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I'm trying to prove the following particular case of Nakayama's lemma. Let $R$ be a commutative ring and $a\in R$ be nilpotent (let's suppose $a^{k-1}\not=0$, $a^k=0$). Then $aM=M \Rightarrow M=\{0\}$.

I have done this. Since $R$ is commutative, $aM\subset M$ is a submodule.

Let $m\in M$. Then $m=an$ for some $n\in M$. If I act by $a^{k-1}$ I get: $a^{k-1}m=a^kn=0$

So I get that $0\ne a^{k-1} \in Ann(M)$, where $Ann$ denotes the annihilator. I don't know where to go from here.

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    yup, exactly. Feel free to post it as an answer, and I'd vote for it.2011-10-12

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Let $m\in M$, let's see that $m=0$.

Since $M=aM$, then there exists $n_1\in M$ such that $m=an_1\in M$. Since $n_1\in M=aM$, there exists $n_2\in M$ such that $n_1=an_2$, whence $m=an_1=a^2n_2$.

By induction we get to $m=a^{k-1}n_k=a^kn_{k+1}=0$ by nilpotency of $a$.