Product of nilpotent ideal and simple module is zero

I am stuck with trying to show that if an ideal $I$ of a ring $R$ is nilpotent and $M$ is a simple $R$-module, then $IM = 0$.

I have attempted showing this by using the fact that the annihilator of a simple module is the primitive ideal, and I'm guessing trying to show that a nilpotent ideal and a primitive ideal are some how related but i think i am missing some crucial information.

I have tried using properties of maximal ideals but to no conclusion, I'm sure I'm just missing an initial step any help on this will be greatly appreciated

Since the Jacobson radical annihilates simple $R$ modules, so must each nilpotent ideal.
Let $M$ be a non-zero simple module. As $M$ is simple, $IM = 0$ or $IM = M$. If $IM = M$, then note that $IM = I^{n}M$ $\forall n \geq 1$. But as $I$ is nilpotent, $I^k = 0$ for some $k$. This implies that $M = 0$, a contradiction. Notice that a consequence is that Nilpotent ideals are contained in the Jacobson radical.
• @Saurabh Can you quickly explain why $M$ simple $\implies IM = 0$ or $IM = M$ ? – Jango Nov 6 '14 at 19:18
• $IM$ is a submodule of $M$. $M$, being simple has only trivial submodules, namely $0$ and $M$. – SMG Nov 7 '14 at 0:41