The Stacks project

Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $S$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $u_ i = u_{T_ i}$. Note that $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$, see Topologies, Lemma 34.9.8. In particular, for every $x \in X_ T$ there exists an $i \in I$ and an $x_ i \in X_ i$ mapping to $x$. Since $\mathcal{O}_{X_ T, x} \to \mathcal{O}_{X_ i, x_ i}$ is flat, hence faithfully flat (see Algebra, Lemma 10.39.17) we conclude that $(u_ i)_{x_ i}$ is injective, surjective, bijective, or zero if and only if $(u_ T)_ x$ is injective, surjective, bijective, or zero. Whence part (1) of the lemma.

Proof of (2). Assume $f$ quasi-compact and $\mathcal{G}$ of finite type. Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be a directed limit of affine $S$-schemes and assume that $u_ T$ is surjective. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $u_ i = u_{T_ i} : \mathcal{F}_ i = \mathcal{F}_{T_ i} \to \mathcal{G}_ i = \mathcal{G}_{T_ i}$. To prove part (2) we have to show that $u_ i$ is surjective for some $i$. Pick $i_0 \in I$ and replace $I$ by $\{ i \mid i \geq i_0\} $. Since $f$ is quasi-compact the scheme $X_{i_0}$ is quasi-compact. Hence we may choose affine opens $W_1, \ldots , W_ m \subset X$ and an affine open covering $X_{i_0} = U_{1, i_0} \cup \ldots \cup U_{m, i_0}$ such that $U_{j, i_0}$ maps into $W_ j$ under the projection morphism $X_{i_0} \to X$. For any $i \in I$ let $U_{j, i}$ be the inverse image of $U_{j, i_0}$. Setting $U_ j = \mathop{\mathrm{lim}}\nolimits _ i U_{j, i}$ we see that $X_ T = U_1 \cup \ldots \cup U_ m$ is an affine open covering of $X_ T$. Now it suffices to show, for a given $j \in \{ 1, \ldots , m\} $ that $u_ i|_{U_{j, i}}$ is surjective for some $i = i(j) \in I$. Using Properties, Lemma 28.16.1 this translates into the following algebra problem: Let $A$ be a ring and let $u : M \to N$ be an $A$-module map. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras. If $N$ is a finite $A$-module and if $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is surjective, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is surjective. This is Algebra, Lemma 10.127.5 part (2).

Proof of (3). Exactly the same arguments as given in the proof of (2) reduces this to the following algebra problem: Let $A$ be a ring and let $u : M \to N$ be an $A$-module map. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras. If $M$ is a finite $A$-module and if $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is zero, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is zero. This is Algebra, Lemma 10.127.5 part (1).

Proof of (4). Assume $f$ quasi-compact and $\mathcal{F}, \mathcal{G}$ of finite presentation. Arguing in exactly the same manner as in the previous paragraph (using in addition also Properties, Lemma 28.16.2) part (3) translates into the following algebra statement: Let $A$ be a ring and let $u : M \to N$ be an $A$-module map. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras. Assume $M$ is a finite $A$-module, $N$ is a finitely presented $A$-module, and $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is an isomorphism. Then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is an isomorphism. This is Algebra, Lemma 10.127.5 part (3). $\square$

Proof. Part (1) is a special case of More on Algebra, Lemma 15.19.3. Part (2) is a special case of More on Algebra, Lemma 15.19.4. $\square$

Proof. Let $A'$ be an object of $\mathcal{C}$. Set $C' = C \otimes _ A A'$ and $N' = N \otimes _ A A'$ similarly to the definitions of $B'$, $M'$ in Situation 38.20.3. Note that $M' \cong N'$ as $B'$-modules. The assumption that $B \to C$ is finite has two consequences: (a) $\mathfrak m_ C = \sqrt{\mathfrak m_ B C}$ and (b) $B' \to C'$ is finite. Consequence (a) implies that

Suppose $\mathfrak q \subset V(\mathfrak m_{A'}B' + \mathfrak m_ B B')$. Then $M'_{\mathfrak q}$ is flat over $A'$ if and only if the $C'_{\mathfrak q}$-module $N'_{\mathfrak q}$ is flat over $A'$ (because these are isomorphic as $A'$-modules) if and only if for every maximal ideal $\mathfrak r$ of $C'_{\mathfrak q}$ the module $N'_{\mathfrak r}$ is flat over $A'$ (see Algebra, Lemma 10.39.18). As $B'_{\mathfrak q} \to C'_{\mathfrak q}$ is finite by (b), the maximal ideals of $C'_{\mathfrak q}$ correspond exactly to the primes of $C'$ lying over $\mathfrak q$ (see Algebra, Lemma 10.36.22) and these primes are all contained in $V(\mathfrak m_{A'}C' + \mathfrak m_ C C')$ by the displayed equation above. Thus the result of the lemma holds. $\square$

Proof. Let $A'$ be an object of $\mathcal{C}$. Set $C' = C \otimes _ A A'$ and $N' = N \otimes _ A A' = M' \otimes _{B'} C'$ similarly to the definitions of $B'$, $M'$ in Situation 38.20.3. Note that

and similarly for $V(\mathfrak m_{A'}C' + \mathfrak m_ C C')$. The ring map

is faithfully flat, hence $V(\mathfrak m_{A'}C' + \mathfrak m_ C C') \to V(\mathfrak m_{A'}B' + \mathfrak m_ B B')$ is surjective. Finally, if $\mathfrak r \in V(\mathfrak m_{A'}C' + \mathfrak m_ C C')$ maps to $\mathfrak q \in V(\mathfrak m_{A'}B' + \mathfrak m_ B B')$, then $M'_{\mathfrak q}$ is flat over $A'$ if and only if $N'_{\mathfrak r}$ is flat over $A'$ because $B' \to C'$ is flat, see Algebra, Lemma 10.39.9. The lemma follows formally from these remarks. $\square$

Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc¹ covering of schemes over $S$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and denote $\mathcal{F}_ i$ the pullback of $\mathcal{F}$ to $X_ i$. Assume that $\mathcal{F}_ i$ satisfies (38.20.7.1) for all $i$. Pick $t \in T$ and let $\xi _ t \in X_ T$ denote the generic point of $X_ t$. We have to show that $\mathcal{F}$ is free in a neighbourhood of $\xi _ t$. For some $i \in I$ we can find a $t_ i \in T_ i$ mapping to $t$. Let $\xi _ i \in X_ i$ denote the generic point of $X_{t_ i}$, so that $\xi _ i$ maps to $\xi _ t$. The fact that $\mathcal{F}_ i$ is free of rank $p$ in a neighbourhood of $\xi _ i$ implies that $(\mathcal{F}_ i)_{x_ i} \cong \mathcal{O}_{X_ i, x_ i}^{\oplus p}$ which implies that $\mathcal{F}_{T, \xi _ t} \cong \mathcal{O}_{X_ T, \xi _ t}^{\oplus p}$ as $\mathcal{O}_{X_ T, \xi _ t} \to \mathcal{O}_{X_ i, x_ i}$ is flat, see for example Algebra, Lemma 10.78.6. Thus there exists an affine neighbourhood $U$ of $\xi _ t$ in $X_ T$ and a surjection $\mathcal{O}_ U^{\oplus p} \to \mathcal{F}_ U = \mathcal{F}_ T|_ U$, see Modules, Lemma 17.9.4. After shrinking $T$ we may assume that $U \to T$ is surjective. Hence $U \to T$ is a smooth morphism of affines with geometrically irreducible fibres. Moreover, for every $t' \in T$ we see that the induced map

is an isomorphism (since by the same argument as before the module on the right is free of rank $p$). It follows from Lemma 38.10.1 that

is injective for every $t' \in T$. Hence we see the surjection $\alpha $ is an isomorphism. This finishes the proof of (1).

Assume that $\mathcal{F}$ is of finite presentation. Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be a directed limit of affine $S$-schemes and assume that $\mathcal{F}_ T$ satisfies (38.20.7.1). Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and denote $\mathcal{F}_ i$ the pullback of $\mathcal{F}$ to $X_ i$. Let $U \subset X_ T$ denote the open subscheme of points where $\mathcal{F}_ T$ is flat over $T$, see More on Morphisms, Theorem 37.15.1. By assumption every generic point of every fibre is a point of $U$, i.e., $U \to T$ is a smooth surjective morphism with geometrically irreducible fibres. We may shrink $U$ a bit and assume that $U$ is quasi-compact. Using Limits, Lemma 32.4.11 we can find an $i \in I$ and a quasi-compact open $U_ i \subset X_ i$ whose inverse image in $X_ T$ is $U$. After increasing $i$ we may assume that $\mathcal{F}_ i|_{U_ i}$ is flat over $T_ i$, see Limits, Lemma 32.10.4. In particular, $\mathcal{F}_ i|_{U_ i}$ is finite locally free hence defines a locally constant rank function $\rho : U_ i \to \{ 0, 1, 2, \ldots \} $. Let $(U_ i)_ p \subset U_ i$ denote the open and closed subset where $\rho $ has value $p$. Let $V_ i \subset T_ i$ be the image of $(U_ i)_ p$; note that $V_ i$ is open and quasi-compact. By assumption the image of $T \to T_ i$ is contained in $V_ i$. Hence there exists an $i' \geq i$ such that $T_{i'} \to T_ i$ factors through $V_ i$ by Limits, Lemma 32.4.11. Then $\mathcal{F}_{i'}$ satisfies (38.20.7.1) as desired. Some details omitted. $\square$

Proof. Let $x \in X$ be a point over $s \in S$. Then the dimension of the closure of $\{ x\} $ in $X_ s$ is $\text{trdeg}_{\kappa (s)}(\kappa (x))$ by Varieties, Lemma 33.20.3. Conversely, if $Z \subset X_ s$ is a closed subset of dimension $d$, then there exists a point $x \in Z$ with $\text{trdeg}_{\kappa (s)}(\kappa (x)) = d$ (same reference). Therefore the equivalence of (1) and (2) holds (even fibre by fibre). The statement on base change follows from Morphisms, Lemmas 29.25.7 and 29.28.3. $\square$

Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $S$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and denote $\mathcal{F}_ i$ the pullback of $\mathcal{F}$ to $X_ i$. Assume that $\mathcal{F}_ i$ is flat over $T_ i$ in dimensions $\geq n$ for all $i$. Let $t \in T$. Choose an index $i$ and a point $t_ i \in T_ i$ mapping to $t$. Consider the cartesian diagram

As the lower horizontal morphism is flat we see from More on Morphisms, Lemma 37.15.2 that the set $Z_ i \subset X_{t_ i}$ where $\mathcal{F}_ i$ is not flat over $T_ i$ and the set $Z \subset X_ t$ where $\mathcal{F}_ T$ is not flat over $T$ are related by the rule $Z_ i = Z_{\kappa (t_ i)}$. Hence we see that $\mathcal{F}_ T$ is flat over $T$ in dimensions $\geq n$ by Morphisms, Lemma 29.28.3.

Assume that $f$ is quasi-compact and locally of finite presentation and that $\mathcal{F}$ is of finite presentation. In this paragraph we first reduce the proof of (2) to the case where $f$ is of finite presentation. Let $T = \mathop{\mathrm{lim}}\nolimits _{i \in I} T_ i$ be a directed limit of affine $S$-schemes and assume that $\mathcal{F}_ T$ is flat in dimensions $\geq n$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and denote $\mathcal{F}_ i$ the pullback of $\mathcal{F}$ to $X_ i$. We have to show that $\mathcal{F}_ i$ is flat in dimensions $\geq n$ for some $i$. Pick $i_0 \in I$ and replace $I$ by $\{ i \mid i \geq i_0\} $. Since $T_{i_0}$ is affine (hence quasi-compact) there exist finitely many affine opens $W_ j \subset S$, $j = 1, \ldots , m$ and an affine open overing $T_{i_0} = \bigcup _{j = 1, \ldots , m} V_{j, i_0}$ such that $T_{i_0} \to S$ maps $V_{j, i_0}$ into $W_ j$. For $i \geq i_0$ denote $V_{j, i}$ the inverse image of $V_{j, i_0}$ in $T_ i$. If we can show, for each $j$, that there exists an $i$ such that $\mathcal{F}_{V_{j, i_0}}$ is flat in dimensions $\geq n$, then we win. In this way we reduce to the case that $S$ is affine. In this case $X$ is quasi-compact and we can choose a finite affine open covering $X = W_1 \cup \ldots \cup W_ m$. In this case the result for $(X \to S, \mathcal{F})$ is equivalent to the result for $(\coprod W_ j, \coprod \mathcal{F}|_{W_ j})$. Hence we may assume that $f$ is of finite presentation.

Assume $f$ is of finite presentation and $\mathcal{F}$ is of finite presentation. Let $U \subset X_ T$ denote the open subscheme of points where $\mathcal{F}_ T$ is flat over $T$, see More on Morphisms, Theorem 37.15.1. By assumption the dimension of every fibre of $Z = X_ T \setminus U$ over $T$ has dimension $< n$. By Limits, Lemma 32.18.5 we can find a closed subscheme $Z \subset Z' \subset X_ T$ such that $\dim (Z'_ t) < n$ for all $t \in T$ and such that $Z' \to X_ T$ is of finite presentation. By Limits, Lemmas 32.10.1 and 32.8.5 there exists an $i \in I$ and a closed subscheme $Z'_ i \subset X_ i$ of finite presentation whose base change to $T$ is $Z'$. By Limits, Lemma 32.18.1 we may assume all fibres of $Z'_ i \to T_ i$ have dimension $< n$. By Limits, Lemma 32.10.4 we may assume that $\mathcal{F}_ i|_{X_ i \setminus T'_ i}$ is flat over $T_ i$. This implies that $\mathcal{F}_ i$ is flat in dimensions $\geq n$; here we use that $Z' \to X_ T$ is of finite presentation, and hence the complement $X_ T \setminus Z'$ is quasi-compact! Thus part (2) is proved and the proof of the lemma is complete. $\square$

Proof. Part (1) follows from the following statement: If $T' \to T$ is a surjective flat morphism of schemes over $S$, then $\mathcal{F}_{T'}$ is flat over $T'$ if and only if $\mathcal{F}_ T$ is flat over $T$, see More on Morphisms, Lemma 37.15.2. Part (2) follows from Limits, Lemma 32.10.4 after reducing to the case where $X$ and $S$ are affine (compare with the proof of Lemma 38.20.12). $\square$