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

\begin{equation}x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 c}}{2}\end{equation}

can be thought of a symmetry of the situation. Assuming that the roots x1,2 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 = {IdK, φ} acting on K.

* φ φ = IdK, 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 IdK 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/n, 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, is a subring of R, and we say R has characteristic 0 . In the latter case, ℤ/n 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 an ≠ 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 aNxN and bMxM. Then since these are leading coefficients, they are non-zero. But since R is a domain, aNbM ≠ 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, (fg) or (fh).

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.


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 = f1f2f3fn 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)= (xa)g(x).

(2) Given f ∈ K[x], there exists distinct roots ai ∈ K, multiplicities mi ∈ ℕ, and g ∈ K[x] without roots, such that f = g(x)∏(xai)mi. 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 = (xα)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 . Suppose that for some prime p, we have that p ∣ ai for 0 ≤ i < d, but p2 ∤ a0. Then f(x) is irreducible over .

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] = ∑(ai mod  p)xi. Clearly, if f = gh then fpgh. 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(xa) = g(xa)h(xa). So f(xa) 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(xa) may be an Eisenstein polynomial).

(4) Tricks and ingenuity.


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 ↦ ag, we require (ag)h=a(gh)).

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

(1) Consider RG = {a ∈ R : g(a) = a for all g ∈ G}. Then RG 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 KG 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 ∈ KG we have that g(a − 1) = g(a) − 1 = a − 1 and so a − 1 ∈ KG. 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 AutK(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,…,xn] 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 Sn by permuting variables: if f(x1,…,xn) is a polynomial and σ ∈ Sn then fσ(x1,…,xn) = f(x1σ,…,xnσ). (This is a right action; traditionally permutations σ ∈ Sn act on {1, …, n} on the right).

Definition. The ring of symmetric polynomials is the fixed ring K[x1,…,xn]Sn

The following elementary symmetric polynomials are clearly elements of the ring K[x1,…,xn]Sn : $$s_{k}=\sum_{i_{1}<i_{2}<\ldots<i_{k}} \prod_{j=1}^{k} x_{i_{j}}$$ 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,…,xn]Sn has the following structure. K[x1,…,xn]Sn = K[s1,…,sn] In other words, every symmetric polynomial f ∈ K[x1,…,xn]Sn is uniquely expressible as a polynomial of the elementary symmetric polynomials si.