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 problem on finiteness of Ext
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](http://en.wikipedia.org/wiki/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:


1. $Tor\_i(R/I,M)$ is in C for all values of i.
2. $Tor\_i(N,M)$ is in C for all finitely generated $R/I$-modules $N$.
3. $H\_i(P \otimes\_R M)$ is in C for all bounded chain complexes $P$ of free $R$-modules whose homology groups are finitely generated $R/I$-modules.
4. $H\_i(K \otimes\_R M)$ 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](http://en.wikipedia.org/wiki/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:


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


