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Question

Let $M$ be an $R$ module such that for every multiplicative closed set $S\subset R$, the kernel of natural map $M\to S^{-1}M$ is $(0)$ or $M$. Show that $(0)$ is primary in $M$.

Attempt

I have the below proof of the above statement when $M$ is finitely generated.

We need to show the map $\lambda_a : M\to M$ sending $m\in M$ to $a.m$, where $a\in R$ is either injective or nilpotent.

Take $S=\{1,a,a^2. . . .\}$. If the natural map $f : M\to S^{-1}M$ is injective and $a.m=0$, we must have $m/1=0$ in $S^{-1}M$.By injectivity of $f$, we have $m=0$.

If the natural map $f : M\to S^{-1}M$ is zero map, then as $M$ is finitely generated and $m/1=0$ for all $m\in M$, we can choose a large enough $N\in\mathbb{N}$ such that $a^N M=0$. Then $\lambda_a^N$ is identically zero.

My doubt is that does the statement hold for modules not finitely generated. I am not able to give a proof for it if so.

Any help is appreciated.

Eric Wofsey
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Akash Yadav
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1 Answers1

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The statement is not true without assuming $M$ is finitely generated. For instance, let $R=\mathbb{Z}$ and $M=\bigoplus_n \mathbb{Z}/(2^n)$. Then the natural map $M\to S^{-1}M$ is zero if $S$ contains an even number and injective otherwise. However, multiplication by $2$ is neither injective nor nilpotent on $M$.

Eric Wofsey
  • 342,377