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Let $R$ be a commutative ring, and let $I \subseteq R$ be a nilpotent ideal. Let moreover $M$ be an $R$-module, and let $IM$ be the submodule generated by the products $xm$ with $x \in I$ and $m \in M$. It is quite easy to prove that, if $M/IM$ is finitely generated as an $R$-module, then $M$ is finitely generated. Indeed, let $\tilde{m}_1, \ldots , \tilde{m}_k$ be generators of $M/IM$ and lift them to elements $m_1, \ldots, m_k$ in $M$. Call $N$ the submodule generated by $m_1, \ldots, m_k$: then, we see that $$ M= N + IM. $$ Indeed, take $m \in M$. By definition of the $m_i$, we have $r_1,\ldots,r_k \in R$ such that $m - \sum_i r_im_i$ lies in $IM$. Now, since $I$ is nilpotent, we conclude (a variant of Nakayama's lemma if you want) that $M=N$, and we are done.

What about finite presentation? Namely:

Let us assume that $I$ is nilpotent and also finitely presented as an $R$-module. Can I conclude that if $M/IM$ is finitely presented as an $R$-module, then the same is true for $M$? (if necessary, I can also assume that $R$ is a coherent ring).

Improve this question
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  • $\begingroup$ I'm not sure how you apply Nakayama in this case, even though I believe the conclusion is true, i.e. that $M$ is f.g. The way I see it is by multiplying the identity $M=N+IM$ by $I^j$ for $j=1,\dots, k-1$ where $k$ is such that $I^k=0$, and then working backwards. $\endgroup$
    – Chris
    Commented Jun 25, 2020 at 12:50
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    $\begingroup$ It's a "variant" of Nakayama: more or less as you say, $M=N+IM$ is the same as $M/N= I(M/N)$, so if $I$ is nilpotent then $M/N=0$. This works for the finitely generated case. My question is about finite presentation, which I feel is quite more difficult... $\endgroup$
    – Francesco Genovese
    Commented Jun 25, 2020 at 12:53
  • $\begingroup$ By the way: if $R$ is Noetherian, then finitely generated and finitely presented modules coincide, so the result is true. Hence, I believe that the assumption of $R$ being coherent is kind of necessary... I'll perhaps modify the original post. $\endgroup$
    – Francesco Genovese
    Commented Jul 11, 2020 at 18:06
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    $\begingroup$ No. $R=k[X_n\mid n\in\mathbb{N}]/(X_1^2)$ is coherent (if $k$ is a field), and the ideal $I$ $=$ $X_1R$ is nilpotent and f.p. If $J$ $=$ $\Sigma\,_{n>1}\,X_1X_nR$, then $M=R/J$ is not f.p., yet $M/IM$ $=$ $R/(I+J)$ $=$ $R/I$ is. $\endgroup$
    – Matthé van der Lee
    Commented Jul 21, 2020 at 18:01

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