Let $V$ be a smooth $\mathbb{Q}$-variety. Assume that for any non-trivial finite Galois extension $K/\mathbb{Q}$ the $K$-points are Zariski dense in $V$. Must $V$ have a $\mathbb{Q}$-point then?
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3$\begingroup$ Remark: I think this is false if you replace $\mathbf{Q}$ with $\mathbf{R}$. $\endgroup$– David Benjamin LimCommented May 12, 2021 at 5:44
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4$\begingroup$ Varieties of the kind you consider have a $0$-cycle of degree $1$. Whether this (much weaker) condition implies the existence of a $\mathbb{Q}$-point is discussed in mathoverflow.net/questions/33774/… . It is false in general, but is true for some families of varieties. $\endgroup$– Uriya FirstCommented May 12, 2021 at 6:15
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2$\begingroup$ This is likely a very hard question, because we have very few methods for constructing rational points. Both a proof and a counterexample would have to do this. $\endgroup$– R. van Dobben de BruynCommented May 12, 2021 at 14:23
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1$\begingroup$ @Uriya: I am probably being silly, but why does such a variety have a $0$-cycle of degree $1$? $\endgroup$– Damian RösslerCommented May 13, 2021 at 9:24
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1$\begingroup$ @DamianRössler To see that there is a $0$-cycle of degree $1$ in the sense of mathoverflow.net/questions/33774/…, take a quadratic extension $K/{Q}$ and a cubic Galois extension $L/{Q}$ with fixed embeddings into $\overline{ Q}$. Choose $x_1\in V(K)$ and $y_1\in V(L)$ and let $x_2,y_2,y_3\in V(\overline{Q})$ be their Galois conjugates (take $x_2=x_1$ if $x_1$ is fixed by $Gal(\overline{ Q}/{ Q})$ and likewise with $y_2,y_3$). Then $y_1+y_2+y_3-x_1-x_2$ is a $0$-cycle of degree $1$. $\endgroup$– Uriya FirstCommented May 13, 2021 at 19:02
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