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

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

Comment #1936 by Matthieu Romagny on

It may clarify to say what is $e_i$.

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.

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• 1 comment(s) on Section 10.39: Flat modules and flat ring maps

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