# Fibrations with isomorphic fibers, but not Zariski locally trivial

I am looking for examples of fibrations $f:X\to Y$ where the fibers are all isomorphic, but $f$ is not Zariski locally trivial. In particular, I am interested in understanding how much such examples are "rare". (I believe they are not that rare)

First of all, by fibration I mean a proper flat surjective morphism of (complex) varieties. But I am not sure this is the correct definition of fibration in Algebraic Geometry; in that case, any correction is much appreciated.

By $f:X\to Y$ being Zariski locally trivial, I mean that there is a variety $F$ such that every point in the base $Y$ has a Zariski open neighborhood $U$ such that $f^{-1}(U)\to U$ is isomorphic to the projection $F\times U\to U$. Here $F$ is called the fiber of $f$ (in particular, Zariski locally trivial fibrations do have isomorphic fibers).

One example I came up with is that of an étale cover of curves: the fibers are discrete of the same size, hence isomorphic, but it is not Zariski locally trivial in general.

Another example might be the Hirzebruch surface $\mathbb F_n\to \mathbb P^1$, with $n\neq 0$.

As for projective bundles $\mathbb P(E)\to Y$, I do not know whether they are Zariski locally trivial or not.

Probably there are many important examples that I am missing here. I would very much appreciate if you could help me to fill in this picture!

Thank you.

• Brenin, I apologize if this is a naive question. What do you mean by "Zariski locally trivial"? I, in particular, don't know what the "trivial"ness would mean? Something like a birational morphism? Thanks! – Alex Youcis Feb 22 '14 at 10:27
• Oh, you mean a trivial fibration? – Alex Youcis Feb 22 '14 at 10:30
• Sorry Alex. Not naive at all. I planned to write it down but I forgot. Now I added (what I believe is) the correct definition of "Zariski locally trivial". Essentially, locally a product. – Brenin Feb 22 '14 at 10:36
• Projective bundles are by definition Zariski locally trivial, since vector bundles are. In particular, Hirzebruch surfaces are indeed Zariski locally trivial, since they are projectivisations of rank 2 bundles on $\mathbf P^1$. – user64687 Feb 22 '14 at 11:09
• By the way, I don't have a good answer to your question, but a search for "Severi--Brauer variety" might be a useful starting point for the case when the fibres are projective spaces. Alternatively, if you post this question on MathOverflow, I'm sure you'll get a thorough answer from Jason Starr. – user64687 Feb 23 '14 at 11:54

• Thank you for your answer! By smooth family you mean a smooth morphism (in the sense of algebraic geometry)? Also, I think of complex manifolds as smooth objects (when viewed in the algebraic category), so I am confused when you say there are no examples with smooth fiber. By the way, for an étale cover of curves $f:X\to Y$, I just wanted to say that it is not isomorphic to one of the form $F\times U\to U$ for open sets $U\subset Y$ covering $Y$. Nevertheless, the fibers are isomorphic because they all consists of $\deg f$ reduced points. – Brenin Jan 8 '15 at 8:58