Abstract algebra

Section 14.1.

We are now starting to discuss Galois theory. It is named after Évariste Galois, a French mathematician who laid foundations of group theory, determined when a polynomial equation is solvable in radicals (a problem which stood open for 350 years), and died from wounds suffered in a dual at the age of 20.

The main idea of Galois theory is also central to many other branches of mathematics. Broadly speaking, the idea is to study symmetries of various mathematical structures. Previ- ously, in this class we learnt how to construct a splitting field K of a polynomial f(x) ∈ F[x] with coefficients in the field F . The splitting field K is a finite field extension of F , and is defined as the minimal extension of F which contains every root of f(x). In what follows, we will study how the group of permutations of the roots of f(x) is related to the splitting field K.

Let us first fix some notations. Recall that an automorphism of a field K is an isomorphism of K with itself, and the group of automorphisms of K is denoted by Aut(K). An automor- phism σ ∈ Aut(K) is said to fix an element k ∈ K if σ(k) = k. Clearly, any automorphism of K has to fix 0 and 1, and therefore fixes the prime subfield of K. In particular, it means that the automorphism groups of Q and Fp are trivial:

Aut(Q) '{1}' Aut(Fp).

Definition 1. Let K/F be a field extension, we denote by Aut(K/F) the set of automorphism of K which preserve F .

Exercise 2. For any extension K/F, the set Aut(K/F) is a subgroup of Aut(K).

Example. If F is the prime subfield of K, then Aut(K/F) = Aut(K).

Proposition 3. Let K/F be a field extension, and σ ∈ Aut(K/F). Then any polynomial f(x) ∈ F[x] which has α ∈ K as a root also has σ(α) ∈ K as a root:

f(α) = 0 ⇔ f(σ(α)) = 0.

Proof. Let α be a root of the polynomial

f(x) = anx n + · · · + a1x + a0, ai ∈ F,

that is anα

n + · · · + a1α + a0 = 0. Applying the automorphism σ to the above equation, we get

σ(an)σ(α) n + · · · + σ(a1)σ(α) + σ(a0) = 0,

but since σ ∈ Aut(K/F), we also have σ(ai) = ai for all i = 1, . . . ,n, and therefore anσ(α)

n + · · · + a1σ(α) + a0 = 0. The latter implies that f(σ(α)) = 0 which completes the proof. �

Example.

(1) Recall that the minimal polynomial of √

2 over Q is x2 −2. Then any automorphism σ ∈ Aut(Q(

√ 2)/Q) has to permute roots of x2 − 2. Moreover, since

Q( √

2) = { a + b

√ 2 ∣∣∣a,b ∈ Q} ,

and σ fixes Q, the automorphism σ is completely determined by the image of σ( √

2) of √

2, which has to be again a root of x2−2. Therefore, there are only two options for 1

2

σ ∈ Aut(Q( √

2)/Q): either σ( √

2) = √

2, in which case σ is the trivial automorphism, or σ(

√ 2) = −

√ 2. Thus we have

Aut ( Q( √

2)/Q ) ' Z/2Z.

(2) Reasoning as above we see that σ ∈ Aut(Q( 3 √

2)/Q) permutes the roots of x3 − 2. However, the other two roots of x3 − 2 do not lie in Q( 3

√ 2), which implies that

σ( 3 √

2) = 3 √

2. The latter completely determines σ and we conclude that

Aut ( Q( 3 √

2)/Q ) '{1} .

We can now see that for any finite extension K/F, the group Aut(K/F) is finite because it is completely determined by the images of generators of K over F . Moreover, Aut(K/F) is nothing else but the group, which permutes the roots of a finite number of equations. However, previous example shows that not every non-trivial permutation gives rise to a non- trivial automorphism.

Definition 4. Let K/F be a field extension. For any subgroup G ⊂ Aut(K/F), the collection of elements of K fixed by G is called a fixed field of G.

Exercise 5.

(1) For any field extension K/F and a subgroup G ⊂ Aut(K/F), the fixed field of G is indeed a field.

(2) For a pair of subfields F1 ⊂ F2 ⊂ K one has Aut(K/F2) ⊂ Aut(K/F1). Conversely, for any pair of subgroups G2 ⊂ G1 ⊂ Aut(K/F) with fixed fields F2 and F1 respectively, one has F1 ⊂ F2.

In the previous example, we see that the fixed field of Aut ( Q( √

2)/Q )

is Q, while the fixed field of Aut

( Q( 3 √

2)/Q ) '{1} is Q( 3

√ 2). We see that in the second case, there is not enough

automorphisms to pick recover Q. The idea of Galois extension is the one with “enough automorphisms”.

Proposition 6. Let K/F be a finite field extension. Then

|Aut(K/F)|6 |K : F|.

Moreover, if K is the splitting field of a separable polynomial f(x) ∈ F[x], the above formula turns into an equality.

Proof. First, recall that if F(α) is a simple extension of F, where α is a root of an irreducible polynomial p(x), and φ: F → E is a field homomorphism, then the number of ways to extend φ to a homomorphism ψ : F(α) → E is precisely the number of roots of the polynomial pφ(x) in E, where pφ(x) is obtained from p(x) by applying the homomorphism φ to its coefficients. Indeed, a homomorphism ψ if exists is given by the formula

ψ(a0 + a1α + · · · + anαn) = φ(a0) + φ(a1)β + · · · + φ(an)βn,

where β = ψ(α) ∈ E. Applying the above formula to the equation p(α) = 0 we see that pφ(β) = 0. Conversely, if pφ(β) = 0, the previous formula indeed defines a homomorphism ψ : F(α) → E.

Now, a field K can be obtained from the field F be a sequence of simple extensions

F = F0 ⊂ F1 ⊂ ···⊂ Fs = K,

3

where the extension Fi is obtained from Fi−1 by adjoining a root of some irreducible poly- nomial pi(x) ∈ Fi−1[x]. By above, any homomorphism φi−1 : Fi−1 → K extends to a homo- morphism φi : Fi → K in no more than ni = deg pi(x) = |Fi : Fi−1| ways. Therefore, a trivial automorphism of the field F extends to an automorphism σ ∈ Aut(K/F) in no more than n = n1 . . .ns = |K : F| ways. This completes the proof of the first statement.

Now, let f(x) ∈ F [x] be a separable polynomial, and K be its splitting field. Then the polynomials pi(x) in the above constructions are irreducible factors of f(x) in the ring Fi−1[x]. Since, K contains every root of f(x), it also contains every root of pi(x) for each i. Moreover, the fact that f(x) is separable ensures that the number of distinct roots of pi(x) in K is deg(pi) for every i, and the statement of the Proposition follows. �

Definition 7. Let K/F be a finite extension, such that |Aut(K/F)| = |K : F|. Then K is said to be a Galois extension of F , the group Aut(K/F) is called the Galois group, and is denoted by Gal(K/F).

Corollary 8. Let F be a finite field or a field of characteristic 0. Then the splitting field of any polynomial f(x) ∈ F[x] is a Galois extension of F .

Definition 9. If f(x) ∈ F[x] is a separable polynomial over the field F, then the Galois group of f(x) over F is the Galois group of its splitting field.

Example.

(1) Let F be a field of any characteristic other than 2, and K/F be its quadratic extension, that is |K : F | = 2. Then K is a Galois extension of F, and Gal(K/F) ' Z/2Z. On the other hand, the inseparable extension F2[x]/F2[t] where x2 = t is not Galois. Indeed, any automorphism σ ∈ Aut(F2[x]/F2[t]) is determined by the image σ(x), which has to be a root of the polynomial x2 − t. It remains to notice that the latter polynomial has just one root in F2[x], since x = −x in a field of characteristic 2.

(2) The extension Q( 3 √

2)/Q is not Galois. (3) Let K = Q(

√ 2, √

3) and F = Q, then the extension K/F is Galois since K is the splitting field for the polynomial (x2 − 2)(x2 − 3). Now, on one hand, any automor- phism σ ∈ Aut(K/F) is completely determined by the images σ(

√ 2), σ(

√ 3), on the

other, √

2 and √

3 are roots of irreducible polynomials x2 −2 and x2 −3 respectively, and therefore we must have σ(

√ 2) = ±

√ 2 and σ(

√ 3) = ±

√ 3 which leaves us with

4 = |K/F| options. Now let us pick elements σ,τ ∈ Gal(K/F) defined by

σ( √

2) = − √

2, σ( √

3) = √

3, τ( √

3) = − √

3, τ( √

2) = √

2.

Clearly, σ and τ generate Gal(K/F), and Gal(K/F) is isomorphic to the Klein 4-group because σ2 = τ2 = 1. We also have a bijection between the subgroups G ⊂ Gal(K/F) and extensions F ⊂ KG ⊂ K, where KG is the fixed field of G: K{1} = K, K{1,σ} = Q(

√ 3), K{1,τ} = Q(

√ 2), K{1,στ} = Q(

√ 6), and KGal(K/F) = Q.

(4) Let K be the splitting field of x3 − 2 over F = Q. Then |Gal(K/F)| = |K : F| = 6 and is a subgroup of the permutation group S3 which permutes the three roots of x3 − 2. Therefore, Gal(K/F) ' S3. To write down an explicit isomorphism, we need to choose generators of Gal(K/F). Since K = Q( 3

√ 2,ζ3), where ζ3 is the primitive

3rd root of 1 and hence is a root of the cyclotomic polynomial x2 + x + 1. Therefore, any automorphism σ ∈ Gal(K/F) sends 3

√ 2 to one of the three roots of x3 − 2, and

sends ζ3 to one of the two roots of x 2 +x+ 1, that is ζ3 or ζ

2 3 = −1−ζ3. For example,

4

we can choose generators defined by

σ( 3 √

2) = ζ3 3 √

2, σ(ζ3) = ζ3, τ( 3 √

2) = 3 √

2, τ(ζ3) = ζ 2 3.

As in the previous example, we could consider subgroups of Gal(K/F) and try to determine their fixed fields. For example, let us identify the fixed field for the subgroup 〈σ〉⊂ Gal(K/F). The field K is a 6-dimensional vector space over F with the basis {

1, 3 √

2, 3 √

4,ζ3,ζ3 3 √

2,ζ3 3 √

4 } .

Then

σ(1) = 1, σ(ζ3 3 √

2) = ζ23 3 √

2 = − 3 √

2 − ζ3 3 √

2,

σ(ζ3) = ζ3, σ( 3 √

4) = ζ23 3 √

4 = − 3 √

4 − ζ3 3 √

4,

σ( 3 √

2) = ζ3 3 √

2, σ(ζ3 3 √

4) = ζ33 3 √

4 = 3 √

4.

Now, clearly Q(ζ3) is fixed by σ, and one can derive from the above calculations that Q(ζ3) is in fact equal to the fixed field of 〈σ〉.

(5) A natural question, whether a Galois extension of a Galois extension is again Galois, has a negative answer. Let K = Q( 4

√ 2) and F = Q. Then K/F is not Galois, since

any automorphism σ ∈ Aut(K/F) permutes the roots of x4 − 2 which are ± 4 √

2 and ±i 4 √

2, and is determined by the image σ( 4 √

2) of 4 √

2. However, ±i 4 √

2 6∈ K, and therefore we have |Aut(K/F)| = 2 while |K : F| = 4. On the other hand, both extensions Q( 4

√ 2)/Q(

√ 2) and Q(

√ 2)/Q are quadratic and hence Galois.

(6) As we noted before, Fpn/Fp is Galois, since Fpn is the splitting field of the polynomial xp

n −x over the finite field Fp. Recall also that the Frobenius homomorphism

σp : Fpn → Fpn, σp(α) = αp

is an isomorphism. Moreover, σp fixes Fp and we get σp ∈ Gal(Fpn/Fp). At the same time, for any α ∈ Fpn we have σkp(α) = αp

k , which implies that σkp = 1 if and only if

k is a multiple of n. We conclude that Gal(Fpn/Fp) is a cyclic group of order n with cyclic generator σp.