Proof. A colimit of a directed system of flat modules is flat, as taking directed colimits is exact and commutes with tensor product. Hence if $M$ is the colimit of a directed system of free finite modules then $M$ is flat.

For the converse, first recall that any module $M$ can be written as the colimit of a directed system of finitely presented modules, in the following way. Choose a surjection $f: R^ I \to M$ for some set $I$, and let $K$ be the kernel. Let $E$ be the set of ordered pairs $(J, N)$ where $J$ is a finite subset of $I$ and $N$ is a finitely generated submodule of $R^ J \cap K$. Then $E$ is made into a directed partially ordered set by defining $(J, N) \leq (J', N')$ if and only if $J \subset J'$ and $N \subset N'$. Define $M_ e = R^ J/N$ for $e = (J, N)$, and define $f_{ee'}: M_ e \to M_{e'}$ to be the natural map for $e \leq e'$. Then $(M_ e, f_{ee'})$ is a directed system and the natural maps $f_ e: M_ e \to M$ induce an isomorphism $\mathop{\mathrm{colim}}\nolimits _{e \in E} M_ e \xrightarrow {\cong } M$.

Now suppose $M$ is flat. Let $I = M \times \mathbf{Z}$, write $(x_ i)$ for the canonical basis of $R^{I}$, and take in the above discussion $f: R^ I \to M$ to be the map sending $x_ i$ to the projection of $i$ onto $M$. To prove the theorem it suffices to show that the $e \in E$ such that $M_ e$ is free form a cofinal subset of $E$. So let $e = (J, N) \in E$ be arbitrary. By Lemma 10.81.2 there is a free finite module $F$ and maps $h: R^ J/N \to F$ and $g: F \to M$ such that the natural map $f_ e: R^ J/N \to M$ factors as $R^ J/N \xrightarrow {h} F \xrightarrow {g} M$. We are going to realize $F$ as $M_{e'}$ for some $e' \geq e$.

Let $\{ b_1, \ldots , b_ n \}$ be a finite basis of $F$. Choose $n$ distinct elements $i_1, \ldots , i_ n \in I$ such that $i_{\ell } \notin J$ for all $\ell$, and such that the image of $x_{i_{\ell }}$ under $f: R^ I \to M$ equals the image of $b_{\ell }$ under $g: F \to M$. This is possible since every element of $M$ can be written as $f(x_ i)$ for infinitely many distinct $i \in I$ (by our choice of $I$). Now let $J' = J \cup \{ i_1, \ldots , i_ n \}$, and define $R^{J'} \to F$ by $x_ i \mapsto h(x_ i)$ for $i \in J$ and $x_{i_{\ell }} \mapsto b_{\ell }$ for $\ell = 1, \ldots , n$. Let $N' = \mathop{\mathrm{Ker}}(R^{J'} \to F)$. Observe:

1. The square

$\xymatrix{ R^{J'} \ar[r] \ar@{^{(}->}[d] & F \ar[d]^{g} \\ R^{I} \ar[r]_{f} & M }$

is commutative, hence $N' \subset K = \mathop{\mathrm{Ker}}(f)$;

2. $R^{J'} \to F$ is a surjection onto a free finite module, hence it splits and so $N'$ is finitely generated;

3. $J \subset J'$ and $N \subset N'$.

By (1) and (2) $e' = (J', N')$ is in $E$, by (3) $e' \geq e$, and by construction $M_{e'} = R^{J'}/N' \cong F$ is free. $\square$

Comment #5032 by Laurent Moret-Bailly on

In the proof, $R^I$ apparently denotes $\bigoplus_{i\in I}R$, while it usually means $\prod_{i\in I}R$. Is this standard SP notation?

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).