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