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. $\square$

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