Galois Extensions

In this section we give a survey (no proofs) of the basic facts about Galois extensions of $\mathbf {Q}$ that will be needed in the rest of this chapter.

Definition 9.1.1 (Galois)   An extension $K/L$ of number fields is Galois if

$$\char93 \Aut (K/L) = [K:L],$$

where $\Aut (K/L)$ is the group of automorphisms of $K$ that fix $L$. We write

$$\Gal (K/L) = \Aut (K/L).$$

For example, if $K\subset \mathbf{C}$ is a number field embedded in the complex numbers, then $K$ is Galois over $\mathbf {Q}$ if every field homomorphism $K\to \mathbf{C}$ has image $K$. As another example, any quadratic extension $K/L$ is Galois over $L$, since it is of the form $L(\sqrt{a})$, for some $a\in L$, and the nontrivial automorphism is induced by $\sqrt{a}\mapsto -\sqrt{a}$, so there is always one nontrivial automorphism. If $f\in L[x]$ is an irreducible cubic polynomial, and $a$ is a root of $f$, then one proves in a course on Galois theory that $L(a)$ is Galois over $L$ if and only if the discriminant of $f$ is a perfect square in $L$. ``Random'' number fields of degree bigger than $2$ are rarely Galois.

If $K\subset \mathbf{C}$ is a number field, then the Galois closure $K^{\gc }$ of $K$ in $\mathbf{C}$ is the field generated by all images of $K$ under all embeddings in  $\mathbf{C}$ (more generally, if $K/L$ is an extension, the Galois closure of $K$ over $L$ is the field generated by images of embeddings $K\to \mathbf{C}$ that are the identity map on $L$). If $K=\mathbf{Q}(a)$, then $K^{\gc }$ is the field generated by all of the conjugates of $a$, and is hence Galois over  $\mathbf {Q}$, since the image under an embedding of any polynomial in the conjugates of $a$ is again a polynomial in conjugates of $a$.

How much bigger can the degree of $K^{\gc }$ be as compared to the degree of $K=\mathbf{Q}(a)$? There is an embedding of $\Gal (K^{\gc }/\mathbf{Q})$ into the group of permutations of the conjugates of $a$. If $a$ has $n$ conjugates, then this is an embedding $\Gal (K^{\gc }/\mathbf{Q})\hookrightarrow S_n$, where $S_n$ is the symmetric group on $n$ symbols, which has order $n!$. Thus the degree of the $K^{\gc }$ over  $\mathbf {Q}$ is a divisor of $n!$. Also $\Gal (K^{\gc }/\mathbf{Q})$ is a transitive subgroup of $S_n$, which constrains the possibilities further. When $n=2$, we recover the fact that quadratic extensions are Galois. When $n=3$, we see that the Galois closure of a cubic extension is either the cubic extension or a quadratic extension of the cubic extension. One can show that the Galois closure of a cubic extension is obtained by adjoining the square root of the discriminant, which is why an irreducible cubic defines a Galois extension if and only if the discriminant is a perfect square.

For an extension $K$ of $\mathbf {Q}$ of degree $5$, it is ``frequently'' the case that the Galois closure has degree $120$, and in fact it is an interesting problem to enumerate examples of degree $5$ extension in which the Galois closure has degree smaller than $120$. For example, the only possibilities for the order of a transitive proper subgroup of $S_5$ are $5$, $10$, $20$, and $60$; there are also proper subgroups of $S_5$ order $2, 3, 4, 6, 8, 12$, and $24$, but none are transitive.

Let $n$ be a positive integer. Consider the field $K=\mathbf{Q}(\zeta_n)$, where $\zeta_n=e^{2\pi i/n}$ is a primitive $n$th root of unity. If $\sigma:K\to \mathbf{C}$ is an embedding, then $\sigma(\zeta_n)$ is also an $n$th root of unity, and the group of $n$th roots of unity is cyclic, so $\sigma(\zeta_n) = \zeta_n^m$ for some $m$ which is invertible modulo $n$. Thus $K$ is Galois and $\Gal (K/\mathbf{Q})\hookrightarrow (\mathbf{Z}/n\mathbf{Z})^*$. However, $[K:\mathbf{Q}]=\varphi (n)$, so this map is an isomorphism. (Remark: Taking a limit using the maps $\Gal (\overline{\mathbf{Q}}/\mathbf{Q})\to \Gal (\mathbf{Q}(\zeta_{p^r})/\mathbf{Q})$, we obtain a homomorphism $\Gal (\overline{\mathbf{Q}}/\mathbf{Q})\to \mathbf{Z}_p^*$, which is called the $p$-adic cyclotomic character.)

Compositums of Galois extensions are Galois. For example, the biquadratic field $K=\mathbf{Q}(\sqrt{5},\sqrt{-1})$ is a Galois extension of $\mathbf {Q}$ of degree $4$, which is the compositum of the Galois extensions $\mathbf{Q}(\sqrt{5})$ and $\mathbf{Q}(\sqrt{-1})$ of $\mathbf {Q}$.

Fix a number field $K$ that is Galois over a subfield $L$. Then the Galois group $G=\Gal (K/L)$ acts on many of the object that we have associated to $K$, including:

• the integers $\O_K$,
• the units $U_K$,
• the group of fractional ideals of $\O_K$,
• the class group $\Cl (K)$, and
• the set $S_\mathfrak{p}$ of prime ideals lying over a given nonzero prime ideal $\mathfrak{p}$ of $\O_L$, i.e., the prime divisors of $\mathfrak{p}\O_K$.

In the next section we will be concerned with the action of $\Gal (K/L)$ on $S_\mathfrak{p}$, though actions on each of the other objects, especially $\Cl (K)$, are also of great interest. Understanding the action of $\Gal (K/L)$ on $S_{\mathfrak{p}}$ will enable us to associate, in a natural way, a holomorphic $L$-function to any complex representation $\Gal (K/L) \to \GL _n(\mathbf{C})$.