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Let $A$ to be a commutative ring with an identity $1$. $M$ is a f.g. $A$--module, Does

$$M/IM=0 \text{ for all maximal } I \Longleftrightarrow M=0$$

hold? It seems like that using localization could solve this (not sure). Is there any other way to get the goal?

For a general $A$--module $M$, a counterexample is $A=\mathbb{Z},\ M=\mathbb{Q}.$

One might get $M = IM$ for all maximal $I$ $\implies$ $M=\mathfrak{R}M$ $\implies$ $M=0$ by Nakayama's lemma. However, the first arrow is incorrect (when $M$ does not need to be a f.g. $A$--module) or not evident.

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    Do you know Nakayama's lemma? – 2017-01-14
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    @EricWofsey Yes. – 2017-01-14

1 Answers 1

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Rather than using Nakayama's lemma with the Jacobson radical of $A$, first localize and then use Nakayama. To show that $M=0$, it suffices to show that the localization $M_I$ is $0$ for all maximal ideals $I$. Now for any $I$, you have $M_I/I_IM_I\cong M/IM=0$ and so $M_I=0$ by Nakayama's lemma over the local ring $A_I$.

Alternatively, if you want to avoid localizing, you can use a stronger version of Nakayama, which states that if $I$ is an ideal and $M$ is a finitely generated module such that $IM=M$, then there exists $i\in I$ such that $(1+i)M=0$. If $M\neq 0$, the annihilator of $M$ is a proper ideal; let $I$ be a maximal ideal containing it. There is then $i\in I$ such that $1+i$ annihilates $M$, which means $1+i\in I$ so $1\in I$, which is a contradiction.