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

Let $A \to B$ be a map of rings such that $B$ is a finitely generated $A$module. Suppose $A$ is semilocal. Then $B$ is also semilocal. 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 semilocal. In particular it has finite length as a $B$module and so $B$ is semilocal. This argument works even if the rings are noncommutative. 


Thanks a lot Doctor Andarkov. By the way, i just find an article where the autor proves that given a semilocal ring $R,$ every left $R$module is semilocal. The article is On semilocal modules and rings Christian Lomp COMMUNICATIONS IN ALGEBRA, 27(4), 19211935 (1999) The result is in Theorem 3.5 

