# showing that the Krull dimension of an extension module is zero

Let $(R,m,k)$ be a Cohen-Macaulay ring of dimension $d>0$ and let $M,C$ be CM $R$-modules such that $\dim M = 0, \dim C = d$.

In the proof of Proposition 3.3.3-b(ii) in Bruns & Herzog, the authors write that it is trivial that $\dim \operatorname{Ext}_R^d(M,C)=0$.

After thinking about it for some time i came up with the following argument, which is however far from trivial.

My argument: Since $\operatorname{height} \operatorname{ann}(M) = d$, there exist elements $x_1,\dots,x_d$ in $\operatorname{ann}(M)$ such that $\operatorname{height}(x_1,\dots,x_i) = i$. Since $\dim C = d$, the $x_i$ give rise to a $C$-regular sequence, denoted $\underline{x}$, which lies in $\operatorname{ann}(M)$. Then by Lemma 1.2.4 in the same book we have that $\operatorname{Ext}_R^d(M,C) \cong \operatorname{Hom}_R(M,C/\underline{x}C)$. Then by exercise 1.2.27 in the same book we have that $\operatorname{Ass}_R \left[\operatorname{Hom}_R(M,C/\underline{x}C)\right] = \operatorname{Supp}_R M \cap \operatorname{Ass}_R C/\underline{x}C = \left\{m\right\}$. Hence $m$ is a minimal prime of $\operatorname{Ext}_R^d(M,C)$ and so this latter module must have Krull dimension equal to zero.

Question 1: Any comments on my argument?

Question 2: Any idea what Bruns&Herzog were having in mind as "trivial" argument?

This is easily shown once you notice that $\operatorname{Supp}\operatorname{Ext}_R^i(M,N)\subseteq\operatorname{Supp} M\cap\operatorname{Supp}N$.