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 10.68.3. Let $R \to R'$ be a flat ring map. Let $M$ be an $R$-module. Suppose that $f_1, \ldots , f_ r \in R$ form an $M$-quasi-regular sequence. Then the images of $f_1, \ldots , f_ r$ in $R'$ form a $M \otimes _ R R'$-quasi-regular sequence.

Proof. Set $J = (f_1, \ldots , f_ r)$, $J' = JR'$ and $M' = M \otimes _ R R'$. We have to show the canonical map $\mu : R'/J'[X_1, \ldots X_ n] \otimes _{R'/J'} M'/J'M' \to \bigoplus (J')^ nM'/(J')^{n + 1}M'$ is an isomorphism. Because $R \to R'$ is flat the sequences $0 \to J^ nM \to M$ and $0 \to J^{n + 1}M \to J^ nM \to J^ nM/J^{n + 1}M \to 0$ remain exact on tensoring with $R'$. This first implies that $J^ nM \otimes _ R R' = (J')^ nM'$ and then that $(J')^ nM'/(J')^{n + 1}M' = J^ nM/J^{n + 1}M \otimes _ R R'$. Thus $\mu $ is the tensor product of (10.68.0.1), which is an isomorphism by assumption, with $\text{id}_{R'}$ and we conclude. $\square$


Comments (2)

Comment #920 by JuanPablo on

I think it wasn't entirely clear how the flatness hypothesis is used. Here flatness hypothesis is used for the following:

If is an ideal of and then . This is seen by tensoring the exact sequence . In this lemma for . The lemma follows tensoring the equation in the definition of quasi-regular sequences.


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