1

1

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...)?

flag

1 Answer

1

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.

link|flag
Why does the map $Q \to S$ have no rational sections? – Pistorious Jun 9 2011 at 7:53

Your Answer

Get an OpenID
or

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