Hi, i would like to know if weather or not a Galois extension of a commutative semi-local ring is also a semilocal ring.
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$\begingroup$ What is a Galois extension of something which is not a field? $\endgroup$– Filippo Alberto EdoardoCommented Sep 29, 2012 at 5:54
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$\begingroup$ A finite étale morphism $Y \to X$ which is Galois. $\endgroup$– user19475Commented Sep 29, 2012 at 8:58
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$\begingroup$ Where Galois means that $\mathrm{Aut}(Y/X)$ acts (simply) transitively on the fibres. $\endgroup$– user19475Commented Sep 29, 2012 at 9:03
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$\begingroup$ 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. $\endgroup$– Karl SchwedeCommented Sep 29, 2012 at 13:47
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$\begingroup$ @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... $\endgroup$– Filippo Alberto EdoardoCommented Oct 1, 2012 at 0:10
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2 Answers
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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.
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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