Lemma 10.103.2. Notation and assumptions as above. If $g$ is good with respect to $(M, f_1, \ldots , f_ d)$, then (a) $g$ is a nonzerodivisor on $M$, and (b) $M/gM$ is Cohen-Macaulay with maximal regular sequence $f_1, \ldots , f_{d - 1}$.

**Proof.**
We prove the lemma by induction on $d$. If $d = 0$, then $M$ is finite and there is no case to which the lemma applies. If $d = 1$, then we have to show that $g : M \to M$ is injective. The kernel $K$ has support $\{ \mathfrak m\} $ because by assumption $\dim \text{Supp}(M) \cap V(g) = 0$. Hence $K$ has finite length. Hence $f_1 : K \to K$ injective implies the length of the image is the length of $K$, and hence $f_1 K = K$, which by Nakayama's Lemma 10.20.1 implies $K = 0$. Also, $\dim \text{Supp}(M/gM) = 0$ and so $M/gM$ is Cohen-Macaulay of depth $0$.

Assume $d > 1$. Observe that $g$ is good for $(M/f_1M, f_2, \ldots , f_ d)$, as is easily seen from the definition. By induction, we have that (a) $g$ is a nonzerodivisor on $M/f_1M$ and (b) $M/(g, f_1)M$ is Cohen-Macaulay with maximal regular sequence $f_2, \ldots , f_{d - 1}$. By Lemma 10.68.4 we see that $g, f_1$ is an $M$-regular sequence. Hence $g$ is a nonzerodivisor on $M$ and $f_1, \ldots , f_{d - 1}$ is an $M/gM$-regular sequence. $\square$

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## Comments (2)

Comment #6015 by Qilin,Yang on

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