Let $Q\to S$ be a quadric fibration over a rational base $S$, over an algebraically closed field of non zero characteristic. Is it true the following?

$Q$ is rational if and only if $Q \to S$ has a rational section.

If not, may it be true under some assumptions (bounds on the dimensions, on the associated Clifford algebras, working over $\mathbb{C}$, etc...)?


In one direction the implication is evident --- if there is a section then $Q$ is rational.

In the other direction the implication is false. For example, consider any projective space $P(V)$, let $S = P(S^2V^*)$, and $Q$ be the universal quadric, that is the canonical divisor of bidegree $(2,1)$ on $P(V)\times P(S^2V^*)$. First, it is clear that $Q$ is rational (because the projection $Q \to P(V)$ is a projectivization of a vector bundle). On the other hand, the map $Q \to S = P(S^2V^*)$ has no rational sections.

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  • $\begingroup$ Why does the map $Q \to S$ have no rational sections? $\endgroup$ – Pistorious Jun 9 '11 at 7:53

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