AATA Field Automorphisms

Section 23.1 Field Automorphisms

Our first task is to establish a link between group theory and field theory by examining automorphisms of fields.

Proposition 23.1.

The set of all automorphisms of a field

F

is a group under composition of functions.

Proof.

σ

and

τ

are automorphisms of

F,

then so are

σ τ

and

σ^{- 1} .

The identity is certainly an automorphism; hence, the set of all automorphisms of a field

F

is indeed a group.

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Proposition 23.2.

🔗

Let

E

be a field extension of

F .

Then the set of all automorphisms of

E

that fix

F

elementwise is a group; that is, the set of all automorphisms

σ : E \to E

such that

σ (α) = α

for all

α \in F

is a group.

Proof.

We need only show that the set of automorphisms of

E

that fix

F

elementwise is a subgroup of the group of all automorphisms of

E .

Let

σ

and

τ

be two automorphisms of

E

such that

σ (α) = α

and

τ (α) = α

for all

α \in F .

Then

σ τ (α) = σ (α) = α

and

σ^{- 1} (α) = α .

Since the identity fixes every element of

E,

the set of automorphisms of

E

that leave elements of

F

fixed is a subgroup of the entire group of automorphisms of

E .

🔗

Let

E

be a field extension of

F .

We will denote the full group of automorphisms of

E

Aut (E) .

We define the Galois group of

E

over

F

to be the group of automorphisms of

E

that fix

F

elementwise; that is,

G (E / F) = {σ \in Aut (E) : σ (α) = α for all α \in F} .

🔗

f (x)

is a polynomial in

F [x]

and

E

is the splitting field of

f (x)

over

F,

then we define the Galois group of

f (x)

to be

G (E / F) .

🔗

Example 23.3.

🔗

Complex conjugation, defined by

σ : a + b i \mapsto a - b i,

is an automorphism of the complex numbers. Since

σ (a) = σ (a + 0 i) = a - 0 i = a,

🔗

the automorphism defined by complex conjugation must be in

G (C / R) .

🔗

Example 23.4.

🔗

Consider the fields

Q \subset Q (\sqrt{5}) \subset Q (\sqrt{3}, \sqrt{5}) .

Then for

a, b \in Q (\sqrt{5}),

σ (a + b \sqrt{3}) = a - b \sqrt{3}

🔗

is an automorphism of

Q (\sqrt{3}, \sqrt{5})

leaving

Q (\sqrt{5})

fixed. Similarly,

τ (a + b \sqrt{5}) = a - b \sqrt{5}

🔗

is an automorphism of

Q (\sqrt{3}, \sqrt{5})

leaving

Q (\sqrt{3})

fixed. The automorphism

μ = σ τ

moves both

\sqrt{3}

and

\sqrt{5} .

It will soon be clear that

{id, σ, τ, μ}

is the Galois group of

Q (\sqrt{3}, \sqrt{5})

over

Q .

The following table shows that this group is isomorphic to

Z_{2} \times Z_{2} .

\begin{array}{ccccc} id & σ & τ & μ \\ id & id & σ & τ & μ \\ σ & σ & id & μ & τ \\ τ & τ & μ & id & σ \\ μ & μ & τ & σ & id \end{array}

🔗

We may also regard the field

Q (\sqrt{3}, \sqrt{5})

as a vector space over

Q

that has basis

{1, \sqrt{3}, \sqrt{5}, \sqrt{15}} .

It is no coincidence that

| G (Q (\sqrt{3}, \sqrt{5}) / Q) | = [Q (\sqrt{3}, \sqrt{5}) : Q)] = 4 .

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Proposition 23.5.

🔗

Let

E

be a field extension of

F

and

f (x)

be a polynomial in

F [x] .

Then any automorphism in

G (E / F)

defines a permutation of the roots of

f (x)

that lie in

E .

Proof.

Let

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}

and suppose that

α \in E

is a zero of

f (x) .

Then for

σ \in G (E / F),

\begin{aligned} 0 & = σ (0) \\ = σ (f (α)) \\ = σ (a_{0} + a_{1} α + a_{2} α^{2} + \dots + a_{n} α^{n}) \\ = a_{0} + a_{1} σ (α) + a_{2} [σ (α)]^{2} + \dots + a_{n} [σ (α)]^{n}; \end{aligned}

therefore,

σ (α)

is also a zero of

f (x) .

🔗

Let

E

be an algebraic extension of a field

F .

Two elements

α, β \in E

are conjugate over

F

if they have the same minimal polynomial. For example, in the field

Q (\sqrt{2})

the elements

\sqrt{2}

and

- \sqrt{2}

are conjugate over

Q

since they are both roots of the irreducible polynomial

x^{2} - 2 .

🔗

A converse of the last proposition exists. The proof follows directly from Lemma 21.32.

🔗

Proposition 23.6.

🔗

α

and

β

are conjugate over

F,

there exists an isomorphism

σ : F (α) \to F (β)

such that

σ

is the identity when restricted to

F .

🔗

Theorem 23.7.

🔗

Let

f (x)

be a polynomial in

F [x]

and suppose that

E

is the splitting field for

f (x)

over

F .

f (x)

has no repeated roots, then

| G (E / F) | = [E : F] .

Proof.

We will use mathematical induction on

[E : F] .

[E : F] = 1,

then

E = F

and there is nothing to show. If

[E : F] > 1,

let

f (x) = p (x) q (x),

where

p (x)

is irreducible of degree

d .

We may assume that

d > 1;

otherwise,

f (x)

splits over

F

and

[E : F] = 1 .

Let

α

be a root of

p (x) .

ϕ : F (α) \to E

is any injective homomorphism, then

ϕ (α) = β

is a root of

p (x),

and

ϕ : F (α) \to F (β)

is a field automorphism. Since

f (x)

has no repeated roots,

p (x)

has exactly

d

roots

β \in E .

By Proposition 23.5, there are exactly

d

isomorphisms

ϕ : F (α) \to F (β_{i})

that fix

F,

one for each root

β_{1}, \dots, β_{d}

p (x)

(see Figure 23.8).

\begin{matrix} E & \overset{ψ}{\to} & E \\ ↓ & ↓ \\ F (α) & \overset{ϕ}{\to} & F (β) \\ ↓ & ↓ \\ F & \overset{identity}{\to} & F \end{matrix}

Figure 23.8. Commutative diagram for Theorem 23.7

Since

E

is a splitting field of

f (x)

over

F,

it is also a splitting field over

F (α) .

Similarly,

E

is a splitting field of

f (x)

over

F (β) .

Since

[E : F (α)] = [E : F] / d,

induction shows that each of the

d

isomorphisms

ϕ

has exactly

[E : F] / d

extensions,

ψ : E \to E,

and we have constructed

[E : F]

isomorphisms that fix

F .

Finally, suppose that

σ

is any automorphism fixing

F .

Then

σ

restricted to

F (α)

ϕ

for some

ϕ : F (α) \to F (β) .

🔗

Corollary 23.9.

🔗

Let

F

be a finite field with a finite extension

E

such that

[E : F] = k .

Then

G (E / F)

is cyclic of order

k .

Proof.

Let

p

be the characteristic of

E

and

F

and assume that the orders of

E

and

F

are

p^{m}

and

p^{n},

respectively. Then

n k = m .

We can also assume that

E

is the splitting field of

x^{p^{m}} - x

over a subfield of order

p .

Therefore,

E

must also be the splitting field of

x^{p^{m}} - x

over

F .

Applying Theorem 23.7, we find that

| G (E / F) | = k .

To prove that

G (E / F)

is cyclic, we must find a generator for

G (E / F) .

Let

σ : E \to E

be defined by

σ (α) = α^{p^{n}} .

We claim that

σ

is the element in

G (E / F)

that we are seeking. We first need to show that

σ

is in

Aut (E) .

α

and

β

are in

E,

σ (α + β) = (α + β)^{p^{n}} = α^{p^{n}} + β^{p^{n}} = σ (α) + σ (β)

by Lemma 22.3. Also, it is easy to show that

σ (α β) = σ (α) σ (β) .

Since

σ

is a nonzero homomorphism of fields, it must be injective. It must also be onto, since

E

is a finite field. We know that

σ

must be in

G (E / F),

since

F

is the splitting field of

x^{p^{n}} - x

over the base field of order

p .

This means that

σ

leaves every element in

F

fixed. Finally, we must show that the order of

σ

k .

By Theorem 23.7, we know that

σ^{k} (α) = α^{p^{n k}} = α^{p^{m}} = α

is the identity of

G (E / F) .

However,

σ^{r}

cannot be the identity for

1 \leq r < k;

otherwise,

x^{p^{n r}} - x

would have

p^{m}

roots, which is impossible.

🔗

Example 23.10.

🔗

We can now confirm that the Galois group of

Q (\sqrt{3}, \sqrt{5})

over

Q

in Example 23.4 is indeed isomorphic to

Z_{2} \times Z_{2} .

Certainly the group

H = {id, σ, τ, μ}

is a subgroup of

G (Q (\sqrt{3}, \sqrt{5}) / Q);

however,

H

must be all of

G (Q (\sqrt{3}, \sqrt{5}) / Q),

since

| H | = [Q (\sqrt{3}, \sqrt{5}) : Q] = | G (Q (\sqrt{3}, \sqrt{5}) / Q) | = 4 .

🔗

Example 23.11.

🔗

Let us compute the Galois group of

f (x) = x^{4} + x^{3} + x^{2} + x + 1

🔗

over

Q .

We know that

f (x)

is irreducible by Exercise 17.5.20 in Chapter 17. Furthermore, since

(x - 1) f (x) = x^{5} - 1,

we can use DeMoivre’s Theorem to determine that the roots of

f (x)

are

ω^{i},

where

i = 1, \dots, 4

and

ω = \cos (2 π / 5) + i \sin (2 π / 5) .

🔗

Hence, the splitting field of

f (x)

must be

Q (ω) .

We can define automorphisms

σ_{i}

Q (ω)

σ_{i} (ω) = ω^{i}

for

i = 1, \dots, 4 .

It is easy to check that these are indeed distinct automorphisms in

G (Q (ω) / Q) .

Since

[Q (ω) : Q] = | G (Q (ω) / Q) | = 4,

🔗

the

σ_{i}

’s must be all of

G (Q (ω) / Q) .

Therefore,

G (Q (ω) / Q) ≅ Z_{4}

since

ω

is a generator for the Galois group.

🔗

Subsection Separable Extensions

🔗

Many of the results that we have just proven depend on the fact that a polynomial

f (x)

F [x]

has no repeated roots in its splitting field. It is evident that we need to know exactly when a polynomial factors into distinct linear factors in its splitting field. Let

E

be the splitting field of a polynomial

f (x)

F [x] .

Suppose that

f (x)

factors over

E

f (x) = (x - α_{1})^{n_{1}} (x - α_{2})^{n_{2}} \dots (x - α_{r})^{n_{r}} = \prod_{i = 1}^{r} (x - α_{i})^{n_{i}} .

🔗

We define the multiplicity of a root

α_{i}

f (x)

to be

n_{i} .

A root with multiplicity 1 is called a simple root. Recall that a polynomial

f (x) \in F [x]

of degree

n

is separable if it has

n

distinct roots in its splitting field

E .

Equivalently,

f (x)

is separable if it factors into distinct linear factors over

E [x] .

An extension

E

F

is a separable extension of

F

if every element in

E

is the root of a separable polynomial in

F [x] .

Also recall that

f (x)

is separable if and only if

gcd (f (x), f^{'} (x)) = 1

(Lemma 22.5).

🔗

Proposition 23.12.

🔗

Let

f (x)

be an irreducible polynomial over

F .

If the characteristic of

F

0,

then

f (x)

is separable. If the characteristic of

F

p

and

f (x) \neq g (x^{p})

for some

g (x)

F [x],

then

f (x)

is also separable.

Proof.

First assume that

char F = 0 .

Since

\deg f^{'} (x) < \deg f (x)

and

f (x)

is irreducible, the only way

gcd (f (x), f^{'} (x)) \neq 1

is if

f^{'} (x)

is the zero polynomial; however, this is impossible in a field of characteristic zero. If

char F = p,

then

f^{'} (x)

can be the zero polynomial if every coefficient of

f^{'} (x)

is a multiple of

p .

This can happen only if we have a polynomial of the form

f (x) = a_{0} + a_{1} x^{p} + a_{2} x^{2 p} + \dots + a_{n} x^{n p} .

🔗

Certainly extensions of a field

F

of the form

F (α)

are some of the easiest to study and understand. Given a field extension

E

F,

the obvious question to ask is when it is possible to find an element

α \in E

such that

E = F (α) .

In this case,

α

is called a primitive element. We already know that primitive elements exist for certain extensions. For example,

Q (\sqrt{3}, \sqrt{5}) = Q (\sqrt{3} + \sqrt{5})

🔗

and

Q (\sqrt[3]{5}, \sqrt{5} i) = Q (\sqrt[6]{5} i) .

🔗

Corollary 22.12 tells us that there exists a primitive element for any finite extension of a finite field. The next theorem tells us that we can often find a primitive element.

🔗

Theorem 23.13. Primitive Element Theorem.

🔗

Let

E

be a finite separable extension of a field

F .

Then there exists an

α \in E

such that

E = F (α) .

Proof.

We already know that there is no problem if

F

is a finite field. Suppose that

E

is a finite extension of an infinite field. We will prove the result for

F (α, β) .

The general case easily follows when we use mathematical induction. Let

f (x)

and

g (x)

be the minimal polynomials of

α

and

β,

respectively. Let

K

be the field in which both

f (x)

and

g (x)

split. Suppose that

f (x)

has zeros

α = α_{1}, \dots, α_{n}

K

and

g (x)

has zeros

β = β_{1}, \dots, β_{m}

K .

All of these zeros have multiplicity

1,

since

E

is separable over

F .

Since

F

is infinite, we can find an

a

F

such that

a \neq \frac{α_{i} - α}{β - β_{j}}

for all

i

and

j

with

j \neq 1 .

Therefore,

a (β - β_{j}) \neq α_{i} - α .

Let

γ = α + a β .

Then

γ = α + a β \neq α_{i} + a β_{j};

hence,

γ - a β_{j} \neq α_{i}

for all

i, j

with

j \neq 1 .

Define

h (x) \in F (γ) [x]

h (x) = f (γ - a x) .

Then

h (β) = f (α) = 0 .

However,

h (β_{j}) \neq 0

for

j \neq 1 .

Hence,

h (x)

and

g (x)

have a single common factor in

F (γ) [x];

that is, the minimal polynomial of

β

over

F (γ)

must be linear, since

β

is the only zero common to both

g (x)

and

h (x) .

β \in F (γ)

and

α = γ - a β

is in

F (γ) .

Hence,

F (α, β) = F (γ) .

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