# Is there an odd degree unirational parametrization of a cubic threefold?

A cubic threefold is a smooth degree $$3$$ hypersurface in $$\mathbb{P}^4$$. Is there a cubic threefold $$X$$ over any field $$k$$ (possibly of positive characteristic) and an odd degree rational map $$\mathbb{P}^3\rightarrow X?$$

Background:

Over an algebraically closed field, all cubic threefolds have unirational parametrizations of degree $$2$$. On the other hand, Clemens-Griffiths showed that cubic threefolds are never rational, i.e., do not have unirational parametrizations of degree $$1$$. (They showed this only in characteristic zero, but IIRC it is not too hard to generalize their argument to positive characteristic.)

There is a folklore conjecture that if a variety has unirational parametrizations of degree $$m$$ and $$n$$, then it has a unirational parametrization of degree $$\operatorname{gcd}(m,n)$$. This is maybe due to Brendan Hassett, but I'm not sure whether he conjectured it or just posed it as a question. If you believe this conjecture (which I personally don't see any strong evidence for), then the answer to my question above should be "no".

One obstruction to odd degree parametrizations in this case would be nonexistence of a decomposition of the diagonal. But it is known that, conditional on the Tate conjecture for divisors on surfaces, there exist cubic threefolds in positive characteristic which do admit decompositions of the diagonal, see https://arxiv.org/abs/1602.06767.

• I believe that this is an open problem. – Jonathan Frink Dec 17 '18 at 20:24
• @JonathanFrink Do you know of a written source? – dhy Dec 18 '18 at 18:48
• Not for this particular question. But see 7.2.2 in arxiv.org/pdf/1612.02415.pdf – Jonathan Frink Dec 18 '18 at 21:24
• Didn’t you have some idea about this in char 3 at an AIM workshop a couple years ago? Did it end up not working out? – Daniel Litt Jan 19 at 5:49