# On finitely generated torsion $R$-modules $M$ when $R$ is commutative.

Let $R$ be a commutative ring and let $M$ be a finitely generated torsion $R$-module. If for any $x\in M$, we have that $R/\operatorname{ann}(x)$ is simple, I want to prove that $$M\cong \bigoplus_{i=1}^n M/l_iM$$ for some maximal ideals $l_1,\ldots, l_n\subset R$.

All I know is that if $M$ is a free module finitely generated, are all it's submodule finitely generated. Again, if $R$ is a PID, then there exists nonzero elements $v_1,\ldots,v_n\in M$ such that $$M=Rv_1\oplus Rv_2\oplus\cdots\oplus Mv_n.$$ However, in this case, $R$ is not necessarily a PID.

Take a finite generating system $x_1, \dots, x_n$ of $M$.
Since this is a generating system, we get that $\operatorname{Ann}_R(M)= \bigcap_{i=1} \operatorname{Ann}_R(x_i)$. Assume wlog that the ideals $\operatorname{Ann}_R(x_i)$ are all distinct by throwing out duplicates.
Now since $R/\operatorname{Ann}_R(x_i)$ is simple, $\operatorname{Ann}_R(x_i)$ must be a maximal ideal. Set $l_i=\operatorname{Ann}_R(x_i)$.
We get that $R/\operatorname{Ann}_R(M) \cong \bigoplus_{i=1}^n R/\operatorname{Ann}_R(x_i) = \bigoplus_{i=1}^n R/l_i$ by the Chinese remainder theorem, since distinct maximal ideals are coprime. Tensoring this isomorphism with $M$ over $R$ gives us
$M \cong M/0 = M/\operatorname{Ann}_R(M)M \cong R/\operatorname{Ann}_R(M) \otimes_R M \cong (\bigoplus_{i=1}^n R/l_i) \otimes_R M \cong \bigoplus_{i=1}^n (R/l_i \otimes_R M) \cong \bigoplus_{i=1}^n M/l_iM$