Let us work over an algebraically closed field $k$ and suppose $\pi:S\rightarrow C$ where $S$ is a surface, $C$ is a smooth curve, and the fibers over closed points are all isomorphic to a fixed smooth projective curve $D$. Must there exist an open subscheme $U\subseteq C$ such that $\pi^{-1}(U)\cong U\times D$? More strongly can I choose $U$ to contain any 2 points of $C$ of my choice?

If $D=\mathbb{P}^1$, then by Prop V.2.2 in Hartshorne, $S$ is isomorphic over $C$ to the projectivization of $\mathcal{E}$, a locally free sheaf of rank 2. Taking an affine open $U$ containing two points of your choice trivializes the vector bundle, and hence $\pi^{-1}(U)\cong U\times\mathbb{P}^1$. Does the same hold (although clearly by a different argument) for $D$ an arbitrary smooth projective curve?


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    $\begingroup$ I don't know the answer, but it seems worth pointing out that Hartshorne makes a substantial simplification in his definition of a geometrically ruled surface, in that he assumes the map $\pi$ has a section. Without this assumption, the proof is much longer, using Tsen's theorem, or, as in Beauville's book, the Noether-Enriques theorem. I imagine whether you take this for granted or not may affect the answer to your question.... $\endgroup$ – Andrew Mar 22 '13 at 22:00
  • $\begingroup$ @Andrew Yes sorry for not mentioning the proof in Hartshorne sidestepped the issue of a section by saying that one has to exist in the ruled surface case at least. I was looking at some cases of an elliptically fibered surface without a section for a possible counterexample, but at least based on a cursory look the elliptic surfaces without a section I found didn't have constant fiber. $\endgroup$ – user16544 Mar 22 '13 at 22:54
  • $\begingroup$ The local-triviality theorem of Fischer-Grauert said that a smooth family of compact complex manifolds is locally trivial if and only if all fibers are analytically isomorphic. I don't know the case for other algebraically closed field. $\endgroup$ – Yuchen Liu Mar 23 '13 at 5:14

This doesn't hold in higher genus.

Consider the $C=\mathbb A^1\setminus \{0\}$ with coordinate $t$, let $S\to C$ be the familly of elliptic curves defined by $$y^2z=x^3+tz^3$$ (suppose $k=\mathbb C$). Each fiber is isomorphic to $D: y^2z=x^3+z^3$ (by the obvious change of variables), but, as you can't take the $6$-th root of $t$ in $O_C(C)$, you see easily that $S\times_C k(C)$ is not isomorphism $D\times_k k(C)$.

What happens in general is there exists a finite étale cover $C'\to C$ such that $S\times_C C'\to D\times_k C'$ ($C'$ can be obtained by adjoining the $n$-torsions of $\mathrm{Pic}^0_{S/C}$ for some $n\ge 3$ prime to the characteristic of $k$). In the above example, $C'=\mathrm{Spec}k[t^{1/3}]$.

The relative curves $S\to C$ under your hypothesis are classified by $H^1_{ét}(C, \mathrm{Aut}(D))$. The point is that the curve $S\times_C k(C)$ over $k(C)$ is a twist of $D\times_k k(C)$ for some finite extension $k(C')/k(C)$ with finite étale $C'\to C$.

  • $\begingroup$ Thank you for the reply. I'm missing some subtlety about this. Pulling back via the generic point $Spec(k(C))\rightarrow C$ gives a twist of $D$ base changed to $k(C)$. These are classified by $H^1(k(C),Aut(D))$. Given such a twist we have some finite extension $L/k(C)$ over which they become isomorphic. $L$ is the function field of a curve $C'$ over $k$. Now this is where I'm confused: why is $C'\rightarrow C$ étale? Also why does every twist in $H^1(k(C),Aut(D))$ correspond to such an $S$ (why can't the fibers over finitely many points be bad?) $\endgroup$ – user16544 Mar 25 '13 at 5:23
  • $\begingroup$ Also related: if I replace the base $C$ with a generic $k$-scheme $T$ will $H^1(k(T),Aut(D))$ or $H^1_{ét}(T,Aut(D)$ (the first doesn't make sense when $T$ isn't irreducible) continue to classify such things? If it's not a curve the above correspondence only works birationally so one at least needs a different argument. $\endgroup$ – user16544 Mar 25 '13 at 5:27
  • $\begingroup$ @Ryan: yes you are right, I edited the answer. In higher dimension I think the current approach doesn't give a classification as you suspected: an element of $H^1_{ét}$ gives a $S\to T$ such that $S\times_T T'$ is birational to $D\times_k T'$ for some finite étale $T'\to T$. I don't see why they would be isomorphic. $\endgroup$ – user18119 Mar 25 '13 at 8:54

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