TITLE: Ext of an $\mathfrak{m}$-primary ideal
QUESTION [8 upvotes]: Let $(A,\mathfrak m,k)$ be a Noetherian local ring, $M$ a finitely generated $A$-module, and $I$ an $\mathfrak{m}$-primary ideal. If $\operatorname{Ext}^{i}_{A}(A/\mathfrak{m},M)=0$ then $\operatorname{Ext}^{i}_{A}(A/I,M)=0$?

REPLY [8 votes]: We will use the following result found as Lemma 3.1.11 on page 93 of Cohen-Macaulay Rings by Bruns and Herzog. It says: Let $R$ be a Noetherian ring, $M$ an $R$-module, $N$ a finite $R$-module and $n>0$ an integer. Suppose that $\operatorname{Ext}_R^n(R/P, M)=0$ for all $P\in \operatorname{Supp} N$. Then $\operatorname{Ext}_R^n(N,M)=0.$
Going back to the probem at hand, take $N=A/I$. Since $I$ is $\mathfrak m$-primary, $\operatorname{Supp}N=\{\mathfrak m\}$. Now the conclusion follows by the above mentioned result.