Lemma 54.16.4. Let $X$ be a Noetherian scheme. Let $E \subset X$ be an exceptional curve of the first kind. If there exists a morphism $f : X \to Y$ such that

1. $Y$ is Noetherian,

2. $f$ is proper,

3. $f$ maps $E$ to a point $y$ of $Y$,

4. $f$ is quasi-finite at every point not in $E$,

Then there exists a contraction of $E$ and it is the Stein factorization of $f$.

Proof. We apply More on Morphisms, Theorem 37.53.4 to get a Stein factorization $X \to X' \to Y$. Then $X \to X'$ satisfies all the hypotheses of the lemma (some details omitted). Thus after replacing $Y$ by $X'$ we may in addition assume that $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and that the fibres of $f$ are geometrically connected.

Assume that $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and that the fibres of $f$ are geometrically connected. Note that $y \in Y$ is a closed point as $f$ is closed and $E$ is closed. The restriction $f^{-1}(Y \setminus \{ y\} ) \to Y \setminus \{ y\}$ of $f$ is a finite morphism (More on Morphisms, Lemma 37.44.1). Hence this restriction is an isomorphism since $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ since finite morphisms are affine. To prove that $\mathcal{O}_{Y, y}$ is regular of dimension $2$ we consider the isomorphism

$\mathcal{O}_{Y, y}^\wedge \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}/\mathfrak m_ y^ n), \mathcal{O})$

of Cohomology of Schemes, Lemma 30.20.7. Let $E_ n = nE$ as in Lemma 54.16.3. Observe that

$E_ n \subset X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}/\mathfrak m_ y^ n)$

because $E \subset X_ y = X \times _ Y \mathop{\mathrm{Spec}}(\kappa (y))$. On the other hand, since $E = f^{-1}(\{ y\} )$ set theoretically (because the fibres of $f$ are geometrically connected), we see that the scheme theoretic fibre $X_ y$ is scheme theoretically contained in $E_ n$ for some $n > 0$. Namely, apply Cohomology of Schemes, Lemma 30.10.2 to the coherent $\mathcal{O}_ X$-module $\mathcal{F} = \mathcal{O}_{X_ y}$ and the ideal sheaf $\mathcal{I}$ of $E$ and use that $\mathcal{I}^ n$ is the ideal sheaf of $E_ n$. This shows that

$X \times _ Y \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}/\mathfrak m_ y^ m) \subset E_{nm}$

Thus the inverse limit displayed above is equal to $\mathop{\mathrm{lim}}\nolimits H^0(E_ n, \mathcal{O}_ n)$ which is a regular two dimensional local ring by Lemma 54.16.3. Hence $\mathcal{O}_{Y, y}$ is a two dimensional regular local ring because its completion is so (More on Algebra, Lemma 15.43.4 and 15.43.1).

We still have to prove that $f : X \to Y$ is the blowup $b : Y' \to Y$ of $Y$ at $y$. We encourage the reader to find her own proof. First, we note that Lemma 54.16.3 also implies that $X_ y = E$ scheme theoretically. Since the ideal sheaf of $E$ is invertible, this shows that $f^{-1}\mathfrak m_ y \cdot \mathcal{O}_ X$ is invertible. Hence we obtain a factorization

$X \to Y' \to Y$

of the morphism $f$ by the universal property of blowing up, see Divisors, Lemma 31.32.5. Recall that the exceptional fibre of $E' \subset Y'$ is an exceptional curve of the first kind by Lemma 54.3.1. Let $g : E \to E'$ be the induced morphism. Because for both $E'$ and $E$ the conormal sheaf is generated by (pullbacks of) $a$ and $b$, we see that the canonical map $g^*\mathcal{C}_{E'/Y'} \to \mathcal{C}_{E/X}$ (Morphisms, Lemma 29.31.3) is surjective. Since both are invertible, this map is an isomorphism. Since $\mathcal{C}_{E/X}$ has positive degree, it follows that $g$ cannot be a constant morphism. Hence $g$ has finite fibres. Hence $g$ is a finite morphism (same reference as above). However, since $Y'$ is regular (and hence normal) at all points of $E'$ and since $X \to Y'$ is birational and an isomorphism away from $E'$, we conclude that $X \to Y'$ is an isomorphism by Varieties, Lemma 33.17.3. $\square$

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