$I$ be a nilpotent ideal in a commutative ring $R$ , $M/IM$ finitely generated over $R/I$ 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 
 A: 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 $M=N$.
A: Unfortunately, there is no hypothesis that $M$ is finitely generated over $R$ -- in fact, that is what we are trying to prove -- so Nakayama's Lemma does not apply; however, as user26857 correctly observed, there is a more elementary proof. We can view this as a sort of Nakayama's Lemma in the case that $M$ is not finitely generated.
Lemma. Given an ideal $I$ of a commutative (unital) ring $R$ and $R$-modules $N \subseteq M$ such that $M = N + IM,$ we have that $M = N + I^2 M.$ Even more, we have that $M = N + I^{2k} M$ for all integers $k \geq 1.$
Proof. By hypothesis, we have that $$M = N + IM = N + I(N + IM) = N + IN + I^2 M = N + I^2 M.$$ But applying this same argument $k$-fold to $M = N + I^2 M$ shows that $M = N + I^{2k} M.$ QED.
Claim. Given an ideal $I$ of a commutative (unital) ring $R$ such that $I^i = 0$ for some integer $i \gg 0$ and an $R$-module $M$ such that $M / IM$ is finitely generated over $R/I,$ then $M$ is finitely generated over $R.$
Proof. Like Mr. Chip proved, we have that $M = N + IM,$ where $N = R \langle m_1, \dots, m_n \rangle$ is the $R$-submodule of $M$ generated by the elements $m_i$ in $M$ that generate $M/IM$ over $R/I.$ By the lemma, we have that $$M = N + IM = N + I^2 M = N + I^4 M = \cdots = N + I^{2k} M = N$$ for some integer $k \gg 0$ (e.g., take $k$ such that $2k \geq i$). Consequently, $M$ is finitely generated over $R.$ QED.
