# If $X\times X$ is rational, must $X$ also be rational?

Is there an example of a smooth projective variety $X$ such that $X$ is irrational, but $X\times X$ is rational?

For instance, is $X\times X$ irrational for a smooth cubic threefold $X$?

• No such example is known, and this is probably out of reach of our present knowledge. Same about the second question.
– abx
Apr 12, 2017 at 13:01
• Naive question: Why can't the Clemens-Griffiths approach to proving the irrationality of cubic threefolds $X$ via the intermediate Jacobian be used to show irrationality of $X \times X$? Is the problem that it only applies to threefolds? Apr 12, 2017 at 13:38
• @abx Is there at least a candidate for a counterexample? For instance, the variety in jstor.org/stable/1971174?seq=1#page_scan_tab_contents ? @ DanielLoughran: Yes.
– byu
Apr 12, 2017 at 13:47
• @Daniel Loughran: yes, the intermediate Jacobian method applies only to threefolds. byu: Why not, but why yes???
– abx
Apr 12, 2017 at 15:46

For the first question I am not that pessimistic. At least there are candidates as follows: Recall that $Z$ is stably rational if there is $n\ge0$ such that $Z\times\mathbf A^n$ is rational. Now suppose there is such a $Z$ such that the minimal $n$ is $\ge2$. I would be extremely surprised if that didn't exist. Then put $Y=Z\times\mathbf A^{n-2}$ and $X=Y\times\mathbf A^1$. By assumption, $X$ is not rational but $X\times\mathbf A^1$ is. Let $d=\dim X\ge1$. Then $$X\times X\cong Y\times \mathbf A^1\times X\cong Y\times\mathbf A^{d+1}\cong X\times\mathbf A^1\times\mathbf A^{d-1}\cong\mathbf A^{d+1}\times\mathbf A^{d-1}\cong\mathbf A^{2d}$$ A candidate for $Y$ would be the non-rational $3$-fold constructed by Beauville et al. for which Shepherd-Barron proved that $Y\times\mathbf A^2$ is rational. I don't know whether $Y\times\mathbf A^1$ is rational or not.