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An rr function (i.e. rational rational function) is a quotient

$$ \frac fg\,:\, \Bbb Q\ \to\ \Bbb Q\cup\{\infty\} $$

such that $\ f,g\,\in\,\Bbb Z[X],\ $ where $\ g\ne 0.$

QUESTION Do there exist rr functions $\ \phi\ \psi\ $ such that set

$$ \{(\phi(x)\ \ \psi(x))\,:\, x\in\Bbb Q\}\ \subseteq \ (\Bbb Q\cup\{\infty\})^2 $$

is dense in a non-empty open subset of $\ \Bbb Q^2\,?$

I'd guess -- NO.

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    $\begingroup$ Even with $\phi ,\psi$ in $\mathbb{C}(X)$, your map extends to an algebraic map $\mathbb{P}^1\rightarrow \mathbb{P}^2$, whose image is an algebraic curve, certainly not dense in $\mathbb{P}^2$. $\endgroup$
    – abx
    Commented Mar 24, 2020 at 8:46
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    $\begingroup$ Is $f\ g$ just the product of $f$ and $g$? The space confuses me. $\endgroup$
    – Wojowu
    Commented Mar 24, 2020 at 9:04
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    $\begingroup$ Thanks for clarification. $\endgroup$
    – Wojowu
    Commented Mar 24, 2020 at 9:21
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    $\begingroup$ comas may be eyesores, but commas are sights for sore eyes. I'm putting 'em in. $\endgroup$
    – Gerry Myerson
    Commented Mar 24, 2020 at 11:57
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    $\begingroup$ Remember elimination theory from algebraic geometry. You just use a resultant to eliminate the parameter. See my notes on Concrete Algebra (on github) for explicit algorithm. So the image is an algebraic curve. $\endgroup$
    – Ben McKay
    Commented Mar 24, 2020 at 12:25

1 Answer 1

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Use the resultant to eliminate the variable $X$. Since the resultant is computed over the rationals, the resultant is a rational coefficient polynomial in the two variables of the plane, satisfied on the image of the parameterized curve. See my (undergraduate!) lecture notes Concrete Algebra on github for complete (and elementary) details.

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  • $\begingroup$ This sounds good. I'll check your link. (+1 for now). U'r more or less translating parametrization into equation (roughly speaking). $\endgroup$
    – Wlod AA
    Commented Mar 24, 2020 at 19:56

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