# Atiyah–Macdonald exercise 3.14

I am trying to understand the hint in exercise 3.14 in Atiyah–Macdonald.

Let $$M$$ be an $$A$$-module and $$\mathfrak a$$ be an ideal of $$A$$. Suppose that $$M_{\mathfrak m} = 0$$ for all maximal ideals $$\mathfrak m \supseteq \mathfrak a$$. Prove that $$M = \mathfrak a M$$.

Hint. Pass to the $$A / \mathfrak a$$-module $$M / \mathfrak a M$$ and use (3.8).

Where (3.8) refers to

$$M = 0 \Leftrightarrow M_{\mathfrak m} = 0 \; \forall \mathfrak m$$ maximal.

I don't quite understand why the hint suggests to regard $$M / \mathfrak a M$$ as $$A / \mathfrak a$$-module, rather than as $$A$$-module. I think we even have as $$A$$-modules

• $$M / \mathfrak a M = 0 \Leftrightarrow (M / \mathfrak a M)_{\mathfrak m} = 0$$ for all $$\mathfrak m \subset A$$ maximal.

• But $$(M / \mathfrak a M)_{\mathfrak m} \simeq M_{\mathfrak m} / (\mathfrak a M)_{\mathfrak m}$$ which is 0 by assumption.

I think that passing to the condition for $$A / \mathfrak a$$-modules is more complicated. Is the above argument correct?

• Your second step is not correct. That module is zero by assumption only for maximal ideals containing $\mathfrak{a}$. What about the other maximal ideals? Jun 28, 2017 at 9:59
• @Crostul Thank you for pointing out the error. (If it were true for all maximal ideals, I should have concluded that M itself is 0.) Jun 28, 2017 at 10:03
• Also, don't forget that (i) The maximal ideals in $A/\mathfrak a$ are all of the form $\mathfrak m / \mathfrak a$ for $\mathfrak m$ maximal in $A$ with $\mathfrak m \supseteq a$; (ii) localization and quotienting commute, so $(M/\mathfrak aM)_{\mathfrak m / \mathfrak a} \cong M_{\mathfrak m} / (\mathfrak aM_{\mathfrak m})$. Jun 28, 2017 at 10:09

The exercise says it is enough to look only at maximal ideals containing $\mathfrak a$. To see why, answer this question:
What is $\mathfrak{a_m}$ if $\mathfrak a\not\subset \mathfrak m$? Hence what is $(\mathfrak a M)_{\mathfrak m}$?
• @Bernard Could you provide some more information? As far as I understand $\mathfrak{a}_\mathfrak{m}$ consists of $a/s$ where $s\notin \mathfrak{m}$. This gives us that $s/s=1\in \mathfrak{a}_\mathfrak{m}$. If $\mathfrak{a}_\mathfrak{m}$ is indeed an ideal then it must be all of $M_\mathfrak{m}$? Then $(\mathfrak{a} M)_\mathfrak{m}=\mathfrak{a}_\mathfrak{m} M_\mathfrak{m}$? So any $\mathfrak{m}$ which does not contain $\matfrak{a}$ will automatically give $(M/\mathfrak{a}M)_\mathfrak{m}=M_\mathfrak{m}/\mathfrak{a}M_\mathfrak{m}=M_\mathfrak{m}/M_\mathfrak{m}=0$. And we are done Feb 6, 2021 at 18:24
• I see. We also could write $M/\mathfrak aM\simeq M\otimes_A A/\mathfrak a$, so that $(M/\mathfrak a M)_{\mathfrak m}\simeq M_{\mathfrak m}\otimes_{A_{\mathfrak m}}(A/\mathfrak a)_{\mathfrak m}= M_{\mathfrak m}\otimes_{A_{\mathfrak m}}0$ Feb 6, 2021 at 20:53