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I am working on the following problem.

Let $R$ be a nilpotent ring (there exists a positive integer $n$ such that the product of $m$ elements of $R$ is $0$). Let $M$ be an $R$-module and $N$ any submodule of $M$. Prove that if $N\not=0$, then $RN\not=N$.

What I have tried: I wanted to prove this by contrapositive. Assume that $RN=N$. We want to show that $N=0$. Let $a\in N$. Then $a\in RN$ and so $a=r_1n_1+\cdots+r_tn_t$ for some $r_i\in R$ and $n_i\in N$.

I know I should probably use that $R$ is nilpotent somewhere, but I am stuck on how to proceed to show that $a=0$. Any clues on how to proceed, or should I try a different approach?

Matt
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  • A commutative ring? – rschwieb Dec 27 '16 at 21:47
  • The problem does not state that $R$ is commutative. – Matt Dec 27 '16 at 21:48
  • And $N$ is not finitely generated or anything? – rschwieb Dec 27 '16 at 21:50
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    I am not quite sure I understand the problem correctly. but it seems to me you could now replace each $n_i$, by such a sum. If you then expand the product, you'd have elements of $N$ preceded by a product of $2$ elements of $R$. Continue like this to have a product of $m$ elements, which is then $0$. – quid Dec 27 '16 at 21:52
  • Yes, that absolutely works. – Matt Dec 27 '16 at 21:58
  • I was thinking this looked a lot like an application of Nakayama's lemma, but maybe the nilpotency assumption makes things a lot easier. – rschwieb Dec 27 '16 at 22:14

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