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I checked if anything regarding this question has been asked on MSE but I didn't find anything. If the specifics of my question are already addressed somewhere please direct me. This is Exercise 14 from Chapter 2 of Introduction to Commutative Algebra by M.F. Atiyah and I.G. Macdonald [AM].

Below I state the exercise, then offer my proof. Although I am convinced by my argument, I am skeptical about some assertions I make. Anything after the ($\dagger$) in my proof is where I think an error could be lurking. Specifically, I feel like I am changing the way I am viewing objects, which may not be permissible. I'm flopping around between viewing things as rings, then as modules, then as modules over a different ring.

3.14: Let $M$ be an $A$-module and $\mathfrak{a}$ be an ideal of $A$. Suppose the $M_{\mathfrak{m}}=0$ for all maximal ideals such that $\mathfrak{a} \subseteq \mathfrak{m}$. Prove that $M = \mathfrak{a}M$. [Pass to the $A/\mathfrak{a}$-module $M/\mathfrak{a}M$ and use (3.8)]

Where (3.8) simply states that being 0 is a local property.

Proof: Consider the exact sequence $$\mathfrak{a} \to A \to A/\mathfrak{a} \to 0.$$

Since tensoring is right exact (Proposition 2.18 in [AM]) we get a new exact sequence $$\mathfrak{a} \otimes_A M \to A \otimes_A M \to A/\mathfrak{a} \otimes_A M \to 0.$$

Now let $\mathfrak{m}$ be any maximal ideal containing $\mathfrak{a}$, and take $S = A - \mathfrak{m}$. Since localizing is exact (Proposition 3.3 in [AM]), then we another exact sequence $$S^{-1}(\mathfrak{a} \otimes_A M) \to S^{-1}(A \otimes_A M) \to S^{-1}(A/\mathfrak{a} \otimes_A M) \to 0. \tag{1}$$

$(\dagger)$ Now since $A \otimes_A M \cong M$ (Proposition 2.14, i in [AM]), then $S^{-1}(A \otimes_A M) \cong S^{-1}M$ which is 0 by hypothesis and since the sequence (1) is exact, then since $S^{-1}(A \otimes_A M) = 0$ we must have that $S^{-1}(A/\mathfrak{a} \otimes_A M) = 0$. By Exercise 2, Chapter 2 we have that $A/\mathfrak{a} \otimes_A M \cong M/\mathfrak{a}M$, thus $S^{-1}(M/\mathfrak{a}M) = 0$. But since $\mathrm{Ann}(M/\mathfrak{a}M)=\mathfrak{a}$, $M/\mathfrak{a}M$ is naturally a faithful $A/\mathfrak{a}$ module, and the maximal ideals of $A/\mathfrak{a}$ are in one to one correspondance with the maximal ideals of $A$ containing $\mathfrak{a}$. Since being 0 is a local property and $S^{-1}(M/\mathfrak{a}M) = 0$ for all maximal ideals of $A/\mathfrak{a}$, then we conclude $M/\mathfrak{a}M = 0$ then $M = \mathfrak{a}M$.

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since $\mathrm{Ann}(M/\mathfrak{a}M) = \mathfrak{a}$

This is wishful thinking. It would only be true if $\mathrm{Ann}(M) \subseteq \mathfrak{a}$ or something.

For instance if

$$A = \mathbb{Z}, M = \mathbb{Z}/2\mathbb{Z}, \mathfrak{a} = 4\mathbb{Z}$$

then $M/\mathfrak{a}M = \mathbb{Z}/2\mathbb{Z}$ and $\mathrm{Ann}(M/\mathfrak{a}M) = 2\mathbb{Z}$.

You also don't need this to be true to apply (3.8) which applies to all modules, not just faithful modules. The rest of the proof is fine.

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  • $\begingroup$ Thank you for your comment and for pointing out that the assertion regarding the annihilator was incorrect. I was actually not making the assertion about being faithful in order to apply to apply (3.8), I was actually just mimicking a proof in the text to conclude that $M$ was an $A/\mathfrak{a}$ module. The proof I refer to is on page 22, the paragraph under Corollary 2.7 where they conclude $M/\mathfrak{a}M$ is a $k$-vector space. Can you explain the relevance of the annihilator claim in the argument? $\endgroup$ – Prince M Apr 24 '17 at 5:25
  • $\begingroup$ @Prince I assume you are talking about the annihilator claim in the book. AM says that "$M/\mathfrak{m}M$ is annihilated by $\mathfrak{m}$" This means only that $\mathfrak{m}(M/\mathfrak{m}M) = 0$ not that $\mathfrak{m} = \mathrm{Ann}(M/\mathfrak{m}M)$. Another way of saying this is $\mathfrak{m} \subseteq \mathrm{Ann}(M/\mathfrak{m}M)$. You need this to get an $A/\mathfrak{m}$ module structure on $M/\mathfrak{m}M$. Recall for modules that $0m = 0$ so if you have $x = 0$ (i.e. $x \in \mathfrak{m}$) then you had better have $xm = 0$ for all $m \in M/\mathfrak{m}M$. $\endgroup$ – Trevor Gunn Apr 24 '17 at 5:35

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