The Stacks project

Proof. We may and do view $f : X \to Y$ as a morphism of algebraic space over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Spaces, Definition 65.16.2). Note that a finite morphism is affine and universally closed, see Morphisms of Spaces, Lemma 67.45.7. By Morphisms of Spaces, Lemma 67.9.8 we see that $Y$ is a separated algebraic space. As $f$ is surjective and $X$ is quasi-compact we see that $Y$ is quasi-compact.

By Lemma 70.11.3 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ a$ with each $X_ a \to Y$ finite and of finite presentation. By Lemma 70.5.10 we see that $X_ a$ is affine for $a$ large enough. Hence we may and do assume that $f : X \to Y$ is finite, surjective, and of finite presentation.

By Proposition 70.8.1 we may write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ as a directed limit of algebraic spaces of finite presentation over $\mathbf{Z}$. By Lemma 70.7.1 we can find $0 \in I$ and a morphism $X_0 \to Y_0$ of finite presentation such that $X_ i = X_0 \times _{Y_0} Y_ i$ for $i \geq 0$ and such that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. By Lemma 70.6.7 we see that $X_ i \to Y_ i$ is finite for $i$ large enough. By Lemma 70.6.4 we see that $X_ i \to Y_ i$ is surjective for $i$ large enough. By Lemma 70.5.10 we see that $X_ i$ is affine for $i$ large enough. Hence for $i$ large enough we can apply Cohomology of Spaces, Lemma 69.17.3 to conclude that $Y_ i$ is affine. This implies that $Y$ is affine and we conclude. $\square$

Proof. We may and do view $f : X \to Y$ as a morphism of algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Spaces, Definition 65.16.2). By Morphisms of Spaces, Lemma 67.9.8 we see that $Y$ is a separated algebraic space. Then by Morphisms of Spaces, Lemma 67.20.11 we find that $f$ is affine. Whereupon by Morphisms of Spaces, Lemma 67.45.7 we see that $f$ is integral.

By the preceding paragraph, we may assume $f : X \to Y$ is surjective and integral, $X$ is affine, and $Y$ is separated. Since $f$ is surjective and $X$ is quasi-compact we also deduce that $Y$ is quasi-compact.

Consider the sheaf $\mathcal{A} = f_*\mathcal{O}_ X$. This is a quasi-coherent sheaf of $\mathcal{O}_ Y$-algebras, see Morphisms of Spaces, Lemma 67.11.2. By Lemma 70.9.1 we can write $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$ as a filtered colimit of finite type $\mathcal{O}_ Y$-modules. Let $\mathcal{A}_ i \subset \mathcal{A}$ be the $\mathcal{O}_ Y$-subalgebra generated by $\mathcal{F}_ i$. Since the map of algebras $\mathcal{O}_ Y \to \mathcal{A}$ is integral, we see that each $\mathcal{A}_ i$ is a finite quasi-coherent $\mathcal{O}_ Y$-algebra. Hence

is a finite morphism of algebraic spaces. Here $\underline{\mathop{\mathrm{Spec}}}$ is the construction of Morphisms of Spaces, Lemma 67.20.7. It is clear that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. Hence by Lemma 70.5.10 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to Y$ factors through each $X_ i$ we see that $X_ i \to Y$ is surjective. Hence we conclude that $Y$ is affine by Lemma 70.15.1. $\square$

Proof. Let $U' \subset X_{red}$ be an open affine subscheme. Let $U \subset X$ be the open subspace corresponding to the open $|U'| \subset |X_{red}| = |X|$. Then $U' \to U$ is surjective and integral. Hence $U$ is affine by Proposition 70.15.2. Thus every point is contained in an open subscheme of $X$, i.e., $X$ is a scheme. $\square$

Proof. An integral morphism is representable by definition, hence if $Y$ is a scheme, so is $X$. Conversely, assume that $X$ is a scheme. Let $U \subset X$ be an affine open. An integral morphism is closed and $|f|$ is bijective, hence $|f|(|U|) \subset |Y|$ is open as the complement of $|f|(|X| \setminus |U|)$. Let $V \subset Y$ be the open subspace with $|V| = |f|(|U|)$, see Properties of Spaces, Lemma 66.4.8. Then $U \to V$ is integral and surjective, hence $V$ is an affine scheme by Proposition 70.15.2. This concludes the proof. $\square$

Proof. Assumptions (1) and (2) imply that $B_{red} = B'_{red}$. Set $X' = X \times _ B B'$. Then $X' \to X$ is closed immersion and $X'_{red} = X_{red}$. Let $U \to B$ be an étale morphism with $U$ affine. Then $X' \times _ B U \to X \times _ B U$ is a closed immersion of algebraic spaces inducing an isomorphism on underlying reduced spaces. Since $X' \times _ B U$ is a scheme (as $B' \to B$ and $X' \to B'$ are representable) so is $X \times _ B U$ by Lemma 70.15.3. Hence $X \to B$ is representable too. Thus we reduce to the case of schemes, see Morphisms, Lemma 29.45.7. $\square$