Galois Extensions

Suppose $K\subset \mathbf{C}$ is a number field. Then $K$ is if every field homomorphism $K\to \mathbf{C}$ has image $K$, or equivalently, $\char93 \Aut (K) = [K:\mathbf{Q}]$. More generally, we have the following definition.

Definition 13.1.1 (Galois)   An extension $K/L$ of number fields is 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, $\mathbf{Q}$ is Galois (over itself), any quadratic extension $K/L$ is Galois, since it is of the form $L(\sqrt{a})$, for some $a\in L$, and the nontrivial embedding 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 in 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 (I will not justify this claim further in this course).

If $K/\mathbf{Q}$ is a number field, then the Galois closure $K^{\gc }$ of $K$ 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 generated by each 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 a natural embedding of $\Gal (K^{\gc }/\mathbf{Q})$ into the group of permutations of the conjugates of $a$. If there are $n$ conjugates of $a$, 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 the Galois group 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. It turns out that that Galois closure of a cubic extension is obtained by adjoining the square root of the discriminant. For an extension $K$ of degree $5$, it is ``frequently'' the case that the Galois closure has degree $120$, and in fact it is a difficult and interesting problem to find examples of degree $5$ extension in which the Galois closure has degree smaller than $120$ (according to : the only possibilities for the order of a transitive proper subgroup of $S_5$ are $5$, $10$, $20$, and $60$; there are five transitive subgroups of $S_5$ out of the total of $19$ subgroups of $S_5$).

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, $[K:\mathbf{Q}]=n$, so this map is an isomorphism. (Side note: Taking a $p$-adic limit and 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 $4$.

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 nonzero fractional ideals of $\O _K$,
• the class group $\Cl (K)$, and
• the set $S_\mathfrak{p}$ of prime ideals $P$ lying over a given prime $\mathfrak{p}$ of $\O _L$.

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 $\Cl (K)$, will be of further interest.