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Suppose $R$ is commutative ring with unity. For ideals $I$, $J \subseteq R$, the ideal quotient $(J:I)$ is $$(J:I) := \{x\in R \, : \, xI \subseteq J\}$$ Let $S\subset R$ be a multiplicative set. When does localization at $S$ commute with taking quotients, i.e. when does the equality $$(S^{-1} J : S^{-1} I) = S^{-1} (J:I)$$ hold? More generally, when is it true that $$S^{-1} \text{Ann}_{R} (M) = \text{Ann}_{S^{-1} R} (S^{-1} M)$$ for an $R$-module $M$?

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Atiyah-Macdonald 's famous book; 3.14 and 3.15:

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    $\begingroup$ A complete answer should add some counterexamples for the case when the ideal/module is not finitely generated. $\endgroup$ – user26857 Aug 10 '15 at 19:46
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Going off user26857's comment, we provide a counterexample for Proposition 3.14 in the case that $M$ is not finitely generated. Hopefully you can use this to construct a counterexample for Corollary 3.15 as well.

Take $A = \mathbb{Z}$, and let $M$ be the direct sum of $\mathbb{Z}/k\mathbb{Z}$ as $k$ ranges through $\mathbb{N}$. This is not finitely generated as a $\mathbb{Z}$-module (there are infinitely many nonzero summands). Now, what is the annihilator of this module? Well, it is just $0$, since for any integer $n$ different from $0$, $n$ does not act by $0$ on the $\mathbb{Z}/(n+1)\mathbb{Z}$ factor, so $n$ does not act by $0$ on $M$. (If $n$ is negative, look at $\mathbb{Z}/(1-n)\mathbb{Z}$. Strictly, this is not necessary since any ideal of $\mathbb{Z}$ is generated by a positive number, but oh well.)

Now, let us localize at $S = \mathbb{Z} \setminus \{0\}$, so $S^{-1}A = \mathbb{Q}$ and $S^{-1}\text{Ann}(M) = S^{-1}0 = 0$.

But what is $S^{-1}M$? We claim that it is $0$. This is not hard to show. Let $e_k$ be the generator of the copy of $\mathbb{Z}/k\mathbb{Z}$ in $M$. Then the set $\{e_k\}$ is a $\mathbb{Z}$-generating set for $M$ in the sense that every element of $M$ is a finite linear combination of the elements $e_k$ with coefficients in $\mathbb{Z}$, so $\{e_k/1\}$ is a $\mathbb{Q}$-generating set for $S^{-1}M$ (in this same sense).

But for each $k$, we have $k \cdot e_k = 0$ in $M$, which shows that $e_k/1 = 0$ in $S^{-1}M$. To be a little more explicit, we want to show that $e_k/1 = 0/1$ in $S^{-1}M$. By definition, this happens if and only if there is some $s \in S$ such that $s(1 \cdot e_k - 0 \cdot 1) = 0$. Now take $s = k$.

So now, we have that $\{0\}$ is a $\mathbb{Q}$-generating set for $S^{-1}M$. Thus, $S^{-1}M = 0$, so $\text{Ann}(S^{-1}M) = \mathbb{Q}$, which is different from $S^{-1}\text{Ann}(M) = 0$. (Here, we need the crucial fact that there exist nonzero rational numbers, i.e., that $\mathbb{Q}$ is not equal to $0$.)

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    $\begingroup$ Perfect answer, Kevin: bravo! $\endgroup$ – Georges Elencwajg Aug 31 '15 at 20:08
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    $\begingroup$ @GeorgesElencwajg Why is this a perfect answer? It doesn't answer the first question. $\endgroup$ – user26857 Mar 4 '16 at 10:38
  • $\begingroup$ Can you kindly give an argument or at least an example for the ideal quotient part? $\endgroup$ – Brozovic Aug 27 '19 at 19:46
  • $\begingroup$ @Brozovic Take $\mathbb{a} =0$. Then, $(\mathbb{a}:\mathbb{b}) = Ann(\mathbb{b})$. So, it is enough. $\endgroup$ – hew Jul 9 '20 at 13:37

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