# 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

The connection you are looking for is that nilpotent ideals are all contained in the Jacobson radical. This is easy to see since the primitive ideals of a ring are prime, and hence each one has to contain all nilpotent ideals. Thus their intersection (the Jacobson radical) contains all nilpotent ideals.

Since the Jacobson radical annihilates simple $R$ modules, so must each nilpotent ideal.

• ah of course thanks for the help! Nov 6, 2014 at 3:07

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.

• thank you! really appreciate it Nov 6, 2014 at 3:06
• @Saurabh Can you quickly explain why $M$ simple $\implies IM = 0$ or $IM = M$ ? Nov 6, 2014 at 19:18
• $IM$ is a submodule of $M$. $M$, being simple has only trivial submodules, namely $0$ and $M$.
– SMG
Nov 7, 2014 at 0:41