4
$\begingroup$

Let $X = Q_1\cap Q_2$ be a complete intersection of two smooth quadrics, over a field $K$, in $\mathbb{P}^4$ with homogeneous coordinates $y_0,y_1,y_2,y_3,y_4$.

Set $Q_1 = \{F_1 = 0\}$ and $Q_2 = \{F_2 = 0\}$ and assume that the monomial $y_0^2$ does not appear in $F_1$ so that $Q_1$ is rational and that the monimial $y_0y_1$ does not appear in $F_2$.

Under these hypotheses could we conclude anything about the unirationality over $K$ of $X$?

What if $X$ is a complete intersection of two quadrics in $\mathbb{P}^n$ with the same properties for $n\geq 5$?

$\endgroup$

1 Answer 1

7
$\begingroup$

No.

Let $Q_1,Q_2$ be arbitrary quadrics. Let $a$ be the coefficient of $y_0^2$ in $F_1$, $b$ the coefficient of $y_0^2$ in $F_2$, $c$ the coefficient of $y_0 y_1$ in $F_1$, $d$ the coefficient of $y_0 y_1$ in $F_2$.

Then the coefficient of $y_0^2$ in $b F_1 - a F_2$ and the coefficient of $y_0 y_1$ in $d F_1 - c F_2$ both vanish. If $ad-bc \neq 0$ (the generic case), then $b F_1 - a F_2$ and $d F_1 - c F_2$ is $Q_1 \cap Q_2$.

So any intersection of two quadrics that satisfies a mild genericity property can be written in this form, and thus the unirationality problem is as hard as for a general intersection of two quadrics.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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