## 76.23 Critère de platitude par fibres

Let $S$ be a scheme. Consider a commutative diagram of algebraic spaces over $S$

$\xymatrix{ X \ar[rr]_ f \ar[dr]_ g & & Y \ar[dl]^ h \\ & Z }$

and a quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$. Given a point $x \in |X|$ we consider the question as to whether $\mathcal{F}$ is flat over $Y$ at $x$. If $\mathcal{F}$ is flat over $Z$ at $x$, then the theorem below states this question is intimately related to the question of whether the restriction of $\mathcal{F}$ to the fibre of $X \to Z$ over $g(x)$ is flat over the fibre of $Y \to Z$ over $g(x)$. To make sense out of this we offer the following preliminary lemma.

Lemma 76.23.1. In the situation above the following are equivalent

1. Pick a geometric point $\overline{x}$ of $X$ lying over $x$. Set $\overline{y} = f \circ \overline{x}$ and $\overline{z} = g \circ \overline{x}$. Then the module $\mathcal{F}_{\overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{F}_{\overline{x}}$ is flat over $\mathcal{O}_{Y, \overline{y}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{Y, \overline{y}}$.

2. Pick a morphism $x : \mathop{\mathrm{Spec}}(K) \to X$ in the equivalence class of $x$. Set $z = g \circ x$, $X_ z = \mathop{\mathrm{Spec}}(K) \times _{z, Z} X$, $Y_ z = \mathop{\mathrm{Spec}}(K) \times _{z, Z} Y$, and $\mathcal{F}_ z$ the pullback of $\mathcal{F}$ to $X_ z$. Then $\mathcal{F}_ z$ is flat at $x$ over $Y_ z$ (as defined in Morphisms of Spaces, Definition 67.31.2).

3. Pick a commutative diagram

$\xymatrix{ & & & U \ar[llld]_ a \ar[rr] \ar[dr] & & V \ar[llld]_>>>>>>>b \ar[dl] \\ X \ar[rr]_ f \ar[dr]_ g & & Y \ar[dl]^ h & & W \ar[llld]_ c \\ & Z }$

where $U, V, W$ are schemes, and $a, b, c$ are étale, and a point $u \in U$ mapping to $x$. Let $w \in W$ be the image of $u$. Let $\mathcal{F}_ w$ be the pullback of $\mathcal{F}$ to the fibre $U_ w$ of $U \to W$ at $w$. Then $\mathcal{F}_ w$ is flat over $V_ w$ at $u$.

Proof. Note that in (2) the morphism $x : \mathop{\mathrm{Spec}}(K) \to X$ defines a $K$-rational point of $X_ z$, hence the statement makes sense. Moreover, the condition in (2) is independent of the choice of $\mathop{\mathrm{Spec}}(K) \to X$ in the equivalence class of $x$ (details omitted; this will also follow from the arguments below because the other conditions do not depend on this choice). Also note that we can always choose a diagram as in (3) by: first choosing a scheme $W$ and a surjective étale morphism $W \to Z$, then choosing a scheme $V$ and a surjective étale morphism $V \to W \times _ Z Y$, and finally choosing a scheme $U$ and a surjective étale morphism $U \to V \times _ Y X$. Having made these choices we set $U \to W$ equal to the composition $U \to V \to W$ and we can pick a point $u \in U$ mapping to $x$ because the morphism $U \to X$ is surjective.

Suppose given both a diagram as in (3) and a geometric point $\overline{x} : \mathop{\mathrm{Spec}}(k) \to X$ as in (1). By Properties of Spaces, Lemma 66.19.4 we can choose a geometric point $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ such that $\overline{x} = a \circ \overline{u}$. Denote $\overline{v} : \mathop{\mathrm{Spec}}(k) \to V$ and $\overline{w} : \mathop{\mathrm{Spec}}(k) \to W$ the induced geometric points of $V$ and $W$. In this setting we know that $\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{U, u}^{sh}$ and similarly for $Y$ and $Z$, see Properties of Spaces, Lemma 66.22.1. In the same vein we have

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

see Properties of Spaces, Lemma 66.29.4. Note that the stalk of $\mathcal{F}_ w$ at $u$ is given by

$(\mathcal{F}_ w)_ u = (a^*\mathcal{F})_ u/\mathfrak m_ w(a^*\mathcal{F})_ u$

and the local ring of $V_ w$ at $v$ is given by

$\mathcal{O}_{V_ w, v} = \mathcal{O}_{V, v}/\mathfrak m_ w\mathcal{O}_{V, v}.$

Since $\mathfrak m_{\overline{z}} = \mathfrak m_ w \mathcal{O}_{Z, \overline{z}} = \mathfrak m_ w \mathcal{O}_{W, w}^{sh}$ we see that

\begin{align*} \mathcal{F}_{\overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{F}_{\overline{x}} & = (a^*\mathcal{F})_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{X, \overline{x}} \\ & = (\mathcal{F}_ w)_ u \otimes _{\mathcal{O}_{U_ w, u}} \mathcal{O}_{U, u}^{sh}/\mathfrak m_ w\mathcal{O}_{U, u}^{sh} \\ & = (\mathcal{F}_ w)_ u \otimes _{\mathcal{O}_{U_ w, u}} \mathcal{O}_{U_ w, \overline{u}}^{sh} \\ & = (\mathcal{F}_ w)_{\overline{u}} \end{align*}

the penultimate equality by Algebra, Lemma 10.156.4 and the last equality by Properties of Spaces, Lemma 66.29.4. The same arguments applied to the structure sheaves of $V$ and $Y$ show that

$\mathcal{O}_{V_ w, \overline{v}}^{sh} = \mathcal{O}_{V, v}^{sh}/\mathfrak m_ w \mathcal{O}_{V, v}^{sh} = \mathcal{O}_{Y, \overline{y}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{Y, \overline{y}}.$

OK, and now we can use Morphisms of Spaces, Lemma 67.31.1 to see that (1) is equivalent to (3).

Finally we prove the equivalence of (2) and (3). To do this we pick a field extension $\tilde K$ of $K$ and a morphism $\tilde x : \mathop{\mathrm{Spec}}(\tilde K) \to U$ which lies over $u$ (this is possible because $u \times _{X, x} \mathop{\mathrm{Spec}}(K)$ is a nonempty scheme). Set $\tilde z : \mathop{\mathrm{Spec}}(\tilde K) \to U \to W$ be the composition. We obtain a commutative diagram

$\xymatrix{ & & & U_ w \times _ w \tilde z \ar[llld]_ a \ar[rr] \ar[dr] & & V_ w \times _ w \tilde z \ar[llld]_>>>>>>>b \ar[dl] \\ X_ z \ar[rr]_ f \ar[dr]_ g & & Y_ z \ar[dl]^ h & & \tilde z \ar[llld]_ c \\ & z }$

where $z = \mathop{\mathrm{Spec}}(K)$ and $w = \mathop{\mathrm{Spec}}(\kappa (w))$. Now it is clear that $\mathcal{F}_ w$ and $\mathcal{F}_ z$ pull back to the same module on $U_ w \times _ w \tilde z$. This leads to a commutative diagram

$\xymatrix{ X_ z \ar[d] & U_ w \times _ w \tilde z \ar[l] \ar[d] \ar[r] & U_ w \ar[d] \\ Y_ z & V_ w \times _ w \tilde z \ar[l] \ar[r] & V_ w }$

both of whose squares are cartesian and whose bottom horizontal arrows are flat: the lower left horizontal arrow is the composition of the morphism $Y \times _ Z \tilde z \to Y \times _ Z z = Y_ z$ (base change of a flat morphism), the étale morphism $V \times _ Z \tilde z \to Y \times _ Z \tilde z$, and the étale morphism $V \times _ W \tilde z \to V \times _ Z \tilde z$. Thus it follows from Morphisms of Spaces, Lemma 67.31.3 that

$\mathcal{F}_ z\text{ flat at }x\text{ over }Y_ z \Leftrightarrow \mathcal{F}|_{U_ w \times _ w \tilde z} \text{ flat at }\tilde x\text{ over }V_ w \times _ w \tilde z \Leftrightarrow \mathcal{F}_ w\text{ flat at }u\text{ over }V_ w$

and we win. $\square$

Definition 76.23.2. 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 $\mathcal{O}_ X$-module. Let $x \in |X|$ be a point and denote $z \in |Z|$ its image.

1. We say the restriction of $\mathcal{F}$ to its fibre over $z$ is flat at $x$ over the fibre of $Y$ over $z$ if the equivalent conditions of Lemma 76.23.1 are satisfied.

2. We say the fibre of $X$ over $z$ is flat at $x$ over the fibre of $Y$ over $z$ if the equivalent conditions of Lemma 76.23.1 hold with $\mathcal{F} = \mathcal{O}_ X$.

3. We say the fibre of $X$ over $z$ is flat over the fibre of $Y$ over $z$ if for all $x \in |X|$ lying over $z$ the fibre of $X$ over $z$ is flat at $x$ over the fibre of $Y$ over $z$

With this definition in hand we can state a version of the criterion as follows. The Noetherian version can be found in Section 76.24.

Theorem 76.23.3. Let $S$ be a scheme. Let $f : X \to Y$ and $Y \to Z$ be morphisms of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

1. $X$ is locally of finite presentation over $Z$,

2. $\mathcal{F}$ an $\mathcal{O}_ X$-module of finite presentation, and

3. $Y$ is locally of finite type over $Z$.

Let $x \in |X|$ and let $y \in |Y|$ and $z \in |Z|$ be the images of $x$. If $\mathcal{F}_{\overline{x}} \not= 0$, then the following are equivalent:

1. $\mathcal{F}$ is flat over $Z$ at $x$ and the restriction of $\mathcal{F}$ to its fibre over $z$ is flat at $x$ over the fibre of $Y$ over $z$, and

2. $Y$ is flat over $Z$ at $y$ and $\mathcal{F}$ is flat over $Y$ at $x$.

Moreover, the set of points $x$ where (1) and (2) hold is open in $\text{Supp}(\mathcal{F})$.

Proof. Choose a diagram as in Lemma 76.23.1 part (3). It follows from the definitions that this reduces to the corresponding theorem for the morphisms of schemes $U \to V \to W$, the quasi-coherent sheaf $a^*\mathcal{F}$, and the point $u \in U$. Thus the theorem follows from the corresponding result for schemes which is More on Morphisms, Theorem 37.16.2. $\square$

Lemma 76.23.4. Let $S$ be a scheme. Let $f : X \to Y$ and $Y \to Z$ be a morphism of algebraic spaces over $S$. Assume

1. $X$ is locally of finite presentation over $Z$,

2. $X$ is flat over $Z$,

3. for every $z \in |Z|$ the fibre of $X$ over $z$ is flat over the fibre of $Y$ over $z$, and

4. $Y$ is locally of finite type over $Z$.

Then $f$ is flat. If $f$ is also surjective, then $Y$ is flat over $Z$.

Proof. This is a special case of Theorem 76.23.3. $\square$

Lemma 76.23.5. Let $S$ be a scheme. Let $f : X \to Y$ and $Y \to Z$ be morphisms of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

1. $X$ is locally of finite presentation over $Z$,

2. $\mathcal{F}$ an $\mathcal{O}_ X$-module of finite presentation,

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

4. $Y$ is locally of finite type over $Z$.

Then the set

$A = \{ x \in |X| : \mathcal{F} \text{ flat at }x \text{ over }Y\} .$

is open in $|X|$ and its formation commutes with arbitrary base change: If $Z' \to Z$ is a morphism of algebraic spaces, and $A'$ is the set of points of $X' = X \times _ Z Z'$ where $\mathcal{F}' = \mathcal{F} \times _ Z Z'$ is flat over $Y' = Y \times _ Z Z'$, then $A'$ is the inverse image of $A$ under the continuous map $|X'| \to |X|$.

Proof. One way to prove this is to translate the proof as given in More on Morphisms, Lemma 37.16.4 into the category of algebraic spaces. Instead we will prove this by reducing to the case of schemes. Namely, choose a diagram as in Lemma 76.23.1 part (3) such that $a$, $b$, and $c$ are surjective. It follows from the definitions that this reduces to the corresponding theorem for the morphisms of schemes $U \to V \to W$, the quasi-coherent sheaf $a^*\mathcal{F}$, and the point $u \in U$. The only minor point to make is that given a morphism of algebraic spaces $Z' \to Z$ we choose a scheme $W'$ and a surjective étale morphism $W' \to W \times _ Z Z'$. Then we set $U' = W' \times _ W U$ and $V' = W' \times _ W V$. We write $a', b', c'$ for the morphisms from $U', V', W'$ to $X', Y', Z'$. In this case $A$, resp. $A'$ are images of the open subsets of $U$, resp. $U'$ associated to $a^*\mathcal{F}$, resp. $(a')^*\mathcal{F}'$. This indeed does reduce the lemma to More on Morphisms, Lemma 37.16.4. $\square$

Lemma 76.23.6. Let $S$ be a scheme. Let $f : X \to Y$ and $Y \to Z$ be a morphism of algebraic spaces over $S$. Assume

1. $X$ is locally of finite presentation over $Z$,

2. $X$ is flat over $Z$, and

3. $Y$ is locally of finite type over $Z$.

Then the set

$\{ x \in |X| : X\text{ flat at }x \text{ over }Y\} .$

is open in $|X|$ and its formation commutes with arbitrary base change $Z' \to Z$.

Proof. This is a special case of Lemma 76.23.5. $\square$

Lemma 76.23.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite presentation. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $x \in |X|$ with image $y \in |Y|$. If $\mathcal{F}$ is flat at $x$ over $Y$, then the following are equivalent

1. $(\mathcal{F}_{\overline{y}})_{\overline{x}}$ is a flat $\mathcal{O}_{X_{\overline{y}}, \overline{x}}$-module,

2. $(\mathcal{F}_{\overline{y}})_{\overline{x}}$ is a free $\mathcal{O}_{X_{\overline{y}}, \overline{x}}$-module,

3. $\mathcal{F}_{\overline{y}}$ is finite free in an étale neighbourhood of $\overline{x}$ in $X_{\overline{y}}$, and

4. $\mathcal{F}$ is finite free in an étale neighbourhood of $x$ in $X$.

Here $\overline{x}$ is a geometric point of $X$ lying over $x$ and $\overline{y} = f \circ \overline{x}$.

Proof. Pick a commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

where $U$ and $V$ are schemes and the vertical arrows are étale such that there is a point $u \in U$ mapping to $x$. Let $v \in V$ be the image of $u$. Applying Lemma 76.23.1 to $\text{id} : X \to X$ over $Y$ we see that (1) translates into the condition “$\mathcal{F}|_{U_ v}$ is flat over $U_ v$ at $u$”. In other words, (1) is equivalent to $(\mathcal{F}|_{U_ v})_ u$ being a flat $\mathcal{O}_{U_ v, u}$-module. By the case of schemes (More on Morphisms, Lemma 37.16.7), we find that this implies that $\mathcal{F}|_ U$ is finite free in an open neighbourhood of $u$. In this way we see that (1) implies (4). The implications (4) $\Rightarrow$ (3) and (2) $\Rightarrow$ (1) are immediate. For the implication (3) $\Rightarrow$ (2) use the description of local rings and stalks in Properties of Spaces, Lemmas 66.22.1 and 66.29.4. $\square$

Lemma 76.23.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite presentation. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module flat over $Y$. Then the set

$\{ x \in |X| : \mathcal{F}\text{ free in an étale neighbourhood of }x\}$

is open in $|X|$ and its formation commutes with arbitrary base change $Y' \to Y$.

Proof. Openness holds trivially. Let $Y' \to Y$ be a morphism of algebraic spaces, set $X' = Y' \times _ Y X$, and let $x' \in |X'|$ be a point lying over $x \in |X|$. By Lemma 76.23.7 we see that $x$ is in our set if and only if $(\mathcal{F}_{\overline{y}})_{\overline{x}}$ is a flat $\mathcal{O}_{X_{\overline{y}}, \overline{x}}$-module. Simiarly, $x'$ is in the analogue of our set for the pullback $\mathcal{F}'$ of $\mathcal{F}$ to $X'$ if and only if $(\mathcal{F}'_{\overline{y}'})_{\overline{x}'}$ is a flat $\mathcal{O}_{X'_{\overline{y}'}, \overline{x}'}$-module (with obvious notation). These two assertions are equivalent by Lemma 76.23.1 applied to the morphism $\text{id} : X \to X$ over $Y$. Thus the statement on base change holds. $\square$

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