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. $\square$