$N(R) \subset \operatorname{ann}_R (M)$ if M is simple left module

$N(R) \subset \operatorname{ann}_R (M)$ if M is simple left module

How do I prove this?

Know that if $a$ is in $N(R)$, then $a^n=0$ for some element. However, don't see how to go from this.

$N(R)=\sum \{ I\}$ where I is nilpotent ideal of R.

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It would not hurt if you edited the question to add an explanation of what $N(R)$ is... – Mariano Suárez-Alvarez Nov 24 '11 at 3:55
Could you say what $N(R)$ is? It's my understanding that the nilpotent elements might not form an ideal in the non-commutative setting. – Dylan Moreland Nov 24 '11 at 3:56
@Dylan, in the quotient $A$ of the free algebra $k\langle x,y\rangle$ by the bilateral ideal $(x^2,y^2)$, the (classes of) $x$ and $y$ are nilpotent, but their sum isn't. – Mariano Suárez-Alvarez Nov 24 '11 at 3:59
@Mariano Ah, thanks! Good to have an example. – Dylan Moreland Nov 26 '11 at 18:30

From your definition of $N(R)$ it is enough to show that $I \subset \text{ann}_R(M)$ for every nilpotent ideal $I$. So let $I$ be nilpotent and suppose $IM \neq \{0\}$. As $IM$ is a submodule of $M$ and $M$ is simple you get $IM = M$. By induction you get $I^nM = M$ for all $n$ but with $I$ being nilpotent you get $M = \{0\}$ a contradiction to $IM \neq{0}$.