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*}

15.17 Flattening over an Artinian ring

A universal flattening exists when the base ring is an Artinian local ring. It exists for an arbitrary module. Hence, as we will see later, a flatting stratification exists when the base scheme is the spectrum of an Artinian local ring.

Lemma 15.17.1. Let $R$ be an Artinian ring. Let $M$ be an $R$-module. Then there exists a smallest ideal $I \subset R$ such that $M/IM$ is flat over $R/I$.

Proof. This follows directly from Lemma 15.16.1 and the Artinian property. $\square$

This ideal has the following universal property.

Lemma 15.17.2. Let $R$ be an Artinian ring. Let $M$ be an $R$-module. Let $I \subset R$ be the smallest ideal $I \subset R$ such that $M/IM$ is flat over $R/I$. Then $I$ has the following universal property: For every ring map $\varphi : R \to R'$ we have

\[ R' \otimes _ R M\text{ is flat over }R' \Leftrightarrow \text{we have }\varphi (I) = 0. \]

Proof. Note that $I$ exists by Lemma 15.17.1. The implication $\Rightarrow $ follows from Algebra, Lemma 10.38.7. Let $\varphi : R \to R'$ be such that $M \otimes _ R R'$ is flat over $R'$. Let $J = \mathop{\mathrm{Ker}}(\varphi )$. By Algebra, Lemma 10.100.7 and as $R' \otimes _ R M = R' \otimes _{R/J} M/JM$ is flat over $R'$ we conclude that $M/JM$ is flat over $R/J$. Hence $I \subset J$ as desired. $\square$


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