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Hi, i would like to know if weather or not a Galois extension of a commutative semi-local ring is also a semilocal ring.

edited Sep 29 at 5:53

Filippo Alberto Edoardo
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asked Sep 28 at 17:39

Hector
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	What is a Galois extension of something which is not a field? – Filippo Alberto Edoardo Sep 29 at 5:54
	A finite étale morphism $Y \to X$ which is Galois. – Timo Keller Sep 29 at 8:58
	Where Galois means that $\mathrm{Aut}(Y/X)$ acts (simply) transitively on the fibres. – Timo Keller Sep 29 at 9:03
	As Konstantin points out in an answer below, any finite map of rings satisfies this property. See for example Atiyah-Macdonald and the going-up / down theorems. – Karl Schwede Sep 29 at 13:47
	@Timo: thanks, for some stupid reason, if you write "Galois morphism of schemes" I see what you mean and if one writes "Galois morphism of rings" I don't. Bad sign... – Filippo Alberto Edoardo Oct 1 at 0:10

Konstantin Ardakov · Answer 1 · 2012-09-29 09:03:37Z UTC

Let $A \to B$ be a map of rings such that $B$ is a finitely generated $A$-module. Suppose $A$ is semi-local. Then $B$ is also semi-local. To see this, note that every simple $B$-module is finitely generated over $A$ hence killed by the Jacobson radical $J(A)$ of $A$ by Nakayama's Lemma. So $J(A) \subseteq J(B)$ which means that $B / J(B)$ is finitely generated as a module over $A/J(A)$ and hence has finite length as an $A$-module since $A$ is semi-local. In particular it has finite length as a $B$-module and so $B$ is semi-local.

This argument works even if the rings are non-commutative.

Hector Pinedo · Answer 2 · 2012-09-30 12:09:35Z UTC

Thanks a lot Doctor Andarkov. By the way, i just find an article where the autor proves that given a semi-local ring $R,$ every left $R$-module is semi-local.

The article is On semilocal modules and rings Christian Lomp COMMUNICATIONS IN ALGEBRA, 27(4), 1921-1935 (1999)

The result is in Theorem 3.5

Galois extension of a semi-local ring

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