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Let $R$ be a commutative ring and $S$ be an $R$-algebra. Assume that $S$ is finitely generated as an $R$-module. Let $M$ and $N$ be finitely generated $S$-modules and $\mathfrak{m}$ a maximal ideal in $R$.

Under what condition does the following equality hold? $$ Ext^{i}_{C}(M,N)_{\mathfrak{m}}=Ext^{i}_{C_{\mathfrak{m}}}(M_{\mathfrak{m}},N_{\mathfrak{m}}) \ \forall i, $$ where the subscript $_{\mathfrak{m}}$ means the localization with respect to $\mathfrak{m}$.

Thank you for your assistance.

2 Answers 2

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It holds always since you can show that $$ S^{-1} H_i(C) \cong H_i (S^{-1}C)$$

where $H_i$ denotes the $i$-th homology group and $C$ the chain complex you're taking homology of.

To see that taking homology commutes with taking fractions let $$ A \xrightarrow{f} B \xrightarrow{g} C$$ be a sequence of $R$-modules such that $gf = 0$ and $S \subset R$ is multiplicative. We claim that $S^{-1}(\mathrm{ker}g / \mathrm{im} f) \cong (S^{-1} \mathrm{ker}g )/ (S^{-1} \mathrm{im} f) \cong (\mathrm{S^{-1} ker}g )/ (\mathrm{S^{-1} im} f)$ from which our claim above will follow.

We know that $S^{-1}$ is an exact functor (Atiyah-Macdonald, p. 39, Prop. 3.3). Also, the following sequence is exact: $$ 0 \to \mathrm{ker} g \hookrightarrow B \xrightarrow{g} C$$ and hence the following sequence is also exact $$ 0 \to S^{-1}\mathrm{ker} g \hookrightarrow S^{-1}B \xrightarrow{S^{-1}g} S^{-1}C$$

so that we get $S^{-1}\mathrm{ker} g \cong \mathrm{ker} S^{-1}g$.

Similarly, $$ A \xrightarrow{f} \mathrm{im} f \to 0$$ is exact so that $$ S^{-1}A \xrightarrow{S^{-1}f} S^{-1} \mathrm{im} f \to 0$$ is also exact and hence $\mathrm{im} S^{-1}f \cong S^{-1} \mathrm{im}f$.

So that we have $$ (S^{-1} \mathrm{ker}g )/ (S^{-1}\mathrm{im}f) \cong (\mathrm{ker} S^{-1}g ) / (\mathrm{im} S^{-1} f) $$

To finish the proof you use that $$ 0 \to \mathrm{im}f \to \mathrm{ker}g \to (\mathrm{ker}g) / ( \mathrm{im}f) \to 0$$ is exact so that $$ 0 \to S^{-1}\mathrm{im}f \to S^{-1} \mathrm{ker}g \to S^{-1}( (\mathrm{ker}g) / ( \mathrm{im}f) ) \to 0$$ is exact and hence $$ S^{-1}( (\mathrm{ker}g) / ( \mathrm{im}f) ) \cong (S^{-1} \mathrm{ker}g) / (S^{-1}\mathrm{im}f) \cong (\mathrm{ker} S^{-1} g) / (\mathrm{im}S^{-1} f)$$

which proves the claim.

  • 0
    Thank you for your reply. I should have emphasized that $S$ may not be commutative. Does localization still commute with taking (co)homology?2012-08-23
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    Dear @M.K., I just noticed that confusingly, I called my multiplicative set $S$, introducing notation that might be confusing given that you are using $S$ for something else in your question. As for your question: I don't know the answer in the non-commutative case.2012-08-23
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    No worry; I was not confused by your notation. Do you know where I can find a proof of your statement in commutative case?2012-08-23
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    @M.K. No, sorry I don't. I had to remind myself of the proof which given in my commutative algebra lecture. I added it to my answer.2012-08-23
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    One could have a quick thought about where commutativity is used. Unfortunately, I should be doing something else right now. I hope someone else can help.2012-08-23
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    Dear Clark: I think there is a counterexample to your claim in Bourbaki (Alg. Com., II.2, Exercise 4).2012-08-23
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    Dear @Pierre-YvesGaillard, it appears as if I cannot access [the relevant page in Bourbaki](http://books.google.com/books?id=Bb30CjGW7EAC&lpg=PP1&hl=de&pg=PA123#v=onepage&q&f=false).2012-08-23
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    Dear Clark: Bourbaki's exercise (which I haven't tried to solve) shows the following. Let $S$ be a multiplicative subset of the commutative ring $A$, and $M$ an $A$-module. Then the natural $A$-algebra morphism $$S^{-1}\text{End}_A(M)\to\text{End}_{S^{-1}A}(S^{-1}M)$$ is not bijective in general.2012-08-23
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    It is really easy when you see condition for which this holds. Take $S = \{s^i, i \in \mathbb{N} \}$ infinite, $M = A^\mathbb{N}$ and $\phi : S^{-1}M \to S^{-1}M$ with $\phi(a_i) = s^{-i}a_i$2016-06-10
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    A comment for future readers: you're actually proving that any exact functor commutes with homology.2017-03-13
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According to Weibel the following is true (Prop 3.3.10):

Let $A$ be a finitely generated module over a commutative Noetherian ring $R$. Then for every multiplicative set $S$, all modules $B$ and all $n$, $$S^{-1} \text{Ext}_R^n(A,B) \simeq \text{Ext}^n_{S^{-1}R}(S^{-1}A,S^{-1}B)$$

Noetherian-ness is important here, for every finitely generated module over a Noetherian ring is finitely presented. The proof then follows from Lemma 3.3.8:

If $A$ is finitely presented as an $R$-module then for every central multiplicative set $S$ in $R$, there is an isomorphism $$S^{-1} \text{Hom}_R(A,B) \simeq \text{Hom}_{S^{-1}R}(S^{-1}A,S^{-1}B)$$