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If R is a commutative noetherian ring, M and N are modules with M finite. It is well known in commutative algebra that $AssHom_R(M,N)=Supp(M)\cap Ass(N)$. But I want to know whether there is a formular for $AssExt_R^i(M,N) ?$ Thanks!

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Are you looking for an equality or would a containment $\subseteq$ be good enough? – Karl Schwede Jul 16 '10 at 11:34
Equalities are going to be hard to find, given that there are lots of examples with, say, even Exts vanishing and odd Exts non-zero. – Graham Leuschke Jul 16 '10 at 12:25
@Karl Schwede: An equality or a containment is welcome. – TmobiusX Jul 17 '10 at 1:12
Pardon my ignorance, but are the associated primes of Ext related to the associated primes of Hom in any way? – Jack Huizenga Jul 17 '10 at 3:04
up vote 2 down vote accepted

The short answer is no, as hinted at in the comments by Karl and Graham. I would argue that even the question of understanding the minimal primes of $\text{Ext}^i(M,N)$ (which is the minimal set of the associated primes, hence an easier question) is intractable. Let's assume that $R$ is Noetherian and $M,N$ are finitely generated.

Since $\text{Ext}$ localizes, one essentially needs to understand when $\text{Ext}^i(M,N)$ vanishes over a local ring $R$. There is no good answer in general. Even in the very nice case when $N$ is the canonical module (assuming it exists), these $\text{Ext}$ are Matlis dual to the local cohomology modules of $M$, and while there are good bounds on the indices when they vanish (below the depth and beyond the dimension), there are no general result which can tell you exactly when.

Here is one more way to see the complexity of the problem even when $i=1$. Take a free cover of $M$ to get an exact sequence $0 \to M_1 \to F \to M \to 0$. Applying $\text{Hom}(-,N)$ to get: $$0 \to \text{Hom}(M,N) \to \text{Hom}(F,N) \to \text{Hom}(M_1,N) \to \text{Ext}^1(M,N) \to 0$$

Assuming some mild condition on $M,N$ (reflexive for example), this shows that the set of associated prime of $\text{Ext}^1(M,N)$ is a subset of the set of primes $p$ such that $\text{depth}(\text{Hom}(M,N)_p) = 2$. Again, this set is not very well understood in general.

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