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 $.



relation between Ass Ext(M,N) and Ass M ,Ass N
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](http://books.google.com/books?id=g2mI0cHQKNQC&lpg=PA193&ots=rvGG2AE770&dq=modules%20over%20unramified&lr&pg=PA237#v=onepage&q=modules%20over%20unramified&f=false). Sadly enough, [the paper he referred to seems to be mysteriously lost](https://mathoverflow.net/questions/72205/a-missing-paper-by-auslander).


Shameless plug: Together with Ryo Takahashi, we accidentallly managed to [recover](https://arxiv.org/abs/1202.5605) 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.


