This week I learned an interesting fact about uniform annihilators of high degree Ext modules from a paper by Iyengar and Takahashi “The Jacobian ideal…”. I dare say there are many other places in the literature to read about it. In fact, I wouldn’t mind at all if you emailed me references where I could learn more about it.

The general gist of the results is that given a “good” Noetherian ring S and an ideal I ⊂ S cutting out the singular locus, then there exist integers m and i_0 such that for all i > i_0 the modules Ext^i_S(M, N) are annihilated by I^m. The key here is that m does not depend on M, N.

Iyengar and Takahashi show that this an essential ingredient if you want to prove strong generation for the category of modules and the derived category D^b_{Coh}(S). I would guess that conversely strong generation of D^b_{Coh}(S) will imply some uniform vanishing of Ext’s but I didn’t try to prove it; have you?

Let me explain a strategy to get a result like this. It only works if you can write S as the quotient of a regular ring and you have plenty of derivations. For example for finite type algebras over perfect fields the proposition below proves what I said above (but you can also find it in the literature of course). If you are still reading, the strategy is given in the proof of the proposition; I suggest skipping the details.

Let R be a regular ring. Let R —> S = R/J be a quotient. Assume we have f_1, …, f_c in J and derivations D_1, …, D_c : R —> R as well as an element z’ in R such that z’ J is contained in (f_1, …, f_c) + J^2. Let

z = det(D_i(f_j))

be as in the previous blog post. Finally, let n be the integer found in the first lemma of the previous post (this integer was found in a nonconstructive manner, but I hope somebody can tell me how to make it effective in some way).

**Proposition:** Let d = dim(R) < ∞. For any finite S-modules M, N we have z^{2n + 1} (z’)^{2n} annihilates Ext^{d + 1}_S(M, N).

**Proof.** Denote i_* : D(S) —> D(R) the pushforward and denote i* : D(R) —> D(S) the pullback. We have Ext^{d + 1}_R(i_*M, i_*N) = 0 because R is regular of dimension d. Thus we have Ext^{d + 1}_S(i*i_*M, N) = 0. But in the previous post we have seen that up to multiplication by z^{2n + 1} (z’)^{2n} the module M is a summand of i*i_*M. This concludes the proof. **EndProof.**

**Remark** If S is a finite type k algebra for some perfect field k and we choose a surjection R = k[x_1, …, x_t] —> S with kernel J then for choices f_1, …, f_c in J and derivations D_1, …, D_c on R, the elements z^{2n}(z’)^{2n + 1} we obtain in S generate an ideal whose vanishing locus is the singular set of Spec(S).