## 37.16 Critère de platitude par fibres

Consider a commutative diagram of schemes (left hand diagram)

\[ \xymatrix{ X \ar[rr]_ f \ar[dr] & & Y \ar[dl] \\ & S } \quad \xymatrix{ X_ s \ar[rr]_{f_ s} \ar[rd] & & Y_ s \ar[dl] \\ & \mathop{\mathrm{Spec}}(\kappa (s)) } \]

and a quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$. Given a point $x \in X$ lying over $s \in S$ with image $y = f(x)$ we consider the question: Is $\mathcal{F}$ flat over $Y$ at $x$? If $\mathcal{F}$ is flat over $S$ at $x$, then the theorem states this question is intimately related to the question of whether the restriction of $\mathcal{F}$ to the fibre

\[ \mathcal{F}_ s = (X_ s \to X)^*\mathcal{F} \]

is flat over $Y_ s$ at $x$. Below you will find a “Noetherian” version, a “finitely presented” version, and earlier we treated a “nilpotent” version, see Lemma 37.10.2.

Theorem 37.16.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in X$. Set $y = f(x)$ and $s \in S$ the image of $x$ in $S$. Assume $S$, $X$, $Y$ locally Noetherian, $\mathcal{F}$ coherent, and $\mathcal{F}_ x \not= 0$. Then the following are equivalent:

$\mathcal{F}$ is flat over $S$ at $x$, and $\mathcal{F}_ s$ is flat over $Y_ s$ at $x$, and

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

**Proof.**
Consider the ring maps

\[ \mathcal{O}_{S, s} \longrightarrow \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} \]

and the module $\mathcal{F}_ x$. The stalk of $\mathcal{F}_ s$ at $x$ is the module $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ and the local ring of $Y_ s$ at $y$ is $\mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$. Thus the implication (1) $\Rightarrow $ (2) is Algebra, Lemma 10.99.15. If (2) holds, then the first ring map is faithfully flat and $\mathcal{F}_ x$ is flat over $\mathcal{O}_{Y, y}$ so by Algebra, Lemma 10.39.4 we see that $\mathcal{F}_ x$ is flat over $\mathcal{O}_{S, s}$. Moreover, $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ is the base change of the flat module $\mathcal{F}_ x$ by $\mathcal{O}_{Y, y} \to \mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$, hence flat by Algebra, Lemma 10.39.7.
$\square$

Here is the non-Noetherian version.

Theorem 37.16.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

$X$ is locally of finite presentation over $S$,

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

$Y$ is locally of finite type over $S$.

Let $x \in X$. Set $y = f(x)$ and let $s \in S$ be the image of $x$ in $S$. If $\mathcal{F}_ x \not= 0$, then the following are equivalent:

$\mathcal{F}$ is flat over $S$ at $x$, and $\mathcal{F}_ s$ is flat over $Y_ s$ at $x$, and

$Y$ is flat over $S$ 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.**
Consider the ring maps

\[ \mathcal{O}_{S, s} \longrightarrow \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} \]

and the module $\mathcal{F}_ x$. The stalk of $\mathcal{F}_ s$ at $x$ is the module $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ and the local ring of $Y_ s$ at $y$ is $\mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$. Thus the implication (1) $\Rightarrow $ (2) is Algebra, Lemma 10.128.9. If (2) holds, then the first ring map is faithfully flat and $\mathcal{F}_ x$ is flat over $\mathcal{O}_{Y, y}$ so by Algebra, Lemma 10.39.4 we see that $\mathcal{F}_ x$ is flat over $\mathcal{O}_{S, s}$. Moreover, $\mathcal{F}_ x/\mathfrak m_ s \mathcal{F}_ x$ is the base change of the flat module $\mathcal{F}_ x$ by $\mathcal{O}_{Y, y} \to \mathcal{O}_{Y, y}/\mathfrak m_ s \mathcal{O}_{Y, y}$, hence flat by Algebra, Lemma 10.39.7.

By Morphisms, Lemma 29.21.11 the morphism $f$ is locally of finite presentation. Consider the set

37.16.2.1
\begin{equation} \label{more-morphisms-equation-open} U = \{ x \in X \mid \mathcal{F} \text{ flat at }x \text{ over both }Y\text{ and }S\} . \end{equation}

This set is open in $X$ by Theorem 37.15.1. Note that if $x \in U$, then $\mathcal{F}_ s$ is flat at $x$ over $Y_ s$ as a base change of a flat module under the morphism $Y_ s \to Y$, see Morphisms, Lemma 29.25.7. Hence at every point of $U \cap \text{Supp}(\mathcal{F})$ condition (1) is satisfied. On the other hand, it is clear that if $x \in \text{Supp}(\mathcal{F})$ satisfies (1) and (2), then $x \in U$. Thus the open set we are looking for is $U \cap \text{Supp}(\mathcal{F})$.
$\square$

These theorems are often used in the following simplified forms. We give only the global statements – of course there are also pointwise versions.

Lemma 37.16.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Assume

$S$, $X$, $Y$ are locally Noetherian,

$X$ is flat over $S$,

for every $s \in S$ the morphism $f_ s : X_ s \to Y_ s$ is flat.

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

**Proof.**
This is a special case of Theorem 37.16.1.
$\square$

Lemma 37.16.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Assume

$X$ is locally of finite presentation over $S$,

$X$ is flat over $S$,

for every $s \in S$ the morphism $f_ s : X_ s \to Y_ s$ is flat, and

$Y$ is locally of finite type over $S$.

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

**Proof.**
This is a special case of Theorem 37.16.2.
$\square$

Lemma 37.16.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

$X$ is locally of finite presentation over $S$,

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

$\mathcal{F}$ is flat over $S$, and

$Y$ is locally of finite type over $S$.

Then the set

\[ U = \{ x \in X \mid \mathcal{F} \text{ flat at }x \text{ over }Y\} . \]

is open in $X$ and its formation commutes with arbitrary base change: If $S' \to S$ is a morphism of schemes, and $U'$ is the set of points of $X' = X \times _ S S'$ where $\mathcal{F}' = \mathcal{F} \times _ S S'$ is flat over $Y' = Y \times _ S S'$, then $U' = U \times _ S S'$.

**Proof.**
By Morphisms, Lemma 29.21.11 the morphism $f$ is locally of finite presentation. Hence $U$ is open by Theorem 37.15.1. Because we have assumed that $\mathcal{F}$ is flat over $S$ we see that Theorem 37.16.2 implies

\[ U = \{ x \in X \mid \mathcal{F}_ s \text{ flat at }x \text{ over }Y_ s\} . \]

where $s$ always denotes the image of $x$ in $S$. (This description also works trivially when $\mathcal{F}_ x = 0$.) Moreover, the assumptions of the lemma remain in force for the morphism $f' : X' \to Y'$ and the sheaf $\mathcal{F}'$. Hence $U'$ has a similar description. In other words, it suffices to prove that given $s' \in S'$ mapping to $s \in S$ we have

\[ \{ x' \in X'_{s'} \mid \mathcal{F}'_{s'} \text{ flat at }x' \text{ over }Y'_{s'}\} \]

is the inverse image of the corresponding locus in $X_ s$. This is true by Lemma 37.15.2 because in the cartesian diagram

\[ \xymatrix{ X'_{s'} \ar[d] \ar[r] & X_ s \ar[d] \\ Y'_{s'} \ar[r] & Y_ s } \]

the horizontal morphisms are flat as they are base changes by the flat morphism $\mathop{\mathrm{Spec}}(\kappa (s')) \to \mathop{\mathrm{Spec}}(\kappa (s))$.
$\square$

Lemma 37.16.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Assume

$X$ is locally of finite presentation over $S$,

$X$ is flat over $S$, and

$Y$ is locally of finite type over $S$.

Then the set

\[ U = \{ x \in X \mid X\text{ flat at }x \text{ over }Y\} . \]

is open in $X$ and its formation commutes with arbitrary base change.

**Proof.**
This is a special case of Lemma 37.16.5.
$\square$

The following lemma is a variant of Algebra, Lemma 10.99.4. Note that the hypothesis that $(\mathcal{F}_ s)_ x$ is a flat $\mathcal{O}_{X_ s, x}$-module means that $(\mathcal{F}_ s)_ x$ is a free $\mathcal{O}_{X_ s, x}$-module which is always the case if $x \in X_ s$ is a generic point of an irreducible component of $X_ s$ and $X_ s$ is reduced (namely, in this case $\mathcal{O}_{X_ s, x}$ is a field, see Algebra, Lemma 10.25.1).

Lemma 37.16.7. Let $f : X \to S$ be a morphism of schemes of finite presentation. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $x \in X$ with image $s \in S$. If $\mathcal{F}$ is flat at $x$ over $S$ and $(\mathcal{F}_ s)_ x$ is a flat $\mathcal{O}_{X_ s, x}$-module, then $\mathcal{F}$ is finite free in a neighbourhood of $x$.

**Proof.**
If $\mathcal{F}_ x \otimes \kappa (x)$ is zero, then $\mathcal{F}_ x = 0$ by Nakayama's lemma (Algebra, Lemma 10.20.1) and hence $\mathcal{F}$ is zero in a neighbourhood of $x$ (Modules, Lemma 17.9.5) and the lemma holds. Thus we may assume $\mathcal{F}_ x \otimes \kappa (x)$ is not zero and we see that Theorem 37.16.2 applies with $f = \text{id} : X \to X$. We conclude that $\mathcal{F}_ x$ is flat over $\mathcal{O}_{X, x}$. Hence $\mathcal{F}_ x$ is free, see Algebra, Lemma 10.78.5 for example. Choose an open neighbourhood $x \in U \subset X$ and sections $s_1, \ldots , s_ r \in \mathcal{F}(U)$ which map to a basis in $\mathcal{F}_ x$. The corresponding map $\psi : \mathcal{O}_ U^{\oplus r} \to \mathcal{F}|_ U$ is surjective after shrinking $U$ (Modules, Lemma 17.9.5). Then $\mathop{\mathrm{Ker}}(\psi )$ is of finite type (see Modules, Lemma 17.11.3) and $\mathop{\mathrm{Ker}}(\psi )_ x = 0$. Whence after shrinking $U$ once more $\psi $ is an isomorphism.
$\square$

Lemma 37.16.8. Let $f : X \to S$ be a morphism of schemes which is locally of finite presentation. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module flat over $S$. Then the set

\[ \{ x \in X : \mathcal{F}\text{ free in a neighbourhood of }x\} \]

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

**Proof.**
Openness holds trivially. Let $x \in X$ mapping to $s \in S$. By Lemma 37.16.7 we see that $x$ is in our set if and only if $\mathcal{F}|_{X_ s}$ is flat at $x$ over $X_ s$. Clearly this is also equivalent to $\mathcal{F}$ being flat at $x$ over $X$ (because this statement is implied by freeness of $\mathcal{F}_ x$ and implies flatness of $\mathcal{F}|_{X_ s}$ at $x$ over $X_ s$). Thus the base change statement follows from Lemma 37.16.5 applied to $\text{id} : X \to X$ over $S$.
$\square$

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