I throwed a look into C. Liedtke's paper SEMI-STABLE REDUCTION FOR CURVES and have some problems to understand an argument (page 1):
Let $S$ be a connected Dedekind scheme. We denote by $K=K(S)$ it's function field $\mathcal{O}_{S, \eta}$.
Definition 2.
An arithmetic surface is an integral, projective, and flat $S$-scheme $\pi: X \to S$ of relative dimension $1$, where $S$ is as before.
Clearly, the generic fiber
$X_{\eta}$
of an arithmetic surface is an integral and
projective curve over the function field $K(S)$ of $S$.
Conversely, if $S = Spec R$
for a DVR $R$
, then an arithmetic surface over $S$ is a model of its generic fiber.
Given a normal proper and geometrically connected curve $C$
over the function field $K(S)$ of $S$ , we seek “good” models over $S$.
For instance, if we choose a projective embedding of
$C$ into some $\mathbb{P}^n_{K(S)}$ then its Zariski closure in $\mathbb{P}^n_{S}$
is an arithmetic surface (where the $S$ -flatness uses that
$S$ is Dedekind). Hence, we can always find an arithmetic surface with generic
fiber $C (=X_{\eta}$. Since $S$ is normal and $C$ is reduced and geometrically connected over
$K(S)$ , so $H^0(C, \mathcal{O}_C)$ is a finite purely inseparable extension field of
$K(S)$, it
follows from considerations with Stein factorization (whose fibers are always
geometrically connected) that any such arithmetic surface has geometrically
connected fibers over $S$, (???) i.e. all fibers $X_s \cong f^{-1}(s)$ for all $s \in S$ are geometrically connected.
Question:
Why the assumption that $H^0(C, \mathcal{O}_C)= H^0(X_{\eta}, \mathcal{O}_{X_{\eta}})$ is a finite purely inseparable extension field of
$K(S)$ imply using Stein factorisation that every fiber $X_s$ is geometrically connected? I don't understand the argument.
We can assume that $S= Spec(A)$ is affine and $A$ is a dvr with generic point $\eta$ and closed $s \in S$ and we need to show that $X_s$ at first is connected. First of all by Stein factorisation $\pi: X \to S$ splits into $ X \xrightarrow{f} Y' \xrightarrow{g} S$ with $Y'=Spec(\pi_*\mathcal{O}_X).$ Then $L:=H^0(X_{\eta}, \mathcal{O}_{X_{\eta}})$ is a finite purely inseparable extension field of
$K(S)$.
Because $f$ has by definition geometrically connected fibers, it suffice to show that $g$ is bijective. And this is exactly the problem.
A classical result tells that if $X$ is an algebraic variety over $k$, and let $K/k$ be an
algebraic purely inseparable extension, then the canonical map $X_K \to X$ is a homeomorphism (see for example Liu's Algebraic Geometry page 89, Prop. 3.2.7 ). In Liu (page 350, Corollary 8.3.6(b) is suggested to use this result to conclude that $g$ is bijective, but I don't understand how this mentioned classical result can be here applied.
Rmk: I know that another way to prove the claim using formal functions theorem works as well. The motivation behind this question is how to use here the assumption on purely inseparability of $L$ over $K(S)$ as guided in Liu's book.
Best Answer
You're exactly on the correct track with writing $\pi$ as a composite $X\stackrel{f}\to Y'\stackrel{g}\to S$ and then noting that we wish to show that $g$ is bijective. We may observe that $g:Y'\to S$ is a finite morphism of integral schemes and $S$ is locally noetherian and normal, plus $K(X)=K(Y')\supset K(S)$ is a purely inseparable extension (as $f_*\mathcal{O}_X = \mathcal{O}_{Y'}$). This means that in the terminology of Liu, it's a finite purely inseparable morphism and per exercise 5.3.9(a) it will be a homeomorphism and thus bijective. Here's a previous post of mine which gives the details of showing that a finite purely inseparable map is a homeomorphism.