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Let $M$ be a f.g. $R$-module, and suppose there exists a non-zero $r\in R$ such that $rM=0$. Does $M$ have finite length?

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Did you want to have $r\neq 0$? – Zev Chonoles Feb 4 '13 at 6:30
This is definitively true over the rings (PIDs) where we have a good handle on f.g. modules, of course. – Alex Youcis Feb 4 '13 at 6:33
@ZevChonoles Yes – aliakbar Feb 4 '13 at 6:36
up vote 1 down vote accepted

Let $R=k[x,y]/(y^2)$ where $k$ is a field. Let $M=(y)=\{yf\mid f\in k[x]\}$, which is a finitely generated $R$-module; indeed, it is generated by the element $y$. We also have that $yM=0$. However, there is an infinite chain of submodules $$(y)\supseteq (xy)\supseteq (x^2y)\supseteq\cdots$$ so $M$ is of infinite length.

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Your module is better identified as the ideal $(y)$, I think; at least the notation is clearer. – Mariano Suárez-Alvarez Feb 4 '13 at 6:38
Good suggestion, thanks! – Zev Chonoles Feb 4 '13 at 6:40

Pick a ring $S$ which has a f.g. module $M$ which is not of finite length. Let $R=S[X]$, and turn $M$ into an $R$-module by declaring that elements of $S$ act as they should and $X\cdot m=0$ for all $m\in M$.

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For example, pick them as in Zev's answer :-) – Mariano Suárez-Alvarez Feb 4 '13 at 6:37

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