Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
What is a Galois extension of something which is not a field? –  Filippo Alberto Edoardo Sep 29 '12 at 5:54
    
A finite étale morphism $Y \to X$ which is Galois. –  Timo Keller Sep 29 '12 at 8:58
    
Where Galois means that $\mathrm{Aut}(Y/X)$ acts (simply) transitively on the fibres. –  Timo Keller Sep 29 '12 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 '12 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 '12 at 0:10
add comment

2 Answers 2

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.

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.