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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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Multiply M by a^k, and use aM = M about k times. –  Jack Schmidt Oct 12 '11 at 15:14
1  
@Jack Ah, I see. Using elements as I was doing, it's $m=an_1\in M \Rightarrow m=a^2n_2\in M \Rightarrow \dots \Rightarrow m=a^kn_k=0$. Thank you. –  Bruno Stonek Oct 12 '11 at 15:27
    
yup, exactly. Feel free to post it as an answer, and I'd vote for it. –  Jack Schmidt Oct 12 '11 at 16:20

1 Answer 1

up vote 2 down vote accepted

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 $a$.

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