most recent 30 from http://mathoverflow.net 2013-05-24T15:21:13Z http://mathoverflow.net/feeds/question/32137 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32137/the-associated-prime-ideals-of-exti-rm-n TmobiusX 2010-07-16T09:05:11Z 2010-07-17T02:37:13Z

<p>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!</p>

http://mathoverflow.net/questions/32137/the-associated-prime-ideals-of-exti-rm-n/32249#32249 Hailong Dao 2010-07-17T02:24:13Z 2010-07-17T02:37:13Z

<p>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. </p> <p>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. </p> <p>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: $$0 \to \text{Hom}(M,N) \to \text{Hom}(F,N) \to \text{Hom}(M_1,N) \to \text{Ext}^1(M,N) \to 0$$</p> <p>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 <code>$\text{depth}(\text{Hom}(M,N)_p) = 2$</code>. Again, this set is not very well understood in general. </p>