The Stacks project

Proof. Given $\mathcal{F}/X/S$ and $U \subset S$ with hypotheses as in the lemma, denote $P$ the property “$\mathcal{F}$ is flat over $S$ at all generic points of fibres”. It is clear that $P$ is preserved under strict transform, see Divisors, Lemma 31.33.3 and Morphisms, Lemma 29.25.7. It is also clear that $P$ is local on $S$. Hence any and all observations of Remark 38.30.1 apply to the problem posed by the lemma.

Consider the function $r : U \to \mathbf{Z}_{\geq 0}$ which assigns to $u \in U$ the integer

where $\xi _ u$ is the generic point of the fibre $X_ u$. By More on Morphisms, Lemma 37.16.7 and the fact that the image of an open in $X_ S$ in $S$ is open, we see that $r(u)$ is locally constant. Accordingly $U = U_0 \amalg U_1 \amalg \ldots \amalg U_ c$ is a finite disjoint union of open and closed subschemes where $r$ is constant with value $i$ on $U_ i$. By Divisors, Lemma 31.34.5 we can find a $U$-admissible blowup to decompose $S$ into the disjoint union of two schemes, the first containing $U_0$ and the second $U_1 \cup \ldots \cup U_ c$. Repeating this $c - 1$ more times we may assume that $S$ is a disjoint union $S = S_0 \amalg S_1 \amalg \ldots \amalg S_ c$ with $U_ i \subset S_ i$. Thus we may assume the function $r$ defined above is constant, say with value $r$.

By Remark 38.30.1 we see that we may assume that we have an effective Cartier divisor $D \subset S$ whose support is $S \setminus U$. Another application of Remark 38.30.1 combined with Divisors, Lemma 31.13.2 tells us we may assume that $S = \mathop{\mathrm{Spec}}(R)$ and $D = \mathop{\mathrm{Spec}}(R/(f))$ for some nonzerodivisor $f \in R$. This case is handled by Lemma 38.30.2. $\square$