Let $X \to B$ be a smooth, proper, dominant map of schemes over $\text{Spec }k$ an algebraically closed field of characteristic zero with $B$ integral. We have the generic fibre $\overline{F}$ defined over $\text{Spec }\overline{K(B)}$ and by base-changing along $\text{Spec }\overline{K(B)} \to \text{Spec }k$, we obtain a map $\overline{X} \to \overline{B}$ such that we can now write down the sequence $\overline{F} \to \overline{X} \to \overline{B}$. To what extent is this a fibre bundle? To ask a definite question, is there some etale map $\overline{B}' \to \overline{B}$ such that further pulling back will yield an isomorphism $\overline{X}' \simeq \overline{F}' \times \overline{B}'$? 


This question is closely related to [Flatness in Algebraic Geometry vs. Fibration in Topology](https://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology) and [Is an algebraic geometer's fibration also an algebraic topologist's fibration?](https://mathoverflow.net/questions/28162/is-an-algebraic-geometers-fibration-also-an-algebraic-topologists-fibration). In particular, it is motivated by (1) Ehresmann's theorem that locally analytically such a morphism should be a (topological) fibre bundle and (2) the fuzzy thinking that "locally analytically" should mean "after an etale base change", but I feel like the answer to the question I posed it above is probably in the negative. For example, it seems unlikely to me that two smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$ which are (automatically) diffeomorphic but not isomorphic should suddenly become isomorphic after an etale base change. However, I don't know of any weaker way to algebro-geometrically state the condition that some map be a fibre bundle, however -- is there anything then that we can say algebro-geometrically with respect to the above maps, or do we have to be content with the differential-geometric statement that it's a fibre bundle in that category?



Is a fibration in algebraic geometry a fibre bundle?
The answer here is a resounding no. I think the most important point is that you're applying the wrong topological intuition here. A variety shouldn't be thought of as like a manifold, but as like a complex manifold, and the corresponding theorem to the "submersion=fiber bundle" theorem in smooth manifolds is just false for complex manifolds. Just as an example, all elliptic curves are topologically the same, so the solutions to $x(x-1)(x-a)=y^2$ are a smooth fiber bundle over (most of) $\mathbb{C}$ with coordinate $a$, but all the fibers which aren't in the same orbit of $SL\_2(\mathbb Z)$ on $\mathbb{C}$ are not isomorphic as complex manifolds. 


The other way of saying this is that complex structures can exist in families; they have moduli. Moduli spaces exactly measure how theorems like "submersion=fiber bundle" fail since they measure continuous variation of structure. 


