TITLE: A question related to fiber bundle
QUESTION [5 upvotes]: Let $f:\mathbb{C}^3 \to \mathbb{C}$ be a morphism of varieties such that it is a smooth fiber bundle. Can I say that the fiber is $\mathbb{C}^2$?

REPLY [2 votes]: Note that $f$ is smooth as a map of varieties. If $f$ comes from a direct product $\mathbb{C}^3\approx S\times \mathbb{C}$ then $S\approx \mathbb{C}^2$. See Section 5.1 of $\mathbb{A}^1$-homotopy theory and contractible varieties: a survey
I do not see why $f$ must come from a direct product so this is not a complete answer.