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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?

Community
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Earthliŋ
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    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? – Crostul Jun 28 '17 at 09:59
  • @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.) – Earthliŋ Jun 28 '17 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 $(M/\mathfrak aM){\mathfrak m / \mathfrak a} \cong M{\mathfrak m} / (\mathfrak aM_{\mathfrak m})$. – Kenny Wong Jun 28 '17 at 10:09

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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
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  • Took me some time to wrap my head around, but it's a nice hint, thank you. – Earthliŋ Jun 29 '17 at 14:08
  • @Earthliŋ: I prefer to give hint as it's more educational, in my opinion. But you may ask for more information if it's too cryptic! – Bernard Jun 29 '17 at 15:01
  • Thanks, but it was just the right level of crypticity =) – Earthliŋ Jun 29 '17 at 18:29
  • @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

     – Mark Murray Feb 06 '21 at 18:24
  • More exactly, $\mathfrak{a_m}$ is all of $A_{\mathfrak m}$, so that $(\mathfrak {a}M){\mathfrak m}=\mathfrak{a_m}M{\mathfrak m}=A_{\mathfrak m}M_{\mathfrak m}=M_{\mathfrak m}$. – Bernard Feb 06 '21 at 18:33
  • Ah yes. I am still a little confused as it doesn't seem that we've used an $A/\mathfrak{a}$ module structure like the hint suggests – Mark Murray Feb 06 '21 at 20:09
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    I see. We also could write $M/\mathfrak aM\simeq M\otimes_A A/\mathfrak a$, so that $(M/\mathfrak a M){\mathfrak m}\simeq M{\mathfrak m}\otimes_{A_{\mathfrak m}}(A/\mathfrak a){\mathfrak m}= M{\mathfrak m}\otimes_{A_{\mathfrak m}}0$ – Bernard Feb 06 '21 at 20:53