# 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 $\overline A$.

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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
Ok; to ensure that "separable" is a mistake, there exists an "Errata" for this book? – Fabio Lucchini Jan 12 '12 at 18:54
Sorry, I don't think there is a list of errata for this book (although certainly Lang's books would do well with lists of errata!). You should be able to check this proof on your own and tell whether separability is ever used. – darij grinberg Jan 12 '12 at 19:06
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