Lemma 29.43.12. Let $f : X \to S$ be a quasi-projective morphism with $S$ quasi-compact and quasi-separated. Then $f$ factors as $X \to X' \to S$ where $X \to X'$ is an open immersion and $X' \to S$ is projective.

**Proof.**
Let $\mathcal{L}$ be $f$-ample. Since $f$ is of finite type and $S$ is quasi-compact $\mathcal{L}^{\otimes n}$ is $f$-very ample for some $n > 0$, see Lemma 29.39.5. Replace $\mathcal{L}$ by $\mathcal{L}^{\otimes n}$. Write $\mathcal{F} = f_*\mathcal{L}$. This is a quasi-coherent $\mathcal{O}_ S$-module by Schemes, Lemma 26.24.1 (quasi-projective morphisms are quasi-compact and separated, see Lemma 29.40.4). By Properties, Lemma 28.22.7 we can find a directed set $I$ and a system of finite type quasi-coherent $\mathcal{O}_ S$-modules $\mathcal{E}_ i$ over $I$ such that $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{E}_ i$. Consider the compositions $\psi _ i : f^*\mathcal{E}_ i \to f^*\mathcal{F} \to \mathcal{L}$. Choose a finite affine open covering $S = \bigcup _{j = 1, \ldots , m} V_ j$. For each $j$ we can choose sections

which generate $\mathcal{L}$ over $f^{-1}V_ j$ and define an immersion

see Lemma 29.39.1. Choose $i$ such that there exist sections $e_{j, t} \in \mathcal{E}_ i(V_ j)$ mapping to $s_{j, t}$ in $\mathcal{F}$ for all $j = 1, \ldots , m$ and $t = 1, \ldots , n_ j$. Then the map $\psi _ i$ is surjective as the sections $f^*e_{j, t}$ have the same image as the sections $s_{j, t}$ which generate $\mathcal{L}|_{f^{-1}V_ j}$. Whence we obtain a morphism

over $S$ such that over $V_ j$ we have a factorization

of the immersion given above. It follows that $r_{\mathcal{L}, \psi _ i}|_{V_ j}$ is an immersion, see Lemma 29.3.1. Since $S = \bigcup V_ j$ we conclude that $r_{\mathcal{L}, \psi _ i}$ is an immersion. Note that $r_{\mathcal{L}, \psi _ i}$ is quasi-compact as $X \to S$ is quasi-compact and $\mathbf{P}(\mathcal{E}_ i) \to S$ is separated (see Schemes, Lemma 26.21.14). By Lemma 29.7.7 there exists a closed subscheme $X' \subset \mathbf{P}(\mathcal{E}_ i)$ such that $i$ factors through an open immersion $X \to X'$. Then $X' \to S$ is projective by definition and we win. $\square$

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