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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 $n\geq 5$?

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1 Answer 1

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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.

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