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Is there an example of a quasiprojective variety $X$ defined over ℚ such that

  1. $X$ is unirational over all finite fields, and
  2. $X$ is unirational over $\mathbb{R}$, and
  3. $X$ is not unirational over $\mathbb{Q}$, and
  4. $X(\mathbb{Q})\neq \emptyset$.

It would be easy to produce trivial examples if we do not demand (3), for example smooth cubic surfaces which violate the Hasse principle.

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    $\begingroup$ I think you also want to assume $X$ is not unirational? $\endgroup$
    – Will Sawin
    Commented Jun 11, 2015 at 18:07
  • $\begingroup$ Yes I am sorry, in fact that was the point of my question (edited). $\endgroup$
    – Dr. Pi
    Commented Jun 11, 2015 at 19:20
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    $\begingroup$ You can take the (quite singular) projective cone over a smooth cubic surface violating the Hasse principle. The only rational point will be the vertex of the cone. $\endgroup$
    – Jason Starr
    Commented Jun 12, 2015 at 0:12
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    $\begingroup$ That is a good example, could you please explain a bit why there would be no other $\mathbb{Q}$ points ? Thanks! $\endgroup$
    – Dr. Pi
    Commented Jun 12, 2015 at 0:22

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