The Stacks project

First proof. This proof is longer but does not use the existence of a complete dévissage. The problem is local around $x$ and $s$, hence we may assume that $X$ and $S$ are affine. During the proof we will finitely many times replace $S$ by an elementary étale neighbourhood of $(S, s)$. The goal is then to find (after such a replacement) an open $V \subset X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ containing $x$ such that $\mathcal{F}|_ V$ is flat over $S$ and finitely presented. Of course we may also replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ at any point of the proof, i.e., we may assume $S$ is a local scheme. We will prove the proposition by induction on the integer $n = \dim _ x(\text{Supp}(\mathcal{F}_ s))$.

We can choose

1. elementary étale neighbourhoods $g : (X', x') \to (X, x)$, $e : (S', s') \to (S, s)$,

2. a commutative diagram

   $$\xymatrix{ X \ar[dd]_ f & X' \ar[dd] \ar[l]^ g & Z' \ar[l]^ i \ar[d]^\pi \\ & & Y' \ar[d]^ h \\ S & S' \ar[l]_ e & S' \ar@{=}[l] }$$

3. a point $z' \in Z'$ with $i(z') = x'$, $y' = \pi (z')$, $h(y') = s'$,

4. a finite type quasi-coherent $\mathcal{O}_{Z'}$-module $\mathcal{G}$,

as in Lemma 38.3.2. We are going to replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$, see remarks in first paragraph of the proof. Consider the diagram

Here we have base changed the schemes $X', Z', Y'$ over $S'$ via $\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}) \to S'$ and the scheme $X$ over $S$ via $\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}) \to S$. It is still the case that $g$ is étale, see Lemma 38.2.2. After replacing $X$ by $X_{\mathcal{O}_{S', s'}}$, $X'$ by $X'_{\mathcal{O}_{S', s'}}$, $Z'$ by $Z'_{\mathcal{O}_{S', s'}}$, and $Y'$ by $Y'_{\mathcal{O}_{S', s'}}$ we may assume we have a diagram as Lemma 38.3.2 where in addition $S = S'$ is a local scheme with closed point $s$. By Lemmas 38.3.3 and 38.3.4 the result for $Y' \to S$, the sheaf $\pi _*\mathcal{G}$, and the point $y'$ implies the result for $X \to S$, $\mathcal{F}$ and $x$. Hence we may assume that $S$ is local and $X \to S$ is a smooth morphism of affines with geometrically irreducible fibres of dimension $n$.

The base case of the induction: $n = 0$. As $X \to S$ is smooth with geometrically irreducible fibres of dimension $0$ we see that $X \to S$ is an open immersion, see Descent, Lemma 35.25.2. As $S$ is local and the closed point is in the image of $X \to S$ we conclude that $X = S$. Thus we see that $\mathcal{F}$ corresponds to a finite flat $\mathcal{O}_{S, s}$ module. In this case the result follows from Algebra, Lemma 10.78.5 which tells us that $\mathcal{F}$ is in fact finite free.

The induction step. Assume the result holds whenever the dimension of the support in the closed fibre is $< n$. Write $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ and $\mathcal{F} = \widetilde{N}$ for some $B$-module $N$. Note that $A$ is a local ring; denote its maximal ideal $\mathfrak m$. Then $\mathfrak p = \mathfrak mB$ is the unique minimal prime lying over $\mathfrak m$ as $X \to S$ has geometrically irreducible fibres. Finally, let $\mathfrak q \subset B$ be the prime corresponding to $x$. By Lemma 38.10.1 we can choose a map

such that $\kappa (\mathfrak p)^{\oplus r} \to N \otimes _ B \kappa (\mathfrak p)$ is an isomorphism. Moreover, as $N_{\mathfrak q}$ is $A$-flat the lemma also shows that $\alpha$ is injective and that $\mathop{\mathrm{Coker}}(\alpha )_{\mathfrak q}$ is $A$-flat. Set $Q = \mathop{\mathrm{Coker}}(\alpha )$. Note that the support of $Q/\mathfrak mQ$ does not contain $\mathfrak p$. Hence it is certainly the case that $\dim _{\mathfrak q}(\text{Supp}(Q/\mathfrak mQ)) < n$. Combining everything we know about $Q$ we see that the induction hypothesis applies to $Q$. It follows that there exists an elementary étale morphism $(S', s) \to (S, s)$ such that the conclusion holds for $Q \otimes _ A A'$ over $B \otimes _ A A'$ where $A' = \mathcal{O}_{S', s'}$. After replacing $A$ by $A'$ we have an exact sequence

(here we use that $\alpha$ is injective as mentioned above) of finite $B$-modules and we also get an element $g \in B$, $g \not\in \mathfrak q$ such that $Q_ g$ is finitely presented over $B_ g$ and flat over $A$. Since localization is exact we see that

is still exact. As $B_ g$ and $Q_ g$ are flat over $A$ we conclude that $N_ g$ is flat over $A$, see Algebra, Lemma 10.39.13, and as $B_ g$ and $Q_ g$ are finitely presented over $B_ g$ the same holds for $N_ g$, see Algebra, Lemma 10.5.3. $\square$

Second proof. We apply Proposition 38.5.7 to find a commutative diagram

of pointed schemes such that the horizontal arrows are elementary étale neighbourhoods and such that $g^*\mathcal{F}/X'/S'$ has a complete dévissage at $x$. (In particular $S'$ and $X'$ are affine.) By Morphisms, Lemma 29.25.13 we see that $g^*\mathcal{F}$ is flat at $x'$ over $S$ and by Lemma 38.2.3 we see that it is flat at $x'$ over $S'$. Via Remark 38.6.5 we deduce that

has a complete dévissage at the prime of $\Gamma (X', \mathcal{O}_{X'})$ corresponding to $x'$. We may base change this complete dévissage to the local ring $\mathcal{O}_{S', s'}$ of $\Gamma (S', \mathcal{O}_{S'})$ at the prime corresponding to $s'$. Thus Lemma 38.10.2 implies that

is flat over $\mathcal{O}_{S', s'}$ and of finite presentation over $\Gamma (X', \mathcal{O}_{X'}) \otimes _{\Gamma (S', \mathcal{O}_{S'})} \mathcal{O}_{S', s'}$. In other words, the restriction of $\mathcal{F}$ to $X' \times _{S'} \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ is of finite presentation and flat over $\mathcal{O}_{S', s'}$. Since the morphism $X' \times _{S'} \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}) \to X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ is étale (Lemma 38.2.2) its image $V \subset X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ is an open subscheme, and by étale descent the restriction of $\mathcal{F}$ to $V$ is of finite presentation and flat over $\mathcal{O}_{S', s'}$. (Results used: Morphisms, Lemma 29.36.13, Descent, Lemma 35.7.3, and Morphisms, Lemma 29.25.13.) $\square$