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Hi @all,

imagine a projective, noetherian and flat family of curves $C\rightarrow S$ (i.e. every geometric fiber is a curve, this is a integral, non singular, proper scheme of dimension 1 over an algebraically closed field) which is smooth. By the stein-factorization there is a scheme $S'$, such that $C\rightarrow S$ decomposes to $C \rightarrow S' \rightarrow S$ where $C\rightarrow S'$ is a projective morphism with connected fibers and $S'\rightarrow S$ is a morphism of finite type.

The paper I'm reading now states, that $S'$ and $S$ are isomorphic on an open subset.

  1. Is this the case?

  2. Why is this the case, especially, which of the many nice properties from above gives us this isomorphism?



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Probably you want $S' \to S$ to be a finite morphism, not just a map of finite type. – Karl Schwede Nov 13 '12 at 12:30
You write that geometric fibres of $C\to S$ are integral, hence connected ; thus $S'=S$ does the job, doesn't it ? – Matthieu Romagny Nov 13 '12 at 13:27
Matthieu, you are right. I missed the integral implies connected... – Karl Schwede Nov 13 '12 at 15:13

If you just want $S'$ with the property that $C\to S'$ has connected fibers, then $S'=S$ suits perfectly as said Matthieu in the comments.

In general, let $f : C\to S$ be a proper surjective morphism of noetherian schemes. Usually, the Stein factorization refers to $S'=\mathrm{Spec}(f_*O_X)$ (it has the advantage to be canonical). This $S'$ is not always isomorphic to $S$: consider $S'\to S$ the normalization morphism of an integral curve with a cusp over $\mathbb C$ and $C=S'\times_{\mathbb C} \mathbb P^1_{\mathbb C}$. However it is true that $S'\to S$ is an isomorphism on an open subset under mild hypothesis.

(a) If $S$ is reduced and the fiber of $f$ at a generic point $\xi$ of $S$ is geometrically reduced and geometrically connected, then $S'\to S$ is an isomorphism above an open neighborhood of $\xi$.

Proof: As the question is local on $S$, we can suppose $S=\mathrm{Spec}(R)$ is affine. As $S'\to S$ is finite, $S'=\mathrm{Spec}(R')$. Because $S$ is reduced at $\xi$, $O_{S,\xi}$ is equal to the residue field $k(\xi)$ of $S$ at $\xi$. So the canonical map $R\to O_{S,\xi}=k(\xi)$ is a localization hence flat and we have
$$R'\otimes_R O_{S,\xi}= H^0(C, O_C)\otimes_R k(\xi)=H^0(C_{\xi}, O_{C_{\xi}})=k(\xi)=O_{S,\xi}.$$ So the finite morphism $S'\to S$ is an isomorphism when localized at $\xi$. By standard arguments, this isomorphism propagates above an open neighborhood of $\xi$.

(b) Suppose $f$ is flat and some geometric fiber $C_{\bar{s}}$ is connected and reduced. Then $S'\to S$ is an isomorphism above an open neighborhood of $s$.

Proof. We have $H^0(C_s, O_{C_s})=k(s)$. Then the statement is just EGA III, 7.8.8.

As a corollary of (b):

(c) If $f$ is flat with reduced and connected geometric fibers, then $S'\to S$ is an isomorphism.

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in fact (a) is a special case of (b) because f is flat over $\xi$. – Qing Liu Nov 15 '12 at 7:52

According to Hartshorne chapter III. corollary 11.5 the stein factorization gives a finite morphism from $S' \longrightarrow S$. Karls comment seems to imply, that this would fix the problem. Right Karl?

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