The finiteness of the associated primes of $Ext^i_R(R/I, M)$ Let $(R,m)$ be a Noetherian local ring and $M$ an $R$-module. If $M$ is a finite $R$-module, then we know that $Ext^i_R(R/I,M)$ is a finite $R$-module for all $i\geq 0$; Now suppose $supp(M)\subseteq V(I)$ and $Ass_R(M)$ is a finite set, then $Ass_R(Hom(R/I,M))=Ass_R(M)$ is finite. For $i>0$, is $Ass_R(Ext_R^i(R/I,M))$ still a finite set?
 A: Take $R = \mathbb{Z}[X]$, so $\mathbb{Z} = R/(X)$. Let $I = (X)$. For each prime number $p$ we consider the module $M_p := (p,X)/(X^2)$. We have $\mathrm{Supp}M_p = V(I)$ and $\mathrm{Ass}_R(M_p) = (X)$. Consider the short exact sequence
$$0 \to M_p \to R/(X^2) \to R/M_p \cong \mathbb{Z}/p\mathbb{Z} \to 0.$$
Applying the $\mathrm{Ext}^i_R(R/(X),-)$ to the above sequence with note that
$$\mathrm{Hom}_R(R/(X), M_p) \cong (X)/(X^2) \cong \mathrm{Hom}_R(R/(X), R/(X^2)$$
we have
$$0 \to \mathrm{Hom}_R(R/(X), \mathbb{Z}/p\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z} \to \mathrm{Ext}^1_R(R/(X),M_p) \to \cdots$$
Thus $(p,X) \in \mathrm{Ass}_R(\mathrm{Ext}^1_R(R/(X),M_p))$. Now set
$$M = \bigoplus_{p: prime} M_p$$
 we have $\mathrm{Supp}(M) = V(I)$, $\mathrm{Ass}_R(M) = (X)$ but 
$$\mathrm{Ass}_R(\mathrm{Ext}^1_R(R/(X),M)) = \{(p,X): p \text{ is prime}\}$$
is an infinite set.
A: One can show that if $R$ is a commutative ring, $S$ is a multiplicative system, and $M$ and $N$ are $R$-modules such that $M$ is finitely generated, then the natural map $S^{-1}Hom_R(M,N)\rightarrow Hom_{S^{-1}R}(S^{-1}M,S^{-1}N)$ is an isomorphism. We need to use the finite generation of $M$ twice, both for the injectivity and the surjectivity. Now, Write a projective resolution $P$ for $R/I$ over $R$. Since $R$ is Noetherian, all the terms in the resolution and all their submodules will be finitely generated. Let $p$ be a prime of $R$ and let $S = R-p$. Then $S^{-1}P$ is a projective resolution for $S^{-1}R/I$ over $S^{-1}R$ (we use here the fact that localization is exact). We can then show, using the above isomorphism, that $S^{-1}Ext^i_R(R/I,M)\cong Ext^i_{S^{-1}R}(S^{-1}R/I,S^{-1}M)$. Since $Ass(M)$ is finite, the last group will be zero for almost all primes $p$. But this means that the number of associated primes of $Ext^i_R(R/I,M)$ is also finite.
