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Let $A$ be a commutative ring with unity, $N$ be the nilradical of $A$, $M$ be an $A$-module. Is it always true that $NM$ is a proper submodule of $M$?

If $M$ is finitely generated then by Nakayamma Lemma $NM$ must be a proper submodule. Are there any counter examples if it is not true in general for arbitarary $A$-modules?

rschwieb
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bharath
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  • In the future, please remember to put question marks on your questions! punctuation is more than just writing rules: it makes sentences easier and faster to parse and digest. – rschwieb Jun 03 '14 at 11:09

1 Answers1

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Let $R$ be $F[x^{\frac12},x^{\frac14},x^{\frac18}\ldots]/(x)$ for a field $F$. Then $R$ has a nilradical $N=(x^{\frac12},x^{\frac14},x^{\frac18}\ldots)$ , and furthermore $N^2=N$. This is an example for your situation where $N=M$.

rschwieb
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