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Let $A$ be a ring, $I\subset A$ an ideal, $M$, $N$ $A$-modules such that $IM=0$ and $IN=0$. Then the modules extend to $A/I$-modules, and we have $$\operatorname{Hom}_A(M,N)=\operatorname{Hom}_{A/I}(M,N).$$ But is this true for higher homological dimensions? i.e., is $$\operatorname{Ext}_A^i(M,N)=\operatorname{Ext}^i_{A/I}(M,N)$$ true for all $i\geq0$?

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Let $A=k[x]/x^2$, $I=(x)$, $M,N=k$, so that $A/I=k$. Then all higher ext groups for $A/I$ vanish, but $\operatorname{Ext}_A(k,k) \cong k[y]$ with $y$ in degree one.

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