Let $R$ be a commutative ring with unity and $I$ be an ideal of $R$ such that $I^n=\{0\}$ for some $n \in \mathbb N$ . If $M$ is an $R$-module such that $M/IM$ is finitely generated over $R/I$, then is it true that $M$ is also finitely generated over $R$ ?
I know that it is enough to show that $IM$ is a finitely generated submodule of $M$ , but I am unable to show this . I can see that $I^{n-1}M$ is an $R/I$ module due to $I^n=0$ , but that doesn't help much .
Please help . Thanks in advance
Take a set of generators $m_1 + IM, \dots, m_n + IM$ for $M/IM$ and let $N$ denote the submodule of $M$ generated by the $m_i$. Then $M = N + IM$: indeed, for any $m \in M$, we have $$m + IM = r_1 m_1 + \dots + r_n m_n + IM$$ for some $r_i \in R$, so $m \in N + IM$.
From $M=N+IM$ we get $M/N=I(M/N)$. Multiplying $n-1$ times by $I$ we obtain $M/N=I^n(M/N)$. Since $I^n=0$ we get $M/N=0$, that is $M=N$.