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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?

flag	asked Jan 12 2012 at 14:56 f.lucchini 97●5

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 2012 at 21:11

darij grinberg · Accepted Answer · 2012-01-12 16:08:31Z UTC

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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 $\overline A$.

answered Jan 12 2012 at 16:08

darij grinberg
12.6k●1●24●75

	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? – f.lucchini Jan 12 2012 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 2012 at 18:25
	Ok; to ensure that "separable" is a mistake, there exists an "Errata" for this book? – f.lucchini Jan 12 2012 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 2012 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. – f.lucchini Jan 12 2012 at 19:30

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A proof in Lang - Algebraic number theory

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