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 $i\ge 0$?

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it would be nice if you add some context of this question. – Martin Brandenburg May 4 '10 at 12:13

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:

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 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.

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Here is what seems to me like a counterexample to the converse. Take $(R, \mathfrak{m})$ local, $I=\mathfrak m$, and $M = E_R(R/\mathfrak{m})$ the injective hull of the residue field. Then $\mathrm{Ext}^i$ is $R/\mathfrak{m}$ for $i=0$ and $0$ otherwise, while $\mathrm{Tor}_0$ is infinitely generated since $E$ is infinitely generated. Does this miss something important? – Graham Leuschke May 4 '10 at 16:00
Graham, E/mE=0. – Hailong Dao May 4 '10 at 16:47
Right. For any x in the maximal ideal the injective module you list will be x-divisible and so the tensor product will actually be zero. – Tyler Lawson May 4 '10 at 16:58
(Assuming x is not a zero divisor.) – Tyler Lawson May 4 '10 at 17:19
Ah, right. Thanks. – Graham Leuschke May 4 '10 at 17:43