# relation between Ass Ext(M,N) and Ass M ,Ass N

Let $R$ be a noetherian ring and $M$ , $N$ be finitely generated $R$-modules. Then what is the relation between $Ass\ Ext^i_R(M,N)$ and $Ass\ M, Ass\ N$?

$Ass$ means set of associated prime ideals.
It's well known that $Ass\ Hom_R(M,N) \subseteq Supp\ M \cap Ass\ N$.

• It would be most helpful if you give some motivation and/or background. The way it phrased now makes it look like you want us to do your homework... Feb 28 '12 at 15:06
• You would really think about me. Feb 28 '12 at 20:56
• This is probably not a homework question. Feb 29 '12 at 3:09
• Actually there's something stronger true than what you say in the last line. Namely, if $R$ is any commutative ring, $M$ is a finitely presented module, and $N$ is any module, then $Ass Hom_R(M,N) = Supp M \cap Ass N$. This is indeed well-known. For instance, it appears as an exercise in the book by Bruns and Herzog. Mar 3 '12 at 21:22
• Please, I want to know the proof of the fact AssHomR(M,N)=SuppM∩AssN if M,N are finitely generated R modules given R is noetherian.
– user40948
Oct 6 '13 at 13:50

The strict answer to your question is no in general. Take a very special case, $R=k[x,y]$, $M=m$ some maximal ideal of $R$, $N=R$. Then $Ass(M) = Ass(N) = \{(0)\}$, but it is not hard to see $Ext^1(M,N) \cong R/m$, so $Ass(Ext^1(M,N)) = \{m\}$.
However, the general question of understanding the associated primes of Ext is harder and I happen to think about it recently. When the ring is regular, one can get a complete (but complicated) description of the support of $Ext^i(M,N)$ based only on the depth of the modules $M,N$ locally at the primes in $Spec(R)$. This was announced by Auslander at the end of his ICM 1962 speech. Sadly enough, the paper he referred to seems to be mysteriously lost.