(I posted this same question on [MSE](https://math.stackexchange.com/questions/685867/fibrations-with-isomorphic-fibers-but-not-zariski-locally-trivial). Sorry if it is too elementary.)


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 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. 


*Remark*. Sometimes a fibration is required to have connected fibers; if this was the correct definition of a fibration, my example would not be an example.


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.



Fibrations with isomorphic fibers, but not Zariski locally trivial
They are not rare. A general construction goes as follows. Start with a finite (étale) group (scheme) $G$ which acts on varieties $F$ and $\tilde Y$, where the second action has no fixed points. 
Let $Y =\tilde Y/G$. Then $(F\times \tilde Y)/G\to Y$ has all its fibres isomorphic to $F$. It is locally trivial in the étale topology, but not usually in the Zariski topology. This includes the above examples, and many others.


