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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 $\mathbb{Q}(i, \sqrt{2})$ 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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