hcju/stacksqa huggingface dataset:
There is an immediate reduction to the case $S = \Spec(R)$ where $R$ is a Nagata ring by assumption (1). We have to show that the integral closure $A$ of $R$ in $\Gamma(X, \mathcal{O}_X)$ is finite over $R$. Since $f$ is quasi-compact by assumption (2) we can write $X = \bigcup_{i = 1, \ldots, n} U_i$ with each $U_i$ affine. Say $U_i = \Spec(B_i)$. Each $B_i$ is reduced by assumption (5) and has finitely many minimal primes $\mathfrak q_{i1}, \ldots, \mathfrak q_{im_i}$ by assumption (3) and Algebra, Lemma \ref{algebra-lemma-irreducible}. We have $$ \Gamma(X, \mathcal{O}_X) \subset B_1 \times \ldots \times B_n \subset \prod\nolimits_{i = 1, \ldots, n} \prod\nolimits_{j = 1, \ldots, m_i} (B_i)_{\mathfrak q_{ij}} $$ the second inclusion by Algebra, Lemma \ref{algebra-lemma-reduced-ring-sub-product-fields}. We have $\kappa(\mathfrak q_{ij}) = (B_i)_{\mathfrak q_{ij}}$ by Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}. Hence the integral closure $A$ of $R$ in $\Gamma(X, \mathcal{O}_X)$ is contained in the product of the integral closures $A_{ij}$ of $R$ in $\kappa(\mathfrak q_{ij})$. Since $R$ is Noetherian it suffices to show that $A_{ij}$ is a finite $R$-module for each $i, j$. Let $\mathfrak p_{ij} \subset R$ be the image of $\mathfrak q_{ij}$. As $\kappa(\mathfrak p_{ij}) \subset \kappa(\mathfrak q_{ij})$ is a finitely generated field extension by assumption (4), we see that $R \to \kappa(\mathfrak q_{ij})$ is essentially of finite type. Thus $R \to A_{ij}$ is finite by Algebra, Lemma \ref{algebra-lemma-nagata-in-reduced-finite-type-finite-integral-closure}.
This is a special case of Lemma \ref{lemma-nagata-normalization-finite-general}. Namely, (2) holds as the finite type morphism $f$ is quasi-compact by definition and quasi-separated by Lemma \ref{lemma-finite-type-Noetherian-quasi-separated}. Condition (3) holds because $X$ is locally Noetherian by Lemma \ref{lemma-finite-type-noetherian}. Finally, condition (4) holds because a finite type morphism induces finitely generated residue field extensions.
We can replace $X$ by an affine neighbourhood of the image of $x'$. Hence we may assume $X = \Spec(A)$ with $A$ Nagata. By Lemma \ref{lemma-nagata-normalization-finite} the morphism $X' \to X$ is finite. Hence we can write $X' = \Spec(A')$ for a finite $A$-algebra $A'$. By Lemma \ref{lemma-normalization-is-normalization} after replacing $X$ by $X'$ we reduce to the case described in the next paragraph. \medskip\noindent The case $X = X' = \Spec(A)$ with $A$ Noetherian. Let $\mathfrak p \subset A$ be the prime ideal corresponding to our point $x'$. Choose $g \in \mathfrak p$ not contained in any minimal prime of $A$ (use prime avoidance and the fact that $A$ has finitely many minimal primes, see Algebra, Lemmas \ref{algebra-lemma-silly} and \ref{algebra-lemma-Noetherian-irreducible-components}). Set $Z = f^{-1}V(g) \subset Y$; it is a closed subscheme of $Y$. Then $f(Z)$ does not contain any generic point by choice of $g$ and does not contain $x'$ because $x'$ is not in the image of $f$. The closure of $f(Z)$ is the set of specializations of points of $f(Z)$ by Lemma \ref{lemma-reach-points-scheme-theoretic-image}. Thus the closure of $f(Z)$ does not contain $x'$ because the condition $\dim(\mathcal{O}_{X', x'}) = 1$ implies only the generic points of $X = X'$ specialize to $x'$. In other words, after replacing $X$ by an affine open neighbourhood of $x'$ we may assume that $f^{-1}V(g) = \emptyset$. Thus $g$ maps to an invertible global function on $Y$ and we obtain a factorization $$ A \to A_g \to \Gamma(Y, \mathcal{O}_Y) $$ Since $X = X'$ this implies that $A$ is equal to the integral closure of $A$ in $A_g$. By Algebra, Lemma \ref{algebra-lemma-integral-closure-localize} we conclude that $A_\mathfrak p$ is the integral closure of $A_\mathfrak p$ in $A_\mathfrak p[1/g]$. By our choice of $g$, since $\dim(A_\mathfrak p) = 1$ and since $A$ is reduced we see that $A_\mathfrak p[1/g]$ is a finite product of fields (the product of the residue fields of the minimal primes contained in $\mathfrak p$). Hence $A_\mathfrak p$ is normal (Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}) and the proof is complete. Some details omitted.
Let $f : Y \to X$ be the morphism (\ref{equation-generic-points}). We get a factorization $Y \to X_{red} \to X$ of $f$ from Schemes, Lemma \ref{schemes-lemma-map-into-reduction}. By Lemma \ref{lemma-characterize-normalization} we obtain a canonical morphism $X^\nu \to X_{red}$ and that $X^\nu$ is the normalization of $X_{red}$ in $Y$. The lemma follows as $Y \to X_{red}$ is identical to the morphism (\ref{equation-generic-points}) constructed for $X_{red}$.
Minimal primes correspond to irreducible components (Algebra, Lemma \ref{algebra-lemma-irreducible}), hence we have (1) by assumption. Then $(0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_t$ because $A$ is reduced (Algebra, Lemma \ref{algebra-lemma-Zariski-topology}). Then we have $Q(A) = \prod A_{\mathfrak q_i} = \prod \kappa(\mathfrak q_i)$ by Algebra, Lemmas \ref{algebra-lemma-total-ring-fractions-no-embedded-points} and \ref{algebra-lemma-minimal-prime-reduced-ring}. This proves (2). Part (3) follows from Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}, or Lemma \ref{lemma-normalization-in-disjoint-union}. Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism $$ \Spec\left(\prod \kappa(\mathfrak q_i)\right) \longrightarrow \Spec(A) $$ where $f : Y \to X$ is the morphism (\ref{equation-generic-points}).
After replacing $X$ by an affine open neighbourhood of $x$ we may assume that $X$ has finitely many irreducible components and that $x$ is contained in each of them. Then the stalk $(\nu_*\mathcal{O}_{X^\nu})_x$ is the integral closure of $A = \mathcal{O}_{X, x}$ in the product $L$ of the residue fields of the minimal primes of $A$. This follows from the construction of the normalization and Lemma \ref{lemma-integral-closure}. Alternatively, you can use Lemma \ref{lemma-description-normalization} and the fact that normalization commutes with localization (Algebra, Lemma \ref{algebra-lemma-integral-closure-localize}). Since $A_{red}$ has finitely many minimal primes (because these correspond exactly to the generic points of the irreducible components of $X$ passing through $x$) we see that $L$ is the total ring of fractions of $A_{red}$ (Algebra, Lemma \ref{algebra-lemma-total-ring-fractions-no-embedded-points}). Thus our ring is also the integral closure of $A$ in the total ring of fractions of $A_{red}$.
Let $f : Y \to X$ be as in (\ref{equation-generic-points}). The scheme $X^\nu$ is a disjoint union of normal integral schemes because $Y$ is normal and every affine open of $Y$ has finitely many irreducible components, see Lemma \ref{lemma-normal-normalization}. This proves (1). Alternatively one can deduce (1) from Lemmas \ref{lemma-normalization-reduced} and \ref{lemma-description-normalization}. \medskip\noindent The morphism $\nu$ is integral by Lemma \ref{lemma-characterize-normalization}. By Lemma \ref{lemma-normal-normalization} the morphism $Y \to X^\nu$ induces a bijection on irreducible components, and by construction of $Y$ this implies that $X^\nu \to X$ induces a bijection on irreducible components. By construction $f : Y \to X$ is dominant, hence also $\nu$ is dominant. Since an integral morphism is closed (Lemma \ref{lemma-integral-universally-closed}) this implies that $\nu$ is surjective. This proves (2). \medskip\noindent Suppose that $\alpha : X' \to X$ is as in (3). It is clear that $X'$ satisfies the assumptions under which the normalization is defined. Let $f' : Y' \to X'$ be the morphism (\ref{equation-generic-points}) constructed starting with $X'$. As $\alpha$ is birational it is clear that $Y' = Y$ and $f = \alpha \circ f'$. Hence the factorization $X^\nu \to X' \to X$ exists and $X^\nu \to X'$ is the normalization of $X'$ by Lemma \ref{lemma-characterize-normalization}. This proves (3). \medskip\noindent Let $g : Z \to X$ be a morphism whose domain is a normal scheme and such that every irreducible component dominates an irreducible component of $X$. By Lemma \ref{lemma-normalization-reduced} we have $X^\nu = X_{red}^\nu$ and by Schemes, Lemma \ref{schemes-lemma-map-into-reduction} $Z \to X$ factors through $X_{red}$. Hence we may replace $X$ by $X_{red}$ and assume $X$ is reduced. Moreover, as the factorization is unique it suffices to construct it locally on $Z$. Let $W \subset Z$ and $U \subset X$ be affine opens such that $g(W) \subset U$. Write $U = \Spec(A)$ and $W = \Spec(B)$, with $g|_W$ given by $\varphi : A \to B$. We will use the results of Lemma \ref{lemma-description-normalization} freely. Let $\mathfrak p_1, \ldots, \mathfrak p_t$ be the minimal primes of $A$. As $Z$ is normal, we see that $B$ is a normal ring, in particular reduced. Moreover, by assumption any minimal prime $\mathfrak q \subset B$ we have that $\varphi^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence if $x \in A$ is a nonzerodivisor, i.e., $x \not \in \bigcup \mathfrak p_i$, then $\varphi(x)$ is a nonzerodivisor in $B$. Thus we obtain a canonical ring map $Q(A) \to Q(B)$. As $B$ is normal it is equal to its integral closure in $Q(B)$ (see Algebra, Lemma \ref{algebra-lemma-normal-ring-integrally-closed}). Hence we see that the integral closure $A' \subset Q(A)$ of $A$ maps into $B$ via the canonical map $Q(A) \to Q(B)$. Since $\nu^{-1}(U) = \Spec(A')$ this gives the canonical factorization $W \to \nu^{-1}(U) \to U$ of $\nu|_W$. We omit the verification that it is unique.
We may assume $X$ is reduced, see Lemma \ref{lemma-normalization-reduced}. Then the lemma follows either from the local description in Lemma \ref{lemma-description-normalization} or from Lemma \ref{lemma-normalization-normal} part (3) because $\coprod Z_i \to X$ is integral and birational (as $X$ is reduced and has locally finitely many irreducible components).
The normalization induces a bijection of irreducible components by Lemma \ref{lemma-normalization-normal}. Let $\eta \in X$ be a generic point of an irreducible component of $X$ and let $\eta^\nu \in X^\nu$ be the generic point of the corresponding irreducible component of $X^\nu$. Then $\eta^\nu \mapsto \eta$ and to finish the proof we have to show that $\mathcal{O}_{X, \eta} \to \mathcal{O}_{X^\nu, \eta^\nu}$ is an isomorphism, see Definition \ref{definition-birational}. Because $X$ and $X^\nu$ are reduced, we see that both local rings are equal to their residue fields (Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}). On the other hand, by the construction of the normalization as the normalization of $X$ in $Y = \coprod \Spec(\kappa(\eta))$ we see that we have $\kappa(\eta) \subset \kappa(\eta^\nu) \subset \kappa(\eta)$ and the proof is complete.
Note that a Nagata scheme is locally Noetherian, thus Definition \ref{definition-normalization} does apply. The lemma is now a special case of Lemma \ref{lemma-nagata-normalization-finite-general} but we can also prove it directly as follows. Write $X^\nu \to X$ as the composition $X^\nu \to X_{red} \to X$. As $X_{red} \to X$ is a closed immersion it is finite. Hence it suffices to prove the lemma for a reduced Nagata scheme (by Lemma \ref{lemma-composition-finite}). Let $\Spec(A) = U \subset X$ be an affine open. By Lemma \ref{lemma-description-normalization} we have $\nu^{-1}(U) = \Spec(\prod A_i')$ where $A_i'$ is the integral closure of $A/\mathfrak q_i$ in its fraction field. As $A$ is a Nagata ring (see Properties, Lemma \ref{properties-lemma-locally-nagata}) each of the ring extensions $A/\mathfrak q_i \subset A'_i$ are finite. Hence $A \to \prod A'_i$ is a finite ring map and we win.
Suppose $f$ is quasi-finite at $x$. Let $x \in U = \Spec(A) \subset X$, $V = \Spec(R) \subset S$ be affine opens as in Definition \ref{definition-quasi-finite}. By either Theorem \ref{theorem-main-theorem} above or Algebra, Lemma \ref{algebra-lemma-quasi-finite-open}, the set of primes $\mathfrak q$ at which $R \to A$ is quasi-finite is open in $\Spec(A)$. Since these all correspond to points of $X$ where $f$ is quasi-finite we get the first statement. The second statement is obvious.
As in the proof of Lemma \ref{lemma-image-nowhere-dense-finite} this reduces immediately to the case where the base $X$ is affine. In this case $Y = \bigcup_{i = 1, \ldots, n} Y_i$ is a finite union of affine opens (as $f$ is quasi-compact). Since each $T \cap Y_i$ is nowhere dense, and since a finite union of nowhere dense sets is nowhere dense (see Topology, Lemma \ref{topology-lemma-nowhere-dense}), it suffices to prove that the image $f(T \cap Y_i)$ is nowhere dense in $X$. This reduces us to the case where both $X$ and $Y$ are affine. At this point we apply Lemma \ref{lemma-quasi-finite-affine} above to get a diagram $$ \xymatrix{ Y \ar[rd]_f \ar[rr]_j & & Z \ar[ld]^\pi \\ & X & } $$ with $Z$ affine, $\pi$ finite and $j$ an open immersion. Set $\overline{T} = \overline{j(T)} \subset Z$. By Topology, Lemma \ref{topology-lemma-image-nowhere-dense-open} we see $\overline{T}$ is nowhere dense in $Z$. Since $f(T) \subset \pi(\overline{T})$ the lemma follows from the corresponding result in the finite case, see Lemma \ref{lemma-image-nowhere-dense-finite}.
The implication (2) $\Rightarrow$ (1) is trivial. The other implication holds because if the image of $\Spec(k) \to Y$ is $y$, then $X_k = \Spec(k) \times_{\Spec(\kappa(y))} X_y$. By definition the fibres of $f$ being universally bounded means that some $n$ exists. Finally, suppose that $X_k = \Spec(A)$. Then $\dim_k A = n$. Hence $A$ is Artinian, all prime ideals are maximal ideals $\mathfrak m_i$, and $A$ is the product of the localizations at these maximal ideals. See Algebra, Lemmas \ref{algebra-lemma-finite-dimensional-algebra} and \ref{algebra-lemma-artinian-finite-length}. Then $\mathfrak m_i$ corresponds to $x_i$, we have $A_{\mathfrak m_i} = \mathcal{O}_{X_k, x_i}$ and hence there is a surjection $A \to  \bigoplus \kappa(\mathfrak m_i) = \bigoplus \kappa(x_i)$ which implies the inequality in the statement of the lemma by linear algebra.
Let $f : X \to Y$ and $g : Y \to Z$ have universally bounded fibres. Say that $\deg(X_y/\kappa(y)) \leq n$ for all $y \in Y$, and that $\deg(Y_z/\kappa(z)) \leq m$ for all $z \in Z$. Let $z \in Z$ be a point. By assumption the scheme $Y_z$ is finite over $\Spec(\kappa(z))$. In particular, the underlying topological space of $Y_z$ is a finite discrete set. The fibres of the morphism $f_z : X_z \to Y_z$ are the fibres of $f$ at the corresponding points of $Y$, which are finite discrete sets by the reasoning above. Hence we conclude that the underlying topological space of $X_z$ is a finite discrete set as well. Thus $X_z$ is an affine scheme (this is a nice exercise; it also follows for example from Properties, Lemma \ref{properties-lemma-maximal-points-affine} applied to the set of all points of $X_z$). Write $X_z = \Spec(A)$, $Y_z = \Spec(B)$, and $k = \kappa(z)$. Then $k \to B \to A$ and we know that (a) $\dim_k(B) \leq m$, and (b) for every maximal ideal $\mathfrak m \subset B$ we have $\dim_{\kappa(\mathfrak m)}(A/\mathfrak mA) \leq n$. We claim this implies that $\dim_k(A) \leq nm$. Note that $B$ is the product of its localizations $B_{\mathfrak m}$, for example because $Y_z$ is a disjoint union of $1$-point schemes, or by Algebra, Lemmas \ref{algebra-lemma-finite-dimensional-algebra} and \ref{algebra-lemma-artinian-finite-length}. So we see that $\dim_k(B) = \sum_{\mathfrak m} \dim_k(B_{\mathfrak m})$ and $\dim_k(A) = \sum_{\mathfrak m} \dim_k(A_{\mathfrak m})$ where in both cases $\mathfrak m$ runs over the maximal ideals of $B$ (not of $A$). By the above, and Nakayama's Lemma (Algebra, Lemma \ref{algebra-lemma-NAK}) we see that each $A_{\mathfrak m}$ is a quotient of $B_{\mathfrak m}^{\oplus n}$ as a $B_{\mathfrak m}$-module. Hence $\dim_k(A_{\mathfrak m}) \leq n \dim_k(B_{\mathfrak m})$. Putting everything together we see that $$ \dim_k(A) = \sum\nolimits_{\mathfrak m} \dim_ta	(A_{\mathfrak m}) \leq \sum\nolimits_{\mathfrak m} n \dim_k(B_{\mathfrak m}) = n \dim_k(B) \leq nm $$ as desired.
Since $X$ is quasi-compact, there exists a finite affine open covering $X = \bigcup_{i = 1, \ldots, n} U_i$ and affine opens $V_i \subset Y$, $i = 1, \ldots, n$ such that $f(U_i) \subset V_i$. Because of the local nature of ``local quasi-finiteness'' (see Lemma \ref{lemma-quasi-finite-at-point-characterize} part (4)) we see that the morphisms $f|_{U_i} : U_i \to V_i$ are locally quasi-finite morphisms of affines, hence quasi-finite, see Lemma \ref{lemma-quasi-finite-locally-quasi-compact}. For $y \in Y$ it is clear that $X_y = \bigcup_{y \in V_i} (U_i)_y$ is an open covering. Hence it suffices to prove the lemma for a quasi-finite morphism of affines (namely, if $n_i$ works for the morphism $f|_{U_i} : U_i \to V_i$, then $\sum n_i$ works for $f$). \medskip\noindent Assume $f : X \to Y$ is a quasi-finite morphism of affines. By Lemma \ref{lemma-quasi-finite-affine} we can find a diagram $$ \xymatrix{ X \ar[rd]_f \ar[rr]_j & & Z \ar[ld]^\pi \\ & Y & } $$ with $Z$ affine, $\pi$ finite and $j$ an open immersion. Since $j$ has universally bounded fibres (Lemma \ref{lemma-immersion-universally-bounded}) this reduces us to showing that $\pi$ has universally bounded fibres (Lemma \ref{lemma-composition-universally-bounded}). \medskip\noindent This reduces us to a morphism of the form $\Spec(B) \to \Spec(A)$ where $A \to B$ is finite. Say $B$ is generated by $x_1, \ldots, x_n$ over $A$ and say $P_i(T) \in A[T]$ is a monic polynomial of degree $d_i$ such that $P_i(x_i) = 0$ in $B$ (a finite ring extension is integral, see Algebra, Lemma \ref{algebra-lemma-finite-is-integral}). With these notations it is clear that $$ \bigoplus\nolimits_{0 \leq e_i < d_i, i = 1, \ldots n} A \longrightarrow B, \quad (a_{(e_1, \ldots, e_n)}) \longmapsto \sum a_{(e_1, \ldots, e_n)} x_1^{e_1} \ldots x_n^{e_n} $$ is a surjective $A$-module map. Thus for any prime $\mathfrak p \subset A$ this induces a surjective map $\kappa(\mathfrak p)$-vector spaces $$ \kappa(\mathfrak p)^{\oplus d_1 \ldots d_n} \longrightarrow B \otimes_A \kappa(\mathfrak p) $$ In other words, the integer $d_1 \ldots d_n$ works in the definition of a morphism with universally bounded fibres.
Assume $g$ has universally bounded fibres, and $f$ is surjective and flat. Say the degree of the fibres of $g$ is bounded by $n \in \mathbf{N}$. We claim $n$ also works for $h$. Let $z \in Z$. Consider the morphism of schemes $X_z \to Y_z$. It is flat and surjective. By assumption $X_z$ is a finite scheme over $\kappa(z)$, in particular it is the spectrum of an Artinian ring (by Algebra, Lemma \ref{algebra-lemma-finite-dimensional-algebra}). By Lemma \ref{lemma-Artinian-affine} the morphism $X_z \to Y_z$ is affine in particular quasi-compact. It follows from Lemma \ref{lemma-fpqc-quotient-topology} that $Y_z$ is a finite discrete as this holds for $X_z$. Hence $Y_z$ is an affine scheme (this is a nice exercise; it also follows for example from Properties, Lemma \ref{properties-lemma-maximal-points-affine} applied to the set of all points of $Y_z$). Write $Y_z = \Spec(B)$ and $X_z = \Spec(A)$. Then $A$ is faithfully flat over $B$, so $B \subset A$. Hence $\dim_k(B) \leq \dim_k(A) \leq n$ as desired.
As $S$ is integral it is irreducible (see Properties, Lemma \ref{properties-lemma-characterize-integral}) and any nonempty open is dense. Hence we may replace $S$ by an affine open of $S$ and assume that $S = \Spec(A)$ is affine. As $S$ is integral we see that $A$ is a domain. 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 $X_i = \Spec(B_i)$. Then $B_i$ is a finite type $A$-algebra, see Lemma \ref{lemma-locally-finite-type-characterize}. 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 \ref{properties-lemma-finite-type-module}. 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 $X_i = \Spec(B_i)$. 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. \medskip\noindent At this point we apply Algebra, Lemma \ref{algebra-lemma-generic-flatness} the pairs $(A \to B_i, M_{ij})$ and to the pairs $(A \to B_i, M_i)$. Thus we obtain nonzero $f_{ij}, f_i \in A$ such that (a) $A_{f_{ij}} \to B_{i, f_{ij}}$ 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 (b) $B_{i, f_i}$ 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 $f = (\prod f_i) (\prod f_{ij})$. We claim that taking $U = D(f)$ works. \medskip\noindent To prove our claim we may replace $A$ by $A_f$, i.e., perform the base change by $U = \Spec(A_f) \to S$. 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 \ref{lemma-locally-finite-presentation-characterize} and Properties, Lemma \ref{properties-lemma-finite-presentation-module}. 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 \ref{algebra-lemma-support-finite-presentation-constructible}. Hence we see that $X \to S$ is quasi-separated by Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated}. This proves the proposition.