# What is the $Ass(Ext^p_R(M,R))$?

Let $$R$$ be a Noetherian commutative local ring, $$M$$ a finitely generated $$R$$-module with $$p=pd M<\infty$$ (projective dimension of $$M$$). What is the relation between $$Ass(Ext^p_R(M,R))$$ and $$Ass(M)$$? Thanks.

• Could you please clarify your question. What do you want to know about the set of associated primes. These primes are contained in the support of $M$. Are you asking about (upper) bounds on the heights of these primes? – Jason Starr Mar 6 at 11:46
• I want to know the relation of $Ass(Ext^p_R(M,R))$ with $Ass M.$ – Tri Nguyen Mar 6 at 13:50
• Unfortunately, that does not clarify the question. Certainly all associated primes of $\text{Ext}^p_R(M,R)$ contain the annihilator ideal of $M$. Thus, each contains one of the minimal associated primes of $M$. Beyond that, what is your expectation? Consider, for instance, the special case that $R$ equals $k[x_0,\dots,x_n]$ and $M$ equals the maximal ideal $\langle x_0,\dots,x_n \rangle$. Then $p$ equals $n>0$, and $\text{Ext}^n_R(M,R)$ equals $R/M=k.$ The unique associated prime of $R/M$ equals $M$. The unique associated prime of $M$ equals $\{0\}$. Direct sums give more examples. – Jason Starr Mar 6 at 14:09
• Do we have $Ass(M)\subseteq Ass(Ext^p_R(M,R))$ ? – Tri Nguyen Mar 6 at 14:24
• No, you do not have that. Again, consider the example that $M$ is the maximal ideal $\langle x_0,\dots,x_n \rangle$ in the polynomial ring $k[x_0,\dots,x_n]$ for $n>0$. Then $\text{Ass}(M)$ equals the singleton set consisting only of the zero ideal $\{0\}$. Yet $\text{Ass}(\text{Ext}^n_R(M,R))$ equals the singleton set consisting only of the ideal $M$. – Jason Starr Mar 6 at 14:33