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Let $M$ be some $A$-module and $f \in A$. Why do we have an isomorphism $$\varinjlim_n \hom_A(f^n A,M) \cong M_f \text{ ?}$$ Background. Let $X$ be a scheme, $U$ an open subscheme, and $F,G$ quasi-coherent sheaves on $X$. There is a canonical homomorphism $$\varinjlim_{J \subseteq \mathcal{O}_X , J|_{U} = \mathcal{O}_U} \hom_{\mathcal{O}_X}(JF,G) \to \hom_{\mathcal{O}_U}(F|_U,G|_U).$$ Deligne has shows that this is an isomorphism when $X$ is noetherian and $F$ is of finite type (see EGA I (1970), Prop. 6.9.17). In particular, we have the following explicit description of the quasi-coherent sheaf $\tilde{M}$ on $\mathrm{Spec}(A)$ for a noetherian ring $A$ and some $A$-module $M$: $$\tilde{M}(U) = \varinjlim_{J \subseteq A, \tilde{J}|_U = \mathcal{O}_U} \hom_A(J,M)$$ Note that, when $U$ is the complement of $V(I)$, the condition $\tilde{J}|_U = \mathcal{O}_U$ means $I \subseteq \sqrt{J}$, or equivalently $I^n \subseteq J$ for some $n$. This gives an even more explicit formula $$\tilde{M}(\mathrm{Spec}(A) \setminus V(I)) = \varinjlim_n \hom_A(I^n,M).$$ In the paper "Sections of quasi-coherent sheaves" it is claimed that this formula also holds when $U$ is a basic-open subset i.e. $I$ is principal and $A$ is not assumed to be noetherian. But I cannot prove this. The colimit becomes $$\cong \varinjlim_n ~\hom_A(f^n A,M) \cong \varinjlim_n ~\{m \in M : \mathrm{Ann}(f^n) \subseteq \mathrm{Ann}(m)\}.$$ The transition maps multiply with $f$. This colimit has a canonical injection to $\varinjlim_n ~ M \cong M_f = \tilde{M}(D(f))$. But I don't see why it should be surjective. Even when $A$ is noetherian, I don't see a direct argument, without repeating Deligne's proof (which essentially uses the Artin-Rees Lemma).

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  • $\begingroup$ Theorem 2.2.16, Brodmann and Sharp, Local Cohomology. $\endgroup$ – user26857 Oct 5 '13 at 14:50
  • $\begingroup$ I don't have access to this book, and some (essential) pages in the google preview are missing. Could you please sketch the argument? I think it has to be very simple ... $\endgroup$ – Martin Brandenburg Oct 5 '13 at 16:12
  • $\begingroup$ Brodmann and Sharp suppose all rings to be noetherian. However, their proof works if $A$ has the property that the $f$-torsion functor preserves injectivity of $A$-modules. $\endgroup$ – Fred Rohrer Oct 5 '13 at 19:32
  • $\begingroup$ I also expect that it doesn't hold if $A$ is arbitrary. Somehow I find errors in published papers every day (why?!). And this is not a pedantic one, since the authors really emphasize that, whereas the general case is treated by Deligne for noetherian rings, the case of basic open sets holds for arbitrary rings. $\endgroup$ – Martin Brandenburg Oct 5 '13 at 19:38
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This is worked out for finitely generated ideals $I \subset A$ in Lemma Tag 01PM with a counter example to your question when $I$ is generated by $2$ elements in Example Tag 01PN. Moreover, there it is shown that the correct thing is to look at the system of modules $\text{Hom}_A(I^n, M/M_n)$ where $M_n = M[I^n]$ is the $I^n$-torsion of $M$.

To get a counter example for one element, we could try the ring $A = k[x, y, z_1, z_2, \ldots]/(x^i z_i)$ and module $M = A$. Then the element $y/x$ of $M_x = \widetilde{M}(D(x))$ doesn't seem to be in the image. Namely, any corresponding map $I^n = (x^n) \to A$ should send $x^n$ to $x^{n - 1} y$, I think. But $x^n$ is annihilated by $z_n$ whereas $x^{n - 1} y$ is not.

OK? I will try to add this to the example in the Stacks project.

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