# smooth morphism on schemes

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?

Thanks

Frank

-
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.

-
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?