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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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4  
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? – Filippo Alberto Edoardo Apr 16 '12 at 16:23
3  
$M$, not $\mathfrak{m}$, should be finitely generated to avoid Filippo's counterexample. – Graham Leuschke Apr 16 '12 at 16:41
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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 – Karl Schwede Apr 16 '12 at 17:13
    
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. – Charles Staats Apr 16 '12 at 18:09
    
Yes, I am sorry, I meant $M$.... – Filippo Alberto Edoardo Apr 16 '12 at 23:58

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