The Stacks project

Proof. By Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with each $X_ i$ of finite type over $\mathbf{Z}$ and with transition morphisms $f_{ii'} : X_ i \to X_{i'}$ affine. Consider the commutative diagram

Note that $X_ i$ is of finite presentation over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Morphisms, Lemma 29.21.9. Hence the base change $X_{i, S} \to S$ is of finite presentation by Morphisms, Lemma 29.21.4. Thus it suffices to show that the arrow $X \to X_{i, S}$ is an immersion for $i$ sufficiently large.

To do this we choose a finite affine open covering $X = V_1 \cup \ldots \cup V_ n$ such that $f$ maps each $V_ j$ into an affine open $U_ j \subset S$. Let $h_{j, a} \in \mathcal{O}_ X(V_ j)$ be a finite set of elements which generate $\mathcal{O}_ X(V_ j)$ as an $\mathcal{O}_ S(U_ j)$-algebra, see Morphisms, Lemma 29.15.2. By Lemmas 32.4.11 and 32.4.13 (after possibly shrinking $I$) we may assume that there exist affine open coverings $X_ i = V_{1, i} \cup \ldots \cup V_{n, i}$ compatible with transition maps such that $V_ j = \mathop{\mathrm{lim}}\nolimits _ i V_{j, i}$. By Lemma 32.4.7 we can choose $i$ so large that each $h_{j, a}$ comes from an element $h_{j, a, i} \in \mathcal{O}_{X_ i}(V_{j, i})$. Thus the arrow in

is a closed immersion. Since $\bigcup (V_{j, i})_{U_ j}$ forms an open of $X_{i, S}$ and since the inverse image of $(V_{j, i})_{U_ j}$ in $X$ is $V_ j$ it follows that $X \to X_{i, S}$ is an immersion. $\square$