Lemma 54.17.1. Let $f : X \to Y$ be a proper birational morphism between integral Noetherian schemes regular of dimension $2$. Then $f$ is a sequence of blowups in closed points.

**Proof.**
Let $V \subset Y$ be the maximal open over which $f$ is an isomorphism. Then $V$ contains all codimension $1$ points of $V$ (Varieties, Lemma 33.17.3). Let $y \in Y$ be a closed point not contained in $V$. Then we want to show that $f$ factors through the blowup $b : Y' \to Y$ of $Y$ at $y$. Namely, if this is true, then at least one (and in fact exactly one) component of the fibre $f^{-1}(y)$ will map isomorphically onto the exceptional curve in $Y'$ and the number of curves in fibres of $X \to Y'$ will be strictly less that the number of curves in fibres of $X \to Y$, so we conclude by induction. Some details omitted.

By Lemma 54.4.3 we know that there exists a sequence of blowing ups

in closed points lying over the fibre $f^{-1}(y)$ and a morphism $X' \to Y'$ such that

is commutative. We want to show that the morphism $X' \to Y'$ factors through $X$ and hence we can use induction on $n$ to reduce to the case where $X' \to X$ is the blowup of $X$ in a closed point $x \in X$ mapping to $y$.

Let $E \subset X'$ be the exceptional fibre of the blowing up $X' \to X$. If $E$ maps to a point in $Y'$, then we obtain the desired factorization by Lemma 54.16.1. We will prove that if this is not the case we obtain a contradiction. Namely, if $f'(E)$ is not a point, then $E' = f'(E)$ must be the exceptional curve in $Y'$. Picture

Arguing as before $f'$ is an isomorphism in an open neighbourhood of the generic point of $E'$. Hence $g : E \to E'$ is a finite birational morphism. Then the inverse of $g$ (a rational map) is everywhere defined by Morphisms, Lemma 29.42.5 and $g$ is an isomorphism. Consider the map

of Morphisms, Lemma 29.31.3. Since the source and target are invertible modules of degree $1$ on $E = E' = \mathbf{P}^1_\kappa $ and since the map is nonzero (as $f'$ is an isomorphism in the generic point of $E$) we conclude it is an isomorphism. By Morphisms, Lemma 29.32.18 we conclude that $\Omega _{X'/Y'}|_ E = 0$. This means that $f'$ is unramified at every point of $E$ (Morphisms, Lemma 29.35.14). Hence $f'$ is quasi-finite at every point of $E$ (Morphisms, Lemma 29.35.10). Hence the maximal open $V' \subset Y'$ over which $f'$ is an isomorphism contains $E'$ by Varieties, Lemma 33.17.3. This in turn implies that the inverse image of $y$ in $X'$ is $E'$. Hence the inverse image of $y$ in $X$ is $x$. Hence $x \in X$ is in the maximal open over which $f$ is an isomorphism by Varieties, Lemma 33.17.3. This is a contradiction as we assumed that $y$ is not in this open. $\square$

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