A problem on finiteness of Ext If $R$ is a commutative noetherian ring and $I$ is an ideal of $R$, $M$ is an $R$-module. Does $Tor_i^R(R/I, M)$ is finitely generated for $i\ge 0$ imply $Ext^i_R(R/I, M)$ is finitely generated for $i\ge 0$?
 A: In fact, the two are equivalent.  My apologies for the length of this argument - if someone else has a shorter one, I'd be happy to hear it.
Let $(a_1,\ldots,a_k) = I$, and let $K_\bullet$ be the Koszul complex  associated to this set of generators.  Note that its zero'th homology group is $R/I$, and all the homology groups are finitely generated $R/I$-modules because $R$ is Noetherian.
Let C be a Serre class of R-modules (i.e. one such that for any $0 \to A' \to A \to A'' \to 0$ exact, $A$ is in $C$ if and only if $A'$ and $A''$ are both in C).  The result you ask is obtained by letting C be the class of finitely generated $R$-modules (which is only a Serre class because $R$ is Noetherian).  We have that the following are equivalent for an R-module M:


*

*$Tor_i(R/I,M)$ is in C for all values of i.

*$Tor_i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.

*$H_i(P \otimes_R M)$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.

*$H_i(K \otimes_R M)$ is in C for all i.


The implication 1 => 2 follows inductively by writing $N$ in a short exact sequence $0 \to J \to \oplus R/I \to N \to 0$ and applying the long exact sequence of Tor.
The implication 2 => 3 follows from a hyperhomology spectral sequence.
The implication 3 => 4 is immediate from the definition of the Koszul complex.
The implication 4 => 1 is proved inductively.  The hyperhomology spectral sequence
$$
E^2_{p,q} = Tor_p(H_q(K), M) \Rightarrow H_{p+q}(K \otimes_R M)
$$
first shows $E^2_{0,0} = Tor_0(R/I,M)$ is in C.  If $Tor_i(R/I,M)$ is in C for $0 \leq i \leq m$, the above argument implies that $Tor_i(N,M)$ is in C for all finitely generated $N$, which forces $E^2_{p,q}$ to be in C for all $p \leq m$.  As the abutment is in C, this forces $E^2_{m+1,0} = Tor_{m+1}(R/I,M)$ to be in C.
Now, there is an exactly analogous string of implications in Ext.  The following are equivalent:


*

*$Ext^i(R/I,M)$ is in C for all values of i.

*$Ext^i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.

*$H^i(\underline{Hom}_R(P,M))$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.

*$H^i(\underline{Hom}_R(K,M))$ is in C for all i.


However, the Koszul complex has self-duality; the tensor product complex $K \otimes_R M$ visibly has the same homology groups as a shift of the Hom-complex $\underline{Hom}_R(K,M)$.  Therefore, the two versions of statement (4) are equivalent.
