4
$\begingroup$

Let $f(u,v,z)\in \mathbb{Q}[u,v,z]$ be a polynomial in three variables such that $X_{\mathbb{R}}\subset \mathbb{R}^{3} $ (the associated surface of real solution) is smooth. Suppose that the set of rational solutions $X_{\mathbb{Q}}$ is dense in $X_{\mathbb{R}}$. My question is the following:

Is there an open set $U\subset X_{\mathbb{R}}$ such that locally we have a diffeomorphism

$$ g: V\subset \mathbb{R}^{2}\longrightarrow U $$ such that

$$ g(x,y)= (\frac{P_{1}(x,y)}{Q_{1}(x,y)},\frac{P_{2}(x,y)}{Q_{2}(x,y)}, \frac{P_{3}(x,y)}{Q_{3}(x,y)})$$ where $P_{i}$ and $Q_{i}$ are in $ \mathbb{Q}[x,y]$

$\endgroup$

1 Answer 1

2
$\begingroup$

The answer to your question is negative. To discuss it, let me make in what follows the additional assumption that the associated projective variety $Y\subset\mathbb{P}^3_{\mathbb{R}}$ is smooth, and let me denote by $d$ the degree of $f$.

Then, the existence of $g$ implies that $Y_{\mathbb{C}}$ is a unirational complex variety. This cannot happen if $d\geq 4$ because $Y_{\mathbb{C}}$ carries a non-trivial $2$-differential form.

When $d\geq 5$, it is expected (but unknown and difficult), that the rational points cannot be dense (even for the Zariski topology). Thus conjecturally, you cannot hope for a counter-example if $d\geq 5$.

On the other hand, when $d=4$, $Y$ is a K3 surface, and it is really possible that the rational points are dense in the real points (for the Zariski topology or even, as you ask, for the euclidean topology). A concrete example is given in [Elkies, On $A^4+B^4+C^4=D^4$], showing that a counterexample to your question is given by $u^4+v^4+z^4-1$.

Finally, let me indicate that when $d=3$ (and if we insist, in your question, that the whole of $Y$ is smooth), the question has a positive answer, going back to [Segre, A note on arithmetical properties of cubic surfaces] : a cubic surface that has a rational point is unirational. This is also the case, and easy, if $d=2$ or $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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