Let $C$ be a finitely generated $R$-algebra where $R$ is unital commutative ring. Let $S$ be a simple $C$-module. Show that the annihilator of $S$ in $R$ is a maximal ideal.
What I have tried is: When $C=R$ then it is a direct consequence of the simplicity of $S$ as an $R$-module.
Further, since $S$ is simple then $S$ is cyclic (i.e. $S=Cy$ for nonzero $y\in S$). Consider the map $$f: C \to S$$ $$c \mapsto cy$$
Then $\ker(f)$ is the annihilator of $S$ in $C$ and $\ker(f) \cap R$ is the annihilator of $S$ in $R$. I have shown that it is an ideal of $R$ but I have no idea to continue showing the maximality. Is it a true direction or is there anything else I should consider to show the maximality?
Best Answer
I got an idea : If we restrict the map $f: C \to S$ on $R$. Then we have an isomorphism between $f(R)=Ry$ and the quotient $R/(\ker(f) \cap R)$, where $\ker(f) \cap R$ is the annihilator of $S$ in $R$.
So to show $R/(\ker(f) \cap R)$ is maximal, it is enough to show that $Ry$ is simple as an $R$-module. Let $ry \in Ry$ be a nonzero element. Then clearly $Ry$ contains $\langle ry \rangle$ as $R$-modules. Conversely, since $ry \in Ry \subset S$ is nonzero, $\langle ry \rangle = S \supset Ry$. So $Ry = \langle ry \rangle$ for any nonzero $ry \in Ry$. Hnece $Ry$ is simple and $R/(\ker(f) \cap R)$ is maximal.
Please let me know if there is ambiguity in my answer.