The Stacks project

Lemma 15.29.2. An $M$-regular sequence is $M$-Koszul-regular. A regular sequence is Koszul-regular.

Proof. Let $R$ be a ring and let $M$ be an $R$-module. It is immediate that an $M$-regular sequence of length $1$ is $M$-Koszul-regular. Let $f_1, \ldots , f_ r$ be an $M$-regular sequence. Then $f_1$ is a nonzerodivisor on $M$. Hence

\[ 0 \to K_\bullet (f_2, \ldots , f_ r) \otimes M \xrightarrow {f_1} K_\bullet (f_2, \ldots , f_ r) \otimes M \to K_\bullet (\overline{f}_2, \ldots , \overline{f}_ r) \otimes M/f_1M \to 0 \]

is a short exact sequence of complexes where $\overline{f}_ i$ is the image of $f_ i$ in $R/(f_1)$. By Lemma 15.28.8 the complex $K_\bullet (R, f_1, \ldots , f_ r)$ is isomorphic to the cone of multiplication by $f_1$ on $K_\bullet (f_2, \ldots , f_ r)$. Thus $K_\bullet (R, f_1, \ldots , f_ r) \otimes M$ is isomorphic to the cone on the first map. Hence $K_\bullet (\overline{f}_2, \ldots , \overline{f}_ r) \otimes M/f_1M$ is quasi-isomorphic to $K_\bullet (f_1, \ldots , f_ r) \otimes M$. As $\overline{f}_2, \ldots , \overline{f}_ r$ is an $M/f_1M$-regular sequence in $R/(f_1)$ the result follows from the case $r = 1$ and induction. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.29: Koszul 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Your email address will not be published.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 062F. Beware of the difference between the letter 'O' and the digit '0'.