Finite-dimensionality of Ext¹ of simple modules For simplicity let $\Bbbk$ be a field of characteristic $0$ and let $A$ be a finitely generated unital associative $\Bbbk$-algebra.
Is it true that for any two simple $A$-modules $S_1, S_2$, we have that $\operatorname{Ext}^1_A (S_1, S_2)$ is finite-dimensional?
If not, what would be a simple counterexample, and what sort of conditions do we need to ensure this?
(As far as I understand, the Weyl algebra $\Bbbk \langle x, y \rangle / (xy - yx - 1)$ has only infinite-dimensional simple modules, but their first extension groups are still finite-dimensional.)
 A: Actually, your claim about the Weyl algebra is only true for simple holonomic modules. It turns out that there’s a simple non-holonomic module $M$ over the 2nd Weyl algebra $A=A_2(\Bbb C)$ for which $\operatorname{Ext}_A^1(M,M)$ is infinite dimensional. This is Corollary 1.3 of “Non-holonomic modules over Weyl algebras and enveloping algebras” by Stafford.
A: It's true if $A$ is commutative. We can compute $\text{Ext}^1(S_1, S_2)$ using the first three terms of a free resolution of $S_1$; since $S_1$ is simple it's cyclic so the first term of this resolution is $A$ and the second term of this resolution is any finite set of generators of the kernel of $A \to S_1$, which is finitely generated because $A$ is Noetherian; Noetherianness also implies that we can take the third term (and in fact every term) to be finite free. So we get a levelwise finite free resolution
$$\cdots \to A^n \to A^m \to A \to S_1 \to 0$$
which computes $\text{Ext}^1(S_1, S_2)$ as a subquotient of $\text{Hom}(A^m, S_2) \cong S_2^m$; by the Nullstellensatz $\dim_k S_2$ is finite so $\dim_k \text{Ext}^1(S_1, S_2)$ must also be finite. More generally $\dim_k \text{Ext}^n(S_1, S_2)$ is finite.
If $A$ is noncommutative $\dim_k S_2$ may be infinite and the kernel of $A \to S_1$ may be infinitely generated (if $A$ is not Noetherian) so this argument fails in two places.
