Lemma 33.17.3. Let $X$ be a Noetherian scheme. Let $f : Y \to X$ be a birational proper morphism of schemes with $Y$ reduced. Let $U \subset X$ be the maximal open over which $f$ is an isomorphism. Then $U$ contains

1. every point of codimension $0$ in $X$,

2. every $x \in X$ of codimension $1$ on $X$ such that $\mathcal{O}_{X, x}$ is a discrete valuation ring,

3. every $x \in X$ such that the fibre of $Y \to X$ over $x$ is finite and such that $\mathcal{O}_{X, x}$ is normal, and

4. every $x \in X$ such that $f$ is quasi-finite at some $y \in Y$ lying over $x$ and $\mathcal{O}_{X, x}$ is normal.

Proof. Part (1) follows from Morphisms, Lemma 29.51.6. Part (2) follows from part (3) and Lemma 33.17.2 (and the fact that finite morphisms have finite fibres).

Part (3) follows from part (4) and Morphisms, Lemma 29.20.7 but we will also give a direct proof. Let $x \in X$ be as in (3). By Cohomology of Schemes, Lemma 30.21.2 we may assume $f$ is finite. We may assume $X$ affine. This reduces us to the case of a finite birational morphism of Noetherian affine schemes $Y \to X$ and $x \in X$ such that $\mathcal{O}_{X, x}$ is a normal domain. Since $\mathcal{O}_{X, x}$ is a domain and $X$ is Noetherian, we may replace $X$ by an affine open of $x$ which is integral. Then, since $Y \to X$ is birational and $Y$ is reduced we see that $Y$ is integral. Writing $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ we see that $A \subset B$ is a finite inclusion of domains having the same field of fractions. If $\mathfrak p \subset A$ is the prime corresponding to $x$, then $A_\mathfrak p$ being normal implies that $A_\mathfrak p \subset B_\mathfrak p$ is an equality. Since $B$ is a finite $A$-module, we see there exists an $a \in A$, $a \not\in \mathfrak p$ such that $A_ a \to B_ a$ is an isomorphism.

Let $x \in X$ and $y \in Y$ be as in (4). After replacing $X$ by an affine open neighbourhood we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $A \subset \mathcal{O}_{X, x}$, see Properties, Lemma 28.29.8. Then $A$ is a domain and hence $X$ is integral. Since $f$ is birational and $Y$ is reduced it follows that $Y$ is integral too. Consider the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$. This is a ring map which is essentially of finite type, the residue field extension is finite, and $\dim (\mathcal{O}_{Y, y}/\mathfrak m_ x\mathcal{O}_{Y, y}) = 0$ (to see this trace through the definitions of quasi-finite maps in Morphisms, Definition 29.20.1 and Algebra, Definition 10.122.3). By Algebra, Lemma 10.124.2 $\mathcal{O}_{Y, y}$ is the localization of a finite $\mathcal{O}_{X, x}$-algebra $B$. Of course we may replace $B$ by the image of $B$ in $\mathcal{O}_{Y, y}$ and assume that $B$ is a domain with the same fraction field as $\mathcal{O}_{Y, y}$. Then $\mathcal{O}_{X, x} \subset B$ have the same fraction field as $f$ is birational. Since $\mathcal{O}_{X, x}$ is normal, we conclude that $\mathcal{O}_{X, x} = B$ (because finite implies integral), in particular, we see that $\mathcal{O}_{X, x} = \mathcal{O}_{Y, y}$. By Morphisms, Lemma 29.42.4 after shrinking $X$ we may assume there is a section $X \to Y$ of $f$ mapping $x$ to $y$ and inducing the given isomorphism on local rings. Since $X \to Y$ is closed (by Schemes, Lemma 26.21.11) necessarily maps the generic point of $X$ to the generic point of $Y$ it follows that the image of $X \to Y$ is $Y$. Then $Y = X$ and we've proved what we wanted to show. $\square$

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