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:

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.

William Stein 2004-05-06