$N(R) \subset \operatorname{ann}_R (M)$ 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.
$N(R) \subset \operatorname{ann}_R (M)$ 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.
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 $IM \neq{0}$.