This is a brief reminder of the main ideas of Galois theory. Any proofs purported here are meant to be suggestive. I learned Galois theory out of Dummit and Foote, which I thought was pretty good. I also have Classical Galois Theory by Gaal on my shelf. This book is in essence one giant worksheet. I have not completed many of the exercises, but I suspect anyone who did would gain a remarkable intuition as to how the theory hangs together.

At any rate, this is mostly for me, since I seem to need to be reminded of the basics of Galois theory every few years.

It may be useful to review Field Extensions and Number Fields before continuing.

Automorphisms of Field Extensions

We want to work in a bit of generality here, so we assume $K | k$ is an extension of number fields. Little generality is lost at this point if you take $k = \mathbb Q$.

Definition

An isomorphism of K is called an automorphism. The set of automorphisms of $K,$ $\mathrm{Aut}(K)$, forms a group under composition. An automorphism σ is said to fix $k$ if $\sigma \gamma=\gamma$ for all $\gamma \in k$. The set of automorphisms of $K$ which fix $k$ is denoted $\mathrm{Aut}(K|k)$ and is a subgroup of $\mathrm{Aut}(K)$.

$\mathrm{Aut}(K|k)$ is a finite group of degree at most $d = [K:k]$. We will use this fact, though the proof would take us too far afield.

Automorphisms of $K$ which preserve $k$ permute roots of polynomials with coefficients in $k$ and roots in $K$.

Proposition

Suppose $g(x) \in k[x]$ is irreducible and there exists $\beta \in K$ such that $g(\beta) = 0$. Then $g( \sigma \beta) = 0 \quad \mbox{for all} \quad \sigma \in \mathrm{Aut}(K|k)$.

Proof

Suppose $g(x) = \sum_m b_m x^m$, then $\sigma b_m = b_m$, and hence $$0= \sigma g(\beta) = \sum_m \sigma b_m (\sigma \beta)^m = \sum_m b_m (\sigma \beta)^m = g(\sigma \beta). \qquad \square$$

One particularly important automorphism is complex conjugation. Suppose that $K | \mathbb Q$ is a number field, and $K \cong \mathbb Q(\alpha)$ for some non-real $\alpha$ (that is the minimal polynomial of $K$ has a non-real root in $\mathbb C$). Then, since complex conjugation is an automorphism of $\mathbb C | \mathbb R$, we have it is also an isomorphism on $K(\alpha) | \mathbb Q$. It follows that if $\beta \in \mathbb Q(\alpha)$ then $\overline \beta \in \mathbb Q(\alpha)$ as well and hence, as sets, $\mathbb Q(\alpha)=\mathbb Q(\overline \alpha)$, and algebraic operations in one of these embeddings can be found from the other by complex conjugation.

Let us distinguish the real and complex roots of $f(x)$ by setting $\alpha_1, \ldots, \alpha_{r_1}$ to be the real roots and $\beta_1, \overline{\beta_1}, \ldots, \beta_{r_2}, \overline{\beta_{r_2}}$ be the non-real complex roots. Clearly $r_1 + r_2 = d$. Then the embeddings $\mathbb Q(\alpha_1), \ldots, \mathbb Q(\alpha_{r_1}) , \mathbb Q(\beta_1) , \ldots, \mathbb Q(\beta_{r_2})$ are called the archimedean embeddings of $K$.

Splitting Fields

Galois theory is concerned about the zeros of rational polynomials and how their zeroes are permuted by the automorphisms of certain extensions of $\mathbb Q$ (which will come to be called Galois extensions). We already noted, that the automorphisms in $\mathrm{Aut}(K|k)$ preserve the set of zeroes in $K$ of any given polynomial $g(x) \in k[x]$. However, the construction $K$ makes no guarantee that a generic polynomial $g(x)$ will have a zero in $K$, and even for the minimal polynomial $f(x)$, the construction of $K$ only guarantees the existence of a single zero of $f(x)$.

In general, if we wanted an extension of $k$ that contains all the zeros of $f(x)$, we would first compute $K = k[x]/ f(x) \mathbb Q[x]$. $K$ contains at least one zero of $f(x)$, and if we factor it in $K[x]$ there will be a linear factor for each of those zeroes. We can then sequentially extend $K$ by constructing field extensions from the remaining irreducible factors of $f(x)$. Each time we extend fields by another irreducible factor, we add another zero of $f(x)$ to the resulting field extension. The process terminates after at most $d$ steps to produce the splitting field of $f(x)$. The splitting field of $f(x)$ has degree bounded by $d!$.

It is possible that the degree of the splitting field is as small as $d$, since it is possible, depending on the nature of $f(x)$, that $K[x]/f(x)\mathbb Q(x)$ itself contains $d$ zeros of $f(x)$.

Example

Suppose $p$ is prime and consider the $p$th cyclotomic polynomial $$\Phi_p(x) = x^{p-1} + x^{p-2} + \cdots + x + 1.$$ Suppose $\zeta$ is a zero of $\Phi_p(x)$ in $\mathbb Q[x]/\Phi_p(x) \mathbb Q[x]$, then it is easily verified that $\zeta^p = 1$.It follows that, if $\ell = 1, \ldots, p-1$, \begin{eqnarray}\Phi_p(\zeta^{\ell}) &=& \zeta^{\ell(p-1)} + \zeta^{\ell(p-2)} + \cdots + \zeta^{\ell} + 1 \\ &=& \zeta^{p-1} + \zeta^{p-2} + \cdots + \zeta + 1 \\ &=& 0.\end{eqnarray} It follows that $\zeta, \zeta^2, \ldots, \zeta^{p-1}$ are all $p-1$ roots of $\Phi_p(x)$ and hence $K = \mathbb Q[x]/\Phi_p(x)\mathbb Q[x]$ is the splitting field of $\Phi_p(x)$.

Galois Theory

Definition

If $K | k$ is a splitting field for a polynomial $g(x) \in k[x]$, then $K$ is said to be Galois over $k,$ and the group of automorphisms of $K$ which fix $k$ is called the Galois group and denoted $\mathrm{Gal}(K|k)$.

Claim

$K | k$ is Galois if and only if $\# \mathrm{Aut}(K|k) = [K : k]$.

We won’t prove this claim (though the only if direction is easy) because it is a bit fiddly with separability and involves a diversion into character theory. Some (most?) authors give this as the definition of Galois and prove that it implies the splitting field definition.

The main result in Galois Theory is a correspondence between intermediate fields of $K | k$ and subgroups of $\mathrm{Gal}(K|k)$. Let us write $G = \mathrm{Gal}(K|k)$ and suppose $H < G$ is a subgroup. Define $$K_H = \{ \gamma \in K : \sigma(\gamma) = \gamma \mbox{ for all } \sigma \in H \}.$$ It is easily verified that $K_H$ is a field, and $k \subset K_H \subset K$ (which we might abbreviate $K | K_H | k$). It will turn out that $H \leftrightarrow K_H$ will be a bijection (called the Galois correspondence) between subgroups of $G$ and intermediate fields of $K | k$.

This correspondence goes beyond a bijection, because there is an interpretation for $H$ and $G/H$ (as a subgroup in the case where $H$ is normal, but to some extent even as a set of cosets in the non-normal case) in terms of the groups of automorphisms $\mathrm{Gal}(K|K_H)$ and $\mathrm{Aut}(K_H|k)$. I hope you objected to the notational switch between $\mathrm{Gal}$ and $\mathrm{Aut}$ in the previous sentence, but it is correct. The fact that $K$ is a splitting field for a polynomial in $k[x]$ means that it is also the splitting field for a polynomial in $K_H[x]$ (namely any one of the irreducible factors of the original polynomial in $k[x]$) and hence $K | K_H$ is Galois and we use the notation $\mathrm{Gal}(K | K_H)$ for the group of automorphisms of $K$ preserving $K_H$. This is, unsurprisingly, equal to $H$. On the other hand, just because $K$ is the splitting field of a polynomial in $k[x]$ doesn’t imply that an intermediate field, such as $K_H$, must be a splitting field for that or any other polynomial in $k[x]$. Thus, in general we need to refer to the automorphism group of $K_H | k$ by $\mathrm{Aut}(K_H | k)$. It will turn out that when $H$ is normal in $G$ then $K_H | k$ is Galois, and $\mathrm{Gal}(K_H | k) \cong G/H$. This will all be enumerated in the Fundamental Theorem of Galois Theory, but we need to develop a few results first.

Given $\gamma \in K$, we call $\sigma \gamma; \sigma \in G$ the Galois conjugates of $\gamma$. Moreover, if $L$ is any intermediate field extension, $K | L | k$, then $\sigma$ gives an isomorphism from $L$ onto $\sigma(L)$ (which fixes $k$). In particular $K_H$ is isomorphic to its image $\sigma K_H$. Notice then that if $\psi \in \mathrm{Aut}(K_H | k)$, then $\sigma \psi \sigma^{-1}$ is an element of $\mathrm{Aut}(\sigma K_H | k)$.

Indeed, $\sigma H \sigma^{-1} = \mathrm{Aut}(\sigma K_H | k)$. We can make this more evocative by denoting the action by conjugation of $G$ on $H$ as $\sigma \cdot \psi = \sigma \psi \sigma^{-1}$, in which case, $$\sigma \cdot \mathrm{Aut}(K_H | k) = \mathrm{Aut}(\sigma K_H | k).$$ If $\sigma K_H = K_H$ for all $\sigma \in G$, then $GHG^{-1} = H$, that is $H$ is normal in $G$. On the other hand, if $H$ is normal in $G$, then $\sigma \psi \sigma^{-1} \in H$ and $\sigma K_H = K_H$ for all $\sigma \in G$.

Now, suppose $H$ is normal and $g(x) \in k[x]$ is a polynomial so that $K_H = k[x]/g(x)k[x]$. From the previous discussion, $x + g(x) k[x]$ is a zero of $g(x)$ in $K_H$, as are $\sigma (x + g(x) k[x])$ for all $\sigma \in G$. To establish $K_H | k$ is Galois, we need to show that the orbit of $x + g(x) k[x]$ under $G$ is equal to $[K_H : k]$. We know for $\sigma \in H$, $\sigma(x + g(x) k[x]) = x + g(x)k[x]$. On the other hand, if $\sigma(x + g(x) k[x]) = x + g(x)k[x]$ then $\sigma \in H$ because $\sigma$ is completely determined by its action on $x + g(x) k[x]$. Thus, the automorphisms of $\mathrm{Aut}(K_H|k)$ are in correspondence with $G/H$. We thus have $[K : K_H] = \#H$, $[K : k] = \#G$ and $[K_H : k] = \#G/H$. It follows that $\# \mathrm{Aut}(K_H|k) = [K_H : k]$ and $K_H | k$ is thus Galois.

To be sure, we have glossed over many details. However, many important observations are captured in the Fundamental Theorem of Galois Theory.

Fundamental Theorem of Galois Theory

Suppose $K | k$ is Galois and $G = \mathrm{Gal}(K|k)$.

CORRESPONDENCE

There is an inclusion reversing correspondence between intermediate fields of $K|k$ and subgroups of $H$.

Normality $\leftrightarrow$ Galois

$H$ is normal in $G$ if and only if $L|k$ is Galois. In this situation $\mathrm{Gal}(L|k) \cong G/H$.

The Correspondence Preserves Lattices

Suppose $H_1 \leftrightarrow L_1$ and $H_2 \leftrightarrow L_2$ for $H_1, H_2 \leq G$ and $L_1, L_2$ intermediate fields of $K|k$. Then $\langle H_1, H_2 \rangle \leftrightarrow L_1 \cap L_2$ and $H_1 \cap H_2 \leftrightarrow L_1 L_2$. (Here $\langle H_1, H_2 \rangle$ is the smallest subgroup of $G$ containing both $H_1$ and $H_2$ and $L_1 L_2$ is the smallest field containing both $L_1$ and $L_2$). Moreover the inclusions (e.g. $L_1 \cap L_2 \subset L_1 \subset L_1 L_2$) are reversed under the correspondence.

Here we write arrows for the inclusion map. The correspondence reverses inclusion.

The correspondence between subgroups of $\mathrm{Gal}(K|k)$ and subfields of $K|k$ is complete as the subfields of $\mathbb{Q}(i, \sqrt[8]{2})$ and subgroups of $G = \langle \sigma, \tau : \sigma^8 = \tau^2 = 1, \sigma \tau = \tau \sigma^3 \rangle$. This example was cribbed from Abstract Algebra, second edition by Dummit and Foote.

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