# The Stacks Project

## Tag 062F

Lemma 15.27.2. 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 15.26.8 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. $\square$

The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5824–5828 (see updates for more information).

\begin{lemma}
\label{lemma-regular-koszul-regular}
An $M$-regular sequence is $M$-Koszul-regular.
A regular sequence is Koszul-regular.
\end{lemma}

\begin{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 \ref{lemma-cone-koszul}
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.
\end{proof}

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