The Stacks project

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