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.
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2$\begingroup$ 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? $\endgroup$– Jason StarrCommented Mar 6, 2019 at 11:46
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1$\begingroup$ I want to know the relation of $Ass(Ext^p_R(M,R))$ with $Ass M.$ $\endgroup$– Tri NguyenCommented Mar 6, 2019 at 13:50
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4$\begingroup$ 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. $\endgroup$– Jason StarrCommented Mar 6, 2019 at 14:09
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$\begingroup$ Do we have $Ass(M)\subseteq Ass(Ext^p_R(M,R))$ ? $\endgroup$– Tri NguyenCommented Mar 6, 2019 at 14:24
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2$\begingroup$ 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 $M$. $\endgroup$– Jason StarrCommented Mar 6, 2019 at 14:33
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