Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $i\geq0$?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.