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Let $(R,\mathfrak{m})$ be a commutative Noetherian local ring and $M$ is an $R$-module such that $\mathfrak{m}^tM=0$ for some non-negative integer $t$. Then the length of $M$ is finite.

Is that right?

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    $\begingroup$ Do you want may be $\mathfrak{m}$ to be finitely generated? What about infinitely many copies of $R/\mathfrak{m}$ which is killed by $\mathfrak{m}$ but has infinite length? $\endgroup$ Apr 16, 2012 at 16:23
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    $\begingroup$ $M$, not $\mathfrak{m}$, should be finitely generated to avoid Filippo's counterexample. $\endgroup$ Apr 16, 2012 at 16:41
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    $\begingroup$ But then yes, if it is finitely generated there is a surjection $R^n \to M$. But then $M$ is also a quotient of $R^n/\mathfrak{m}^t$, which is certainly of finitely length $\endgroup$ Apr 16, 2012 at 17:13
  • $\begingroup$ A slightly different approach: under the hypotheses, $M$ is a module over the Artinian ring $A/\mathfrak{m}^tA$. Thus, $M$ is of finite length iff it is finitely generated. $\endgroup$ Apr 16, 2012 at 18:09
  • $\begingroup$ Yes, I am sorry, I meant $M$.... $\endgroup$ Apr 16, 2012 at 23:58

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