# Localization and Extension of modules

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.

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## 2 Answers

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.

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Thank you for your reply. I should have emphasized that $S$ may not be commutative. Does localization still commute with taking (co)homology? – M. K. Aug 23 '12 at 7:04
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. – Rudy the Reindeer Aug 23 '12 at 7:24
No worry; I was not confused by your notation. Do you know where I can find a proof of your statement in commutative case? – M. K. Aug 23 '12 at 7:58
@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. – Rudy the Reindeer Aug 23 '12 at 8:34
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. – Rudy the Reindeer Aug 23 '12 at 8:36

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 $$S^{-1} \text{Hom}_R(A,B) \simeq \text{Hom}_{S^{-1}R}(S^{-1}A,S^{-1}B)$$

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