The Stacks project

Lemma 10.69.6. Let $(R, \mathfrak m)$ be a local Noetherian ring. Let $M$ be a nonzero finite $R$-module. Let $f_1, \ldots , f_ c \in \mathfrak m$ be an $M$-quasi-regular sequence. Then $f_1, \ldots , f_ c$ is an $M$-regular sequence.

Proof. Set $J = (f_1, \ldots , f_ c)$. Let us show that $f_1$ is a nonzerodivisor on $M$. Suppose $x \in M$ is not zero. By Krull's intersection theorem there exists an integer $r$ such that $x \in J^ rM$ but $x \not\in J^{r + 1}M$, see Lemma 10.51.4. Then $f_1 x \in J^{r + 1}M$ is an element whose class in $J^{r + 1}M/J^{r + 2}M$ is nonzero by the assumed structure of $\bigoplus J^ nM/J^{n + 1}M$. Whence $f_1x \not= 0$.

Now we can finish the proof by induction on $c$ using Lemma 10.69.5. $\square$

Comments (2)

Comment #2656 by Ko Aoki on

I think "By Krull's intersection theorem" is more straightforward than "By the Artin-Rees lemma". Actually the lemma cited in the sentence is Krull's intersection theorem.

There are also:

  • 2 comment(s) on Section 10.69: Quasi-regular sequences

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Your email address will not be published.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 061S. Beware of the difference between the letter 'O' and the digit '0'.