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*}

Proof. Assume $M$ is flat and let $\sum f_ i x_ i = 0$ be a relation in $M$. Let $I = (f_1, \ldots , f_ n)$, and let $K = \mathop{\mathrm{Ker}}(R^ n \to I, (a_1, \ldots , a_ n) \mapsto \sum _ i a_ i f_ i)$. So we have the short exact sequence $0 \to K \to R^ n \to I \to 0$. Then $\sum f_ i \otimes x_ i$ is an element of $I \otimes _ R M$ which maps to zero in $R \otimes _ R M = M$. By flatness $\sum f_ i \otimes x_ i$ is zero in $I \otimes _ R M$. Thus there exists an element of $K \otimes _ R M$ mapping to $\sum e_ i \otimes x_ i \in R^ n \otimes _ R M$ where $e_ i$ is the $i$th basis element of $R^ n$. Write this element as $\sum k_ j \otimes y_ j$ and then write the image of $k_ j$ in $R^ n$ as $\sum a_{ij} e_ i$ to get the result.

Assume every relation is trivial, let $I$ be a finitely generated ideal, and let $x = \sum f_ i \otimes x_ i$ be an element of $I \otimes _ R M$ mapping to zero in $R \otimes _ R M = M$. This just means exactly that $\sum f_ i x_ i$ is a relation in $M$. And the fact that it is trivial implies easily that $x$ is zero, because

\[ x = \sum f_ i \otimes x_ i = \sum f_ i \otimes \left(\sum a_{ij}y_ j\right) = \sum \left(\sum f_ i a_{ij}\right) \otimes y_ j = 0 \]
$\square$


Comments (4)

Comment #1936 by Matthieu Romagny on

It may clarify to say what is .

Comment #2459 by Oliver Eivind Anderson on

Is the sentence "a relation in M" formally defined anywhere in the stacks project?

Comment #2460 by on

Dear Oliver Eivind Anderson. We sometimes allow ourselves to use a discussion directly preceding in a lemma. Please click on the section link (immediately above the lemma) to check out the text immediately preceding. A while ago I put in an issue for the stacks-website project --- namely https://github.com/stacks/stacks-website/issues/67 --- that we show this text surrounding a lemma or definition in the website. Maybe that would help.

There are also:

  • 1 comment(s) on Section 10.38: Flat modules and flat ring maps

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