Timeline for if $I$ is finitely presented nilpotent and $M/IM$ is finitely presented, then $M$ is finitely presented

7 events

when toggle format	what		by	license	comment
Jul 21, 2020 at 18:01	comment	added	Matthé van der Lee		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.
Jul 11, 2020 at 18:06	comment	added	Francesco Genovese		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.
Jun 25, 2020 at 12:54	history	edited	Francesco Genovese	CC BY-SA 4.0	added 33 characters in body
Jun 25, 2020 at 12:53	comment	added	Francesco Genovese		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...
Jun 25, 2020 at 12:50	comment	added	Chris		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.
Jun 25, 2020 at 12:28	history	edited	YCor		edited tags
Jun 25, 2020 at 12:00	history	asked	Francesco Genovese	CC BY-SA 4.0