The Stacks project

Proof. For each $S$ we will prove the statement for all $p \geq 0$ concurrently. The functor $H_ p$ is a sheaf for the fppf topology by Lemma 38.20.8. Hence combining Descent, Lemma 35.39.1, More on Morphisms, Lemma 37.57.1 , and Descent, Lemma 35.24.1 we see that the question is local for the étale topology on $S$. In particular, the question is Zariski local on $S$.

For $s \in S$ denote $\xi _ s$ the unique generic point of the fibre $X_ s$. Note that for every $s \in S$ the restriction $\mathcal{F}_ s$ of $\mathcal{F}$ is locally free of some rank $p(s) \geq 0$ in some neighbourhood of $\xi _ s$. (As $X_ s$ is irreducible and smooth this follows from generic flatness for $\mathcal{F}_ s$ over $X_ s$, see Algebra, Lemma 10.118.1 although this is overkill.) For future reference we note that

In particular $H_{p(s)}(s)$ is nonempty and $H_ q(s)$ is empty if $q \not= p(s)$.

Let $U \subset X$ be an open subscheme. As $f : X \to S$ is smooth, it is open. It is immediate from (38.20.7.2) that the functor $H_ p$ for the pair $(f|_ U : U \to f(U), \mathcal{F}|_ U)$ and the functor $H_ p$ for the pair $(f|_{f^{-1}(f(U))}, \mathcal{F}|_{f^{-1}(f(U))})$ are the same. Hence to prove the existence of $S_ p$ over $f(U)$ we may always replace $X$ by $U$.

Pick $s \in S$. There exists an affine open neighbourhood $U$ of $\xi _ s$ such that $\mathcal{F}|_ U$ can be generated by at most $p(s)$ elements. By the arguments above we see that in order to prove the statement for $H_{p(s)}$ in an neighbourhood of $s$ we may assume that $\mathcal{F}$ is generated by $p(s)$ elements, i.e., that there exists a surjection

In this case it is clear that $H_{p(s)}$ is equal to $F_{iso}$ (38.20.1.1) for the map $u$ (this follows immediately from Lemma 38.19.1 but also from Lemma 38.12.1 after shrinking a bit more so that both $S$ and $X$ are affine.) Thus we may apply Theorem 38.23.3 to see that $H_{p(s)}$ is representable by a closed immersion in a neighbourhood of $s$.

The result follows formally from the above. Namely, the arguments above show that locally on $S$ the function $s \mapsto p(s)$ is bounded. Hence we may use induction on $p = \max _{s \in S} p(s)$. The functor $H_ p$ is representable by a closed immersion $S_ p \to S$ by the above. Replace $S$ by $S \setminus S_ p$ which drops the maximum by at least one and we win by induction hypothesis.

Assume $\mathcal{F}$ is of finite presentation. Then $S_ p \to S$ is locally of finite presentation by Lemma 38.20.8 part (2) combined with Limits, Remark 32.6.2. Then we redo the induction argument in the paragraph to see that each $S_ p$ is quasi-compact when $S$ is affine: first if $p = \max _{s \in S} p(s)$, then $S_ p \subset S$ is closed (see above) hence quasi-compact. Then $U = S \setminus S_ p$ is quasi-compact open in $S$ because $S_ p \to S$ is a closed immersion of finite presentation (see discussion in Morphisms, Section 29.22 for example). Then $S_{p - 1} \to U$ is a closed immersion of finite presentation, and so $S_{p - 1}$ is quasi-compact and $U' = S \setminus (S_ p \cup S_{p - 1})$ is quasi-compact. And so on. $\square$

Proof. Let $T \to S$ be any morphism. Then $h_ T : X'_ T \to X_ T$ is étale as a base change of the étale morphism $g$. For $t \in T$ denote $Z \subset X_ t$ the set of points where $\mathcal{F}_ T$ is not flat over $T$, and similarly denote $Z' \subset X'_ t$ the set of points where $\mathcal{F}'_ T$ is not flat over $T$. As $\mathcal{F}'_ T = h_ T^*\mathcal{F}_ T$ we see that $Z' = h_ t^{-1}(Z)$, see Morphisms, Lemma 29.25.13. Hence $Z' \to Z$ is an étale morphism, so $\dim (Z') \leq \dim (Z)$ (for example by Descent, Lemma 35.21.2 or just because an étale morphism is smooth of relative dimension $0$). This implies that $F_ n \subset F_ n'$.

Finally, suppose that $h(X') \supset \text{Ass}_{X/S}(\mathcal{F})$ and that $T \to S$ is a morphism such that $F_ n'(T)$ is nonempty, i.e., such that $\mathcal{F}'_ T$ is flat in dimensions $\geq n$ over $T$. Pick a point $t \in T$ and let $Z \subset X_ t$ and $Z' \subset X'_ t$ be as above. To get a contradiction assume that $\dim (Z) \geq n$. Pick a generic point $\xi \in Z$ corresponding to a component of dimension $\geq n$. Let $x \in \text{Ass}_{X_ t}(\mathcal{F}_ t)$ be a generalization of $\xi $. Then $x$ maps to a point of $\text{Ass}_{X/S}(\mathcal{F})$ by Divisors, Lemma 31.7.3 and Remark 31.7.4. Thus we see that $x$ is in the image of $h_ T$, say $x = h_ T(x')$ for some $x' \in X'_ T$. But $x' \not\in Z'$ as $x \leadsto \xi $ and $\dim (Z') < n$. Hence $\mathcal{F}'_ T$ is flat over $T$ at $x'$ which implies that $\mathcal{F}_ T$ is flat at $x$ over $T$ (by Morphisms, Lemma 29.25.13). Since this holds for every such $x$ we conclude that $\mathcal{F}_ T$ is flat over $T$ at $\xi $ by Theorem 38.26.1 which is the desired contradiction. $\square$

Proof. As $X$ is affine and $\mathcal{F}$ is quasi-coherent of finite presentation we know that $\mathcal{F}$ can be generated by $c \geq 0$ elements. Then $\dim _{\kappa (x)}(\mathcal{F}_ x \otimes \kappa (x))$ in any point $x \in X$ never exceeds $c$. In particular $H_ p = \emptyset $ for $p > c$. Moreover, note that there certainly is an inclusion $\coprod H_ p \to F_ d$. Having said this the content of the lemma is that, if a base change $\mathcal{F}_ T$ is flat in dimensions $\geq d$ over $T$ and if $t \in T$, then $\mathcal{F}_ T$ is free of some rank $r$ in an open neighbourhood $U \subset X_ T$ of the unique generic point $\xi $ of $X_ t$. Namely, then $H_ r$ contains the image of $U$ which is an open neighbourhood of $t$. The existence of $U$ follows from More on Morphisms, Lemma 37.16.7. $\square$

Proof. A preliminary remark is that $X$, $S$ are affine schemes and that it suffices to prove $F_ d$ is representable by a monomorphism of finite presentation $Z_ d \to S$ on the category of affine schemes over $S$. (Of course we do not require $Z_ d$ to be affine.) Hence throughout the proof of the lemma we work in the category of affine schemes over $S$.

Let $(Z_ k, Y_ k, i_ k, \pi _ k, \mathcal{G}_ k, \alpha _ k)_{k = 1, \ldots , n}$ be a complete dévissage of $\mathcal{F}/X/S$ over $s$, see Definition 38.5.1. We will use induction on the length $n$ of the dévissage. Recall that $Y_ k \to S$ is smooth with geometrically irreducible fibres, see Definition 38.4.1. Let $d_ k$ be the relative dimension of $Y_ k$ over $S$. Recall that $i_{k, *}\mathcal{G}_ k = \mathop{\mathrm{Coker}}(\alpha _ k)$ and that $i_ k$ is a closed immersion. By the definitions referenced above we have $d_1 = \dim (\text{Supp}(\mathcal{F}_ s))$ and

for $k = 2, \ldots , n$. It follows that $d_1 > d_2 > \ldots > d_ n \geq 0$ because $\alpha _ k$ is an isomorphism in the generic point of $(Y_ k)_ s$.

Note that $i_1$ is a closed immersion and $\mathcal{F} = i_{1, *}\mathcal{G}_1$. Hence for any morphism of schemes $T \to S$ with $T$ affine, we have $\mathcal{F}_ T = i_{1, T, *}\mathcal{G}_{1, T}$ and $i_{1, T}$ is still a closed immersion of schemes over $T$. Thus $\mathcal{F}_ T$ is flat in dimensions $\geq d$ over $T$ if and only if $\mathcal{G}_{1, T}$ is flat in dimensions $\geq d$ over $T$. Because $\pi _1 : Z_1 \to Y_1$ is finite we see in the same manner that $\mathcal{G}_{1, T}$ is flat in dimensions $\geq d$ over $T$ if and only if $\pi _{1, T, *}\mathcal{G}_{1, T}$ is flat in dimensions $\geq d$ over $T$. The same arguments work for “flat in dimensions $\geq d + 1$” and we conclude in particular that $\pi _{1, *}\mathcal{G}_1$ is flat over $S$ in dimensions $\geq d + 1$ by our assumption on $\mathcal{F}$.

Suppose that $d_1 > d$. It follows from the discussion above that in particular $\pi _{1, *}\mathcal{G}_1$ is flat over $S$ at the generic point of $(Y_1)_ s$. By Lemma 38.12.1 we may replace $S$ by an affine neighbourhood of $s$ and assume that $\alpha _1$ is $S$-universally injective. Because $\alpha _1$ is $S$-universally injective, for any morphism $T \to S$ with $T$ affine, we have a short exact sequence

and still the first arrow is $T$-universally injective. Hence the set of points of $(Y_1)_ T$ where $\pi _{1, T, *}\mathcal{G}_{1, T}$ is flat over $T$ is the same as the set of points of $(Y_1)_ T$ where $\mathop{\mathrm{Coker}}(\alpha _1)_ T$ is flat over $S$. In this way the question reduces to the sheaf $\mathop{\mathrm{Coker}}(\alpha _1)$ which has a complete dévissage of length $n - 1$ and we win by induction.

If $d_1 < d$ then $F_ d$ is represented by $S$ and we win.

The last case is the case $d_1 = d$. This case follows from a combination of Lemma 38.27.3 and Lemma 38.27.1. $\square$

Proof. The functor $F_ n$ is a sheaf for the fppf topology by Lemma 38.20.12. Observe that a monomorphism of finite presentation is separated and quasi-finite (Morphisms, Lemma 29.20.15). Hence combining Descent, Lemma 35.39.1, More on Morphisms, Lemma 37.57.1 , and Descent, Lemmas 35.23.31 and 35.23.13 we see that the question is local for the étale topology on $S$.

In particular the situation is local for the Zariski topology on $S$ and we may assume that $S$ is affine. In this case the dimension of the fibres of $f$ is bounded above, hence we see that $F_ n$ is representable for $n$ large enough. Thus we may use descending induction on $n$. Suppose that we know $F_{n + 1}$ is representable by a monomorphism $Z_{n + 1} \to S$ of finite presentation. Consider the base change $X_{n + 1} = Z_{n + 1} \times _ S X$ and the pullback $\mathcal{F}_{n + 1}$ of $\mathcal{F}$ to $X_{n + 1}$. The morphism $Z_{n + 1} \to S$ is quasi-finite as it is a monomorphism of finite presentation, hence Lemma 38.16.4 implies that $\mathcal{F}_{n + 1}$ is pure relative to $Z_{n + 1}$. Since $F_ n$ is a subfunctor of $F_{n + 1}$ we conclude that in order to prove the result for $F_ n$ it suffices to prove the result for the corresponding functor for the situation $\mathcal{F}_{n + 1}/X_{n + 1}/Z_{n + 1}$. In this way we reduce to proving the result for $F_ n$ in case $S_{n + 1} = S$, i.e., we may assume that $\mathcal{F}$ is flat in dimensions $\geq n + 1$ over $S$.

Fix $n$ and assume $\mathcal{F}$ is flat in dimensions $\geq n + 1$ over $S$. To finish the proof we have to show that $F_ n$ is representable by a monomorphism $Z_ n \to S$ of finite presentation. Since the question is local in the étale topology on $S$ it suffices to show that for every $s \in S$ there exists an elementary étale neighbourhood $(S', s') \to (S, s)$ such that the result holds after base change to $S'$. Thus by Lemma 38.5.8 we may assume there exist étale morphisms $h_ j : Y_ j \to X$, $j = 1, \ldots , m$ such that for each $j$ there exists a complete dévissage of $\mathcal{F}_ j/Y_ j/S$ over $s$, where $\mathcal{F}_ j$ is the pullback of $\mathcal{F}$ to $Y_ j$ and such that $X_ s \subset \bigcup h_ j(Y_ j)$. Note that by Lemma 38.27.2 the sheaves $\mathcal{F}_ j$ are still flat over in dimensions $\geq n + 1$ over $S$. Set $W = \bigcup h_ j(Y_ j)$, which is a quasi-compact open of $X$. As $\mathcal{F}$ is pure along $X_ s$ we see that

contains all generalizations of $s$. By More on Morphisms, Lemma 37.25.5 $E$ is a constructible subset of $S$. We have seen that $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \subset E$. By Morphisms, Lemma 29.22.4 we see that $E$ contains an open neighbourhood of $s$. Hence after shrinking $S$ we may assume that $E = S$. It follows from Lemma 38.27.2 that it suffices to prove the lemma for the functor $F_ n$ associated to $X = \coprod Y_ j$ and $\mathcal{F} = \coprod \mathcal{F}_ j$. If $F_{j, n}$ denotes the functor for $Y_ j \to S$ and the sheaf $\mathcal{F}_ i$ we see that $F_ n = \prod F_{j, n}$. Hence it suffices to prove each $F_{j, n}$ is representable by some monomorphism $Z_{j, n} \to S$ of finite presentation, since then

Thus we have reduced the theorem to the special case handled in Lemma 38.27.4. $\square$

Proof. These statements follow immediately from Theorem 38.27.5 applied to $F_0 = F_{flat}$ and the fact that if $f$ is proper then $\mathcal{F}$ is automatically pure over the base, see Lemma 38.17.1. $\square$