The Stacks project

Proof. Assume (1), (2) and (3). Condition (3) means $S = \mathop{\mathrm{Spec}}(R)$ for some ring $R$. Condition (1) means by definition there exists a quasi-coherent $\mathcal{O}_ S$-module $\mathcal{E}$ and an immersion $\alpha : X \to \mathbf{P}(\mathcal{E})$ such that $\mathcal{L} = \alpha ^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)$. Write $\mathcal{E} = \widetilde{M}$ for some $R$-module $M$. Thus we have

Since $\alpha$ is an immersion, and since the topology of $\text{Proj}(\text{Sym}_ R(M))$ is generated by the standard opens $D_{+}(f)$, $f \in \text{Sym}_ R^ d(M)$, $d \geq 1$, we can find for each $x \in X$ an $f \in \text{Sym}_ R^ d(M)$, $d \geq 1$, with $\alpha (x) \in D_{+}(f)$ such that

is a closed immersion. Condition (2) implies $X$ is quasi-compact. Hence we can find a finite collection of elements $f_ j \in \text{Sym}_ R^{d_ j}(M)$, $d_ j \geq 1$ such that for each $f = f_ j$ the displayed map above is a closed immersion and such that $\alpha (X) \subset \bigcup D_{+}(f_ j)$. Write $U_ j = \alpha ^{-1}(D_{+}(f_ j))$. Note that $U_ j$ is affine as a closed subscheme of the affine scheme $D_{+}(f_ j)$. Write $U_ j = \mathop{\mathrm{Spec}}(A_ j)$. Condition (2) also implies that $A_ j$ is of finite type over $R$, see Lemma 29.15.2. Choose finitely many $x_{j, k} \in A_ j$ which generate $A_ j$ as a $R$-algebra. Since $\alpha |_{U_ j}$ is a closed immersion we see that $x_{j, k}$ is the image of an element

Finally, choose $n \geq 1$ and elements $y_0, \ldots , y_ n \in M$ such that each of the polynomials $f_ j, f_{j, k} \in \text{Sym}_ R(M)$ is a polynomial in the elements $y_ t$ with coefficients in $R$. Consider the graded ring map

Denote $F_ j$, $F_{j, k}$ the elements of $R[Y_0, \ldots , Y_ n]$ such that $\psi (F_ j) = f_ j$ and $\psi (F_{j, k}) = f_{j, k}$. By Constructions, Lemma 27.11.1 we obtain an open subscheme

and a morphism $r_\psi : U(\psi ) \to \mathbf{P}^ n_ R$. This morphism satisfies $r_\psi ^{-1}(D_{+}(F_ j)) = D_{+}(f_ j)$, and hence we see that $\alpha (X) \subset U(\psi )$. Moreover, it is clear that

is still an immersion since $i^\sharp (F_{j, k}/F_ j^{e_{j, k}}) = x_{j, k} \in A_ j = \Gamma (U_ j, \mathcal{O}_ X)$ by construction. Moreover, the morphism $r_\psi$ comes equipped with a map $\theta : r_\psi ^*\mathcal{O}_{\mathbf{P}^ n_ R}(1) \to \mathcal{O}_{\text{Proj}(\text{Sym}_ R(M))}(1)|_{U(\psi )}$ which is an isomorphism in this case (for construction $\theta$ see lemma cited above; some details omitted). Since the original map $\alpha$ was assumed to have the property that $\mathcal{L} = \alpha ^*\mathcal{O}_{\text{Proj}(\text{Sym}_ R(M))}(1)$ we win. $\square$