Here, rings are commutative. The question comes from Commutative Algebra by Matsumura.
$A$ is a local ring with maximal ideal $\mathfrak{m}$. $B$ is a finite $A$-algebra, that is, there exists $b_1, \dotsc, b_n \in B$ with $B = Ab_1 + \dotsb + Ab_n$ as $A$-modules. Then prove that $B$ is semi-local, i.e., $B$ has finite maximal ideals.
I know that $B/\mathfrak{m}B$ is a finite $A/\mathfrak{m}$-algebra, and hence Artinian.
Best Answer
There is a natural map $f \colon A \longrightarrow B$. Since $A$ is local, $\mathrm{Rad}(A) = \mathfrak{m}$ and every proper ideal is contained in $\mathfrak{m}$.
Since $B$ is a finite $A$-module, Nakayama’s lemma tells that for every proper ideal $I$ of $A$, $IB \neq B$. Then this implies that for every maximal ideal $J$ of $B$, the preimage $f^{-1}(J)$ is a maximal ideal of $A$, so $$f^{-1}(J) = \mathfrak{m} \,.$$ Thus, $\mathfrak{m} B \subseteq J$. So we only need to prove that $B/\mathfrak{m}B$ is semi-local. But $B/\mathfrak{m}B$ is Artinian and so is isomorphic to a finite product of Artinian local rings. Therefore, it has finite maximal ideals.