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Suppose given a smooth morphism $f:X\to Y$ between varieties over $\mathbb{C}$ whose fibres are $\mathbb{P}^n$. Then I have an equality of Hodge polynomials $H(X) = H(Y)H(\mathbb{P}^n)$, say because the hyperplane class generates the cohomology of $\mathbb{P}^n$ and hence $f_* \mathbb{Z}_X$ cannot have monodromy.

Is any such fibration in fact Zariski locally trivial? Even if not, do I have the equality in the Grothendieck group of varieties $[X] = [Y][\mathbb{P}^n]$?

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Vivek, since the aut. scheme of $\mathbf{P}^n$ is ${\rm{PGL}}_n$, and any smooth surjective map has etale-local sections, your question is about etale PGL$_n$-torsors. Using the central extension $1 \rightarrow {\mathbf{G}}_m \rightarrow {\rm{GL}}_n \rightarrow {\rm{PGL}}_n \rightarrow 1$, at the generic pt of $Y$ we get a connecting map of pointed sets ${\rm{H}}^1(k(Y), {\rm{PGL}}_n) \rightarrow {\rm{H}}^2(k(Y),\mathbf{G}_m) = {\rm{Br}}(k(Y))$ which is a bijection onto ${\rm{Br}}(k(Y))[n]$ (see Serre's book on Galois cohom). See Grothendieck's Brauer exposes for the finer theory over $Y$. – BCnrd Dec 31 '10 at 6:40
Oops, as usual I made a typo and wrote PGL$_n$ rather than PGL$_{n+1}$, and likewise should use $(n+1)$-torsion in the Brauer group. – BCnrd Dec 31 '10 at 7:00
Thanks, Brian. I guess probably this is answered in those references, but does one generally expect to be able to compute the Br(k(Y)) and/or that map? – Vivek Shende Dec 31 '10 at 7:18
Vivek, Serre describes the bijection onto $(n+1)$-torsion in the Brauer groups as well as the inverse map; look up "Brauer-Severi variety" in the index at the back of the book (I think). As for actually computing the Brauer group, this is subtle beyond the 1-dim'l case handled by Tsen's theorem. As an indication of the difficulties, in dimension 3 nontrivial 2-torsion Brauer classes can fail to be represented by quaternion division algebras (in contrast with number fields, local fields, and function fields of curves over finite fields). There's quite a bit of literature for surfaces. – BCnrd Dec 31 '10 at 8:00
up vote 14 down vote accepted

It is not necessarily trivial in the Zariski topology. Consider for instance the plane quadric $\{x^2+sy^2+tz^2\}\subseteq \mathbb P^2\times\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$ as a family of $\mathbb P^1$'s over $\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$. It is not even isomorphic to $\mathbb P^1$ over the generic point of $\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$ as it doesn't have a rational point. As for the multiplicativity in the Grothendieck group there are examples when it is not true not even in the localised and completed Grothendieck group, see arXiv:0903.3143.

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thanks for the examples! Do you have counter-examples to the perhaps weaker claim that $X$ is in the ring generated by $Y$ and the class of the affine line? – Vivek Shende Dec 31 '10 at 7:09
Yes, the same example gives that. – Torsten Ekedahl Dec 31 '10 at 13:37

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