66.31 Flat modules

In this section we define what it means for a module to be flat at a point. To do this we will use the notion of the stalk of a sheaf on the small étale site $X_{\acute{e}tale}$ of an algebraic space, see Properties of Spaces, Definition 65.19.6.

Lemma 66.31.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $x \in |X|$. The following are equivalent

1. for some commutative diagram

$\xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ X \ar[r]^ f & Y }$

where $U$ and $V$ are schemes, $a, b$ are étale, and $u \in U$ mapping to $x$ the module $a^*\mathcal{F}$ is flat at $u$ over $V$,

2. the stalk $\mathcal{F}_{\overline{x}}$ is flat over the étale local ring $\mathcal{O}_{Y, \overline{y}}$ where $\overline{x}$ is any geometric point lying over $x$ and $\overline{y} = f \circ \overline{x}$.

Proof. During this proof we fix a geometric proof $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ over $x$ and we denote $\overline{y} = f \circ \overline{x}$ its image in $Y$. Given a diagram as in (1) we can find a geometric point $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ with $\overline{x} = a \circ \overline{u}$, see Properties of Spaces, Lemma 65.19.4. Set $\overline{v} = h \circ \overline{u}$ with image $v \in V$. We know that

$\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{U, u}^{sh} \quad \text{and}\quad \mathcal{O}_{Y, \overline{y}} = \mathcal{O}_{V, v}^{sh}$

see Properties of Spaces, Lemma 65.22.1. We obtain a commutative diagram

$\xymatrix{ \mathcal{O}_{U, u} \ar[r] & \mathcal{O}_{X, \overline{x}} \\ \mathcal{O}_{V, v} \ar[u] \ar[r] & \mathcal{O}_{Y, \overline{y}} \ar[u] }$

of local rings. Finally, we have

$\mathcal{F}_{\overline{x}} = (\varphi ^*\mathcal{F})_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}}$

by Properties of Spaces, Lemma 65.29.4. Thus Algebra, Lemma 10.39.9 tells us $(\varphi ^*\mathcal{F})_ u$ is flat over $\mathcal{O}_{V, v}$ if and only if $\mathcal{F}_{\overline{x}}$ is flat over $\mathcal{O}_{V, v}$. Hence the result follows from More on Flatness, Lemma 38.2.5. $\square$

Definition 66.31.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.

1. Let $x \in |X|$. We say $\mathcal{F}$ is flat at $x$ over $Y$ if the equivalent conditions of Lemma 66.31.1 hold.

2. We say $\mathcal{F}$ is flat over $Y$ if $\mathcal{F}$ is flat over $Y$ at all $x \in |X|$.

Having defined this we have the obligatory base change lemma. This lemma implies that formation of the flat locus of a quasi-coherent sheaf commutes with flat base change.

Lemma 66.31.3. Let $S$ be a scheme. Let

$\xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

be a cartesian diagram of algebraic spaces over $S$. Let $x' \in |X'|$ with image $x \in |X|$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$ and denote $\mathcal{F}' = (g')^*\mathcal{F}$.

1. If $\mathcal{F}$ is flat at $x$ over $Y$ then $\mathcal{F}'$ is flat at $x'$ over $Y'$.

2. If $g$ is flat at $f'(x')$ and $\mathcal{F}'$ is flat at $x'$ over $Y'$, then $\mathcal{F}$ is flat at $x$ over $Y$.

In particular, if $\mathcal{F}$ is flat over $Y$, then $\mathcal{F}'$ is flat over $Y'$.

Proof. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Choose a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Choose a scheme $V'$ and a surjective étale morphism $V' \to V \times _ Y Y'$. Then $U' = V' \times _ V U$ is a scheme endowed with a surjective étale morphism $U' = V' \times _ V U \to Y' \times _ Y X = X'$. Pick $u' \in U'$ mapping to $x' \in |X'|$. Then we can check flatness of $\mathcal{F}'$ at $x'$ over $Y'$ in terms of flatness of $\mathcal{F}'|_{U'}$ at $u'$ over $V'$. Hence the lemma follows from More on Morphisms, Lemma 37.15.2. $\square$

The following lemma discusses “composition” of flat morphisms in terms of modules. It also shows that flatness satisfies a kind of top down descent.

Lemma 66.31.4. Let $S$ be a scheme. Let $X \to Y \to Z$ be morphisms of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Let $x \in |X|$ with image $y \in |Y|$.

1. If $\mathcal{F}$ is flat at $x$ over $Y$ and $Y$ is flat at $y$ over $Z$, then $\mathcal{F}$ is flat at $x$ over $Z$.

2. Let $x : \mathop{\mathrm{Spec}}(K) \to X$ be a representative of $x$. If

1. $\mathcal{F}$ is flat at $x$ over $Y$,

2. $x^*\mathcal{F} \not= 0$, and

3. $\mathcal{F}$ is flat at $x$ over $Z$,

then $Y$ is flat at $y$ over $Z$.

3. Let $\overline{x}$ be a geometric point of $X$ lying over $x$ with image $\overline{y}$ in $Y$. If $\mathcal{F}_{\overline{x}}$ is a faithfully flat $\mathcal{O}_{Y, \overline{y}}$-module and $\mathcal{F}$ is flat at $x$ over $Z$, then $Y$ is flat at $y$ over $Z$.

Proof. Pick $\overline{x}$ and $\overline{y}$ as in part (3) and denote $\overline{z}$ the induced geometric point of $Z$. Via the characterization of flatness in Lemmas 66.31.1 and 66.30.8 the lemma reduces to a purely algebraic question on the local ring map $\mathcal{O}_{Z, \overline{z}} \to \mathcal{O}_{Y, \overline{y}}$ and the module $\mathcal{F}_{\overline{x}}$. Part (1) follows from Algebra, Lemma 10.39.4. We remark that condition (2)(b) guarantees that $\mathcal{F}_{\overline{x}}/ \mathfrak m_{\overline{y}} \mathcal{F}_{\overline{x}}$ is nonzero. Hence (2)(a) $+$ (2)(b) imply that $\mathcal{F}_{\overline{x}}$ is a faithfully flat $\mathcal{O}_{Y, \overline{y}}$-module, see Algebra, Lemma 10.39.15. Thus (2) is a special case of (3). Finally, (3) follows from Algebra, Lemma 10.39.10. $\square$

Sometimes the base change happens “up on top”. Here is a precise statement.

Lemma 66.31.5. Let $S$ be a scheme. Let $f : X \to Y$, $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Let $\mathcal{G}$ be a quasi-coherent sheaf on $Y$. Let $x \in |X|$ with image $y \in |Y|$. If $f$ is flat at $x$, then

$\mathcal{G}\text{ flat over }Z\text{ at }y \Leftrightarrow f^*\mathcal{G}\text{ flat over }Z\text{ at }x.$

In particular: If $f$ is surjective and flat, then $\mathcal{G}$ is flat over $Z$, if and only if $f^*\mathcal{G}$ is flat over $Z$.

Proof. Pick a geometric point $\overline{x}$ of $X$ and denote $\overline{y}$ the image in $Y$ and $\overline{z}$ the image in $Z$. Via the characterization of flatness in Lemmas 66.31.1 and 66.30.8 and the description of the stalk of $f^*\mathcal{G}$ at $\overline{x}$ of Properties of Spaces, Lemma 65.29.5 the lemma reduces to a purely algebraic question on the local ring maps $\mathcal{O}_{Z, \overline{z}} \to \mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ and the module $\mathcal{G}_{\overline{y}}$. This algebraic statement is Algebra, Lemma 10.39.9. $\square$

Lemma 66.31.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume $f$ locally finite presentation, $\mathcal{F}$ of finite type, $X = \text{Supp}(\mathcal{F})$, and $\mathcal{F}$ flat over $Y$. Then $f$ is universally open.

Proof. Choose a surjective étale morphism $\varphi : V \to Y$ where $V$ is a scheme. Choose a surjective étale morphism $U \to V \times _ Y X$ where $U$ is a scheme. Then it suffices to prove the lemma for $U \to V$ and the quasi-coherent $\mathcal{O}_ V$-module $\varphi ^*\mathcal{F}$. Hence this lemma follows from the case of schemes, see Morphisms, Lemma 29.25.11. $\square$

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).