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*}

10.80 Characterizing flatness

In this section we discuss criteria for flatness. The main result in this section is Lazard's theorem (Theorem 10.80.4 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 10.38.11. It turns out that this can be massaged into a seemingly much stronger property.

Lemma 10.80.1. Let $M$ be an $R$-module. The following are equivalent:

  1. $M$ is flat.

  2. 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 $x \in \mathop{\mathrm{Ker}}(h)$.

  3. 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 $h'$.

  4. 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 10.80.2. 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 10.80.1. $\square$

Lemma 10.80.3. 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 10.80.2 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 10.80.2, hence is flat. $\square$

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 10.80.2 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. $\square$

[1] In fact, a module map $f : R^ n \to M$ corresponds to a choice of elements $x_1, x_2, \ldots , x_ n$ of $M$ (namely, the images of the standard basis elements $e_1, e_2, \ldots , e_ n$); furthermore, an element $x \in \mathop{\mathrm{Ker}}(f)$ corresponds to a relation between these $x_1, x_2, \ldots , x_ n$ (namely, the relation $\sum _ i f_ i x_ i = 0$, where the $f_ i$ are the coordinates of $x$). The module map $h$ (represented as an $m \times n$-matrix) corresponds to the matrix $(a_{ij})$ from Lemma 10.38.11, and the $y_ j$ of Lemma 10.38.11 are the images of the standard basis vectors of $R^ m$ under $g$.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 058C. Beware of the difference between the letter 'O' and the digit '0'.