The Stacks project

Proof. Since $f$ is surjective and $X$ is quasi-compact we see that $S$ is quasi-compact. Since $X$ is separated and $f$ is surjective and universally closed (Morphisms, Lemma 29.44.7), we see that $S$ is separated (Morphisms, Lemma 29.41.11).

By Lemma 32.9.8 we can write $X = \mathop{\mathrm{lim}}\nolimits _ a X_ a$ with $X_ a \to S$ finite and of finite presentation. By Lemma 32.4.13 we see that $X_ a$ is affine for some $a \in A$. Replacing $X$ by $X_ a$ we may assume that $X \to S$ is surjective, finite, of finite presentation and that $X$ is affine.

By Proposition 32.5.4 we may write $S = \mathop{\mathrm{lim}}\nolimits _{i \in I} S_ i$ as a directed limits of schemes of finite type over $\mathbf{Z}$. By Lemma 32.10.1 we can after shrinking $I$ assume there exist schemes $X_ i \to S_ i$ of finite presentation such that $X_{i'} = X_ i \times _ S S_{i'}$ for $i' \geq i$ and such that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. By Lemma 32.8.3 we may assume that $X_ i \to S_ i$ is finite for all $i \in I$ as well. By Lemma 32.4.13 once again we may assume that $X_ i$ is affine for all $i \in I$. Hence the result follows from the Noetherian case, see Cohomology of Schemes, Lemma 30.13.3. $\square$

Proof. By Morphisms, Lemma 29.41.11 the scheme $S$ is separated. Then by Morphisms, Lemma 29.11.11 we find that $f$ is affine. Whereupon by Morphisms, Lemma 29.44.7 we see that $f$ is integral.

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

By Lemma 32.7.3 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with $X_ i \to S$ finite. By Lemma 32.4.13 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to S$ factors through each $X_ i$ we see that $X_ i \to S$ is surjective. Hence we conclude that $S$ is affine by Lemma 32.11.1. $\square$

Proof. Let $Z_ i \subset X$, $i = 1, \ldots , n$ be affine closed subschemes such that $X = \bigcup Z_ i$ set theoretically. Then $\coprod Z_ i \to X$ is surjective and integral with affine source. Hence $X$ is affine by Proposition 32.11.2. $\square$

Proof. If $\mathcal{L}$ is ample, then $i^*\mathcal{L}$ is ample for example by Morphisms, Lemma 29.37.7. Assume $i^*\mathcal{L}$ is ample. Then $Z$ is quasi-compact (Properties, Definition 28.26.1) and separated (Properties, Lemma 28.26.8). Since $i$ is surjective, we see that $X$ is quasi-compact. Since $i$ is universally closed and surjective, we see that $X$ is separated (Morphisms, Lemma 29.41.11).

By Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a directed limit of finite type schemes over $\mathbf{Z}$ with affine transition morphisms. We can find an $i$ and an invertible sheaf $\mathcal{L}_ i$ on $X_ i$ whose pullback to $X$ is isomorphic to $\mathcal{L}$, see Lemma 32.10.2.

For each $i$ let $Z_ i \subset X_ i$ be the scheme theoretic image of the morphism $Z \to X_ i$. If $\mathop{\mathrm{Spec}}(A_ i) \subset X_ i$ is an affine open subscheme with inverse image of $\mathop{\mathrm{Spec}}(A)$ in $X$ and if $Z \cap \mathop{\mathrm{Spec}}(A)$ is defined by the ideal $I \subset A$, then $Z_ i \cap \mathop{\mathrm{Spec}}(A_ i)$ is defined by the ideal $I_ i \subset A_ i$ which is the inverse image of $I$ in $A_ i$ under the ring map $A_ i \to A$, see Morphisms, Example 29.6.4. Since $\mathop{\mathrm{colim}}\nolimits A_ i/I_ i = A/I$ it follows that $\mathop{\mathrm{lim}}\nolimits Z_ i = Z$. By Lemma 32.4.15 we see that $\mathcal{L}_ i|_{Z_ i}$ is ample for some $i$. Since $Z$ and hence $X$ maps into $Z_ i$ set theoretically, we see that $X_{i'} \to X_ i$ maps into $Z_ i$ set theoretically for some $i' \geq i$, see Lemma 32.4.10. (Observe that since $X_ i$ is Noetherian, every closed subset of $X_ i$ is constructible.) Let $T \subset X_{i'}$ be the scheme theoretic inverse image of $Z_ i$ in $X_{i'}$. Observe that $\mathcal{L}_{i'}|_ T$ is the pullback of $\mathcal{L}_ i|_{Z_ i}$ and hence ample by Morphisms, Lemma 29.37.7 and the fact that $T \to Z_ i$ is an affine morphism. Thus we see that $\mathcal{L}_{i'}$ is ample on $X_{i'}$ by Cohomology of Schemes, Lemma 30.17.5. Pulling back to $X$ (using the same lemma as above) we find that $\mathcal{L}$ is ample. $\square$

Proof. Recall that a scheme is quasi-affine if and only if the structure sheaf is ample, see Properties, Lemma 28.27.1. Hence if $Z$ is quasi-affine, then $\mathcal{O}_ Z$ is ample, hence $\mathcal{O}_ X$ is ample by Lemma 32.11.4, hence $X$ is quasi-affine. A proof of the converse, which can also be seen in an elementary way, is gotten by reading the argument just given backwards. $\square$

Proof. Since $\mathop{\mathrm{Spec}}(A)$ is quasi-compact, we may replace $W$ by a quasi-compact open still containing the image of $\mathop{\mathrm{Spec}}(A) \to X$. Recall that $X$ is quasi-separated and quasi-compact by dint of having an ample invertible sheaf, see Properties, Definition 28.26.1 and Lemma 28.26.7. By Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a limit of a directed system of schemes of finite type over $\mathbf{Z}$ with affine transition morphisms. For some $i$ the ample invertible sheaf $\mathcal{L}$ on $X$ descends to an ample invertible sheaf $\mathcal{L}_ i$ on $X_ i$ and the open $W$ is the inverse image of a quasi-compact open $W_ i \subset X_ i$, see Lemmas 32.4.15, 32.10.3, and 32.4.11. We may replace $X, W, \mathcal{L}$ by $X_ i, W_ i, \mathcal{L}_ i$ and assume $X$ is of finite presentation over $\mathbf{Z}$. Write $A = \mathop{\mathrm{colim}}\nolimits A_ j$ as the colimit of its finite $k$-subalgebras. Then for some $j$ the morphism $\mathop{\mathrm{Spec}}(A) \to X$ factors through a morphism $\mathop{\mathrm{Spec}}(A_ j) \to X$, see Proposition 32.6.1. Since $\mathop{\mathrm{Spec}}(A_ j)$ is finite this reduces the lemma to Properties, Lemma 28.29.6. $\square$