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Let $A$ be a ring with unity, and let $I$ be a two sided ideal of $A$ such that $I^n = 0.$ Let $M$ be a simple $A$-module. I want to show that $IM = 0.$

We know that $IM$ is a submodule of $M$. Suppose $IM = M.$ Then $I^2M = I(IM) = IM = M,$ and so $I^nM = M.$ But $I^nM = \{0\}M = 0,$ but $M$ cannot be $0$ as it is simple, hence nontrivial, a contradiction. So $IM = 0.$

This seems too simple to me (this question was on a past paper, and my solution seems much shorter than the number of marks), so I was wondering if it was correct. Particularly, I was wondering if $I^2M = I(IM)$ actually works? I can't see how it shouldn't work from the definitions, but I can never be too sure about products of ideals and modules.

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    $\begingroup$ The proof works just fine. $I^2M = I(IM)$ follows form the associativity of multiplication (just write down the definition of the module on either side and see what you get). $\endgroup$ – Arthur May 22 '17 at 14:01
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    $\begingroup$ Yes: it even already is confirmed on this site. $\endgroup$ – rschwieb May 23 '17 at 21:12
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Let $A,I,M, n$ be given as in the question. We want to show that $IM=0$.

Claim 1. For all $k\in\mathbb{N}$, $I^{k}M:=\left\{i_{1}m_{1}+\cdots+i_{r}m_{r}:r\in\mathbb{N}, i_{j}\in I^{k}\text{ and }m_{j}\in M\text{ for all }1\leq j\leq r\right\}$ is a submodule of $M$.

Proof. Let $k\in\mathbb{N}$. Clearly $0\in I^{k}M$. Suppose $x,y\in I^{k}M$ and $a\in A$. Write $x=i_{1}m_{1}+\cdots+i_{k}m_{k}$ and $y=j_{1}p_{1}+\cdots +j_{r}p_{r}$, where the $i$'s and $j$'s are in $I^{k}$ and the $m$'s and $p$'s are in $M$. Then, since $M$ is an $A$-module, $$ \begin{aligned}x+ay&=i_{1}m_{1}+\cdots+i_{k}m_{k}+a( j_{1}p_{1}+\cdots +j_{r}p_{r})\\ &=i_{1}m_{1}+\cdots+i_{k}m_{k}+(aj_{1})p_{1}+\cdots+(aj_{r})p_{r} \end{aligned} $$ Since $I$ is an ideal in $A$, $aj_{\ell}\in I$ for all $1\leq \ell\leq r$. Thus, $x+ay\in I^{k}M$, which proves that $I^{k}M$ is a submodule of $M$ by the submodule criterion. //

Claim 2. Assuming that $IM=M$, it follows that $I^{k}M=M$ for all $k\in\mathbb{N}$.

Proof. We proceed by induction. For the base case, $k=1$, the result is true by assumption. So, suppose that for some fixed $k\in\mathbb{N}$, $I^{k}M=M$. We need to show that $I^{k+1}M=M$. Suppose $i_{1},\ldots,i_{k+1}\in I$ and $m\in M$. Then, since $M$ is an $A$-module, $(i_{1}\cdots i_{k+1})m=(i_{1}\cdots i_{k})(i_{k+1}m)$. Since $IM=M$, $i_{k+1}m\in M$, so that $(i_{1}\cdots i_{k+1})m=(i_{1}\cdots i_{k})m_{1}$ for some $m_{1}\in M$. By the inductive hypothesis, $I^{k}M=M$, so there is an $m_{2}\in M$ such that $ (i_{1}\cdots i_{k+1})m=(i_{1}\cdots i_{k})m_{1}=m_{2}$. It follows that each element in $I^{k+1}M$ is a finite sum of elements in $M$, and hence, belongs to $M$. Thus, $I^{k+1}M\subseteq M$. Conversely, given $m'\in M$, we can reverse the above process to obtain $i_{1}',\ldots,i_{k+1}'\in I$ and $m_{1}'\in M$ such that $m'=(i_{1}'\cdots i_{k+1}')m'_{1}\in I^{k+1}M$. Whence, $M\subseteq I^{k+1}M$. The result follows by mutual inclusion. //

Now, since $M$ is a simple $A$-module, it follows by Claim 1 that $IM=0$ or $IM=M$. If $IM=0$, then we are done. So, let's suppose that $IM=M$. But then Claim 2 says that $\{0\}=\{0\}M=I^{n}M=M$, which contradicts that $M$ is simple, since simple modules cannot be trivial.

Note: We only ever invoked Claim 1 with $k=1$.

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