I'm preparing for the qualify exam and coming up with the following exercise:
Question: Let $S \subset R$ be a subring contained in the center of $R$. Suppose that $R$ is finitely generated as a left $S$-module. Prove: $\mathrm{rad}(S) \subset \mathrm{rad}(R)$. Here $\mathrm{rad}(\cdot)$ denotes the Jacobson radical.
My attempts: It looks extremely like an application of Nakayama's lemma, as $R$ is a finitely generated $S$-module and Jacobson radicals involve here. So I'm trying to argue as $$ \mathrm{rad}(S) \cdot (\star) = (\star) \Rightarrow (\star)=0 \Rightarrow \mathrm{rad}(S) \subset \mathrm{rad}(R), $$ for some mysterious object $(\star)$ as a finitely generated left $S$-module. But I have been struggling to find an appropriate $(\star)$ to make things work.
Then "best" candidate for $(\star)$ that I can think of is $\mathrm{rad}(S) \cdot R/\mathrm{rad}(R)$. Then I need to show that this is finitely generated and $$ \mathrm{rad}(S)^2 \cdot R/\mathrm{rad}(R) = \mathrm{rad}(S) \cdot R/\mathrm{rad}(R), $$ yet I got stuck here. As I haven't invoke the condition that $S \subset Z(R)$, it doesn't seem working.
Thank you all for answering and commenting!
Let $M$ be a simple $R$-module. Then it is finitely generated as an $S$-module, so has maximal submodules. Thus $\mathrm{rad}_S(M)<M$, and $\mathrm{rad}(S)M\subseteq\mathrm{rad}_S(M)$.
Next, $R\mathrm{rad}(S)=\mathrm{rad}(S)R$ as ideals in $R$, since $S$ is contained in the centre of $R$. Thus $\mathrm{rad}(S)M$ is a proper $R$-submodule of the simple module $M$, so is zero. This shows that $\mathrm{rad}(S)$ kills every simple $R$-module, and hence is contained in $\mathrm{rad}(R)$.