The Stacks project

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. $\square$