The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.67.9. Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots , f_ r \in R$ and $e_1, \ldots , e_ r > 0$ integers. Then $f_1, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1^{e_1}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence.

Proof. We will prove this by induction on $r$. If $r = 1$ this follows from the following two easy facts: (a) a power of a nonzerodivisor on $M$ is a nonzerodivisor on $M$ and (b) a divisor of a nonzerodivisor on $M$ is a nonzerodivisor on $M$. If $r > 1$, then by induction applied to $M/f_1M$ we have that $f_1, f_2, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1, f_2^{e_2}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence. Thus it suffices to show, given $e > 0$, that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1, \ldots , f_ r$ is an $M$-regular sequence. We will prove this by induction on $e$. The case $e = 1$ is trivial. Since $f_1$ is a nonzerodivisor under both assumptions (by the case $r = 1$) we have a short exact sequence

\[ 0 \to M/f_1M \xrightarrow {f_1^{e - 1}} M/f_1^ eM \to M/f_1^{e - 1}M \to 0 \]

Suppose that $f_1, f_2, \ldots , f_ r$ is an $M$-regular sequence. Then by induction the elements $f_2, \ldots , f_ r$ are $M/f_1M$ and $M/f_1^{e - 1}M$-regular sequences. By Lemma 10.67.8 $f_2, \ldots , f_ r$ is $M/f_1^ eM$-regular. Hence $f_1^ e, f_2, \ldots , f_ r$ is $M$-regular. Conversely, suppose that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence. Then $f_2 : M/f_1^ eM \to M/f_1^ eM$ is injective, hence $f_2 : M/f_1M \to M/f_1M$ is injective, hence by induction(!) $f_2 : M/f_1^{e - 1}M \to M/f_1^{e - 1}M$ is injective, hence

\[ 0 \to M/(f_1, f_2)M \xrightarrow {f_1^{e - 1}} M/(f_1^ e, f_2)M \to M/(f_1^{e - 1}, f_2)M \to 0 \]

is a short exact sequence by Lemma 10.4.1. This proves the converse for $r = 2$. If $r > 2$, then we have $f_3 : M/(f_1^ e, f_2)M \to M/(f_1^ e, f_2)M$ is injective, hence $f_3 : M/(f_1, f_2)M \to M/(f_1, f_2)M$ is injective, and so on. Some details omitted. $\square$


Comments (4)

Comment #787 by Keenan Kidwell on

I see how the snake lemma gives that is regular on , but not how it gives, e.g., regularity of on , etc..

Comment #3793 by Kestutis Cesnavicius on

The statement can be strengthened to an 'if and only if,' see Gabber--Ramero, "Foundations for almost ring theory," Lemma 7.8.8 (ii) in v13 of https://arxiv.org/abs/math/0409584

Comment #3915 by on

One remark on changing lemmas from "if" to "if and only if" is to ask: does one ever really need the "noninteresting variant". Anyway, I changed the argument slightly to prove if and only if in this case. Thanks! Changes are here.


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