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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

thanks in advance

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2 Answers 2

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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.

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    $\begingroup$ ah of course thanks for the help! $\endgroup$
    – Peter A
    Nov 6, 2014 at 3:07
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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.

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

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