# Integral Galois Extensions (Lang)

I have trouble understanding an argument in the proof of Proposition 2.5, p. 342, of Lang's Algebra. The setup of my question is the following:

Let $A$ be integrally closed in its quotient field $K$ and $B$ be its integral closure in a finite Galois extension $L$ of $K$, with group $G$. Let $p$ be a maximal ideal of $A$ and $\beta$ a maximal ideal of $B$ lying above $p$. Denote $\bar{B}=B/\beta$ and $\bar{A}=A/p$.

In his proof, Lang takes an element $\bar{x} \in \bar{B}$ that generates a separable extension of $\bar{A}$. He shows that $\bar{B}$ is normal over $\bar{A}$ and that $[\bar{A}(\bar{x}):\bar{A}] \le [K(x):K] \le [L:K]$, where $x$ is a representative of $\bar{x}$ in $B$. So far so good.

Then he says that "...this implies that the maximal separable subextension of $\bar{A}$ in $\bar{B}$ is of finite degree over $\bar{A}$ (using the primitive element theorem of elementary field theory)".

Question 1: How can we use the P.E.T. to see that? The statement of the primitive element theorem assumes that the underlying field extension is finite.

Question 2: Is $\bar{B}$ over $\bar{A}$ separable? Why is Lang referring to the "Galois group" of $\bar{B}/\bar{A}$?

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Could you clarify when Lang mentions the Galois group of $\bar{B}/\bar{A}$ for those of us without the text handy? You've only mentioned $G$ the Galois group of $L$ over $K$. – Kevin Carlson Jul 31 '12 at 21:41
@KevinCarlson: Let $G_{\beta}$ be the subgroup of $G$ such that $\sigma \beta = \beta, \forall \sigma \in G_{\beta}$. In the statement of the proposition he mentions that there is a homomorphism of $G_{\beta}$ onto the Galois group of $B/\beta$ over $A/p$. – Manos Jul 31 '12 at 21:50

The primitive element theorem implies that every finite separable subextension of $\overline B/\overline A$ is simple, hence its degree over $\overline A$ is at most $[L:K]$ by the previous considerations. Let $M$ be a finite separable subextension of maximal degree. If $x \in \overline B$ is separable then $M(x)/\overline{A}$ is a finite separable extension, hence equal to $M$ by the maximality of the degree of $M$. This shows that $M$ is actually the maximal separable subextension of $\overline B/\overline A$.
Concerning you second question: For every extension of fields $L/K$ (without the assumption of separability or normality) one may consider the group of $K$-automorphisms of $L$, i.e. the automorphisms $L\to L$ that leave every element of $K$ fixed. Usually this group is only called the Galois group of $L/K$ if the extension is Galois. However, in the proof of Proposition 2.5 Lang seems to call this the Galois group even though the extension is not necessarily Galois. In the case where $L/K$ is normal it can be shown (as Lang claims) that any $K$-automorphism of $L$ is uniquely determined by its restriction to the maximal separable subfield.
Why needs $B/A$ to be finite? – Manos Jul 31 '12 at 22:12
You are right, $B/A$ doesn't need to be finite. I edited my answer to replace this by a different argument. – marlu Jul 31 '12 at 23:22