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. $\square$

Proof. By Lemma 32.9.1 above there exists a morphism $Y \to S$ of finite presentation and an immersion $i : X \to Y$ of schemes over $S$. For every point $x \in X$, there exists an affine open $V_ x \subset Y$ such that $i^{-1}(V_ x) \to V_ x$ is a closed immersion. Since $X$ is quasi-compact we can find finitely may affine opens $V_1, \ldots , V_ n \subset Y$ such that $i(X) \subset V_1 \cup \ldots \cup V_ n$ and $i^{-1}(V_ j) \to V_ j$ is a closed immersion. In other words such that $i : X \to X' = V_1 \cup \ldots \cup V_ n$ is a closed immersion of schemes over $S$. Since $S$ is quasi-separated and $Y$ is quasi-separated over $S$ we deduce that $Y$ is quasi-separated, see Schemes, Lemma 26.21.12. Hence the open immersion $X' = V_1 \cup \ldots \cup V_ n \to Y$ is quasi-compact. This implies that $X' \to Y$ is of finite presentation, see Morphisms, Lemma 29.21.6. We conclude since then $X' \to Y \to S$ is a composition of morphisms of finite presentation, and hence of finite presentation (see Morphisms, Lemma 29.21.3). $\square$

Proof. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the quasi-coherent sheaf of ideals defining $X$ as a closed subscheme of $Y$. By Properties, Lemma 28.22.3 we can write $\mathcal{I}$ as a directed colimit $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{I}_ i$ of its quasi-coherent sheaves of ideals of finite type. Let $X_ i \subset Y$ be the closed subscheme defined by $\mathcal{I}_ i$. These form an inverse system of schemes indexed by $I$. The transition morphisms $X_ i \to X_{i'}$ are affine because they are closed immersions. Each $X_ i$ is quasi-compact and quasi-separated since it is a closed subscheme of $Y$ and $Y$ is quasi-compact and quasi-separated by our assumptions. We have $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ as follows directly from the fact that $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{I}_ a$. Each of the morphisms $X_ i \to Y$ is of finite presentation, see Morphisms, Lemma 29.21.7. $\square$

Proof. By Lemma 32.9.3 there is a closed immersion $X \to Y$ with $Y \to S$ of finite presentation. Then $Y$ is quasi-separated by Schemes, Lemma 26.21.12. Since $X$ is quasi-compact, we may assume $Y$ is quasi-compact by replacing $Y$ with a quasi-compact open containing $X$. We see that $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with $X_ i \to Y$ a closed immersion of finite presentation by Lemma 32.9.4. The morphisms $X_ i \to S$ are of finite presentation by Morphisms, Lemma 29.21.3. $\square$

Proof. Apply Lemma 32.9.5 and note that $X_ i \to S$ is separated for large $i$ by Lemma 32.4.17 as we have assumed that $X \to S$ is separated. $\square$

Proof. We may write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as in Lemma 32.9.5. Applying Lemma 32.4.19 we see that $X_ i \to S$ is finite for large enough $i$. $\square$

Proof. We may write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as in Lemma 32.9.5. Applying Lemma 32.4.19 we see that $X_ i \to S$ is finite for large enough $i$. $\square$