# A proof in Lang - Algebraic number theory

I'm reading Proposition 14, page 15, Chapter I, taken in Lang - Algebraic number theory. It states that if $A$ is an integrally closed domain with field of fractions $K$, $L$ be a finite galois extension of $K$, $B$ be the integral closure of $A$ in $L$, $\mathfrak P$ be maximal ideal of $B$ lying over a maximal ideal $\mathfrak p$ of $A$, then $\bar B=B/\mathfrak P$ is a normal extension of $\bar A=A/\mathfrak p$.

The proof shows, actually, only that each finite separable subextension of $\bar B/\bar A$ is normal. This implies that $\bar B$ is normal over $\bar A$? Why?

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Little inaccuracy in your question: The proof does NOT show that "each finite separable subextension of $\overline B / \overline A$ is normal; it shows this for the maximal finite separable subextension. – darij grinberg Jan 12 '12 at 21:11

Lang actually proves that for every $\overline{x}\in \overline B$ (no separability condition required), there exists a polynomial over $\overline A$ which has $\overline{x}$ as a root and splits into linear factors over $\overline B$. This yields that $\overline B$ is normal over $\overline A$.
Lang write: "Let $\bar x$ generate a separable subextension of $\bar B$ over $\bar A$." Why it requires separable even if his argumentation holds without separability conditions? – Fabio Lucchini Jan 12 '12 at 17:50
I think the "separable" in this sentence is just a mistake. The argument doesn't require $\overline x$ to be separable, does it? (The first time Lang needs separability is when he invokes the primitive element theorem, but at that time the normality is already proven.) – darij grinberg Jan 12 '12 at 18:25
I'm sure that the separability condition on $\bar x$ can be dropped to prove the normality of $\bar B$ over $\bar A$. My doubt is that under the hypothesys of the proposition a statement such as "if the maximal separable subextension of $\bar B/\bar A$ is normal, then $\bar B/\bar A$ is normal" holds, hence Lang can assume $\bar x$ separable. – Fabio Lucchini Jan 12 '12 at 19:30