

is a field extension, then for any

α ∈ F

with minimal

polynomial

, we have

[

]

/⟨P

⟩

∼

(

). Since an irreducible

is the minimal

polynomial of its roots, we can view the above lemma as telling us something

about Hom

(K(α), L).

Corollary. Let

L/K

be a field extension,

f ∈ K

[

] irreducible,

deg f >

0. Then

|Hom

(K[t]/⟨f⟩, L)| ≤ deg f.

In particular, if E = K[t]/⟨f ⟩, then

|Aut

(E)| = |Root

(E)| ≤ deg f = [E : K].

So E/K is a Galois extension iff |Root

(E)| = deg f.

Proof. This follows directly from the following three facts:

– |Root

(L)| ≤ deg f

– Aut

(E) = Hom

(E, E)

– deg f = [K(α) : K] = [E : K].

Definition (Splitting field). Let

L/K

be a field extensions,

f ∈ K

[

]. We say

splits over L if we can factor f as

f = a(t − α

) ···(t − α

)

for some

a ∈ K

and

∈ L

. Alternatively, this says that

contains all roots of

We say

is a splitting field of

(

, ··· , α

). This is the smallest

field where f has all its roots.

Example.

– C is the splitting field of t

+ 1 ∈ R[t].

– Q

(

√

2, µ

) is a splitting field of

−

∈ Q

[

], where

is a third root of

unity.

–

By the fundamental theorem of algebra, for any

K ⊆ C

and

f ∈ K

[

there is a splitting field L ⊆ C of f.

Note that the degree of the splitting field need not be (bounded by) the

degree of the polynomial. In the second example, we have [

(

√

2, µ

) :

] = 6,

but t

− 2 only has degree 3.

More generally, we can show that every polynomial has a splitting field, and

this is unique up to isomorphism. This is important, since we would like to talk

about the splitting field of a polynomial all the time.

Theorem. Let K be a field, f ∈ K[t]. Then

(i) There is a splitting field of f.

(ii) The splitting field is unique (up to K-isomorphism).

Proof.

(i)

deg f

= 0, then

is a splitting field of

. Otherwise, we add the roots

of f one by one.

Pick

g | f

[

], where

is irreducible and

deg g >

0. We have the field

extension

K ⊆ K

[

]

/⟨g⟩

. Let

. Then

(

) = 0 which implies that

(

) = 0. Hence we can write

= (

t − α

)

(

)[

]. Note that

deg h < deg f

. So we can repeat the process on

iteratively to get a field

extensions

K ⊆ K

(

, ··· , α

). This

(

, ··· , α

) is a splitting field of

(ii)

Assume

and

′

are both splitting fields of

over

. We want to find a

K-isomorphism from L to L

′

Pick largest

F, F

′

such that

K ⊆ F ⊆ L

and

K ⊆ F

′

⊆ L

′

are field

extensions and there is a

-isomorphism from

F → F

′

. By “largest”,

we mean we want to maximize [F : K].

We want to show that we must have

. Then we are done because

this means that F

′

is a splitting field, and hence F

′

= L

′

So suppose

F 

. We will try to produce a larger

with

-isomorphism

F →

′

⊆ L

′

Since

F 

, we know that there is some

α ∈ Root

(

) such that

α ∈ F

Then there is some irreducible

g ∈ K

[

] with

deg g >

0 such that

(

) = 0

and g | f . Say f = gh.

Now we know there is an isomorphism

[

]

/⟨g⟩ → F

(

) by

t 7→ α

. The

isomorphism

F → F

′

extends to a isomorphism

[

]

→ F

′

[

Then since the coefficients of

are in

, we have

(

) =

(

)

(

)

| f

′

[

]. Since

is irreducible in

[

(

) is irreducible in

′

[

]. So there is some

′

∈ Root

µ(g)

(

′

)

⊆ Root

(

′

) and isomorphism

′

[t]/⟨µ(g)⟩ → F

′

(α

′

Now

induces a

-isomorphism

[

]

/⟨g⟩ → F

′

[

]

/⟨µ

(

)

⟩

, which in turn

induces a

-isomorphism

(

)

→ F

′

(

′

). This contradicts the maximality

of F . So we must have had F = L.

Note that the splitting is unique just up to isomorphism. We could be

quotienting by different polynomials and still get the same splitting field.

Example.

(

√

) is a splitting field of

−

∈ Q

[

]. At the same time,

(

√

)

is also a splitting field of t

+ 3t +

∈ Q[t].

2.5 Algebraic closures

The splitting field gives us the field with the root of one particular polynomial.

We could be greedy and ask for the roots for all polynomials, and get the

algebraic closure. The algebraic closure will not be of much use in this course,

but is a nice thing to know about. The major theorems would be the existence

and uniqueness of algebraic closures.

Definition (Algebraically closed field). A field

is algebraically closed if for all

f ∈ L[t], we have

f = a(t − α

)(t − α

) ···(t − α

)

for some a, α

∈ L. In other words, L contains all roots of its polynomials.

Let L/K be a field extension. We say L is an algebraic closure of K if

– L is algebraic over K

– L is algebraically closed.

Example.

is an algebraically closed field iff (

L ⊆ E

is a finite extension

implies E = L).

This is since if

L ⊆ E

is finite, then

is algebraic over

, and hence must

be L.

Example.

is algebraically closed by the fundamental theorem of algebra, and

is the algebraic closure of R (but not Q).

Before we prove our next theorem, we need the following technical lemma:

Lemma. If

is a commutative ring, then it has a maximal ideal. In particular,

if I is an ideal of R, then there is a maximal ideal that contains I.

Proof. Let

P = {I : I is an ideal of R, I = R}.

⊆ I

⊆ ···

is any chain of

∈ P

, then

∈ P

. By Zorn’s lemma,

there is a maximal element of

(containing

). So

has at least one maximal

ideal (containing I).

Theorem (Existence of algebraic closure). Any field

has an algebraic closure.

Proof. Let

A = {λ = (f, j) : f ∈ K[t] irreducible monic, 1 ≤ j ≤ deg f}.

We can think of

as labelling which root of

we want. For each

λ ∈ A

, we

assign a variable t

. We take

R = K[t

: λ ∈ A]

to be the polynomial ring over

with variables

. This

contains all the

“roots” of the polynomials in

. However, we’ve got a bit too much. For example,

(if

), in

√

and

√

+ 1 would be put down as separate, unrelated

variables. So we want to quotient this R by something.

For every monic and irreducible f ∈ K[t], we define

f = f −

deg f

j=1

(t − t

(f,j)

) ∈ R[t].

If we want the

(f,j)

to be roots of

, then

should vanish for all

. Denote the

coefficient of t

ℓ

f by b

(f,ℓ)

. Then we want b

(f,ℓ)

= 0 for all f, .

To do so, let

I ⊆ R

be the ideal generated by all such coefficients. We now

want to quotient R by I. We first have to check that I = R.

Suppose not. So there are

,ℓ

)

, ··· , b

,ℓ

)

with

, ··· , g

∈ R

such that

,ℓ

)

+ ··· + g

,ℓ

)

= 1. (∗)

We will attempt to reach a contradiction by constructing a homomorphism

that sends each b

,ℓ

)

to 0.

Let E be a splitting field of f

···f

. So in E[t], for each i, we can write

deg f

j=1

(t − α

i,j

Then we define a homomorphism φ : R → E by

(

φ(t

,j)

) = α

i,j

φ(t

) = 0 otherwise

This induces a homomorphism

φ : R[t] → E[t].

Now apply

φ(

) =

φ(f

) −

deg f

j=1

φ(t − t

,j)

)

= f

−

deg f

j=1

(t − α

i,j

)

= 0

So φ(b

,ℓ

)

) = 0 as b

,ℓ

)

is a coefficient of f

Now we apply φ to (∗) to obtain

φ(g

,ℓ

)

+ ··· + g

,ℓ

)

) = φ(1).

But this is a contradiction since the left had side is 0 while the right is 1. Hence

we must have I = R.

We would like to quotient by

, but we have to be a bit more careful, since

the quotient need not be a field. Instead, pick a maximal ideal

containing

, and consider

R/M

. Then

is a field. Moreover, since we couldn’t

have quotiented out anything in

(any ideal containing anything in

would

automatically contain all of

), this is a field extension

L/K

. We want to show

that L is an algebraic closure.

Now we show that

is algebraic over

. This should all work out smoothly,

since that’s how we constructed

. First we pick

α ∈ L

. Since

R/M

and

is generated by the terms t

, there is some (f

, j

), ··· , (f

, j

) such that

α ∈ K(

)

, ··· ,

)

is algebraic over

if each

)

is algebraic over

. To show this, note

that

= 0, since we’ve quotiented out each of its coefficients. So by definition,

0 = f

(t) −

deg f

j=1

(t −

,j)

So f

(

)

) = 0. So done.

Finally, we have to show that

is algebraically closed. Suppose

L ⊆ E

is a

finite (and hence algebraic) extension. We want to show that L = E.

Consider arbitrary

β ∈ E

. Then

is algebraic over

, say a root of

f ∈ L

[

]. Since every coefficient of

can be found in some finite extension

(

)

, ··· ,

)

), there is a finite extension

that contains all coeffi-

cients of

. Since

(

) is a finite extension of

, we know

(

) is a finite and

hence algebraic extension of K. In particular, β is algebraic in K.

Let

be the minimal polynomial of

over

. Since all polynomials in

split over

by construction (

(

) =

(

t −

(f,j)

)), its roots must in

. In

particular, β ∈ L. So L = E.

Theorem (Uniqueness of algebraic closure). Any field

has a unique algebraic

closure up to K-isomorphism.

This is the same proof as the proof that the splitting field is unique — given

two algebraic closures, we take the largest subfield of the algebraic closures that

biject with each other. However, since there could be infinitely many subfields,

we have to apply Zorn’s lemma to obtain the maximal such subfield.

Proof. (sketch) Suppose L, L

′

are both algebraic closures of K. Let

H = {(F, ψ) : K ⊆ F ⊆ L, ψ ∈ Hom

(F, L

′

)}.

We define a partial order on

by (

, ψ

)

≤

(

, ψ

) if

≤ F

and

We have to show that chains have upper bounds. Given a chain

{

(

, ψ

)

}

we define

F =

[

, ψ(x) = ψ

(x) for x ∈ F

Then (

F, ψ

)

∈ H

. Then applying Zorn’s lemma, there is a maximal element of

H, say (F, ψ).

Finally, we have to prove that

, and that

(

) =

′

. Suppose

F 

Then we attempt to produce a larger

and a

-isomorphism

F →

′

⊆ L

′

Since

F 

, there is some

α ∈ L \ F

. Since

is an algebraic extension of

there is some irreducible g ∈ K[t] such that deg g > 0 and g(α) = 0.

Now there is an isomorphism

[

]

/⟨g⟩ → F

(

) defined by

t 7→ α

. The

isomorphism

F → F

′

then extends to an isomorphism

[

]

→ F

′

[

]

and thus to

[

]

/⟨g⟩ → F

′

[

]

/⟨µ

(

)

⟩

. Then if

′

is a root of

(

), then we have

′

[

]

//⟨µ

(

)

⟩

∼

′

(

′

). So this gives an isomorphism

(

)

→ F

(

′

). This

contradicts the maximality of φ.

By doing the argument the other way round, we must have

(

) =

′

. So

done.

2.6 Separable extensions

Here we will define what it means for an extension to be separable. This is

done via defining separable polynomials, and then an extension is separable if

all minimal polynomials are separable.

At first, the definition of separability might seem absurd — surely every

polynomial should be separable. Indeed, polynomials that are not separable

tend to be weird, and our theories often break without separability. Hence it is

important to figure out when polynomials are separable, and when they are not.

Fortunately, we will end up with a result that tells us exactly when a polynomial

is not separable, and this is just a very small, specific class. In particular, in

fields of characteristic zero, all polynomials are separable.

Definition (Separable polynomial). Let

be a field,

f ∈ K

[

] non-zero, and

a splitting field of

. For an irreducible

, we say it is separable if

has no

repeated roots, i.e.

|Root

(

)

deg f

. For a general polynomial

, we say it is

separable if all its irreducible factors in K[t] are separable.

It should be obvious from definition that if

is separable and

Q | P

, then

is also separable.

Note that some people instead define a separable polynomial to be one with

no repeated roots, so (

x−

over

would not be separable under this definition.

Example. Any linear polynomial t − a (with a ∈ K) is separable.

This is, however, not a very interesting example. To get to more interesting

examples, we need even more preparation.

Definition (Formal derivative). Let

be a field,

f ∈ K

[

]. (Formal) differenti-

ation the K-linear map K[t] → K[t] defined by t

7→ nt

n−1

The image of a polynomial f is the derivative of f , written f

′

This is similar to how we differentiate real or complex polynomials (in case

that isn’t obvious).

The following lemma summarizes the properties of the derivative we need.

Lemma. Let K be a field, f, g ∈ K[t]. Then

(i) (f + g)

′

= f

′

+ g

′

, (f g)

′

= fg

′

+ f

′

(ii)

Assume

f 

= 0 and

is a splitting field of

. Then

has a repeated root in

if and only if

and

′

have a common (non-constant) irreducible factor

in K[t] (if and only if f and f

′

have a common root in L).

This will allow us to show when irreducible polynomials are separable.

Proof.

(i) (f + g)

′

= f

′

+ g

′

is true by linearity.

To show that (

)

′

, we use linearity to reduce to the case

where

, g

. Then both sides are (

)

n+m−1

. So this holds.

(ii)

First assume that

has a repeated root. So let

= (

t −α

)

h ∈ L

[

] where

α ∈ L

. Then

′

= 2(

t − α

)

+ (

t − α

)

′

= (

t − α

)(2

+ (

t − α

)

′

). So

(

) =

′

(

) = 0. So

and

′

have common roots. However, we want a

common irreducible factor in

[

], not

[

]. So we let

be the minimal

polynomial of α over K. Then P

| f and P

| f

′

. So done.

Conversely, suppose

is a common irreducible factor of

and

′

[

with deg e > 0. Pick α ∈ Root

(L). Then α ∈ Root

(L) ∩ Root

′

(L).

Since α is a root of f, we can write f = (t − α)q ∈ L[t] for some q. Then

′

= (t − α)q

′

+ q.

Since (t − α) | f

′

, we must have (t − α) | q. So (t − α)

| f.

Recall that the characteristic of a field

char K

is the minimum

such that

p ·

= 0. If no such

exists, we say

char K

= 0. For example,

has

characteristic 0 while Z

has characteristic p.

Corollary. Let K be a field, f ∈ K[t] non-zero irreducible. Then

(i) If char K = 0, then f is separable.

(ii)

char K

p >

0, then

is not separable iff

deg f >

0 and

f ∈ K

[

]. For

example, t

+ 3t

+ 1 is not separable.

Proof.

By definition, for irreducible

is not separable iff

has a repeated

root. So by our previous lemma,

is not separable if and only if

and

′

have a common irreducible factor of positive degree in

[

]. However, since

irreducible, its only factors are 1 and itself. So this can happen if and only if

′

= 0.

To make it more explicit, we can write

f = a

+ ··· + a

t + a

Then we can write

′

= na

n−1

+ ··· + a

Now f

′

= 0 if and only if all coefficients ia

= 0 for all i.

(i)

Suppose

char K

= 0, then if

deg f

= 0, then

is trivially separable. If

deg f >

0, then

is not separable iff

′

= 0 iff

= 0 for all

iff

= 0

for

i ≥

1. But we cannot have a polynomial of positive degree with all its

coefficients zero (apart from the constant term). So f must be separable.

(ii) If deg f = 0, then f is trivially separable. So assume deg f > 0.

Then

is not separable

⇔ f

′

= 0

⇔ ia

= 0 for

i ≥

⇔ a

= 0 for all

i ≥ 1 not multiples of p ⇔ f ∈ K[t

Using this, it should be easy to find lots of examples of separable polynomials.

Definition (Separable elements and extensions). Let

K ⊆ L

be an algebraic

field extension. We say

α ∈ L

is separable over

is separable, where

is the minimal polynomial of α over K.

We say

is separable over

(or

K ⊆ L

is separable) if all

α ∈ L

are

separable.

Example.

–

The extensions

Q ⊆ Q

(

√

) and

R ⊆ C

are separable because

char Q

char R = 0. So we can apply our previous corollary.

–

Let

(

) be the field of rational functions in

over

(which is the

fraction field of

[

]), and

(

). We have

K ⊆ L

, and

(

Since

∈ K

is a root of

− s

∈ K

[

]. So

is algebraic over

and

hence

is algebraic over

. In fact

−s

is the minimal polynomial

of s over K.

Now

−s

= (

t−s

)

since the field has characteristic

. So

Root

−s

(

) =

{s}. So P

is not separable.

As mentioned in the beginning, separable extensions are nice, or at least

non-weird. One particular nice result about separable extensions is that all finite

separable extensions are simple, i.e. if

K ⊆ L

is finite separable, then

(

)

for some

α ∈ L

. This is what we will be working towards for the remaining of

the section.

Example. Consider

Q ⊆ Q

(

√

). This is a separable finite extension. So

we should be able to generate

(

√

) by just one element, not just two. In

fact, we can use α =

√

2 +

√

3, since we have

= 11

√

2 + 9

√

3 = 2

√

2 + 9α.

So since α

∈ Q(α), we know that

√

2 ∈ Q(α). So we also have

√

3 ∈ Q(α).

In general, it is not easy to find an

that works, but we our later result will

show that such an α exists.

Before that, we will prove some results about the K-homomorphisms.

Lemma. Let

L/F/K

be finite extensions, and

E/K

be a field extension. Then

for all α ∈ L, we have

|Hom

(F (α), E)| ≤ [F (α) : F ]|Hom

(F, E)|.

Note that if

is the minimal polynomial of

over

, then [

(

) :

] =

deg P

. So we can interpret this intuitively as follows: for each

ψ ∈ Hom

(

F, E

we can obtain a

-homomorphism in

Hom

(

)

, E

) by sending things in

according to

, and then send

to any root of

. Then there are at

most [

(

) :

]

-homomorphisms generated this way. Moreover, each

homomorphism in

Hom

(

)

, E

) can be created this way. So we get this

result.

Proof.

We show that for each

ψ ∈ Hom

(

F, E

), there are at most [

(

) :

]

-isomorphisms in

Hom

(

)

, E

) that restrict to

. Since each

isomorphism in

Hom

(

)

, E

) has to restrict to something, it follows that

there are at most [

(

) :

]

|Hom

(

F, E

)

| K

-homomorphisms from

(

) to

Now let

be the minimal polynomial for

, and let

ψ ∈ Hom

(

F, E

To extend ψ to a morphism F (α) → E, we need to decide where to send α. So

there should be some sort of correspondence

Root

(E) ←→ {φ ∈ Hom

(F (α), E) : φ|

= ψ}.

Except that the previous sentence makes no sense, since

∈ F

[

] but we are

not told that F is a subfield of E. So we use our ψ to “move” our things to E.

We let

(

)

⊆ E

, and

q ∈ M

[

] be the image of

under the

homomorphism

[

]

→ M

[

] induced by

. As we have previously shown, there

is a one-to-one correspondence

Root

(E) ←→ Hom

(M[t]/⟨q⟩, E).

What we really want to show is the correspondence between

Root

(

) and the

-homomorphisms

[

]

/⟨P

⟩ → E

that restrict to

. Let’s ignore the

quotient for the moment and think: what does it mean for

φ ∈ Hom

(

[

]

, E

) to

restrict to

? We know that any

φ ∈ Hom

(

[

]

, E

) is uniquely determined

by the values it takes on

and

. Hence if

φ|

, then our

must send

(

) =

, and can send

to anything in

. This corresponds exactly to

the

-homomorphisms

[

]

→ E

that does nothing to

and sends

to that

“anything” in E.

The situation does not change when we put back the quotient. Changing

from

[

]

→ E

[

]

/⟨q⟩ → E

just requires that the image of

must be

a root of

. On the other hand, using

[

]

/⟨P

⟩

instead of

[

] requires that

(

)) = 0. But we know that

(

) =

(

) =

. So this just requires

q(t) = 0 as well. So we get the one-to-one correspondence

Hom

(M[t]/⟨q⟩, E) ←→ {φ ∈ Hom

(F [t]/⟨P

⟩, E) : φ|

= ψ}.

Since F [t]/⟨P

⟩ = F (α), there is a one-to-one correspondence

Root

(E) ←→ {φ ∈ Hom

(F (α), E) : φ|

= ψ}.

So done.

Theorem. Let L/K and E/K be field extensions. Then

(i) |Hom

(L, E)| ≤ [L : K]. In particular, |Aut

(L)| ≤ [L : K].

(ii) If equality holds in (i), then for any intermediate field K ⊆ F ⊆ L:

(a) We also have |Hom

(F, E)| = [F : K].

(b) The map Hom

(L, E) → Hom

(F, E) by restriction is surjective.

Proof.

(i) We have previously shown we can find a sequence of field extensions

K = F

⊆ F

⊆ ··· ⊆ F

= L

such that for each

, there is some

such that

i−1

(

). Then by

our previous lemma, we have

|Hom

(L, E)| ≤ [F

: F

n−1

]|Hom

n−1

, E)|

≤ [F

: F

n−1

][F

n−1

: F

n−2

]|Hom

n−2

, E)|

≤ [F

: F

n−1

][F

n−1

: F

n−2

] ···[F

: F

]|Hom

, E)|

= [F

: F

]

= [L : K]

(ii) (a)

If equality holds in (i), then every inequality in the proof above has

to an equality. Instead of directly decomposing

K ⊆ L

as a chain

above, we can first decompose

K ⊆ F

, then

F ⊆ L

, then join them

together. Then we can assume that F = F

for some i. Then we get

|Hom

(L, E)| = [L : F ]|Hom

(F, E)| = [L : K].

Then the tower law says

|Hom

(F, E)| = [F : K].

(b)

By the proof of the lemma, for each

ψ ∈ Hom

(

F, E

), we know that

{φ : Hom

(L, E) : φ|

= ψ} ≤ [L : F ]. (∗)

As we know that

|Hom

(F, E)| = [F : K], |Hom

(L, E)| = [L : K]

we must have had equality in (

∗

), or else we won’t have enough

elements. So in particular

{φ

Hom

(

L, E

) :

φ|

ψ} ≥

1. So the

map is surjective.

With this result, we can prove prove the following result characterizing

separable extensions.

Theorem. Let

L/K

be a finite field extension. Then the following are equivalent:

(i) There is some extension E of K such that |Hom

(L, E)| = [L : K].

(ii) L/K is separable.

(iii) L

(

, ··· , α

) such that

, the minimal polynomial of

over

is separable for all i.

(iv) L

(

, ··· , α

) such that

, the minimal polynomial of

over

K(α

, ··· , α

i−1

) is separable for all i.

Proof.

–

(i)

⇒

(ii): For all

α ∈ L

, if

is the minimal polynomial of

over

then since K(α) is a subfield of L, by our previous theorem, we have

|Hom

(K(α), E)| = [K(α) : K].

We also know that

|Root

(

)

|Hom

(

)

, E

)

, and that [

(

) :

] =

deg P

. So we know that

has no repeated roots in any splitting

field. So P

is a separable. So L/K is a separable extension.

– (ii) ⇒ (iii): Obvious from definition

–

(iii)

⇒

(iv): Since

is a minimal polynomial in

(

, ··· , α

i−1

), we

know that R

| P

. So R

is separable as P

is separable.

–

(iv)

⇒

(i): Let

be the splitting field of

, ··· , P

. We do induction

to show that this satisfies the properties we want. If

= 1, then

L = K(α

). Then we have

|Hom

(L, E)| = |Root

(E)| = deg P

= [K(α

) : K] = [L : K].

We now induct on

. So we can assume that (iv)

⇒

(i) holds for smaller

number of generators. For convenience, we write

(

, ··· , α

Then we have

|Hom

n−1

, E)| = [K

n−1

: K].

We also know that

|Hom

, E)| ≤ [K

: K

n−1

]|Hom

n−1

, E)|.

What we actually want is equality. We now re-do (parts of) the proof of

this result, and see that separability guarantees that equality holds. If

we pick

ψ ∈ Hom

(

n−1

, E

), then there is a one-to-one correspondence

between

{φ ∈ Hom

(

, E

) :

φ|

n−1

ψ}

and

Root

(

), where

q ∈ M

[

]

is defined as the image of

under

n−1

[

]

→ M

[

], and

is the image

of ψ.

Since

∈ K

[

] and

| P

, then

q | P

. So

splits over

. By

separability assumption , we get that

|Root

(E)| = deg q = deg R

= [K

: K

n−1

Hence we know that

|Hom

(L, E)| = [K

: K

n−1

]|Hom

n−1

, E)|

= [K

: K

n−1

][K

n−1

: K]

= [K

: K].

So done.

Before we finally get to the primitive element theorem, we prove the following

lemma. This will enable us to prove the trivial case of the primitive element

theorem, and will also be very useful later on.

Lemma. Let

be a field,

∗

L \ {

}

be the multiplicative group of

. If

is a finite subgroup of L

∗

, then G is cyclic.

Proof.

Since

∗

is abelian,

is also abelian. Then by the structure theorem on

finite abelian groups,

∼

⟨n

⟩

× ··· ×

⟨n

⟩

for some

∈ N

. Let

be the least common multiple of

, ··· , n

, and let

f = t

− 1.

If α ∈ G, then α

= 1. So f(α) = 0 for all α ∈ G. Therefore

|G| = n

···n

≤ |Root

(L)| ≤ deg f = m.

Since

is the least common multiple of

, ··· , n

, we must have

···n

and thus (

, n

) = 1 for all

i 

. Then by the Chinese remainder theorem, we

have

∼

⟨n

⟩

× ··· ×

⟨n

⟩

⟨n

···n

⟩

So G is cyclic.

We now come to the main theorem of the lecture:

Theorem (Primitive element theorem). Assume

L/K

is a finite and separable

extension. Then L/K is simple, i.e. there is some α ∈ L such that L = K(α).

Proof.

At some point in our proof, we will require that

is infinite. So we

first do the finite case first. If

is finite, then

is also finite, which in turns

implies

∗

is finite too. So by the lemma,

∗

is a cyclic group (since it is a finite

subgroup of itself). So there is some

α ∈ L

∗

such that every element in

∗

is a

power of α. So L = K(α).

So focus on the case where

is infinite. Also, assume

K 

. Then since

L/K

is a finite extension, there is some intermediate field

K ⊆ F ⊊ L

such that

(

) for some

. Now

L/K

is separable. So

F/K

is also separable, and

[

]

[

]. Then by induction on degree of extension, we can assume

F/K

is simple. In other words, there is some

λ ∈ F

such that

(

). Now

L = K(λ, β). In the rest of the proof, we will try to replace the two generators

λ, β with just a single generator.

Unsurprisingly, the generator of

will be chosen to be a linear combination

of β and λ. We set

α = β + aλ

for some

a ∈ K

to be chosen later. We will show that

(

) =

. Actually,

almost any choice of

will do, but at the end of the proof, we will see which

ones are the bad ones.

Let

and

be the minimal polynomial of

and

over

respectively.

Consider the polynomial f = P

(α − at) ∈ K(α)[t]. Then we have

f(λ) = P

(α − aλ) = P

(β) = 0.

On the other hand, P

(λ) = 0. So λ is a common root of P

and f .

We now want to pick an

such that

is the only common root of

and

(in

). If so, then the gcd of

and

(

) must only have

as a root.

But since

is separable, it has no double roots. So the gcd must be

t − λ

. In

particular, we must have

λ ∈ K

(

). Since

aλ

, it follows that

β ∈ K

(

)

as well, and so K(α) = L.

Thus, it remains to choose an

such that there are no other common roots.

We work in a splitting field of P

, and write

= (t − β

) ···(t − β

)

= (t − λ

) ···(t − λ

We wlog β

= β and λ

= λ.

Now suppose θ is a common root of f and P

. Then

(

f(θ) = 0

(θ) = 0

⇒

(

(α − aθ) = 0

(θ) = 0

⇒

(

α − aθ = β

θ = λ

for some i, j. Then we know that

α = β

+ aλ

However, by definition, we also know that

α = β + aλ

Now we see how we need to choose

. We need to choose

such that the elements

β + aλ = β

+ aλ

for all i, j. But if they were equal, then we have

a =

λ − λ

− β

and there are only finitely many elements of this form. So we just have to pick

an a not in this list.

Corollary. Any finite extension

L/K

of field of characteristic 0 is simple, i.e.

L = K(α) for some α ∈ L.

Proof.

This follows from the fact that all extensions of fields of characteristic

zero are separable.

We have previously seen that

(

√

)

is a simple extension, but that

is of course true from this theorem. A more interesting example would be one in

which this fails. We will need a field with non-zero characteristic.

Example. Let

(

s, u

), the fraction field of

[

s, u

]. Let

(

, u

We have L/K. We want to show this is not simple.

α ∈ L

, then

∈ K

. So

is a root of

− α

∈ K

[

]. Thus the minimal

polynomial

has degree at most

. So [

(

) :

] =

deg P

≤ p

. On the other

hand, we have [

] =

, since

: 0

≤ i, j < p}

is a basis. So for any

we have

(

)



. So

L/K

is not a simple extension. This then implies

L/K

is not separable.

At this point, one might suspect that all fields with positive characteristic

are not separable. This is not true by considering a rather silly example.

Example. Consider

and

[

]

/⟨s

+ 1

⟩

. We can check manually

that

+ 1 has no roots and hence irreducible. So

is a field. So

L/F

is a

finite extension. Note that L only has 4 elements.

Now if

α ∈ L \ F

, and

is the minimal polynomial of

over

, then

| t

+ t + 1. So P

is separable as a polynomial. So L/F

is separable.

In fact, we have

Proposition. Let

L/K

be an extension of finite fields. Then the extension is

separable.

Proof.

Let the characteristic of the fields be

. Suppose the extension were not

separable. Then there is some non-separable element

α ∈ L

. Then its minimal

polynomial must be of the form P

Now note that the map

K → K

given by

x 7→ x

is injective, hence surjective.

So we can write a

= b

for all i. Then we have





and so P

is not irreducible, which is a contradiction.

2.7 Normal extensions

We are almost there. We will now move on to study normal extensions. Normal

extensions are very closely related to Galois extensions. In fact, we will show

that if an extension is normal and separable, then it is Galois. The advantage

of introducing the idea of normality is that normality is a much more concrete

definition to work with. It is much easier to check if an extension is normal than

to check if

|Aut

(

)

= [

]. In particular, we will shortly prove that the

splitting field of any polynomial is normal.

This is an important result, since we are going to use the splitting field to

study the roots of a polynomial, and since we mostly care about polynomials

over

, this means all these splitting fields are automatically Galois extensions

of Q.

It is not immediately obvious why these extensions are called “normal” (just

like most other names in Galois theory). We will later see that normal extensions

are extensions that correspond to normal subgroups, in some precise sense given

by the fundamental theorem of Galois theory.

Definition (Normal extension). Let

K ⊆ L

be an algebraic extension. We say

L/K

is normal if for all

α ∈ L

, the minimal polynomial of

over

splits over

In other words, given any minimal polynomial, L should have all its roots.

Example. The extension

(

√

)

is not normal since the minimal polynomial

− 2 does not split over Q(

√

2).

In some sense, extensions that are not “normal” are missing something. This

is somewhat similar to how Galois extensions work. Before we go deeper into

this, we need a lemma.

Lemma. Let

L/F/K

be finite extensions, and

is the algebraic closure of

Then any ψ ∈ Hom

(F,

K) extends to some φ ∈ Hom

(L,

K).

Proof.

Let

ψ ∈ Hom

(

). If

, then the statement is trivial. So assume

L = F .

Pick

α ∈ L \ F

. Let

∈ F

[

] be the minimal polynomial of

over

Consider

(

)

∈

[

]. Let

be any root of

, which exists since

algebraically closed. Then as before, we can extend

(

) by sending

β. More explicitly, we send

i=0

7→

ψ(a

)β

which is well-defined since any polynomial relation satisfied by

is also

satisfied by β.

Repeat this process finitely many times to get some element in

Hom

(

We will use this lemma to characterize normal extensions.

Theorem. Let

L/K

be a finite extension. Then

L/K

is a normal extension if

and only if L is the splitting field of some f ∈ K[t].

Proof.

Suppose

L/K

is normal. Since

is finite, let

(

, ··· , α

) for some

∈ L

. Let

be the minimal polynomial of

over

. Take

···P

Since

L/K

is normal, each

splits over

. So

splits over

, and

is a

splitting field of f.

For the other direction, suppose that

is the splitting field of some

f ∈ K

[

First we wlog assume

L ⊆

. This is possible since the natural injection

K →

extends to some

L →

by our previous lemma, and we can replace

with

φ(L).

Now suppose

β ∈ L

, and let

be its minimal polynomial. Let

′

be another

root. We want to show it lives in L.

Now consider

(

). By the proof of the lemma, we can produce an embedding

(

)

→

that sends

′

. By the lemma again, this extends to an

embedding of

into

. But any such embedding must send a root of

to a

root of

. So it must send

. In particular,

(

) =

′

∈ L

. So

splits

over L.

This allows us to identify normal extensions easily. The following theorem

then allows us to identify Galois extensions using this convenient tool.

Theorem. Let L/K be a finite extension. Then the following are equivalent:

(i) L/K is a Galois extension.

(ii) L/K is separable and normal.

(iii) L

(

, ··· , α

) and

, the minimal polynomial of

over

, is

separable and splits over L for all i.

Proof.

–

(i)

⇒

(ii): Suppose

L/K

is a Galois extension. Then by definition, this

means

|Hom

(L, L)| = |Aut

(L)| = [L : K].

To show that

L/K

is separable, recall that we proved that an extension is

separable if and only if there is some

such that

|Hom

(

L, E

)

= [

In this case, just pick

. Then we know that the extension is separable.

To check normality, let

α ∈ L

, and let

be its minimal polynomial over

K. We know that

|Root

(L)| = |Hom

(K[t]/⟨P

⟩, L)| = |Hom

(K(α), L)|.

But since

|Hom

(

L, L

)

= [

] and

(

) is a subfield of

, this implies

|Hom

(K(α), L)| = [K(α) : K] = deg P

Hence we know that

|Root

(L)| = deg P

So P

splits over L.

–

(ii)

⇒

(iii): Just pick

, ··· , α

such that

(

, ··· , α

). Then these

polynomials are separable since the extension is separable, and they split

since

L/K

is normal. In fact, by the primitive element theorem, we can

pick these such that n = 1.

–

(iii)

⇒

(i): Since

(

, ··· , α

) and the minimal polynomials

over

are separable, by a previous theorem, there are some extension

of K such that

|Hom

(L, E)| = [L : K].

To simplify notation, we first replace

with its image inside

under some

-homomorphism

L → E

, which exists since

|Hom

(

L, E

)

= [

]

So we can assume L ⊆ E.

We now claim that the inclusion

Hom

(L, L) → Hom

(L, E)

is a surjection, hence a bijection. Indeed, if

L → E

, then

takes

(

), which is a root of

. Since

splits over

, we know

(

)

∈ L

for all i. Since L is generated by these α

, it follows that φ(L) ⊆ L.

Thus, we have

[L : K] = |Hom

(L, E)| = |Hom

(L, L)|,

and the extension is Galois.

From this, it follows that if

L/K

is Galois, and we have an intermediate field

K ⊆ F ⊆ L, then L/F is also Galois.

Corollary. Let

be a field and

f ∈ K

[

] be a separable polynomial. Then the

splitting field of f is Galois.

This is one of the most crucial examples.

2.8 The fundamental theorem of Galois theory

Finally, we can get to the fundamental theorem of Galois theory. Roughly, given

a Galois extension

K ⊆ L

, the fundamental theorem tell us there is a one-to-one

correspondence between intermediate field extensions

K ⊆ F ⊆ L

and subgroups

of the automorphism group Gal(L/K).

Given an intermediate field

, we can obtain a subgroup of

Gal

(

L/K

) by

looking at the automorphisms that fix

. To go the other way round, given a

subgroup

H ≤ Gal

(

L/K

), we can obtain a corresponding field by looking at the

field of elements that are fixed by everything in

. This is known as the fixed

field, and can in general be defined even for non-Galois extensions.

Definition (Fixed field). Let

L/K

be a field extension,

H ≤ Aut

(

) a

subgroup. We define the fixed field of H as

= {α ∈ L : φ(α) = α for all φ ∈ H}.

It is easy to see that L

is an intermediate field K ⊆ L

⊆ L.

Before we get to the fundamental theorem, we first prove Artin’s lemma.

This in fact proves part of the results in the fundamental theorem, but is also

useful on its own right.

Lemma (Artin’s lemma). Let

L/K

be a field extension and

H ≤ Aut

(

) a

finite subgroup. Then L/L

is a Galois extension with Aut

(L) = H.

Note that we are not assuming that L/K is Galois, or even finite!

Proof. Pick any α ∈ L. We set

{α

, ··· , α

} = {φ(α) : φ ∈ H},

where

are distinct. Here we are allowing for the possibility that

(

) =

(

)

for some distinct φ, ψ ∈ H.

By definition, we clearly have n < |H|. Let

f =

(t − α

) ∈ L[t].

We know that any

φ ∈ H

gives an homomorphism

[

]

→ L

[

], and any such

map fixes

because

just permutes the

. Thus, the coefficients of

are in

, and thus f ∈ L

[t].

Since

id ∈ H

, we know that

(

) = 0. So

is algebraic over

. Moreover,

if q

is the minimal polynomial of α over L

, then q

| f in L

[t]. Hence

(α) : L

] = deg q

≤ deg f ≤ |H|.

Further, we know that

has distinct roots. So

is separable, and so

separable. So it follows that L/L

is a separable extension.

We next show that

L/L

is simple. This doesn’t immediately follow from

the primitive element theorem, because we don’t know it is a finite extension

yet, but we can still apply the theorem cleverly.

Pick

α ∈ L

such that [

(

) :

] is maximal. This is possible since

(α) : L

] is bounded by |H|. The claim is that L = L

(α).

We pick an arbitrary

β ∈ L

, and will show that this is in

(

). By the

above arguments,

⊆ L

(

α, β

) is a finite separable extension. So by the

primitive element theorem, there is some

λ ∈ L

such that

(

α, β

) =

(

Note that we must have

(λ) : L

] ≥ [L

(α) : L

By maximality of [

(

) :

], we must have equality. So

(

) =

(

). So

β ∈ L

(α). So L = L

(α).

Finally, we show it is a Galois extension. Let L = L

(α). Then

[L : L

] = [L

(α) : L

] ≤ |H| ≤ |Aut

(L)|

Recall that we have previously shown that for any extension

L/L

, we have

|Aut

(L)| ≤ [L : L

]. Hence we must have equality above. So

[L : L

] = |Aut

(L)|.

So the extension is Galois. Also, since we know that

H ⊆ Aut

(

), we must

have H = Aut

(L).

Theorem. Let

L/K

be a finite field extension. Then

L/K

is Galois if and only

if L

= K, where H = Aut

(L).

Proof.

(

⇒

) Suppose

L/K

is a Galois extension. We want to show

Using Artin’s lemma (and the definition of H), we have

[L : K] = |Aut

(L)| = |H| = |Aut

(L)| = [L : L

]

So [L : K] = [L : L

]. So we must have L

= K.

(⇐) By the lemma, K = L

⊆ L is Galois.

This is an important theorem. Given a Galois extension

L/K

, this gives us

a very useful test of when elements of

α ∈ L

are in fact in

. We will use this a

lot.

Finally, we get to the fundamental theorem.

Theorem (Fundamental theorem of Galois theory). Assume

L/K

is a (finite)

Galois extension. Then

(i) There is a one-to-one correspondence

H ≤ Aut

(L) ←→ intermediate fields K ⊆ F ⊆ L.

This is given by the maps

H 7→ L

and

F 7→ Aut

(

) respectively.

Moreover, |Aut

(L) : H| = [L

: K].

(ii) H ≤ Aut

(

) is normal (as a subgroup) if and only if

is a normal

extension if and only if L

/K is a Galois extension.

(iii)

H ◁ Aut

(

), then the map

Aut

(

)

→ Aut

(

) by the restriction

map is well-defined and surjective with kernel isomorphic to H, i.e.

Aut

(L)

= Aut

Proof. Note that since L/K is a Galois extension, we know

|Aut

(L)| = |Hom

(L, L)| = [L : K],

By a previous theorem, for any intermediate field

K ⊆ F ⊆ L

, we know

|Hom

(

F, L

)

= [

] and the restriction map

Hom

(

L, L

)

→ Hom

(

F, L

)

is surjective.

(i)

The maps are already well-defined, so we just have to show that the maps

are inverses to each other. By Artin’s lemma, we know that

Aut

(

and since

L/F

is a Galois extension, the previous theorem tells that

Aut

(L)

. So they are indeed inverses. The formula relating the index

and the degree follows from Artin’s lemma.

(ii)

Note that for every

φ ∈ Aut

(

), we have that

ϕHϕ

−1

φL

, since

α ∈ L

ϕHϕ

−1

iff

(

−1

(

))) =

for all

ψ ∈ H

iff

(

−1

(

)) =

−1

(

)

for all ψ ∈ H iff α ∈ φL

. Hence H is a normal subgroup if and only if

φ(L

) = L

for all φ ∈ Aut

(L). (∗)

Assume (

∗

). We want to first show that

Hom

(

, L

) =

Hom

(

, L

Let

ψ ∈ Hom

(

, L

). Then by the surjectivity of the restriction map

Hom

(

L, L

)

→ Hom

(

, L

must be the restriction of some

ψ ∈

Hom

(

L, L

). So

fixes

by (

∗

). So

sends

. So

ψ ∈

Hom

, L

). So we have

|Aut

)| = |Hom

, L

)| = |Hom

, L)| = [L

: K].

So L

/K is Galois, and hence normal.

Now suppose

is a normal extension. We want to show this implies

(

∗

). Pick any

α ∈ L

and

φ ∈ Aut

(

). Let

be the minimal polynomial

over

. So

(

) is a root of

(since

fixes

∈ K

, and hence

maps roots to roots). Since

is normal,

splits over

. This

implies that φ(α) ∈ L

. So φ(L

) = L

Hence,

is a normal subgroup if and only if

(

) =

if and only if

/K is a Galois extension.

(iii)

Suppose

is normal. We know that

Aut

(

) =

Hom

(

L, L

) restricts

Hom

(

, L

) surjectively. To show that we in fact have restriction

Aut

(

), by the proof above, we know that

(

) =

for all

φ ∈ Aut

(

). So this does restrict to an automorphism of

. In other

words, the map

Aut

(

)

→ Aut

(

) is well-defined. It is easy to see

this is a group homomorphism.

Finally, we have to calculate the kernel of this homomorphism. Let

be the kernel. Then by definition,

E ⊇ H

. So it suffices to show that

|E|

|H|

. By surjectivity of the map and the first isomorphism theorem

of groups, we have

|Aut

(L)|

|E|

= |Aut

)| = [L

: K] =

[L : K]

[L : L

]

|Aut

(L)|

|H|

noting that

and

L/K

are both Galois extensions, and

|H|

= [

K] by Artin’s lemma. So |E| = |H|. So we must have E = H.

Example. Let

be an odd prime, and

be a primitive

th root of unity. Given

a (square-free) integer

, when is

√

(

)? We know that

√

n ∈ Q

(

) if and

only if

(

√

)

⊆ Q

(

). Moreover, [

(

√

) :

] = 2, i.e.

(

√

) is a quadratic

extension.

We will later show that

Gal

(

)

∼

(

Z/pZ

)

∗

∼

p−1

. Then by the

fundamental theorem of Galois theory, quadratic extensions contained in

(

)

correspond to index 2-subgroups of

Gal

(

)

). By general group theory,

there is exactly one such subgroup. So there is exactly one square-free

such

that

(

√

)

⊆ Q

(

) (since all quadratic extensions are of the form

(

√

)),

given by the fixed field of the index 2 subgroup of (Z/pZ)

∗

Now we shall try to find some square root lying in

(

). We will not fully

justify the derivation, since we can just square the resulting number to see that

it is correct. We know the general element of Q(ζ

) looks like

p−1

k=0

We know

Gal

(

)

∼

(

Z/pZ

)

∗

acts by sending

7→ ζ

for each

n ∈

(

Z/pZ

)

∗

and the index 2 subgroup consists of the quadratic residues. Thus, if an element

is fixed under the action of the quadratic residues, the quadratic residue powers

all have the same coefficient, and similarly for the non-residue powers.

If we wanted this to be a square root, then the action of the remaining

elements of

Gal

(

)

) should negate this object. Since these elements swap

the residues and the non-residues, we would want to have something like

= 1 if

is a quadratic residue, and

−

1 if it is a non-residue, which is just the Legendre

symbol! So we are led to try to square

τ =

p−1

k=1





It is an exercise in the Number Theory example sheet to show that the square

of this is in fact



−1



So we have

√

p ∈ Q(ζ

) if p ≡ 1 (mod 4), and

√

−p ∈ Q(ζ

) if p ≡ 3 (mod 4).

2.9 Finite fields

We’ll have a slight digression and look at finite fields. We adopt the notation

where

is always a prime number, and

Z/⟨p⟩

. It turns out finite fields are

rather simple, as described in the lemma below:

Lemma. Let K be a finite field with q = |K| element. Then

(i) q = p

for some d ∈ N, where p = char K > 0.

(ii)

Let

− t

. Then

(

) = 0 for all

α ∈ K

. Moreover,

is the splitting

field of f over F

This means that a finite field is completely determined by the number of

elements.

Proof.

(i)

Consider the set

{m·

}

m∈Z

, where 1

is the unit in

and

m·

represents

repeated addition. We can identify this with

. So we have the extension

⊆ K. Let d = [K : F

]. Then q = |K| = p

(ii)

Note that

∗

K \ {

}

is a finite multiplicative group with order

q −

Then by Lagrange’s theorem, α

q−1

= 1 for all α ∈ K

∗

. So α

− α = 0 for

all α = 0. The α = 0 case is trivial.

Now every element in

is a root of

. So we need to check that all roots

are in

. Note that the derivative

′

q−1

−

1 =

−

1 (since

is a

power of the characteristic). So

′

(

) =

−



= 0 for all

α ∈ K

. So

and

′

have no common roots. So

has no repeated roots. So

contains

distinct roots of f. So K is a splitting field.

Lemma. Let q = p

, q

′

= p

′

, where d, d

′

∈ N. Then

(i)

There is a finite field

with exactly

elements, which is unique up to

isomorphism. We write this as F

(ii) We can embed F

⊆ F

′

iff d | d

′

Proof.

(i)

Let

− t

, and let

be a splitting field of

over

. Let

Root

(

). The objective is to show that

. Then we will have

|K|

|L|

|Root

(

)

deg f

, because the proof of the previous

lemma shows that f has no repeated roots.

To show that

, by definition, we have

L ⊆ K

. So we need to show

every element in

is in

. We do so by showing that

itself is a field.

Then since

contains all the roots of

and is a subfield of the splitting

field K, we must have K = L.

It is straightforward to show that L is a field: if α, β ∈ L, then

(α + β)

= α

+ β

= α + β.

So α + β ∈ L. Similarly, we have

(αβ)

= α

= αβ.

So αβ ∈ L. Also, we have

(α

−1

)

= (α

)

−1

= α

−1

So α

−1

∈ L. So L is in fact a field.

Since any field of size

is a splitting field of

, and splitting fields are

unique to isomorphism, we know that K is unique.

(ii)

Suppose

⊆ F

′

. Then let

= [

′

]. So

′

. So

′

. So

d | d

′

On the other hand, suppose

d | d

′

. Let

′

. We let

′

− t

. Then

for any α ∈ F

, we have

f(α) = α

′

− α = α

− α = (···((α

)

···)

− α = α − α = 0.

Since

′

is the splitting field of

, all roots of

are in

′

. So we know

that F

⊆ F

′

Note that if

is the algebraic closure of

, then

⊆

for every

We then have

[

k∈N

because any α ∈

is algebraic over F

, and so belongs to some F

Definition. Consider the extension

, where

is a power of

. The

Frobenius Fr

: F

→ F

is defined by α 7→ α

This is a homomorphism precisely because the field is of characteristic zero.

In fact, Fr

∈ Aut

), since α

= α for all α ∈ F

The following two theorems tells us why we care about the Frobenius.

Theorem. Consider

. Then

is an element of order

as an element

of Aut

Proof.

For all

α ∈ F

, we have

(

) =

. So the order of

divides

If m | n, then the set

{α ∈ F

: Fr

(α) = α} = {α ∈ F

: α

= α} = F

So if m is the order of Fr

, then F

= F

. So m = n.

Theorem. The extension

is Galois with Galois group

Gal

(

) =

Aut

)

∼

Z/nZ, generated by Fr

Proof.

The multiplicative group

∗

\ {

}

is finite. We have previously

seen that multiplicative groups of finite fields are cyclic. So let

be a generator

of this group. Then

(

). Let

be the minimal polynomial of

over

. Then since Aut

) has an element of order n, we get

n ≤ |Aut

)| = |Hom

(α), F

)|.

Since F

(α) is generated by one element, we know

|Hom

(α), F

)| = |Root

So we have

n ≤ |Root

)| ≤ deg P

= [F

: F

] = n.

So we know that

|Aut

)| = [F

: F

] = n.

So F

is a Galois extension.

Since

|Aut

(

)

, it has to be generated by

, since this has order

. In

particular, this group is cyclic.

We see that finite fields are rather nice — there is exactly one field of order

for each

and prime

, and these are all of the finite fields. All extensions

are Galois and the Galois group is a simple cyclic group.

Example. Consider F

. We can write

= {0, 1} ⊆ F

= {0, 1, α, α

where α is a generator of F

∗

. Define φ ∈ Aut

) by φ(α) = α

. Then

Aut

) = {id, φ}

since it has order 2.

Note that we can also define the Frobenius

→

, where

α 7→ α

Then

is the elements of

fixed by

. So we can recover this subfield by

just looking at the Frobenius.

3 Solutions to polynomial equations

We have now proved the fundamental theorem of Galois theory, and this gives a

one-to-one correspondence between (intermediate) field extensions and subgroups

of the Galois group. That is our first goal achieved. Our next big goal is to use

this Galois correspondence to show that, in general, polynomials of degree 5 or

more cannot be solved by radicals.

First of all, we want to make this notion of “solving by radicals” precise.

We all know what this means if we are working over

, but we need to be very

precise when working with arbitrary fields.

For example, we know that the polynomial

−

∈ Q

[

] can be “solved

by radicals”. In this case, we have

Root

√

5, µ

√

5, µ

√

5},

where

= 1

, µ 

= 1. In general fields, we want to properly define the analogues

of µ and

√

These will correspond to two different concepts. The first is cyclotomic

extensions, where the extension adds the analogues of

, and the second is

Kummer extensions, where we add things like

√

Then, we would say a polynomial is soluble by radicals if the splitting field

of the polynomial can be obtained by repeatedly taking cyclotomic and Kummer

extensions.

3.1 Cyclotomic extensions

Definition (Cyclotomic extension). For a field

, we define the

th cyclotomic

extension to be the splitting field of t

− 1.

Note that if

is a field and

is the

th cyclotomic extension, then

Root

−1

(

) is a subgroup of multiplicative group

∗

L \ {

}

. Since this is a

finite subgroup of L

∗

, it is a cyclic group.

Moreover, if

char K

= 0 or 0

< char K ∤ n

, then (

−

′

n−1

and this

has no common roots with

−

1. So

−

1 has no repeated roots. In other

words, t

− 1 has n distinct roots. So as a group,

Root

−1

(L)

∼

Z/nZ.

In particular, this group has at least one element µ of order n.

Definition (Primitive root of unity). The

th primitive root of unity is an

element of order n in Root

−1

(L).

These elements correspond to the elements of the multiplicative group of

units in Z/nZ, written (Z/nZ)

The next theorem tells us some interesting information about these roots

and some related polynomials.

Theorem. For each

d ∈ N

, there exists a

th cyclotomic monic polynomial

∈ Z[t] satisfying:

(i) For each n ∈ N, we have

− 1 =

d|n

(ii) Assume char K = 0 or 0 < char K ∤ n. Then

Root

(L) = {nth primitive roots of unity}.

Note that here we have an abuse of notation, since

is a polynomial in

[

], not

[

], but we can just use the canonical map

[

]

→ K

[

] mapping

1 to 1 and t to t.

Proof.

We do induction on

to construct

. When

= 1, let

t −

1. Then

(i) and (ii) hold in this case, trivially.

Assume now that (i) and (ii) hold for smaller values of n. Let

f =

d|n,d<n

By induction,

f ∈ Z

[

]. Moreover, if

d | n

and

d < n

, then

(

−

1) because

(

−

(

−

1). We would like to say that

also divides

−

1. However, we

have to be careful, since to make this conclusion, we need to show that

and

′

have no common roots for distinct d, d

′

| n (and d, , d

′

< n).

Indeed, by induction, φ

and φ

′

have no common roots because

Root

(L) = {dth primitive roots of unity},

Root

′

(L) = {d

′

th primitive roots of unity},

and these two sets are disjoint (or else the roots would not be primitive).

Therefore

and

′

have no common irreducible factors. Hence

f | t

−

1. So

we can write

− 1 = fφ

where

∈ Q

[

]. Since

is monic,

has integer coefficients. So indeed

∈ Z[t]. So the first part is proven.

To prove the second part, note that by induction,

Root

(L) = {non-primitive nth roots of unit},

since all nth roots of unity are dth primitive roots of unity for some smaller d.

Since

fφ

−

contains the remaining, primitive

th roots of unit.

Since

−

1 has no repeated roots, we know that

does not contain any extra

roots. So

Root

(L) = {nth primitive roots of unity}.

These

are what we use to “build up” the polynomials

−

1. These

will later serve as a technical tool to characterize the Galois group of the

cyclotomic extension of Q.

Before we an reach that, we first take a tiny step, and prove something that

works for arbitrary fields first.

Theorem. Let

be a field with

char K

= 0 or 0

< char K ∤ n

. Let

be the

th cyclotomic extension of

. Then

L/K

is a Galois extension, and there is an

injective homomorphism θ : Gal(L/K) → (Z/nZ)

In addition, every irreducible factor of φ

(in K[t]) has degree [L : K].

The important thing about our theorem is the homomorphism

θ : Gal(L/K) → (Z/nZ)

In general, we don’t necessarily know much about

Gal

(

L/K

), but the group

(

Z/nZ

)

is well-understood. In particular, we now know that

Gal

(

L/K

) is

abelian.

Proof. Let µ be an nth primitive root of unity. Then

Root

−1

(L) = {1, µ, µ

, ··· , µ

n−1

}

is a cyclic group of order

generated by

. We first construct the homomorphism

Aut

(

)

→

(

Z/nZ

)

as follows: for each

φ ∈ Aut

(

is completely

determined by the value of

(

) since

(

). Since

is an automorphism, it

must take an

th primitive root of unity to another

th primitive root of unity.

(

) =

for some

such that (

i, n

) = 1. Now let

(

) =

i ∈

(

Z/nZ

)

. Note

that this is well-defined since if µ

= µ

, then i − j has to be a multiple of n.

Now it is easy to see that if

φ, ψ ∈ Aut

(

) are given by

(

) =

, and

(

) =

, then

φ ◦ ψ

(

) =

(

) =

. So

(

φψ

) =

(

)

(

). So

is a

group homomorphism.

Now we check that

is injective. If

(

) =

(note that (

Z/nZ

)

is a

multiplicative group with unit 1), then φ(µ) = µ. So φ = id.

Now we show that

L/K

is Galois. Recall that

(

), and let

a minimal polynomial of

over

. Since

is a root of

−

1, we know that

| t

−

1. Since

−

1 has no repeated roots,

has no repeated roots. So

is separable. Moreover,

splits over

−

1 splits over

. So the extension

is separable and normal, and hence Galois.

Applying the previous theorem, each irreducible factor

is a minimal

polynomial of some nth primitive root of unity, say λ. Then L = K(λ). So

deg g = deg P

= [K(λ) : K] = [L : K].

Example. We can calculate the following in Q[t].

(i) φ

= t − 1.

(ii) φ

= t + 1 since t

− 1 = φ

(iii) φ

= t

+ t + 1.

(iv) φ

= t

+ 1.

These are rather expected. Now take

. Then 1 =

−

1. So we might be

able to further decompose these polynomials. For example,

+ 1 =

t −

1 in

So we have

= t

+ 1 = t

− 1 = φ

So in

is not irreducible. Similarly, if we have too much time, we can show

that

= (t

+ t + 1)(t

+ t

+ 1).

is not irreducible. However, they are irreducible over the rationals, as we

will soon see.

So far, we know

Gal

(

L/K

) is an abelian group, isomorphic to a subgroup of

(

Z/nZ

)

. However, we are greedy and we want to know more. The following

lemma tells us when this θ is an isomorphism.

Lemma. Under the notation and assumptions of the previous theorem,

irreducible in K[t] if and only if θ is an isomorphism.

Proof.

(

⇒

) Suppose

is irreducible. Recall that

Root

(

) is exactly the

primitive roots of unity. So if

is an

th primitive root of unity, then

, the

minimal polynomial of

over

. In particular, if

is also an

th primitive

root of unity, then

. This implies that there is some

∈ Aut

(

) such

that φ

(µ) = λ.

Now if

i ∈

(

Z/nZ

)

, then taking

, this shows that we have

∈

Aut

(L) such that θ(φ

) =

i. So θ is surjective, and hence an isomorphism.

(

⇔

) Suppose that

is an isomorphism. We will reverse the above argument

and show that all roots have the same minimal polynomial. Let

be a

primitive root of unity, and pick

i ∈

(

Z/nZ

)

, and let

. Since

is an

isomorphism, there is some

∈ Aut

(

) such that

(

) =

, i.e.

(

) =

= λ. Then we must have P

= P

Since every

th primitive root of unity is of the form

(with (

i, n

) = 1), this

implies that all

th primitive roots have the same minimal polynomial. Since

the roots of

are all the

th primitive roots of unity, its irreducible factors are

exactly the minimal polynomials of the primitive roots. Moreover, φ

does not

have repeated roots. So φ

= P

. In particular, φ

is irreducible.

We want to apply this lemma to the case of rational numbers. We want to

show that

is an isomorphism. So we have to show that

is irreducible in

Q[t].

Theorem. φ

is irreducible in Q[t]. In particular, it is also irreducible in Z[t].

Proof.

As before, this can be achieved by showing that all

th primitive roots

have the same minimal polynomial. Moreover, let

be our favorite

th primitive

root. Then all other primitive roots

are of the form

, where (

i, n

) = 1. By

the fundamental theorem of arithmetic, we can write

as a product

···q

Hence it suffices to show that for all primes

q ∤ n

, we have

. Noting

that µ

is also an nth primitive root, this gives

= P

(µ

)

= P

= ··· = P

···q

= P

So we now let

be an

th primitive root,

be its minimal polynomial. Since

µ is a root of φ

, we can write P

| φ

inside Q[t]. So we can write

= P

Since

and

are monic,

is also monic. By Gauss’ lemma, we must have

, R ∈ Z[t].

Note that showing

is the same as showing

is a root of

, since

deg P

. So suppose it’s not. Since

is an

th primitive root of unity,

it is a root of

. So

must be a root of

. Now let

(

). Then

is a

root of S, and so P

| S.

We now reduce mod

. For any polynomial

f ∈ Z

[

], we write the result of

reducing the coefficients mod

. Then we have

R(t

)

R(t)

. Since

divides

(by Gauss’ lemma), we know

and

R(t)

have common roots. But

, and so this implies

has repeated roots. This is impossible since

divides

−

1, and since

q ∤ n

, we know the derivative of

−

1 does not

vanish at the roots. So we are done.

Corollary. Let

and

be the

th cyclotomic extension of

. Then the

injection θ : Gal(L/Q) → (Z/nZ)

is an isomorphism.

Example. Let

be a prime number, and

d ∈ N

. Consider

, a field

with

elements, and let

be the

th cyclotomic extension of

(where

p ∤ n

Then we have a homomorphism θ : Gal(L/F

) → (Z/nZ)

We have previously shown that

Gal

(

L/F

) must be a cyclic group. So if

(

Z/nZ

)

is non-cyclic, then

is not an isomorphism, and

is not irreducible

in F

[t].

For example, take p = q = 7 and n = 8. Then

(Z/8Z)

= {

is not cyclic, because manual checking shows that there is no element of order 4.

Hence

Gal

(

L/F

)

→

(

)

is not an isomorphism, and

is not irreducible

in F

[t].

3.2 Kummer extensions

We shall now consider a more general case, and study the splitting field of

− λ ∈ K

[

]. As we have previously seen, we will need to make use of the

primitive roots of unity.

The definition of a Kummer extension will involve a bit more that it being

the splitting field of

− λ

. So before we reach the definition, we first studying

some properties of an arbitrary splitting field of

−λ

, and use this to motivate

the definition of a Kummer extension.

Definition (Cyclic extension). We say a Galois extension

L/K

is cyclic is

Gal(L/K) is a cyclic group.

Theorem. Let

be a field,

λ ∈ K

non-zero,

n ∈ N

char K

= 0 or 0

< char K ∤

n. Let L be the splitting field of t

− λ. Then

(i) L contains an nth primitive root of unity, say µ.

(ii) L/K

(

) is a cyclic (and in particular Galois) extension with degree [

K(µ)] | n.

(iii) [L : K(µ)] = n if and only if t

− λ is irreducible in K(µ)[t].

Proof.

(i)

Under our assumptions,

− λ

and (

− λ

)

′

n−1

have no common

roots in L. So t

− λ has distinct roots in L, say α

, ··· , α

∈ L.

It then follows by direct computation that

−1

, α

−1

, ··· , α

−1

are

distinct roots of unity, i.e. roots of

−

1. Then one of these, say

must

be an nth primitive root of unity.

(ii)

We know

L/K

(

) is a Galois extension because it is the splitting field of

the separable polynomial t

− λ.

To understand the Galois group, we need to know how this field exactly

looks like. We let

be any root of

−λ

. Then the set of all roots can be

written as

{α, µα, µ

α, ··· , µ

n−1

α}

Then

L = K(α

, ··· , α

) = K(µ, α) = K(µ)(α).

Thus, any element of

Gal

(

L/K

(

)) is uniquely determined by what it sends

to, and any homomorphism must send

to one of the other roots of

− λ, namely µ

α for some i.

Define a homomorphism

Gal

(

L/K

(

))

→ Z/nZ

that sends

to the

corresponding i (as an element of Z/nZ, so that it is well-defined).

It is easy to see that σ is an injective group homomorphism. So we know

Gal

(

L/K

(

)) is isomorphic to a subgroup of

Z/nZ

. Since the subgroup of

any cyclic group is cyclic, we know that

Gal

(

L/K

(

)) is cyclic, and its size

is a factor of

by Lagrange’s theorem. Since

|Gal

(

L/K

(

))

= [

(

)]

by definition of a Galois extension, it follows that [L : K(µ)] divides n.

(iii)

We know that [

(

)] = [

(

µ, α

) :

(

)] =

deg q

. So [

(

)] =

if and only if

deg q

. Since

is a factor of

− λ

deg q

if and

only if

−λ

. This is true if and only if

−λ

is irreducible

(

)[

So done.

Example. Consider

+ 2

∈ Q

[

]. Let

√

−1

, which is a 4th primitive root

of unity. Now

+ 2 = (t − α)(t + α)(t − µα)(t + µα),

where

√

−2

is one of the roots of

+ 2. Then we have the field extension

Q ⊆ Q(µ) ⊆ Q(µ, α), where Q(µ, α) is a splitting field of t

+ 2.

Since

√

−2 ∈ Q

(

), we know that

+ 2 is irreducible in

(

)[

] by looking at

the factorization above. So by our theorem,

(

)

⊆ Q

(

µ, α

) is a cyclic extension

of degree exactly 4.

Definition (Kummer extension). Let

be a field,

λ ∈ K

non-zero,

n ∈ N

char K

= 0 or 0

< char K ∤ n

. Suppose

contains an

th primitive root of

unity, and

is a splitting field of

− λ

. If

deg

[

] =

, we say

L/K

is a

Kummer extension.

Note that we used to have extensions

K ⊆ K

(

)

⊆ L

. But if

already

contains a primitive root of unity, then

(

). So we are left with the cyclic

extension K ⊆ L.

To following technical lemma will be useful:

Lemma. Assume

L/K

is a field extension. Then

Hom

(

L, L

) is linearly in-

dependent. More concretely, let

, ··· , λ

∈ L

and

, ··· , φ

∈ Hom

(

L, L

)

distinct. Suppose for all α ∈ L, we have

(α) + ··· + λ

(α) = 0.

Then λ

= 0 for all i.

Proof. We perform induction on n.

Suppose we have some λ

∈ L and φ

∈ Hom

(L, L) such that

(α) + ··· + λ

(α) = 0.

The

= 1 case is trivial, since

= 0 implies

= 0 (the zero homomorphism

does not fix K).

Otherwise, since the homomorphisms are distinct, pick

β ∈ L

such that

(β) = φ

(β). Then we know that

(αβ) + ··· + λ

(αβ) = 0

for all α ∈ L. Since φ

are homomorphisms, we can write this as

(α)φ

(β) + ··· + λ

(α)φ

(β) = 0.

On the other hand, by just multiplying the original equation by φ

(β), we get

(α)φ

(β) + ··· + λ

(α)φ

(β) = 0.

Subtracting the equations gives

(α)(φ

(β) − φ

(β)) + ··· + λ

n−1

(α)(φ

n−1

(β) − φ

(β)) = 0

for all

α ∈ L

. By induction,

(

)

− φ

(

)) = 0 for all 1

≤ i ≤ n −

1. In

particular, since φ

(β) − φ

(β) = 0, we have λ

= 0. Then we are left with

(α) + ··· + λ

(α) = 0.

Then by induction again, we know that all coefficients are zero.

Theorem. Let

be a field,

n ∈ N

char K

= 0 or 0

< char K ∤ n

. Suppose

contains an

th primitive root of unity, and

L/K

is a cyclic extension of degree

[L : K] = n. Then L/K is a Kummer extension.

This is a rather useful result. If we look at the splitting field of a polynomial

− λ

, even if the ground field includes the right roots of unity, a priori, this

doesn’t have to be a Kummer extension if it doesn’t have degree

. But we

previously showed that the extension must be cyclic. And so this theorem shows

that it is still a Kummer extension of some sort.

This is perhaps not too surprising. For example, if, say,

= 4 and

secretly a square, then the splitting field of

− λ

is just the splitting field of

−

√

λ.

Proof.

Our objective here is to find a clever

λ ∈ K

such that

is the splitting

field of t

− λ. To do so, we will have to hunt for a root β of t

− λ in L.

Pick

a generator of

Gal

(

L/K

). We know that if

were a root of

− λ

then

(

) =

−1

for some primitive

th root of unity

. Thus, we want to find

an element that satisfies such a property.

By the previous lemma, we can find some α ∈ L such that

β = α + µφ(α) + µ

(α) + ··· + µ

n−1

(α) = 0.

Then, noting that

is the identity and

fixes

µ ∈ K

, we see that

trivially

satisfies

φ(β) = φ(α) + µφ

α + ··· + µ

n−1

(α) = µ

−1

β,

In particular, we know that φ(β) ∈ K(β).

Now pick

. Then

(

) =

−n

. So

fixes

. Since

generates Gal(L/K), we know all automorphisms of L/K fixes β

. So β

∈ K.

Now the roots of

− λ

are

β, µβ, ··· , µ

n−1

. Since these are all in

, we

know K(β) is the splitting field of t

− λ.

Finally, to show that

(

) =

, we observe that

id, φ|

K(β)

, . . . ,

K(β)

are

distinct elements of

Aut

(

)) since they do different things to

. Recall our

previous theorem that

[K(β) : K] ≥ |Aut

(K(β))|.

So we know that n = [L : K] = [K(β) : K]. So L = K(β). So done.

Example. Consider

−

∈ Q

[

], and

a third primitive root of unity. Then

we have the extension

Q ⊆ Q

(

)

⊆ Q

(

µ,

√

). Then

Q ⊆ Q

(

) is a cyclotomic

extension of degree 2, and

(

)

⊆ Q

(

µ,

√

) is a Kummer extension of degree 3.

3.3 Radical extensions

We are going to put these together and look at radical extensions, which allows

us to characterize what it means to “solve a polynomial with radicals”.

Definition (Radical extension). A field extension

L/K

is radical if there is

some further extension E/L and with a sequence

K = E

⊆ E

⊆ ··· ⊆ E

= E,

such that each

⊆ E

i+1

is a cyclotomic or Kummer extension, i.e.

i+1

is a

splitting field of t

− λ

i+1

over E

for some λ

i+1

∈ E

Informally, we say

i+1

is obtained by adding the roots “

i+1

” to

Hence we interpret a radical extension as an extension that only adds radicals.

Definition (Solubility by radicals). Let

be a field, and

f ∈ K

[

. We say

f is soluble by radicals if the splitting field of f is a radical extension of K.

This means that f can be solved by radicals of the form

√

Let’s go back to our first lecture and describe what we’ve done in the language

we’ve developed in the course.

Example. As we have shown in lecture 1, any polynomial

f ∈ Q

[

] of degree at

most 4 can be solved by radicals.

For example, assume

deg f

= 3. So

. Let

be the

splitting field of

. Recall we reduced the problem of “solving”

to the case

= 0 by the substitution

x 7→ x −

. Then we found our

β, γ ∈ C

such that

each root

can be written as a linear combination of

and

(and

), i.e.

L ⊆ Q(β, γ, µ).

Then we showed that

{β

, γ

} =

(

−27c ±

(27c)

+ 4 × 27b

)

We now let

λ =

(27c)

+ 4 × 27b

Then we have the extensions

Q ⊆ Q(λ) ⊆ Q(λ, µ) ⊆ Q(λ, µ, β),

and also

Q ⊆ L ⊆ Q(λ, µ, β).

Note that the first extension

Q ⊆ Q

(

) is a Kummer extension since it is a

splitting field of

−λ

. Then

(

)

⊆ Q

(

λ, µ

) is the third cyclotomic extension.

Finally,

(

λ, µ

)

⊆ Q

(

λ, µβ

) is a Kummer extension, the splitting field of

−β

So Q ⊆ L is a radical extension.

Let’s go back to the definition of a radical extension. We said

L/K

is radical

if there is a further extension

E/L

that satisfies certain nice properties. It would

be great if

E/K

is actually a Galois extensions. To show this, we first need a

technical lemma.

Lemma. Let

L/K

be a Galois extension,

char K

= 0,

γ ∈ L

and

the splitting

field of

−γ

over

. Then there exists a further extension

E/F

such that

E/L

is radical and E/K is Galois.

Here we have the inclusions

K ⊆ L ⊆ F ⊆ E,

where

K, L

and

are given and

is what we need to find. The idea of the proof

is that we just add in the “missing roots” to obtain

so that

E/K

is Galois,

and doing so only requires performing cyclotomic and Kummer extensions.

Proof.

Since we know that

L/K

is Galois, we would rather work in

than in

However, our

is in

, not

. Hence we will employ a trick we’ve used before,

where we introduce a new polynomial

, and show that its coefficients are fixed

Gal

(

L/K

), and hence in

. Then we can look at the splitting field of

its close relatives.

Let

f =

ϕ∈Gal(L/K)

− φ(γ)).

Each

φ ∈ Gal

(

L/K

) induces a homomorphism

[

]

→ L

[

]. Since each

φ ∈ Gal

(

L/K

) just rotates the roots of

around, we know that this induced

homomorphism fixes

. Since all automorphisms in

Gal

(

L/K

) fix the coefficients

of f , the coefficients must all be in K. So f ∈ K[t].

Now since

L/K

is Galois, we know that

L/K

is normal. So

is the splitting

field of some

g ∈ K

[

]. Let

be the splitting field of

over

. Then

K ⊆ E

is normal. Since the characteristic is zero, this is automatically separable. So

the extension K ⊆ E is Galois.

We have to show that

L ⊆ E

is a radical extension. We pick our fields as

follows:

– E

= L

– E

= splitting field of t

− 1 over E

– E

= splitting field of t

− γ over E

– E

= splitting field of t

− φ

(γ) over E

– . . .

– E

= E,

where we enumerate Gal(L/K) as {id, φ

, φ

, ···}.

We then have the sequence of extensions

L = E

⊆ E

⊆ ··· ⊆ E

Here

⊆ E

is a cyclotomic extension, and

⊆ E

etc. are

Kummer extensions since they contain enough roots of unity and are cyclic. By

construction, F ⊆ E

. So F ⊆ E.

Theorem. Suppose

L/K

is a radical extension and

char K

= 0. Then there is

an extension E/L such that E/K is Galois and there is a sequence

K = E

⊆ E

⊆ ··· ⊆ E,

where E

⊆ E

i+1

is cyclotomic or Kummer.

Proof. Note that this is equivalent to proving the following statement: Let

K = L

⊆ L

⊆ ···L

be a sequence of cyclotomic or Kummer extensions. Then there exists an

extension

⊆ E

such that

K ⊆ E

is Galois and can be written as a sequence

of cyclotomic or Kummer extensions.

We perform induction on s. The s = 0 case is trivial.

s >

0, then by induction, there is an extension

M/L

s−1

such that

M/K

Galois and is a sequence of cyclotomic and Kummer extensions. Now

is a

splitting field of

−γ

over

s−1

for some

γ ∈ L

s−1

. Let

be the splitting field

− γ

over

. Then by the lemma and its proof, there exists an extension

E/M

that is a sequence of cyclotomic or Kummer extensions, and

E/K

is Galois.

s−1

= L

s−1

(

√

γ)

F = M(

√

γ)

However, we already know that

M/K

is a sequence of cyclotomic and Kummer

extensions. So

E/K

is a sequence of cyclotomic and Kummer extension. So

done.

3.4 Solubility of groups, extensions and polynomials

Let

f ∈ K

[

]. We defined the notion of solubility of

in terms of radical

extensions. However, can we decide whether

is soluble or not without resorting

to the definition? In particular, is it possible to decide whether its soluble by

just looking at

Gal

(

L/K

), where

is the splitting field of

over

? It would

be great if we could do so, since groups are easier to understand than fields.

The answer is yes. It turns out the solubility of

corresponds to the solubility

Gal

(

L/K

). Of course, we will have to first define what it means for a group

to be soluble. After that, we will find examples of polynomials

of degree at

least 5 such that

Gal

(

L/K

) is not soluble. In other words, there are polynomials

that cannot be solved by radicals.

Definition (Soluble group). A finite group

is soluble if there exists a sequence

of subgroups

= {1} ◁ ··· ◁ G

◁ G

= G,

where G

i+1

is normal in G

and G

i+1

is cyclic.

Example. Any finite abelian group is solvable by the structure theorem of finite

abelian groups:

∼

⟨n

⟩

× ··· ×

⟨n

⟩

Example. Let

be the symmetric group of permutations of

letters. We

know that G

is soluble since

{1} ◁ A

◁ S

where S

∼

Z/⟨2⟩ and A

/{0}

∼

Z/⟨3⟩.

is also soluble by

{1} ◁ {e, (1 2)(3 4)} ◁ {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ◁ A

◁ S

We can show that the quotients are

Z/⟨

⟩

Z/⟨

⟩

Z/⟨

⟩

and

Z/⟨

⟩

respectively.

How about

for higher

? It turns out they are no longer soluble for

n ≥

To prove this, we first need a lemma.

Lemma. Let G be a finite group. Then

(i) If G is soluble, then any subgroup of G is soluble.

(ii)

A ◁ G

is a normal subgroup, then

is soluble if and only if

and

G/A

are both soluble.

Proof.

(i) If G is soluble, then by definition, there is a sequence

= {1} ◁ ··· ◁ G

◁ G

= G,

such that G

i+1

is normal in G

and G

i+1

is cyclic.

Let

H ∩ G

. Note that

i+1

is just the kernel of the obvious homo-

morphism

→ G

i+1

. So

i+1

◁ H

. Also, by the first isomorphism

theorem, this gives an injective homomorphism

i+1

→ G

i+1

. So

i+1

is cyclic. So H is soluble.

(ii)

(

⇒

) By (i), we know that

is solvable. To show the quotient is soluble,

by assumption, we have the sequence

= {1} ◁ ··· ◁ G

◁ G

= G,

such that

i+1

is normal in

and

i+1

is cyclic. We construct the

sequence for the quotient in the obvious way. We want to define

as the

quotient

, but since

is not necessarily a subgroup of

, we instead

define

to be the image of quotient map

→ G/A

. Then we have a

sequence

= {1} ◁ ··· ◁ E

= G/A.

The quotient map induces a surjective homomorphism

i+1

→

i+1

, showing that E

i+1

are cyclic.

(⇐) From the assumptions, we get the sequences

= {1} ◁ ··· ◁ A

= A

= A ◁ ··· ◁ F

= G

where each quotient is cyclic. So we get a sequence

= {1} ◁ A

◁ ··· ◁ A

= F

◁ F

n−1

◁ ··· ◁ F

= G,

and each quotient is cyclic. So done.

Example. We want to show that

is not soluble if

n ≥

5. It is a well-known

fact that

is a simple non-abelian group, i.e. it has no non-trivial subgroup.

So A

is not soluble. So S

is not soluble.

The key observation in Galois theory is that solubility of polynomials is

related to solubility of the Galois group.

Definition (Soluble extension). A finite field extension

L/K

is soluble if there

is some extension L ⊆ E such that K ⊆ E is Galois and Gal(E/K) is soluble.

Note that this definition is rather like the definition of a radical extension,

since we do not require the extension itself to be “nice”, but just for there to be

a further extension that is nice. In fact, we will soon see they are the same.

Lemma. Let

L/K

be a Galois extension. Then

L/K

is soluble if and only if

Gal(L/K) is soluble.

This means that the whole purpose of extending to

is just to make it a

Galois group, and it isn’t used to introduce extra solubility.

Proof. (⇐) is clear from definition.

(

⇒

) By definition, there is some

E ⊆ L

such that

E/K

is Galois and

Gal

(

E/K

) is soluble. By the fundamental theorem of Galois theory,

Gal

(

L/K

) is

a quotient of

Gal

(

E/K

). So by our previous lemma,

Gal

(

L/K

) is also soluble.

We now come to the main result of the lecture:

Theorem. Let

be a field with

char K

= 0, and

L/K

is a radical extension.

Then L/K is a soluble extension.

Proof.

We have already shown that if we have a radical extension

L/K

, then

there is a finite extension

K ⊆ E

such that

K ⊆ E

is a Galois extension, and

there is a sequence of cyclotomic or Kummer extensions

= K ⊆ E

⊆ ··· ⊆ E

= E.

Let

Gal

(

E/E

). By the fundamental theorem of Galois theory, inclusion of

subfields induces an inclusion of subgroups

= Gal(E/K) ≥ G

≥ ··· ≥ G

= {1}.

In fact,

▷ G

i+1

because

⊆ E

i+1

are Galois (since cyclotomic and Kummer

extensions are). So in fact we have

= Gal(E/K) ▷ G

▷ ··· ▷ G

= {1}.

Finally, note that by the fundamental theorem of Galois theory,

i+1

= Gal(E

i+1

We also know that the Galois groups of cyclotomic and Kummer extensions are

abelian. Since abelian groups are soluble, our previous lemma implies that

L/K

is soluble.

In fact, we will later show that the converse is also true. So an extension is

soluble if and only if it is radical.

Corollary. Let

be a field with

char K

= 0, and

f ∈ K

[

]. If

can be solved

by radicals, then

Gal

(

L/K

) is soluble, where

is the splitting field of

over

Again, we will later show that the converse is also true. However, to construct

polynomials that cannot be solved by radicals, this suffices. In fact, this corollary

is all we really need.

Proof.

We have seen that

L/K

is a Galois extension. By assumption,

L/K

thus a radical extension. By the theorem,

L/K

is also a soluble extension. So

Gal(L/K) is soluble.

To find an

f ∈ Q

[

] which cannot be solved by radicals, it suffices to find an

such that the Galois group is not soluble. We don’t know many non-soluble

groups so far. So in fact, we will find an f such that Gal(L/Q) = S

To do so, we want to relate Galois groups to permutation groups.

Lemma. Let

be a field,

f ∈ K

[

] of degree

with no repeated roots. Let

be the splitting field of

over

. Then

L/K

is Galois and there exist an

injective group homomorphism

Gal(L/K) → S

Proof.

Let

Root

(

) =

{α

, ··· , α

}

. Let

be the minimal polynomial of

over

. Then

| f

implies that

is separable and splits over

. So

L/K

Galois.

Now each

φ ∈ Gal

(

L/K

) permutes the

, which gives a map

Gal

(

L/K

)

→ S

It is easy to show this is an injective group homomorphism.

Note that there is not a unique or naturally-defined injective group homo-

morphism to

. This homomorphism, obviously, depends on how we decide to

number our roots.

Example. Let

= (

−

2)(

−

∈ Q

[

]. Let

be the splitting field of

over

Q. Then the roots are

Root

(L) = {

√

2, −

√

3, −

√

3}.

We label these roots as

, α

in order. Now note that

(

√

and thus [

] = 4. Hence

|Gal

(

L/Q

)

= 4 as well. We can let

Gal

(

L/Q

) =

{id, ϕ, ψ, λ}, where

id(

√

2) =

√

2 id(

√

3) =

√

ϕ(

√

2) = −

√

2 ϕ(

√

3) =

√

ψ(

√

2) =

√

2 ψ(

√

3) = −

√

λ(

√

2) = −

√

2 λ(

√

3) = −

√

Then the image of Gal(L/Q) → S

is given by

{e, (1 2), (3 4), (1 2)(3 4)}.

What we really want to know is if there are polynomials in which this map is

in fact an isomorphism, i.e. the Galois group is the symmetric group. If so, then

we can use this to produce a polynomial that is not soluble by polynomials.

To find this, we first note a group-theoretic fact.

Lemma. Let p be a prime, and σ ∈ S

have order p. Then σ is a p-cycle.

Proof. By IA Groups, we can decompose σ into a product of disjoint cycles:

σ = σ

···σ

Let σ

have order m

> 1. Again by IA Groups, we know that

p = order of σ = lcm(m

, ··· , m

Since

is a prime number, we know that

for all

. Hence we must have

= 1, since the cycles are disjoint and there are only

elements. So

Hence σ is indeed an p cycle.

We will use these to find an example where the Galois group is the symmetric

group. The conditions for this to happen are slightly awkward, but the necessity

of these will become apparent in the proof.

Theorem. Let

f ∈ Q

[

] be irreducible and

deg f

prime. Let

L ⊆ C

be the

splitting field of f over Q. Let

Root

(L) = {α

, α

, ··· , α

p−2

, α

p−1

, α

Suppose that

, α

, ··· , α

p−2

are all real numbers, but

p−1

and

are not.

In particular,

p−1

¯α

. Then the homomorphism

Gal

(

L/Q

)

→ S

is an

isomorphism.

Proof.

From IA groups, we know that the cycles (1 2

··· p

) and (

p −

)

generate the whole of

. So we show that these two are both in the image of

is irreducible, we know that

, the minimal polynomial of

over Q. Then

p = deg P

= [Q(α

) : Q].

By the tower law, this divides [

], which is equal to

|Gal

(

L/Q

)

since the

extension is Galois. Since

divides the order of

Gal

(

L/Q

), by Cauchy’s theorem

of groups, there must be an element of

Gal

(

L/Q

) that is of order

. This maps

to an element σ ∈ im β of order exactly p. So σ is a p-cycle.

On the other hand, the isomorphism

C → C

given by

z 7→ ¯z

restricted to

gives an automorphism in

Gal

(

L/Q

). This simply permutes

p−1

and

, since

it fixes the real numbers and

p−1

and

must be complex conjugate pairs. So

τ = (p − 1 p) ∈ im β.

Now for every 1

≤ i < p

, we know that

again has order

, and hence

is a

-cycle. So if we change the labels of the roots

, ··· , α

and replace

with

, and then waffle something about combinatorics, we can assume

σ = (1 2 ··· p − 1 p). So done.

Example. Let t

− 4t + 2 ∈ Q[t]. Let L be the splitting field of f over Q.

First note that

deg f

= 5 is a prime. Also, by Eisenstein’s criterion,

irreducible. By finding the local maximum and minimum points, we find that

has exactly three real roots. So by the theorem, there is an isomorphism

Gal(L/Q) → S

. In other words, Gal(L/Q)

∼

We know S

is not a soluble group. So f cannot be solved by radicals.

After spending 19 lectures, we have found an example of a polynomial that

cannot be solved by radicals. Finally.

Note that there are, of course, many examples of

f ∈ Q

[

] irreducible of

degree at least 5 that can be solved by radicals, such as f = t

− 2.

3.5 Insolubility of general equations of degree 5 or more

We now want to do something more interesting. Instead of looking at a particular

example, we want to say there is no general formula for solving polynomial

equations of degree 5 or above. First we want to define certain helpful notions.

Definition (Field of symmetric rational functions). Let

be a field,

(

, ··· , x

), the field of rational functions over

. Then there is an injective

homomorphism S

→ Aut

(L) given by permutations of x

We define the field of symmetric rational functions

to be the fixed

field of S

There are a few important symmetric rational functions that we care about

more.

Definition (Elementary symmetric polynomials). The elementary symmetric

polynomials are e

, e

, ··· , e

defined by

1≤l

<···<l

≤n

ℓ

···x

ℓ

It is easy to see that

= x

+ x

+ ··· + x

= x

+ x

+ ··· + x

n−1

= x

···x

Obviously, e

, ··· , e

∈ F .

Theorem (Symmetric rational function theorem). Let

be a field,

(

, ··· , x

). Let

the field fixed by the automorphisms that permute the

Then

(i) L is the splitting field of

f = t

− e

n−1

+ ··· + (−1)

over F .

(ii) F = L

⊆ L is a Galois group with Gal(L/F ) isomorphic to S

(iii) F = K(e

, ··· , e

Proof.

(i) In L[t], we have

f = (t − x

) ···(t − x

So L is the splitting field of f over F .

(ii) By Artin’s lemma, L/K is Galois and Gal(L/F )

∼

(iii)

Let

(

, ··· , e

). Clearly,

E ⊆ F

. Now

E ⊆ L

is a Galois extension,

since L is the splitting field of f over E and f has no repeated roots.

By the fundamental theorem of Galois theory, since we have the Galois ex-

tensions

E ⊆ F ⊆ L

, we have

Gal

(

L/F

)

≤ Gal

(

L/E

). So

≤ Gal

(

L/E

However, we also know that

Gal

(

L/E

) is a subgroup of

, we must have

Gal(L/E) = Gal(L/F ) = S

. So we must have E = F .

Definition (General polynomial). Let

be a field,

, ··· , u

variables. The

general polynomial over K of degree n is

f = t

+ u

n−1

+ ··· + u

Technically, this is a polynomial in the polynomial ring

(

, ··· , u

)[

]. How-

ever, we say this is the general polynomial over

be cause we tend to think of

these u

as representing actual elements of K.

We say the general polynomial over

of degree

can be solved by radicals

if f can be solved by radicals over K(u

, ··· , u

Example. The general polynomial of degree 2 over Q is

+ u

t + u

This can be solved by radicals because its roots are

−u

− 4u

Theorem. Let

be a field with

char K

= 0. Then the general polynomial

polynomial over K of degree n cannot be solved by radicals if n ≥ 5.

Proof. Let

f = t

+ u

n−1

+ ··· + u

be our general polynomial of degree

n ≥

5. Let

be a splitting field of

over

K(u

, ··· , u

). Let

Root

(N) = {α

, ··· , α

We know the roots are distinct because

is irreducible and the field has charac-

teristic 0. So we can write

f = (t − α

) ···(t − α

) ∈ N[t].

We can expand this to get

= −(α

+ ··· + α

)

= α

+ α

+ ··· + α

n−1

= (−1)

(ith elementary symmetric polynomial in α

, ··· , α

Now let

, ··· , x

be new variables, and

the

th elementary symmetric

polynomial in

, ··· , x

. Let

(

, ··· , x

), and

(

, ··· , e

). We

know that F ⊆ L is a Galois extension with Galois group isomorphic to S

We define a ring homomorphism

θ : K[u

, ··· , u

] → K[e

, ··· , e

] ⊆ K[x

, ··· , x

]

7→ (−1)

This is our equations of u

in terms α

, but with x

instead of α

We want to show that

is an isomorphism. Note that since the homomorphism

just renames

into

, the fact that

is an isomorphism means there are no

“hidden relations” between the

. It is clear that

is a surjection. So it suffices

to show θ is injective. Suppose θ(h) = 0. Then

h(−e

, ··· , (−1)

) = 0.

Since the x

are just arbitrary variables, we now replace x

with α

. So we get

h(−e

(α

, ··· , α

), ··· , (−1)

(α

, ··· , α

))) = 0.

Using our expressions for u

in terms of e

, we have

h(u

, ··· , u

) = 0,

But

(

, ··· , u

) is just

itself. So

= 0. Hence

is injective. So it is an

isomorphism. This in turns gives an isomorphism between

K(u

, ··· , u

) → K(e

, ··· , e

) = F.

We can extend this to their polynomial rings to get isomorphisms between

K(u

, ··· , u

)[t] → F [t].

In particular, this map sends our original f to

f 7→ t

− e

n−1

+ ··· + (−1)

= g.

Thus, we get an isomorphism between the splitting field of

over

(

, ··· , u

)

and the splitting field g over F .

The splitting field of

over

(

, ··· , u

) is just

by definition. From the

symmetric rational function theorem, we know that the splitting field of

over

F is just L, and So N

∼

L. So we have an isomorphism

Gal(N/K(u

, ··· , u

)) → Gal(L/F )

∼

Since S

is not soluble, f is not soluble.

This is our second main goal of the course, to prove that general polynomials

of degree 5 or more are not soluble by radicals.

Recall that we proved that all radical extensions are soluble. We now prove

the converse.

Theorem. Let

be a field with

char K

= 0. If

L/K

is a soluble extension,

then it is a radical extension.

Proof.

Let

L ⊆ E

be such that

K ⊆ E

is Galois and

Gal

(

E/K

) is soluble. We

can replace

with

, and assume that in fact

L/K

is a soluble Galois extension.

So there is a sequence of groups

{0} = G

◁ ··· ◁ G

◁ G

= Gal(L/K)

such that G

i+1

is cyclic.

By the fundamental theorem of Galois theory, we get a sequence of field

extension given by L

= L

K = L

⊆ ··· ⊆ L

= L.

Moreover, we know that

⊆ L

i+1

is a Galois extension with Galois group

Gal(L

i+1

)

∼

i+1

. So Gal(L

i+1

) is cyclic.

Let

= [

]. Recall that we proved a previous theorem that if

Gal

(

i+1

) is cyclic, and

contains a primitive

th root of unity (with

= [

i+1

]), then

⊆ L

i+1

is a Kummer extension. However, we do not

know of

contains the right root of unity. Hence, the trick here is to add an

nth primitive root of unity to each field in the sequence.

Let

th primitive root of unity. Then if we add the

th primitive root

to each item of the sequence, we have

(µ) ··· L

(µ) L

i+1

(µ) ··· L

(µ)

K = L

··· L

i+1

··· L

= L

⊆

⊆ ⊆

⊆

⊆ ⊆ ⊆ ⊆ ⊆

We know that

⊆ L

(

) is a cyclotomic extension by definition. We will now

show that

(

)

⊆ L

i+1

(

) is a Kummer extension for all

. Then

L/K

is radical

since L ⊆ L

(µ).

Before we do anything, we have to show

(

)

⊆ L

i+1

(

) is a Galois extension.

To show this, it suffices to show L

⊆ L

i+1

(µ) is a Galois extension.

Since

⊆ L

i+1

is Galois,

⊆ L

i+1

is normal. So

i+1

is the splitting of

some

over

. Then

i+1

(

) is just the splitting field of (

−

. So

⊆

i+1

(

) is normal. Also,

⊆ L

i+1

(

) is separable since

char K

char L

= 0.

Hence L

⊆ L

i+1

(µ) is Galois, which implies that L

(µ) ⊆ L

i+1

(µ) is Galois.

We define a homomorphism of groups

Gal(L

i+1

(µ)/L

(µ)) → Gal(L

i+1

)

by restriction. This is well-defined because L

i+1

is the splitting field of some h

over

, and hence any automorphism of

i+1

(

) must send roots of

to roots

of h, i.e. L

i+1

to L

i+1

Moreover, we can see that this homomorphism is injective. If

φ 7→ φ|

i+1

then it fixes everything in

i+1

. Also, since it is in

Gal

(

i+1

(

)

(

)), it fixes

(µ). In particular, it fixes µ. So φ must fix the whole of L

i+1

(µ). So φ = id.

By injectivity, we know that

Gal

(

i+1

(

)

(

)) is isomorphic to a subgroup

Gal

(

i+1

). Hence it is cyclic. By our previous theorem, it follows that

(µ) ⊆ L

i+1

(µ) is a Kummer extension. So L/K is radical.

Corollary. Let

be a field with

char K

= 0 and

h ∈ K

[

]. Let

be the

splitting of

over

. Then

can be solved by radicals if and only if

Gal

(

L/K

)

is soluble.

Proof. (⇒) Proved before.

(

⇐

) Since

L/K

is a Galois extension,

L/K

is a soluble extension. So it is a

radial extension. So h can be solved by radicals.

Corollary. Let

be a field with

char K

= 0. Let

f ∈ K

[

] have

deg f ≤

Then f can be solved by radicals.

Proof. Exercise.

Note that in the case where

, we have proven this already by given

explicit solutions in terms of radicals in the first lecture.

4 Computational techniques

In the last three lectures, we will look at some techniques that allow us to

actually compute the Galois group of polynomials (i.e. Galois groups of their

splitting fields).

4.1 Reduction mod p

The goal of this chapter is to see what happens when we reduce a polynomial

f ∈ Z[t] to the corresponding polynomial

f ∈ F

[t].

More precisely, suppose we have a polynomial

f ∈ Z

[

], and

is its splitting

field over

. We then reduce

f ∈ F

[

] by reducing the coefficients mod

and let

E be the splitting field of

f over F

The ultimate goal is to show that under mild assumptions, there is an

injection

Gal(E/F

) → Gal(E/Q).

To do this, we will go through a lot of algebraic fluff to obtain an alternative

characterization of the Galois group, and obtain the result as an easy corollary.

This section will be notationally heavy. First, in the background, we have a

polynomial

of degree

(whose field we shall specify later). Then we will have

three distinct set of variables, namely (

, ··· , x

), (

, ··· , u

), plus a

. They

will play different roles.

–

The

will be placeholders. After establishing our definitions, we will then

map each x

to α

, a root of our f.

– The u

will stay as “general coefficients” all the time.

– t

will be the actual variable we think our polynomial is in, i.e. all polyno-

mials will be variables in

, and

and

will form part of the coefficients.

To begin with, let

L = Q(x

, ··· , x

)

F = Q(e

, ··· , e

where

are variables and

are the symmetric polynomials in the

, ··· , x

We have seen that Gal(L/F )

∼

Now let

B = Z[x

, ··· , x

]

A = Z[e

, ··· , e

It is an exercise on example sheet 4 to show that

B ∩ F = A. (∗)

We will for now take this for granted.

We now add it new variables

, ··· , u

, t

. We previously mentioned that

can act on, say

[

, ··· , u

, t

] by permuting the variables. Here there are two

ways in which this can happen — a permutation can either permute the

, or

permute the u

. We will have to keep this in mind.

Now for each σ ∈ S

, we define the linear polynomial

= t − x

σ(1)

− ··· − x

σ(n)

For example, we have

(1)

= t − x

− ··· − x

As mentioned, an element

ρ ∈ S

can act on

in two ways: it either sends

7→ R

ρσ

or R

7→ R

σρ

−1

It should be clear that the first action permutes the

. What the second

action does is permute the

. To see this, we can consider a simple case where

n = 2. Then the action ρ acting on R

(1)

sends

t − x

− x

7→ t − x

−1

(1)

− x

−2

(2)

= t − x

ρ(1)

− x

ρ(2)

Finally, we define the following big scary polynomial:

R =

σ∈S

∈ B[u

, ··· , u

, t].

We see that this is fixed by any permutation in

σ ∈ S

under both actions.

Considering the first action and using (∗), we see that in fact

R ∈ A[u

, ··· , u

, t].

This is since if we view

as a polynomial over

in the variables

, ··· , u

, t

then its coefficients will be invariant under permuting the

. So the coefficients

must be a function of the e

, i.e. lie in A.

With these definitions in place, we can focus on a concrete polynomial.

Let K be a field, and let

f = t

+ a

n−1

+ ··· + a

∈ K[t]

have no repeated roots. We let E be the splitting field of f over K. Write

Root

(E) = {α

, ··· , α

Note that this is the setting we had at the beginning of the chapter, but with an

arbitrary field K instead of Q and F

We define a ring homomorphism

B → E

7→ α

. This extends to a

ring homomorphism

θ : B[u

, ··· , u

, t] → E[u

, ··· , u

, t].

Note that the ring homomorphism

send

7→

(

−

. So in particular, if

we restrict the homomorphism

, then the image is restricted to the field

generated by

. But we already have

∈ K

. So

(

) =

. In particular, since

R ∈ A[u

, ··· , u

, t], we have

θ(R) ∈ K[u

, ··· , u

, t].

Now let

be an irreducible factor of

(

) in

[

, ··· , u

, t

]. We want to say

each such irreducible polynomial is related to the Galois group

Gal

(

E/K

Since

has no repeated roots, we can consider

as a subgroup of

, where

the elements of

are just the permutations of the roots

. We will then show

that each irreducible polynomial corresponds to a coset of G.

Recall that at the beginning, we said

can act on our polynomial rings

by permuting the

. However, once we have mapped the

to the

and focus on a specific field,

as a whole can no longer act on the

, since

there might be non-trivial relations between the

. Instead, only the subgroup

G ≤ S

can act on α

. On the other hand, S

can still act on u

Recall that

is defined as a product of linear factors

’s. So we can find a

subset Λ ⊆ S

such that

P =

σ∈Λ

We will later see this Λ is just a coset of the Galois group G.

Pick σ ∈ Λ. Then by definition of P ,

| P

[

, ··· , u

, t

]. Now if

ρ ∈ G

, then we can let

act on both sides by

permuting the

(i.e. the

). This does not change

because

has coefficients

in K and the action of G has to fix K. Hence we have

ρσ

| P.

More generally, if we let

H =

ρ∈G

ρσ

∈ E[u

, ··· , u

, t],

then

H | P

by the irreducibility of P .

Since

is also invariant under the action of

, we know

H ∈ K

[

, ··· , u

, t

By the irreducibility of P , we know H = P. Hence, we know

Λ = Gσ.

We have thus proved that the irreducible factors of

(

) in

[

, ··· , u

, t

] are

in one-to-one correspondence with the cosets of

. In particular, if

corresponds to G itself, then

P =

τ∈G

In general, if

corresponds to a coset

Gσ

, we can let

λ ∈ S

act on

permuting the u

’s. Then this sends

P =

ρ∈G

ρσ

7→ Q =

ρ∈G

ρσλ

−1

So this corresponds to the coset

Gσλ

−1

. In particular,

if and only if

Gσ

Gσλ

−1

. So we can use this to figure out what permutations preserve an

irreducible factor. In particular, taking σ = (1), we have

Theorem.

G = {λ ∈ S

: λ preserves the irreducible factor corresponding to G}. (†)

This is the key result of this chapter, and we will apply this as follows:

Theorem. Let

f ∈ Z

[

] be monic with no repeated roots. Let

be the splitting

field of

over

, and take

f ∈ F

[

] be the obvious polynomial obtained by

reducing the coefficients of

mod

. We also assume this has no repeated roots,

and let

E be the splitting field of

Then there is an injective homomorphism

G = Gal(

E/F

) → G = Gal(E/Q).

Moreover, if

factors as a product of irreducibles of length

, n

, ··· , n

, then

Gal(f) contains an element of cycle type (n

, ··· , n

Proof. We apply the previous theorem twice. First, we take K = Q. Then

θ(R) ∈ Z[u

, ··· , u

, t].

Let

be the irreducible factor of

(

) corresponding to the Galois group

Applying Gauss’ lemma, we know P has integer coefficients.

Applying the theorem again, taking

. Denote the ring homomorphism

. Then

(

)

∈ F

[

, ··· , u

, t

]. Now let

be the irreducible factor

(

)

corresponding to

Now note that

(

(1)

)

| P

and

(

(1)

)

| Q

, since the identity is in

and

Also, note that

(

) =

θ(R)

, where the bar again denotes reduction mod

. So

Q |

P .

Considering the second action of

(i.e. permuting the

), we can show

G ⊆ G, using the characterization (†). Details are left as an exercise.

This is incredibly useful for computing Galois groups, as it allows us to

explicitly write down some cycles in Gal(E, Q).

4.2 Trace, norm and discriminant

We are going to change direction a bit and look at traces and norms. These will

help us understand the field better, and perhaps prove some useful facts from

it. They will also lead to the notion of the discriminant, which is again another

tool that can be used to compute Galois groups, amongst many other things.

Definition (Trace). Let

be a field. If

= [

] is an

n × n

matrix over

we define the trace of A to be

tr(A) =

i=1

i.e. we take the sum of the diagonal terms.

It is a well-known fact that if B is an invertible n × n matrix, then

tr(B

−1

AB) = tr(A).

Hence given a finite-dimensional vector space

over

and

V → V

K-linear map, then we can define the trace for the linear map as well.

Definition (Trace of linear map). Let

be a finite-dimensional vector space

over K, and σ : V → V a K-linear map. Then we can define

tr(σ) = tr(any matrix representing σ).

Definition (Trace of element). Let

K ⊆ L

be a finite field extension, and

α ∈ L

. Consider the

-linear map

L → L

given by multiplication with

, i.e.

β 7→ αβ. Then we define the trace of α to be

L/K

(α) = tr(σ).

Similarly, we can consider the determinant, and obtain the norm.

Definition (Norm of element). We define the norm of α to be

L/K

(α) = det(σ),

where σ is, again, the multiplication-by-α map.

This construction gives us two functions

L/K

, N

L/K

L → K

. It is easy to

see from definition that tr

L/K

is additive while N

L/K

is multiplicative.

Example. Let

L/K

be a finite field extension, and

x ∈ K

. Then the matrix of

x is represented by xI, where I is the identity matrix. So

L/K

(x) = x

[L:K]

, tr

L/K

(x) = [L : K]x.

Example. Let

(

). Consider an element

bi ∈ Q

(

), and pick

the basis {1, i} for Q(i). Then the matrix of a + bi is



a −b

b a



So we find that tr

L/K

(a + bi) = 2a and N(a + bi) = a

+ b

= |a + bi|

In general, if

and

(

√

−d

) where

d >

0 is square-free, then

(

√

−d

) =

√

−d|

. However, for other fields, the norm is

not at all related to the absolute value.

In general, computing norms and traces with the definition directly is not

fun. It turns out we can easily find the trace and norm of

from the minimal

polynomial of

, just like how we can find usual traces and determinants from

the characteristic polynomial.

To do so, we first prove the transitivity of trace and norm.

Lemma. Let L/F/K be finite field extensions. Then

L/K

= tr

F/K

◦tr

L/F

, N

L/K

= N

F/K

◦ N

L/F

To prove this directly is not difficult, but involves some confusing notation.

Purely for the sake of notational convenience, we shall prove the following more

general fact:

Lemma. Let

F/K

be a field extension, and

-vector space. Let

V → V

be an F -linear map. Then it is in particular a K-linear map. Then

det

T = N

F/K

(det

T ), tr

T = tr

F/K

(tr

T ).

Taking

to be

and

to be multiplication by

α ∈ F

clearly gives the

original intended result.

Proof.

For

α ∈ F

, we will write

F → F

for multiplication by

map viewed

as a K-linear map.

By IB Groups, Rings and Modules, there exists a basis

}

such that

in rational canonical form, i.e. such that

is block diagonal with each diagonal

looking like







0 0 ··· 0 a

1 0 ··· 0 a

0 1 ··· 0 a

0 0 ··· 1 a

r−1







Since the norm is multiplicative and trace is additive, and

det



A 0

0 B



= det A det B, tr



A 0

0 B



= tr A + tr B,

we may wlog T is represented by a single block as above.

From the rational canonical form, we can read off

det

T = (−1)

r−1

, tr

T = a

r−1

We now pick a basis

}

over

, and then

}

is a basis for

over

K. Then in this basis, the matrix of T over K is given by







0 0 ··· 0 m

1 0 ··· 0 m

0 1 ··· 0 m

0 0 ··· 1 m

r−1







It is clear that this has trace

r−1

) = tr

F/K

r−1

) = tr

F/K

(tr

T ).

Moreover, writing n = [L : K], we have

det







0 0 ··· 0 m

1 0 ··· 0 m

0 1 ··· 0 m

0 0 ··· 1 m

r−1







= (−1)

n(r−1)

det







0 0 ··· 0

1 0 ··· 0

0 1 ··· 0

r−1

0 0 ··· 1







= (−1)

n(r−1)

det

)

= det

((−1)

r−1

)

= N

F/K

((−1)

r−1

)

= N

F/K

(det

T ).

So the result follows.

As a corollary, we have the following very powerful tool for computing norms

and traces.

Corollary. Let

L/K

be a finite field extension, and

α ∈ L

. Let

= [

(

)]

and let P

be the minimal polynomial of α over K, say

= t

+ a

n−1

+ ··· + a

with a

∈ K. Then

L/K

(α) = −ra

n−1

and

L/K

(α) = (−1)

Note how this resembles the relation between the characteristic polynomial

and trace/determinants in linear algebra.

Proof.

We first consider the case

= 1. Write

for the matrix representing

multiplication by

. Then

is the minimal polynomial of

. But since

deg P

dim

(

), it follows that this is also the characteristic polynomial.

So the result follows.

Now if

r 

= 1, we can consider the tower of extensions

L/K

(

)

. Then we

have

L/K

(α) = N

K(α)/K

L/K(α)

(α)) = N

K(α)/K

(α

)

= (N

K(α)/K

(α))

= (−1)

The computation for trace is similar.

It is also instructive to prove this directly. In the case

= 1, we can pick the

basis {1, α, α

, ··· , α

n−1

} of L over K. Then the multiplication map sends

1 7→ α

α 7→ α

n−1

7→ α

= −a

n−1

− ··· − a

So the matrix is just

A =







0 0 ··· −a

1 0 ··· −a

0 1 ··· −a

0 0 ··· −a

n−1







The characteristic polynomial of this matrix is

det(tI − A) = det







t 0 ··· a

−1 t ··· a

0 −1 ··· a

0 0 ··· t + a

n−1







By adding t

multiples of the ith row to the first row for each i, this gives

det(tI − A) = det







0 0 ··· P

−1 t ··· a

0 −1 ··· a

0 0 ··· t + a

n−1







= P

Then we notice that for

r 

= 1, in an appropriate choice of basis, the matrix

looks like

C =







A 0 ··· 0

0 A ··· 0

0 0 ··· A







Theorem. Let

L/K

be a finite but not separable extension. Then

L/K

(

) = 0

for all α ∈ L.

Proof.

Pick

β ∈ L

such that

, the minimal polynomial of

over

, is not

separable. Then by the previous characterization of separable polynomials, we

know p = char K > 0 with P

= q(t

) for some q ∈ K[t].

Now consider

K ⊆ K(β

) ⊆ K(β) ⊆ L.

To show

L/K

= 0, by the previous proposition, it suffices to show

K(β)/K(β

)

Note that the minimal polynomial of

over

because

(

) = 0 and

is irreducible. Then [

(

) :

] =

deg P

p deg q

and

deg

[

(

) :

] =

deg q

So [K(β) : K(β

)] = p.

Now

{

, β, β

, ··· , β

p−1

}

is a basis of

(

) over

(

). Let

be the

minimal polynomial of β

over K(β

). Then

(

t − 1 i = 0

− β

i = 0

We get the second case using the fact that

is a prime number, and hence

(

)(

) =

(

) if 1

≤ i < p

. So [

(

)(

) :

(

)] =

and hence the

minimal polynomial has degree p. Hence tr

K(β)/K(β

)

(β

) = 0 for all i.

Thus, tr

K(β)/K(β

)

= 0. Hence

L/K

= tr

K(β

)/K

◦tr

K(β)/K(β

)

◦tr

L/K(β)

= 0.

Note that if L/K is a finite extension, and char K = 0, then

L/K

(1) = [L : K] = 0.

L/K



= 0. It is in fact true that all separable extensions have

L/K



= 0, not

only when the field has characteristic 0.

Example. We want to show

√

3 ∈ Q

(

√

). Suppose not. Then we have

(

√

) =

(

√

), since both extensions of

have degree 3. Then there

exists some a, b, c ∈ Q such that

√

3 = a + b

√

2 + c

√

We now compute the traces over Q. The minimal polynomials over Q are

√

= t

− 3, P

√

= t

− 2, P

√

= t

− 4.

So we have

L/Q

(

√

3) = a tr

L/Q

(1) + b tr

L/Q

(

√

2) + c tr

L/Q

(

√

4).

Since the minimal polynomials above do not have coefficients in

, the traces of

the cube roots are zero. So we need a = 0. Then we are left with

√

3 = b

√

2 + c

√

We apply the same trick again. We multiply by

√

2 to obtain

√

6 = b

√

4 + 2c.

We note that the minimal polynomial of

√

6 is t

− 6. Taking the trace gives

L/Q

(

√

6) = b tr

L/Q

(

√

4) + 6c.

Again, the traces are zero. So c = 0. So we have

√

3 = b

√

In other words,

which is clearly nonsense. This is a contradiction. So

√

3 ∈ Q(

√

2).

We can obtain another formula for the trace and norm as follows:

Theorem. Let

L/K

be a finite separable extension. Pick a further extension

E/L such that E/K is normal and

|Hom

(L, E)| = [L : K].

Write Hom

(L, E) = {ϕ

, ··· , ϕ

}. Then

L/K

(α) =

i=1

(α), N

L/K

(α) =

i=1

(α)

for all α ∈ L.

Proof.

Let

α ∈ L

. Let

be the minimal polynomial of

over

. Then there

is a one-to-one correspondence between

Hom

(K(α), E) ←→ Root

(E) = {α

, ··· , α

wlog we let α = α

Also, since

|Hom

(L, E)| = [L : K],

we get

|Hom

(K(α), E)| = [K(α) : K] = deg P

Moreover, the restriction map

Hom

(

L, E

)

→ Hom

(

)

, E

) (defined by

ϕ 7→ ϕ|

K(α)

) is surjective and sends exactly [

(

) :

] elements to any particular

element in Hom

(K(α), E).

Therefore

(α) = [L : K(α)]

ψ∈Hom

(K(α),E)

ψ(α) = [L : K(α)]

i=1

Moreover, we can read the sum of roots of a polynomial is the (negative of the)

coefficient of t

d−1

, where

= t

+ a

d−1

+ ··· + a

(α) = [L : K(α)](−a

d−1

) = tr

L/K

(α).

Similarly, we have

(α) =





ψ∈Hom

(K(α),E)

ψ(α)





[L:K(α)]

i=1

[L:K(α)]

= ((−1)

)

[L:K(α)]

= N

L/K

(α).

Corollary. Let

L/K

be a finite separable extension. Then there is some

α ∈ L

such that tr

L/K

(α) = 0.

Proof. Using the notation of the previous theorem, we have

L/K

(α) =

(α).

Similar to a previous lemma, we can show that

, ··· , ϕ

are “linearly indepen-

dent” over

, and hence

cannot be identically zero. Hence there is some

such that

L/K

(α) =

(α) = 0.

Example. Let

⊆ L

, with

is a power of some prime number

By a previous theorem on finite fields, we know L/K is Galois and

Gal(L/K) =

and is generated by the Frobenius ϕ = Fr

To apply the theorem, we had to pick an

such that

E/K

is normal and

Hom

(

L, E

) = [

]. However, since

L/K

is Galois, we can simply pick

E = L.

Then we know

L/K

(α) =

ψ∈Gal(L/K)

ψ(α)

n−1

i=0

(α)

= α + α

+ α

+ ··· + α

n−1

Similarly, the norm is

L/K

(α) =

n−1

i=0

(α) = α · α

· ··· · α

n−1

Recall that when solving quadratic equations

, we defined the

discriminant as

−

. This discriminant then determined the types of roots of

. In general, we can define the discriminant of a polynomial of any degree, in a

scary way.

Definition (Discriminant). Let

be a field and

f ∈ K

[

the splitting field

of f over K. So we have

f = a(t − α

) ···(t − α

)

for some a, α

, ··· , α

∈ L. We define

∆

i<j

(α

− α

), D

= ∆

= (−1)

n(n−1)/2

i=j

(α

− α

We call D

the discriminant of f .

Clearly, D

= 0 if and only if f has no repeated roots.

Theorem. Let

be a field and

f ∈ K

[

is the splitting field of

over

Suppose D

= 0 and char K = 2. Then

(i) D

∈ K.

(ii)

Let

Gal

(

L/K

), and

G → S

be the embedding given by the

permutation of the roots. Then

im θ ⊆ A

if and only if ∆

∈ K

(if and

only if D

is a square in K).

Proof.

(i) It is clear that D

is fixed by Gal(L/K) since it only permutes the roots.

(ii)

Consider a permutation

σ ∈ S

of the form

= (

 m

), and let it act on

the roots. Then we claim that

σ(∆

) = −∆

. (†)

So in general, odd elements in

negate ∆

while even elements fix it.

Thus, ∆

∈ K

iff ∆

is fixed by

Gal

(

L/K

) iff every element of

Gal

(

L/K

)

is even.

To prove (

†

), we have to painstakingly check all terms in the product. We

wlog

 < m

. If

k < , m

. Then this swaps (

−α

ℓ

) with

−α

), which

has no effect. The

k > m

case is similar. If

 < k < m

, then this sends

(

ℓ

− α

)

7→

(

− α

) and (

− α

)

7→

(

ℓ

− α

). This introduces two

negative signs, which has no net effect. Finally, this sends (

− α

) to

its negation, and so introduces a negative sign.

We will later use this result to compute certain Galois groups. Before that,

we see how this discriminant is related to the norm.

Theorem. Let

be a field, and

f ∈ K

[

] be an

-degree monic irreducible

polynomial with no repeated roots. Let

be the splitting field of

over

, and

let α ∈ Root

(L). Then

= (−1)

n(n−1)/2

K(α)/K

′

(α)).

Proof.

Let

Hom

(

)

, L

) =

{ϕ

, ··· , ϕ

}

. Recall these are in one-to-one

correspondence with Root

(L) = {α

, ··· , α

}. Then we can compute

i=j

(α

− α

) =

j=i

(α

− α

Note that since f is just monic, we have

f = (t − α

) ···(t − α

Computing the derivative directly, we find

j=i

(α

− α

) = f

′

(α

So we have

i=j

(α

− α

) =

′

(α

Now since the ϕ

just maps α to α

, we have

i=j

(α

− α

) =

′

(α)) = N

K(α)/K

′

(α)).

Finally, multiplying the factor of (−1)

n(n−1)/2

gives the desired result.

Example. Let

be a field with

char K 

= 2

3. Let

f ∈ K

[

] have degree 3, say

f = t

+ bt + c

where we have gotten rid of the

term as in the first lecture. We further assume

f is irreducible with no repeated roots, and let L be the splitting field of f.

We want to compute the discriminant of this polynomial. Let

α ∈ Root

(

Then

β = f

′

(α) = 3α

+ b.

Then we can see

β = −2b −

Alternatively, we have

α =

−3c

β + 2b

. (∗)

Putting (

∗

) into

bα

= 0, we find the minimal polynomial of

has

constant term −4b

− 27c

. This then gives us the norm, and we get

= −N

K(α)/K

(β) = −4b

− 27c

This is the discriminant of a cubic.

We can take a specific example, where

f = t

− 31t + 62.

Then

is irreducible over

. We can compute

, and find that it is a square.

So the previous theorem says the image of the Galois group

Gal

(

L/K

) is a

subgroup of

. However, we also know

Gal

(

L/K

) has three elements since

deg f = 3. So we know Gal(L/K)

∼