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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 $.

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    $\begingroup$ 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... $\endgroup$
    – Vladimir Dotsenko
    Commented Feb 28, 2012 at 15:06
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    $\begingroup$ You would really think about me. $\endgroup$
    – Stella
    Commented Feb 28, 2012 at 20:56
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    $\begingroup$ This is probably not a homework question. $\endgroup$
    – Hailong Dao
    Commented Feb 29, 2012 at 3:09
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    $\begingroup$ 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. $\endgroup$
    – Neil Epstein
    Commented Mar 3, 2012 at 21:22
  • $\begingroup$ 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. $\endgroup$
    – user40948
    Commented Oct 6, 2013 at 13:50

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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.

Shameless plug: Together with Ryo Takahashi, we accidentallly managed to recover Auslander's Theorem 3 from the speech cited above (which is about support of Tor). I am optimistic that his final paragraph can be deciphered in the near future.

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