The Stacks project

Proof. Let $n$ be the degree of $\mathcal{L}|_ E$ as in Lemma 54.16.7. Observe that $n > 0$ as $\mathcal{L}$ is ample on $E$ (Varieties, Lemma 33.44.14 and Properties, Lemma 28.26.3). After replacing $\mathcal{L}$ by a power we may assume $H^ i(X, \mathcal{L}^{\otimes e}) = 0$ for all $i > 0$ and $e > 0$, see Cohomology of Schemes, Lemma 30.17.1. Finally, after replacing $\mathcal{L}$ by another power we may assume there exist global sections $t_0, \ldots , t_ n$ of $\mathcal{L}$ which define a closed immersion $\psi : X \to \mathbf{P}^ n_ S$, see Morphisms, Lemma 29.39.4.

Set $\mathcal{M} = \mathcal{L}(nE)$. Then $\mathcal{M}|_ E \cong \mathcal{O}_ E$. Since we have the short exact sequence

and since $H^1(X, \mathcal{M}(-E))$ is zero (by Lemma 54.16.7 and the fact that $n > 0$) we can pick a section $s_{n + 1}$ of $\mathcal{M}$ which generates $\mathcal{M}|_ E$. Finally, denote $s_0, \ldots , s_ n$ the sections of $\mathcal{M}$ we get from the sections $t_0, \ldots , t_ n$ of $\mathcal{L}$ chosen above via $\mathcal{L} \subset \mathcal{L}(nE) = \mathcal{M}$. Combined the sections $s_0, \ldots , s_ n, s_{n + 1}$ generate $\mathcal{M}$ in every point of $X$ and therefore define a morphism

over $S$, see Constructions, Lemma 27.13.1.

Below we will check the conditions of Lemma 54.16.4. Once this is done we see that the Stein factorization $X \to X' \to \mathbf{P}^{n + 1}_ S$ of $\varphi $ is the desired contraction which proves (1). Moreover, the morphism $X' \to \mathbf{P}^{n + 1}_ S$ is finite hence $X'$ is proper over $S$ (Morphisms, Lemmas 29.44.11 and 29.41.4). This proves (2). Observe that $X'$ has an ample invertible sheaf. Namely the pullback $\mathcal{M}'$ of $\mathcal{O}_{\mathbf{P}^{n + 1}_ S}(1)$ is ample by Morphisms, Lemma 29.37.7. Observe that $\mathcal{M}'$ pulls back to $\mathcal{M}$ on $X$ (by Constructions, Lemma 27.13.1). Finally, $\mathcal{M} = \mathcal{L}(nE)$. Since in the arguments above we have replaced the original $\mathcal{L}$ by a positive power we conclude that the invertible $\mathcal{O}_{X'}$-module $\mathcal{L}'$ mentioned in (3) of the lemma is ample on $X'$ by Properties, Lemma 28.26.2.

Easy observations: $\mathbf{P}^{n + 1}_ S$ is Noetherian and $\varphi $ is proper. Details omitted.

Next, we observe that any point of $U = X \setminus E$ is mapped to the open subscheme $W$ of $\mathbf{P}^{n + 1}_ S$ where one of the first $n + 1$ homogeneous coordinates is nonzero. On the other hand, any point of $E$ is mapped to a point where the first $n + 1$ homogeneous coordinates are all zero, in particular into the complement of $W$. Moreover, it is clear that there is a factorization

of $\psi |_ U$ where $\text{pr}$ is the projection using the first $n + 1$ coordinates and $\psi : X \to \mathbf{P}^ n_ S$ is the embedding chosen above. It follows that $\varphi |_ U : U \to W$ is quasi-finite.

Finally, we consider the map $\varphi |_ E : E \to \mathbf{P}^{n + 1}_ S$. Observe that for any point $x \in E$ the image $\varphi (x)$ has its first $n + 1$ coordinates equal to zero, i.e., the morphism $\varphi |_ E$ factors through the closed subscheme $\mathbf{P}^0_ S \cong S$. The morphism $E \to S = \mathop{\mathrm{Spec}}(R)$ factors as $E \to \mathop{\mathrm{Spec}}(H^0(E, \mathcal{O}_ E)) \to \mathop{\mathrm{Spec}}(R)$ by Schemes, Lemma 26.6.4. Since by assumption $H^0(E, \mathcal{O}_ E)$ is a field we conclude that $E$ maps to a point in $S \subset \mathbf{P}^{n + 1}_ S$ which finishes the proof. $\square$