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I'm currently reading "Rational surfaces with many nodes" by Dolgachev et al., avaliable here: http://www.math.lsa.umich.edu/~idolga/lisbon.pdf

A "surface" is always smooth and projective and let us work over $\mathbb{C}$ for simplicity.

Let $Z$ be a ruled surface with $h^1(Z, \mathcal{O}_Z) = q(Z) > 0$, and consider its Albanese map: $$ \alpha: Z \longrightarrow A. $$ Then (Beauville "complex algebraic surfaces" V.18) the image of $\alpha$ is a curve, let us write $C$, so that $$ \alpha: Z \longrightarrow C. $$ The curve $C$ is is smooth and of genus $q(Z)$ (Beauville V.15).

Now it is stated in the mentioned paper (proof of 3.2) that the generic fibre is isomorphic to $\mathbb{P}^1$. My question is, why? It seems like magic to me.

My thoughts so far: It seems very likely that it has to do with $Z$ being birational to the surface $D \times \mathbb{P}^1$, where $D$ is a smooth curve of genus $q(Z)$. I have tried to use generic smoothness in characteristic zero (Hartshorne 10.7) so that at least the fibres of $\alpha$ are smooth curves, but this did not bring me anywhere. Moreover, it should not be necessary, since this all should work over any algebraically closed field of characteristic $\neq 2$.

Hope: Is it in general true that if $Z$ is ruled and there is a surjective map $Z \rightarrow C$ to a smooth curve of genus $q(Z)$, the general fibre of this map is $\mathbb{P}^1$?

Lastly, any remarks on unnecessaryness of characteristic zero are highly appreciated!

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Here's an answer that you might find useful. Let me stick to $\mathbb{C}$; maybe others can fill in the details for positive characteristic.

First of all, $\alpha$ has connected fibres in your case, and by Bertini its general fibre $G$ is smooth, so $G$ is an irreducible smooth curve.

On the other hand, since the curve $C$ has genus >0, there cannot be any nonconstant morphism $\mathbb{P}^1 \rightarrow C$, so for any smooth fibre $F$ (and hence any fibre) of the ruling of $Z$, we must have $\alpha(F) = \{pt\}$. In particular, if we take $F$ to be a general fibre of the ruling, it must be contained in a fibre of $\alpha$. Now if $G$ is not a fibre of the ruling, then $G$ intersects $F$. But then $F \subsetneq G$, contradicting the fact that $G$ is smooth and irreducible.

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  • $\begingroup$ Thanks, crystal clear answer! $\endgroup$ – Joachim Apr 4 '13 at 17:50
  • $\begingroup$ You're welcome! $\endgroup$ – user64687 Apr 4 '13 at 17:53
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Here is an attempt. Hopefully I haven't made any errors, the question turned out to be more tricky than I initially suspected.

Suppose that $Z$ is birational to $D\times\mathbb P^1.$ Using this, we can define a morphism $\phi: U'\to C$ where $U'\subseteq D\times\mathbb P^1$ is an open subset isomorphic to some open $U\subseteq Z.$ Since $C$ is projective and $U'$ is dense, for any $(d,x)\in D\times\mathbb P^1$ where $\phi$ is defined, we can extend $\phi$ uniquely to be defined on all of $\{d\}\times\mathbb P^1.$ This follows by considering $\phi$ as a rational map $\phi(d,-):U'\cap(\{d\}\times\mathbb P^1)\to C$ from a quasi-projective curve contained in $\mathbb P^1$ to $C.$ Thus, $\phi$ extends to a morphism $(D\times\mathbb P^1)\setminus(\cup_{i=1}^n D_i)\to C$ where $D_i=\{d_i\}\times\mathbb P^1$ for a finite number of points $d_i\in D.$

On the other hand, given $x\in\mathbb P^1$, the map $\phi$ also defines morphisms $\phi(-,x):D\setminus\{d_1,\ldots, d_n\}\to C,$ and we can use the same trick. Namely, since $D$ minus finitely many points is quasi-projective, we can uniquely extend $\phi$ to be defined over the $d_i$'s. Thus, we can extend $\phi$ to be defined on all of $D\times\mathbb P^1.$ Notice that $\phi$ must be constant on any $\{d\}\times\mathbb P^1$ since $g(C)=q(Z)\ge 1.$ This implies the $\{d\}\times\mathbb P^1$ lie in fibres of $\phi.$ Since $\alpha$ is surjective, these fibres are finite disjoint unions of copies of $\mathbb P^1$.

Now, there exists a surface $S$ and (birational) morphisms $f:S\to Z,g:S\to D\times\mathbb P^1,$ which are the blowups of either $Z$ or $D\times\mathbb P^1$ at finitely many points, and which resolve the birational map $D\times\mathbb P^1\dashrightarrow Z.$ By our arguments above, $\phi\circ g$ has generic fibre equal to a disjoint union of copies of $\mathbb P^1.$ But $\phi\circ g = \alpha\circ f,$ which implies that the same is true for $\alpha\circ f.$ Away from a finite set of points, $f$ is an isomorphism, which implies that generically, the fibres of $\alpha$ are also finite disjoint unions of copies of $\mathbb P^1.$

Now, using the fact that $\alpha$ is the albanese map, and thus has connected fibres, we see that the generic fibre of $\alpha$ must be a $\mathbb P^1.$

So, to answer your last question, in addition I needed to know that $\alpha$ is surjective with connected fibres, but I only needed that $g(C)\ge 1.$

Actually, recall that by the Noether-Enriques theorem, it suffices to show that a single fibre of a smooth map is $\mathbb P^1.$ I was initially trying to prove this, but got the stronger version directly instead!

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  • $\begingroup$ Ah, I see that an answer was posted while I was typing! My difficulty was in justifying the fibres of $\alpha$ are irreducible. I didn't think about Bertini's theorem, which simplifies things substantially! $\endgroup$ – Andrew Apr 4 '13 at 17:58
  • $\begingroup$ I haven't had time to look at your answer in detail, but i am pretty sure that it will contain general reasoning from which i will learn. So in any case thanks a lot Andrew! $\endgroup$ – Joachim Apr 4 '13 at 18:21
  • $\begingroup$ Dear @Joachim, you are most welcome! $\endgroup$ – Andrew Apr 4 '13 at 18:29

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