The Stacks project

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. $\square$


Comments (4)

Comment #6015 by Qilin,Yang on

It is not clear that:"The kernel has support because by assumption

Comment #6017 by Qilin,Yang on

"The kernel has support because by assumption “ follows from EGA IV lemme 16.2.3.1.

Comment #6169 by on

I think it is clear. The support of is contained in the support of and contained in because it is annihilated by . Then the assumption says that the dimension of is . Hence the support of (which is closed) must consist just of the maximal ideal if not empty.

If you are a random reader and you are mystified as to why we have this lemma here, please see the comment #3048 on Lemma 10.103.3.

Comment #9762 by Branislav Sobot on

Sorry, but I don't see why the case can't occur. Simply take for some ideal of definition . I think that this lemma is simply wrong in this case, while the case can't occur in the next lemma.

There are also:

  • 6 comment(s) on Section 10.103: Cohen-Macaulay modules

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00N4. Beware of the difference between the letter 'O' and the digit '0'.