MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...)?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
Why does the map $Q \to S$ have no rational sections? – Pistorious Jun 9 '11 at 7:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.