I am reading Liu's Algebraic Geometry and Arithmetic curves, p.356~p.357 and stuck at some point.
In beggining of p.356~ p.357, section 8.3.3 Contraction, he wrote,
Let $X\to S$ be a normal fibered surface over a Dedekind scheme of dimension $1$. Let $E_1, \dots, E_n$ be irreducible vertical divisors on $X$. The aim of this subsection is to consturct, when that is possible, a fibered surface $Y\to S$ and a birational projective morphism $f:X\to Y$ such that $f(E_i)$ is reduced to a point for each $i\le n$, and that $f$ is an isomorphism outside of $\cup_i E_i$ ( in other words, $\cup_i E_i$ is the exceptional locus of $f$, Definition 7.2.21 ).
Why the bold statement is true? Here, the 'exceptional locus' is as follows :
Lemma 2.20. Let $X$, $Y$ be Noetherian integral schemes. Let $f:X\to Y$ be a separated birational morphism of finite type. (a) There exists a non-empty open subset $V$ of $Y$ such that $f^{-1}(V) \to V$ is an isomorphism. (b) Let $W$ be the union of the open sets $V$ as in $(a)$. Then $x\in f^{-1}(W)$ if and only if $\mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ is an isomorphism.
Definition 2.21. Let us keep the hypotheses and notation of the lemma above. We call the closed subset $E:= X - f^{-1}(W)$ the exceptional locus of $f$
My first attempt to the bold statement is as follows.
Let $\{ V_j \}_{j \in J}$ be the set of all non-empty open subsets of $Y$ such that $f^{-1}(V_j) \to V_j$ is an isomorphism, so that $W = \cup_{j\in J}V_j$ as above in the Lemma 2.20.
Let $E:=\cup_{i\in I}E_i$. We need to show that $E = X - f^{-1}(W)$, or equivalently, $X-E = f^{-1}(W)$, Since $f :X \to Y$ is an isomorphism outside of $E$, by the Lemma 2.20 -(b), $X-E \subseteq f^{-1}(W)$. It remains to show that $f^{-1}(W) \subseteq X-E$, or equivalently, $E \subseteq f^{-1}(W)^{c} = \cap_{j\in J}f^{-1}(V_j^{c})$ Now let $x \in E$. Then there exists $i$, say $1$, such that $x \in E_1$. It suffices to show that $x \in f^{-1}(V_j^{c})$ for all $j$ ; i.e., $f(x) \in V_j^{c}$ for all $j$.
Case 1) $x \in E_1$ is not the generic point of $E_1$ : By the way of contradiction, assume that $f(x) \notin V_j^{c} $ for some $j$ ; i.e., $f(x) \in V_{j}$ for some $j$. Since $f^{-1}(V_j) \to V_j$ is isomorphism (so that bijection) and $f(E_1)$ is reduced to a point, we get $E_1 \cap f^{-1}(V_j) = \{x\}$ ; i.e., $\{x\}$ is open in $E_1$. So $E_1 - \{x\}$ is closed subset. Note that $E_1 = \bar{\{x\}} \cup (E_1 - \{x\})$. Since $x$ is not the generic point of $E_1$, we get contradiction to that $E_1$ is irreducible. So $f(x) \in V_j^{c}$ for all $j$.
Case 2) $x= \xi_{1}$ is the generic point of $E_1$. Like case 1), by the contrary, assume that $f(\xi_{1}) \notin V_j^{c} $ for some $j$. Then by the similar argument as the Case 1), $\xi_{1}$ is open generic point of $E_1$. It is not hard to show that for an integral noetherian scheme, the generic point is open if and only if the scheme is a finite set. ( C.f. When is the generic point of an integral noetherian scheme open (reference)? ) So $E_1$ is finite set. But note that $f(E_1) = \{y_1 \}$ is reduced to a point so that by base change, we can view $E_1$ as a non-empty (?) $k(y_1)$-scheme of finite type. So by the Gortz's Algebraic Geometry Proposition 5.20, $\dim E_1 =0$, which contradicts to fact that $\dim E_1 =1 $ ( C.f. Irreducible vertical divisor on a (normal) fibered surface has dimension $1$ ? ( Liu's Algebraic Geometry ) ).
Correct?
Or is there any other route to show that $f(x) \in V_j^{c}$ for all $j$, .. or $f^{-1}(W) \subseteq X-E$ ? For $x \in E$, $\mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ is not isomorphism? If so, then by the Lemma 2.20 -(b) above, $E \subseteq f^{-1}(W)^{c}$ and we are done.
Can anyone help ? Thanks for reading.
