10.81 Characterizing flatness
In this section we discuss criteria for flatness. The main result in this section is Lazard's theorem (Theorem 10.81.4 below), which says that a flat module is the colimit of a directed system of free finite modules. We remind the reader of the “equational criterion for flatness”, see Lemma 10.39.11. It turns out that this can be massaged into a seemingly much stronger property.
Lemma 10.81.1. Let $M$ be an $R$-module. The following are equivalent:
$M$ is flat.
If $f: R^ n \to M$ is a module map and $x \in \mathop{\mathrm{Ker}}(f)$, then there are module maps $h: R^ n \to R^ m$ and $g: R^ m \to M$ such that $f = g \circ h$ and $x \in \mathop{\mathrm{Ker}}(h)$.
Suppose $f: R^ n \to M$ is a module map, $N \subset \mathop{\mathrm{Ker}}(f)$ any submodule, and $h: R^ n \to R^{m}$ a map such that $N \subset \mathop{\mathrm{Ker}}(h)$ and $f$ factors through $h$. Then given any $x \in \mathop{\mathrm{Ker}}(f)$ we can find a map $h': R^ n \to R^{m'}$ such that $N + Rx \subset \mathop{\mathrm{Ker}}(h')$ and $f$ factors through $h'$.
If $f: R^ n \to M$ is a module map and $N \subset \mathop{\mathrm{Ker}}(f)$ is a finitely generated submodule, then there are module maps $h: R^ n \to R^ m$ and $g: R^ m \to M$ such that $f = g \circ h$ and $N \subset \mathop{\mathrm{Ker}}(h)$.
Proof.
That (1) is equivalent to (2) is just a reformulation of the equational criterion for flatness1. To show (2) implies (3), let $g: R^ m \to M$ be the map such that $f$ factors as $f = g \circ h$. By (2) find $h'': R^ m \to R^{m'}$ such that $h''$ kills $h(x)$ and $g: R^ m \to M$ factors through $h''$. Then taking $h' = h'' \circ h$ works. (3) implies (4) by induction on the number of generators of $N \subset \mathop{\mathrm{Ker}}(f)$ in (4). Clearly (4) implies (2).
$\square$
Lemma 10.81.2. Let $M$ be an $R$-module. Then $M$ is flat if and only if the following condition holds: if $P$ is a finitely presented $R$-module and $f: P \to M$ a module map, then there is a free finite $R$-module $F$ and module maps $h: P \to F$ and $g: F \to M$ such that $f = g \circ h$.
Proof.
This is just a reformulation of condition (4) from Lemma 10.81.1.
$\square$
Lemma 10.81.3. Let $M$ be an $R$-module. Then $M$ is flat if and only if the following condition holds: for every finitely presented $R$-module $P$, if $N \to M$ is a surjective $R$-module map, then the induced map $\mathop{\mathrm{Hom}}\nolimits _ R(P, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, M)$ is surjective.
Proof.
First suppose $M$ is flat. We must show that if $P$ is finitely presented, then given a map $f: P \to M$, it factors through the map $N \to M$. By Lemma 10.81.2 the map $f$ factors through a map $F \to M$ where $F$ is free and finite. Since $F$ is free, this map factors through $N \to M$. Thus $f$ factors through $N \to M$.
Conversely, suppose the condition of the lemma holds. Let $f: P \to M$ be a map from a finitely presented module $P$. Choose a free module $N$ with a surjection $N \to M$ onto $M$. Then $f$ factors through $N \to M$, and since $P$ is finitely generated, $f$ factors through a free finite submodule of $N$. Thus $M$ satisfies the condition of Lemma 10.81.2, hence is flat.
$\square$
Theorem 10.81.4 (Lazard's theorem). Let $M$ be an $R$-module. Then $M$ is flat if and only if it is the colimit of a directed system of free finite $R$-modules.
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:
The square
\[ \xymatrix{ R^{J'} \ar[r] \ar@{^{(}->}[d] & F \ar[d]^{g} \\ R^{I} \ar[r]_{f} & M } \]
is commutative, hence $N' \subset K = \mathop{\mathrm{Ker}}(f)$;
$R^{J'} \to F$ is a surjection onto a free finite module, hence it splits and so $N'$ is finitely generated;
$J \subset J'$ 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.
$\square$
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