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Suppose $R$ is Noetherian and $M$ is a finitely generated $R$-module. IF $g \in \text {End}_R(M,M)$ does there exist a k such that $g^k =g^{k+1}= \ldots$

I'm trying to work with the generators, but not getting anywhere. Any assistance would be appreciated.

user26857
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user92612
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    Hint: Use the ascending chain condition on submodules, which holds in $M$, because a finitely generated module over a noetherian ring is noetherian. But check the assignment: the statement you report is false. – egreg Apr 28 '14 at 23:27
  • @egreg: what? Think about the case that $R$ is a field. – Qiaochu Yuan Apr 28 '14 at 23:28
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    @QiaochuYuan I was completing the comment. – egreg Apr 28 '14 at 23:29

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No, not in general. Let $R=\mathbb{K}$ be a field and let $M$ be a vector space of dimension $2$ over $\mathbb{K}$. Write $M=\mathbb{K}e_1\oplus\mathbb{K}e_2$. Then the map $g$ defined by $e_1 \mapsto e_2$ and $e_2\mapsto e_1$ has $g^k\neq g^{k+1}$ for every $k$.

Seth
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