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Let $A$ be a commutative ring with $1$ and $\mathcal m$ be a maximal ideal. One knows that then there is a canonical isomorphism

$A_{\mathcal m}/{{\mathcal m}A_{\mathcal m}} \simeq A/{\mathcal m}$.

Does one have an isomorphism of extension groups

$\operatorname{Ext}^1_A(A/{\mathcal m}, A/{\mathcal m}) \simeq \operatorname{Ext}^1_{A_{\mathcal m}}(A_{\mathcal m}/{{\mathcal m}A_{\mathcal m}},A_{\mathcal m}/{{\mathcal m}A_{\mathcal m}})$

via $A\rightarrow A_{\mathcal m}?$

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Can you compute both things in an example, say with $A=k[x,y]$ and $m$ the maximal ideal at the origin? – Mariano Suárez-Alvarez Oct 29 '11 at 22:34

Let $A$ be a commutative ring with $1$ and $\mathfrak m$ a maximal ideal. The natural morphism from $A$ to $A_{\mathfrak m}$ inducing an isomorphism of $A/\mathfrak m$ onto $A_{\mathfrak m}/\mathfrak m_{\mathfrak m}$, we may denote these two fields by the single letter $K$. Then $K$ is also an $A$-module and an $A_{\mathfrak m}$-module, and we have $$ \text{Hom}_A(A,K)=\text{Hom}_{A_{\mathfrak m}}(A_{\mathfrak m},K)=K.\tag1 $$ Let $n$ be a nonnegative integer. We claim the existence of a canonical isomorphism $$ \text{Ext}^n_A(K,K)\simeq\text{Ext}^n_{A_{\mathfrak m}} (K,K). $$ To prove this, apply the localization functor, which is exact and commutes with direct sums, to a free resolution of $K$ in the category of $A$-modules, and use (1).

Thank you to Mariano (see his comment) for having pointed out a mistake: in the previous version I incorrectly called "injection" the natural morphism from $A$ to $A_{\mathfrak m}$.

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This proof is the very first one that comes to mind as soon as one tries to actually compuite the two Ext's. – Mariano Suárez-Alvarez Oct 30 '11 at 11:47
Dear @Mariano: thank you very much! – Pierre-Yves Gaillard Oct 30 '11 at 12:13
Cher Pierre, there is no need to add that to the body of the answer! – Mariano Suárez-Alvarez Oct 30 '11 at 12:15
Cher Mariano: It's not by masochism. It's just to let people understand the comments. – Pierre-Yves Gaillard Oct 30 '11 at 12:21

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