**Proof.**
Both cases follow from Lemma 54.16.8 using standard results on ample invertible modules and (quasi-)projective morphisms.

Proof of (1). Projectivity of $f$ means that $f$ is proper and there exists an $f$-ample invertible module $\mathcal{L}$, see Morphisms, Lemma 29.43.13 and Definition 29.40.1. Let $U \subset S$ be an affine open containing the image of $E$. By Lemma 54.16.8 there exists a contraction $c : f^{-1}(U) \to V'$ of $E$ and an ample invertible module $\mathcal{N}'$ on $V'$ whose pullback to $f^{-1}(U)$ is equal to $\mathcal{L}(nE)|_{f^{-1}(U)}$. Let $v \in V'$ be the closed point such that $c$ is the blowing up of $v$. Then we can glue $V'$ and $X \setminus E$ along $f^{-1}(U) \setminus E = V' \setminus \{ v\} $ to get a scheme $X'$ over $S$. The morphisms $c$ and $\text{id}_{X \setminus E}$ glue to a morphism $b : X \to X'$ which is the contraction of $E$. The inverse image of $U$ in $X'$ is proper over $U$. On the other hand, the restriction of $X' \to S$ to the complement of the image of $v$ in $S$ is isomorphic to the restriction of $X \to S$ to that open. Hence $X' \to S$ is proper (as being proper is local on the base by Morphisms, Lemma 29.41.3). Finally, $\mathcal{N}'$ and $\mathcal{L}|_{X \setminus E}$ restrict to isomorphic invertible modules over $f^{-1}(U) \setminus E = V' \setminus \{ v\} $ and hence glue to an invertible module $\mathcal{L}'$ over $X'$. The restriction of $\mathcal{L}'$ to the inverse image of $U$ in $X'$ is ample because this is true for $\mathcal{N}'$. For affine opens of $S$ avoiding the image of $v$, we see that the same is true because it holds for $\mathcal{L}$. Thus $\mathcal{L}'$ is $(X' \to S)$-relatively ample by Morphisms, Lemma 29.37.4 and (1) is proved.

Proof of (2). We can write $X$ as an open subscheme of a scheme $\overline{X}$ projective over $S$ by Morphisms, Lemma 29.43.12. By (1) there is a contraction $b : \overline{X} \to \overline{X}'$ and $\overline{X}'$ is projective over $S$. Then we let $X' \subset \overline{X}$ be the image of $X \to \overline{X}'$; this is an open as $b$ is an isomorphism away from $E$. Then $X \to X'$ is the desired contraction. Note that $X'$ is quasi-projective over $S$ as it has an $S$-relatively ample invertible module by the construction in the proof of part (1).
$\square$

## Comments (2)

Comment #7907 by Laurent Moret-Bailly on

Comment #8165 by Aise Johan de Jong on