Introduction

Galois theory studies the symmetries of the roots of a polynomial equation. The existence of the two solutions

can be thought of a symmetry of the situation. Assuming that the roots x_1,2 are (real or complex but) irrational, we have the following.

* Solutions generate a field extension K = $\mathbb Q$(x₁, x₂) of degree two of the rational field $\mathbb Q$.

* K has a field automorphism (symmetry) defined by φ(x₁) = x₂, giving rise to a group G = {Id_K, φ} acting on K.

* φ ◦ φ = Id_K, so as an abstract group, {e, φ} is of order two, isomorphic to C₂.

* The rational subfield $\mathbb Q$ is precisely the subfield of K fixed by both Id_K and φ.

The main aim of this course is to study groups of automorphisms of field extensions, their fixed subfields, and the relationship between the structure of field extensions and the structure of the associated groups. We will pay special attention to the question whether a general polynomial has roots expressible by a formula consisting of field operations and taking n-th roots, as in formula (1) for the quadratic polynomial. It was known by the 16th century that such a formula exists if deg f ≤ 4 (Tartaglia, Cardano, Ferrari). In contrast, it was discovered in the 19th century that there is usually no such formula for deg f ≥ 5 (Ruffini, Abel, Galois). We will prove these results as an application of our structure theory of field extensions.

1. Rings, Fields and Polynomial Rings

1.1. Rings and domains. Recall that R is a ring if it has addition, additive inverses, an additive identity 0 ∈ R, and it also has multiplication that is distributive over addition. All our rings will also have a multiplicative identity 1 ∈ R, and will have both operations commutative and associative.

Ring homomorphisms f : R → S between rings are required to map the multiplicative identity of R to that of S. For example, $\mathbb Z$ is a ring while $\mathbb N$ is not a ring; the map f : $\mathbb Z→\mathbb Z$ defined by f(n) = 0 is not a ring homomorphism.

A ring R is an integral domain, if there are no zero-divisors, i.e. ab = 0 implies that a = 0 or b = 0.

A subset I of a ring R is said to be an ideal, denoted I $\trianglelefteq$ R, if

(1) for a ∈ R, x ∈ I, we have xa ∈ I;

(2) 0 ∈ I and for a, b ∈ I, we have a + b, -a ∈ I.

Example. Any ideal in the ring of integers is of the form nℤ = {0,  ± n,  ± 2n, …} ⊴ ℤ. The corresponding quotient ring is ℤ/nℤ, which is a domain if and only if it is a field if and only if n is prime.

Given any ring R we can define a ring homomorphism ϕ : ℤ → R such that $$\phi(n)=\left\{\begin{array}{l} \underbrace{1+1+\ldots+1}_{n \text { times }}(n \geq 0) \\ \underbrace{-(1+1+\ldots+1)}_{|n| \text { times }}(n<0) \end{array}\right.$$ Its kernel ker ϕ = {n ∈ ℤ : ϕ(n) = 0} is an ideal in ℤ. So $$\operatorname{ker} \phi=\left\{\begin{array}{c} 0 \\ n \mathbb{Z} \end{array}\right.$$ In the first case, ℤ is a subring of R, and we say R has characteristic 0 . In the latter case, ℤ/nℤ becomes a subring of R, and we say that R has characteristic n. In both cases, we call this the prime subring of R. The characteristic of a domain is 0 or a prime number.

1.2. Fields. A ring K is a field if non-zero elements have multiplicative inverses. A field is automatically an integral domain, and therefore has characteristic 0 or prime p. In the first case, we have $\mathbb Q$ ⊂ K, in the latter case, $\mathbb F$_p ⊂ K, the prime subfield of K.

Proposition. Given an integral domain R, there is a field K, the field of fractions of R, with the following properties: there is a homomorphism ϕ : R → K which is injective and for every x ∈ K, there are a, b ∈ R such that $x=\frac{\phi(a)}{\phi(b)}$.

Proof. Construct K as the set of pairs (a,b), a ∈ R, b ∈ R ∖ {0} under the equivalence relation (a,b) ∼ (a^′,b^′) ⇔ ab^′ = ba^′. Then K has addition (a,b) + (c,d)= (ad+bc,bd) and multiplication (a,b)(c,d) = (ac,bd). This has all the required structure including multiplicative inverses: (a,b)(b,a) = 1 for a, b ≠ 0. Hence K is a field. The homomorphism ϕ : R → K is given by ϕ(a) = (a,1) and $(a, b)=\frac{\phi(a)}{\phi(b)}$ if b ≠ 0.

Lemma. (i) Let K be a field, I ⊴ K. Then either I = {0} or I = K.

(ii) Let K, L be fields. Then if f : K → L is a ring homomorphism, then f is injective.

Proof. (i) For a ≠ 0 ∈ I, a^− 1 ∈ K and so 1 ∈ I. Then for b ∈ K, 1b ∈ I and thus K ⊆ I. So I = K.

Part (ii) follows from (i). ker f = {a ∈ K : f(a) = 0} ⊴ K. So either ker f = K, and so f(1) = 0 which cannot happen as f(1) = 1, or ker f = {0} and thus f is injective. 1.3. Polynomial Rings. Given a ring R, we can form another ring by considering $$R[x]=\left\{\sum_{i=0}^{d} a_{i} x^{i}: a_{i} \in R\right\}$$ the set of all polynomials with coefficients in R. This becomes a ring by extending the addition and multiplication in the usual way ("opening the bracket"). For a nonzero polynomial f ∈ R[x], its degree deg f is the largest n for which a_n ≠ 0. A polynomial f ∈ R[x] is said to be monic if its leading coefficient is 1. Given a polynomial f ∈ K[x], α ∈ K is a root of f if f(α) = 0.

Lemma. If R is a domain, then R[x] is a domain.

Proof. Suppose f, g ≠ 0 ∈ R[x] with leading coefficients a_Nx^N and b_Mx^M. Then since these are leading coefficients, they are non-zero. But since R is a domain, a_Nb_M ≠ 0 and so fg ≠ 0. So R[x] is a domain.

Now let K be a field. We get a ring K[x], the polynomial ring over K, and its field of fractions $$K(x)=\left\{\frac{f(x)}{g(x)}: f, g \in K[x], g \neq 0\right\} .$$ Definition.    (1) A polynomial f ∈ K[x] is said to be irreducible if f = gh only if deg g = 0 or deg h = 0.

(2) A polynomial f ∈ K[x] divides a polynomial g, written f ∣ g, if there exists h ∈ K[x] such that g = fh.

(3) A polynomial f ∈ K[x] is prime, if whenever f ∣ gh, (f∣g) or (f∣h).

An ideal I is said to be prime if ab ∈ I implies a ∈ I or b ∈ I. Hence f is prime in K[x] if and only if (f) is a prime ideal.

Recall

Proposition. K[x] is a Euclidean ring; that is, given two monic polynomials f, g with deg g ≥ 1 then there exist q, r ∈ K[x] wth deg r < deg g, satisfying f = gq + r.

This implies

Theorem. If K is a field, then K[x] is a unique factorization domain. That is,

(1) The irreducible polynomials are precisely the prime polynomials.

(2) Given f ≠ 0 ∈ K[x], it can be expressed as a product f = f₁f₂f₃…f_n of irreducibles in an essentially unique way (up to rearrangement and multiplication by scalars).

Corollary. Let K be a field.

(1) Given a polynomial f ∈ K[x], a ∈ K is a root if f(a) = 0 ⇔ f(x)= (x−a)g(x).

(2) Given f ∈ K[x], there exists distinct roots a_i ∈ K, multiplicities m_i ∈ ℕ, and g ∈ K[x] without roots, such that f = g(x)∏(x−a_i)^m_i. Concerning irreducibility, we have the important and useful Lemma.

(Gauss’ Lemma) Let f ∈ ℤ[x] be monic. Then f is irreducible in ℤ[x], if and only if f is irreducible in $\mathbb Q$[x].

Corollary. Let f be a monic polynomial with integer coefficients and deg f ≤ 3. Then if f has no integral root, then f is irreducible in $\mathbb Q$[x].

Proof. By Gauss’ Lemma, f(x) is reducible in $\mathbb Q$[x] if and only if f(x) is reducible in ℤ[x] if and only if f = f₁f₂ with 1 ≤ deg f₁, 1 ≤ deg f₂. Since we have made the assumption that deg f = 3, it must be that deg f₁ = 1 (wlog) and so f = (x−α)f₂(x). Then α ∈ ℤ is a root.

Proposition. (Eisenstein’s Criterion) Let $$f(x)=x^{d}+\sum_{i=0}^{d-1} a_{i} x^{i}$$ be a monic polynomial over ℤ. Suppose that for some prime p, we have that p ∣ a_i for 0 ≤ i < d, but p² ∤ a₀. Then f(x) is irreducible over ℤ.

We are interested in methods of determining whether monic polynomials are irreducible in ℤ[x] (and hence in $\mathbb Q$[x] ). There are various tools that can be used to do this:

(1) Eisenstein’s criterion.

(2) More generally, reduction modulo p, for p a prime. Given a polynomial f ∈ ℤ[x], we have (f mod  p) ∈ $\mathbb F$_p[x] = ∑(a_i mod  p)xⁱ. Clearly, if f = gh then f≡_pgh. Conversely, if for some p, (f mod  p) is irreducible, so if f. Checking irreducibility in $\mathbb F$_p[x] is a finite (though perhaps cumbersome) task, since we have only finitely many choices for each coefficient in a splitting.

(3) Substitution: if f(x) = g(x)h(x) then f(x−a) = g(x−a)h(x−a). So f(x−a) irreducible implies f(x) is irreducible. There is an example of this on a problem sheet. This might help (say by the fact that f(x−a) may be an Eisenstein polynomial).

(4) Tricks and ingenuity.

GROUP ACTIONS ON RINGS AND FIELDS

Basic notions.

Definition. A group G acts on a ring R (on the left) if for every g ∈ G, we are given a ring automorphism (bijective ring homomorphism) g : R → R mapping a ↦ g(a), such that for a ∈ R and g, h ∈ G, we have h(g(a)) = (hg)(a). (For G to act on R on the right, often written a ↦ ag or a ↦ a^g, we require (ag)h=a(gh)).

Lemma. Let the group G act on the ring R.

(1) Consider R^G = {a ∈ R : g(a) = a for all g ∈ G}. Then R^G is a subring of R.

(2) If R = K is a field, then for g ∈ G and a ∈ K ∖ {0} we have g(1) = 1 and g(a^− 1) = (g(a))^− 1.

(3) If R = K is a field, then K^G is a subfield, containing the prime subfield of K.

Proof. For (1), just use the subring test. For (2), g(1) = g(1²) = g(1)g(1) so g(1) = 1 so 1 = g(1) = g(aa^− 1) = g(a)g(a^− 1). For (3), taking an element a ∈ K^G we have that g(a^− 1) = g(a)^− 1 = a^− 1 and so a^− 1 ∈ K^G. Moreover, 1 is fixed, and the prime subfield consists of elements which are ratios of sums of 1, so the prime subfield must always be fixed.

For K a field, denote by Aut (K) the group of all field automorphisms (bijective field homomorphisms) g : K → K, with the group operation being composition. More generally, for a field extension L/K, let Aut_K(L) = {g ∈ Aut (L) ∣  for all a ∈ K, g(a) = a} denote the group of all field automorphisms of L over K. Using this notation, an action of a group G on a field K is a group homomorphism G → Aut (K). In most of our examples, this homomophism will be injective (different elements of G give different automorphisms of K ) in which case we will call the action faithful.

Symmetric Polynomials. Take K a field and let

R = K[x₁,x₂,…,x_n] be the polynomial ring in n variables. This is a domain, since we can define this iteratively adding one variable at a time. The ring R has an action by the symmetric group S_n by permuting variables: if f(x₁,…,x_n) is a polynomial and σ ∈ S_n then f^σ(x₁,…,x_n) = f(x_1σ,…,x_nσ). (This is a right action; traditionally permutations σ ∈ S_n act on {1, …, n} on the right).

Definition. The ring of symmetric polynomials is the fixed ring K[x₁,…,x_n]^S_n

The following elementary symmetric polynomials are clearly elements of the ring K[x₁,…,x_n]^S_n : $$s_{k}=\sum_{i_{1}<i_{2}<\ldots<i_{k}} \prod_{j=1}^{k} x_{i_{j}}$$ For example. $$\begin{aligned} s_{1} &=\sum_{i} x_{i}, \\ s_{2} &=\sum_{i<j} x_{i} x_{j}, \\ s_{3} &=\sum_{i<j<k} x_{i} x_{j} x_{k}, \end{aligned}$$ and so on.

Theorem. (Theorem on Symmetric Functions) K[x₁,…,x_n]^S_n has the following structure. K[x₁,…,x_n]^S_n = K[s₁,…,s_n] In other words, every symmetric polynomial f ∈ K[x₁,…,x_n]^Sn is uniquely expressible as a polynomial of the elementary symmetric polynomials s_i.