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 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. $\square$