**Proof.**
For the impatient reader: This proof is a repeat of the proof of Proposition 29.27.1 using Algebra, Lemma 10.118.7 instead of Algebra, Lemma 10.118.3.

Since being flat and being of finite presentation is local on the base, see Lemmas 29.25.2 and 29.21.2, we may work affine locally on $S$. Thus we may assume that $S = \mathop{\mathrm{Spec}}(A)$, where $A$ is a reduced ring (see Properties, Lemma 28.3.2). As $f$ is of finite type, it is quasi-compact, so $X$ is quasi-compact. Hence we can find a finite affine open cover $X = \bigcup _{i = 1, \ldots , n} X_ i$. Write $X_ i = \mathop{\mathrm{Spec}}(B_ i)$. Then $B_ i$ is a finite type $A$-algebra, see Lemma 29.15.2. Moreover there are finite type $B_ i$-modules $M_ i$ such that $\mathcal{F}|_{X_ i}$ is the quasi-coherent sheaf associated to the $B_ i$-module $M_ i$, see Properties, Lemma 28.16.1. Next, for each pair of indices $i, j$ choose an ideal $I_{ij} \subset B_ i$ such that $X_ i \setminus X_ i \cap X_ j = V(I_{ij})$ inside $X_ i = \mathop{\mathrm{Spec}}(B_ i)$. Set $M_{ij} = B_ i/I_{ij}$ and think of it as a $B_ i$-module. Then $V(I_{ij}) = \text{Supp}(M_{ij})$ and $M_{ij}$ is a finite $B_ i$-module.

At this point we apply Algebra, Lemma 10.118.7 the pairs $(A \to B_ i, M_{ij})$ and to the pairs $(A \to B_ i, M_ i)$. Thus we obtain dense opens $U(A \to B_ i, M_{ij}) \subset S$ and dense opens $U(A \to B_ i, M_ i) \subset S$ with notation as in Algebra, Equation (10.118.3.2). Since a finite intersection of dense opens is dense open, we see that

\[ U = \bigcap \nolimits _{i, j} U(A \to B_ i, M_{ij}) \quad \cap \quad \bigcap \nolimits _ i U(A \to B_ i, M_ i) \]

is open and dense in $S$. We claim that $U$ is the desired open.

Pick $u \in U$. By definition of the loci $U(A \to B_ i, M_{ij})$ and $U(A \to B, M_ i)$ there exist $f_{ij}, f_ i \in A$ such that (a) $u \in D(f_ i)$ and $u \in D(f_{ij})$, (b) $A_{f_{ij}} \to B_{i, f_{ij}}$ is flat and of finite presentation and $M_{ij, f_{ij}}$ is flat over $A_{f_{ij}}$ and of finite presentation over $B_{i, f_{ij}}$, and (c) $B_{i, f_ i}$ is flat and of finite presentation over $A_ f$ and $M_{i, f_ i}$ is flat and of finite presentation over $B_{i, f_ i}$. Set $f = (\prod f_ i) (\prod f_{ij})$. Now it suffices to prove that $X \to S$ is flat and of finite presentation over $D(f)$ and that $\mathcal{F}$ restricted to $X_{D(f)}$ is flat over $D(f)$ and of finite presentation over the structure sheaf of $X_{D(f)}$.

Hence we may replace $A$ by $A_ f$, i.e., perform the base change by $\mathop{\mathrm{Spec}}(A_ f) \to S$. After this base change we see that each of $A \to B_ i$ is flat and of finite presentation and that $M_ i$, $M_{ij}$ are flat over $A$ and of finite presentation over $B_ i$. This already proves that $X \to S$ is quasi-compact, locally of finite presentation, flat, and that $\mathcal{F}$ is flat over $S$ and of finite presentation over $\mathcal{O}_ X$, see Lemma 29.21.2 and Properties, Lemma 28.16.2. Since $M_{ij}$ is of finite presentation over $B_ i$ we see that $X_ i \cap X_ j = X_ i \setminus \text{Supp}(M_{ij})$ is a quasi-compact open of $X_ i$, see Algebra, Lemma 10.40.8. Hence we see that $X \to S$ is quasi-separated by Schemes, Lemma 26.21.6. This proves the proposition.
$\square$

## Comments (1)

Comment #7088 by Matthieu Romagny on