The Stacks project

Proof. By Limits, Proposition 32.5.4 there exists an affine morphism $X \to X_0$ where $X_0$ is a scheme of finite type over $\mathbf{Z}$. Below we produce a morphism $Y_0 \to X_0$ with all the desired properties. Then setting $Y = X \times _{X_0} Y_0$ and $f$ equal to the projection $f : Y \to X$ we conclude. To see this observe that $f$ is of finite presentation (Morphisms, Lemma 29.21.4), $f$ is proper (Morphisms, Lemma 29.41.5), $f$ is completely decomposed (Lemma 37.78.3). Finally, since $Y \to Y_0$ is affine (as the base change of $X \to X_0$) we see that $Y$ has an ample family of invertible modules by Lemma 37.79.2. This reduces us to the case discussed in the next paragraph.

Assume $X$ is of finite type over $\mathbf{Z}$. In particular $\dim (X) < \infty$. We will argue by induction on $\dim (X)$. If $\dim (X) = 0$, then $X$ is affine and has the resolution property. In general, there exists a dense open $U \subset X$ and a $U$-admissible blowing up $X' \to X$ such that $X'$ has an ample family of invertible modules, see Lemma 37.80.2. Since $f : X' \to X$ is an isomorphism over $U$ we see that every point of $U$ lifts to a point of $X'$ with the same residue field. Let $Z = X \setminus U$ with the reduced induced scheme structure. Then $\dim (Z) < \dim (X)$ as $U$ is dense in $X$ (see above). By induction we find a proper, completely decomposed morphism $W \to Z$ such that $W$ has an ample family of invertible modules. Then it follows that $Y = X' \amalg W \to X$ is the desired morphism. $\square$