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 $\mathop{\mathrm{Spec}}(\kappa (s')) \to \mathop{\mathrm{Spec}}(\kappa (s))$.
$\square$

## Comments (0)

There are also: