The Stacks project

Proof. Choose a $M$-regular sequence $f_1, \ldots , f_ d$ with $d = \dim (\text{Supp}(M))$. If $g$ is good with respect to $(M, f_1, \ldots , f_ d)$ we win by Lemma 10.103.2. In particular the lemma holds if $d = 1$. (The case $d = 0$ does not occur.) Assume $d > 1$. Choose an element $h \in R$ such that (i) $h$ is good with respect to $(M, f_1, \ldots , f_ d)$, and (ii) $\dim (\text{Supp}(M) \cap V(h, g)) = d - 2$. To see $h$ exists, let $\{ \mathfrak q_ j\} $ be the (finite) set of minimal primes of the closed sets $\text{Supp}(M)$, $\text{Supp}(M)\cap V(f_1, \ldots , f_ i)$, $i = 1, \ldots , d - 1$, and $\text{Supp}(M) \cap V(g)$. None of these $\mathfrak q_ j$ is equal to $\mathfrak m$ and hence we may find $h \in \mathfrak m$, $h \not\in \mathfrak q_ j$ by Lemma 10.15.2. It is clear that $h$ satisfies (i) and (ii). From Lemma 10.103.2 we conclude that $M/hM$ is Cohen-Macaulay. By (ii) we see that the pair $(M/hM, g)$ satisfies the induction hypothesis. Hence $M/(h, g)M$ is Cohen-Macaulay and $g : M/hM \to M/hM$ is injective. By Lemma 10.68.4 we see that $g : M \to M$ and $h : M/gM \to M/gM$ are injective. Combined with the fact that $M/(g, h)M$ is Cohen-Macaulay this finishes the proof. $\square$