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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?2011-10-29

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