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Let $f:X \to Y$ be a dominant morphism between two integral proper surfaces (therefore $2$-dimensional, proper $k$-schemes).

Futhermore we assume

  1. for the structure sheaf holds $\mathcal{O}_Y= f_*(\mathcal{O}_X)$; therefore $f$ is a fibration.

  2. $Y$ is a normal surface (therefore the stalks a normal rings)

There is a well known fact (without assumption 2. that $Y$ is normal) that there exist a dense open $U \subset Y$ such that the restriction $f \vert _{f^{-1}(U)} \to U$ is an isomorphism.

My question is why and how to see that if we assump that 2. holds we can the choose the $U$ such that it has following shape: $U = Y \backslash \{z_1, ..., z_s\}$ with $z_i \in Y$ closed points such that $E_i := f^{-1}(z_i)$ are (connected) exceptional curves (so $k$-subschemes of dimension $1$).

My attempts: First remark is that if $z \in Y$ is closed then the fiber $f^{-1}(z)$ is connected since $f$ is a fibration (see: Stein factorization)

My idea is to show that if $z \in Y$ is closed with $f^{-1}(z) = x$ (therefore the fiber is NOT a curve) then $\mathcal{O}_{Y,z} = \mathcal{O}_{X,x} $. Therefore in this case the point $z$ could be "added" to $U$ since then $f$ induces an isomorphism on these stalks.

Therefore let assump that $f^{-1}(z) = x$ is a point. Using the fact that $f(\eta_X)= \eta_Y$ holds for generic points we conclude the induced map on the level of stalks $f^{\#}_z:\mathcal{O}_{Y,z} \to \mathcal{O}_{X,x}$ is a inclusion. Since $z$ and $x$ are closed points the stalks are both $2$-dimensional local rings.

Futhermore, since $f$ is a proper morphism and properness is stable under base change, the corresponding map of $f^{\#}_z$ on Specs - let call it $g_z: Spec(\mathcal{O}_{X,x}) \to Spec(\mathcal{O}_{Y,z})$ - is also proper.

Here occurs the first problem: I intend to show that $g_z$ is a finite morphism. Can I conclude from the properness of $g_z$ that it is also finite. (I know that here I need an extra argument; does anybody have an idea?)

If I succeed in this then I know that $\mathcal{O}_{X,x}$ is an integral extension of $\mathcal{O}_{Y,z}$.

From here I need the last step to win: To show that $\mathcal{O}_{Y,z}$ is normal. I guess that here I could use Serre's criterion for normality. Does anybody see how to get here the $S_2$ condition?

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  • $\begingroup$ That follows from Zariski's Main Theorem, namely, the connectedness part. $\endgroup$ Commented Sep 30, 2018 at 10:08
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    $\begingroup$ @Jason Starr: Do you have this in mind: en.wikipedia.org/wiki/Zariski%27s_connectedness_theorem? If yes, then it just states that fibers under certain conditions for $f$ (e.g.as given in my case) are connected. But the core problem in my "proof" is to show that if $f^{-1}(z) = \{x\}$ is a point then $\mathcal{O}_{Y,z} \cong \mathcal{O}_{X,x}$ must hold. And here the obstacle is to proof that $g_z$ is a finite morphism. I don't see how the connectedness part from ZMT prove this. $\endgroup$
    – user267839
    Commented Sep 30, 2018 at 12:39
  • $\begingroup$ Please read "The Original form" of Zariski's Main Theorem, p. 288, Section III.9 of David Mumford, "The Red Book of Varieties and Schemes", Lecture Notes in Math. 1358 (1988), Springer-Verlag, Heidelberg. $\endgroup$ Commented Sep 30, 2018 at 17:51

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