The units with norm $+1$ in a pure cubic number field $K$ generated
by a cube root of $m = ab^2$, where $a$ and $b$ are coprime and
squarefree integers, correspond to integral points on the torus
$$ R\_{K/\mathbb Q}^{(1)}: 
 X\_1^3 + ab^2X\_2^3 + a^2bX\_3^2 - 3abX\_1X\_2X\_3 = 1. $$
According to Voskresenskii (Algebraic Groups and their birational
Invariants), all tori of dimension $2$ such as the one above
are rational.


I am having problems with finding such a rational parametrization.


The surface has three singular points at infinity, all of them
defined over the normal closure of $K$; the line through the pair
of conjugate singular points is necessarily contained in the surface 
and defined over $K$. 


* Is there a way of finding a parametrization 
from the singular points or the three lines connecting them?


A different idea is looking at the tangent plane in $(1,0,0)$.
It intersects the surface in a singular cubic, which can be 
parametrized via sweeping lines and produces the parametrization
$$ X\_1 = 1, \quad X\_2 = \frac{3t}{b+at^3}, \quad X\_2 = \frac{3t^2}{b+at^3}. $$
By looking at the tangent plane at these rational points I would get a 
2-parameter family of rational points; the calculations are, however,
quite involved. So:


* Is there a slick way of obtaining this family?
* Once I have written down the 2-parameter family of rational points, how
can I show that the parametrization includes all rational points?


An additional question in this connection is the following: 
conics such as $x^2 - my^2 = 1$ can be parametrized by trigonometric
or hyperbolic functions. 


* Are there (periodic) analytic functions that parametrize the 
cubic surface above?


I have always wondered why there are so many books on diophantine equation, but few if any explaining some simple geometric techniques useful for finding rational points on algebraic surfaces. Is there such a book out there?



Parametrization of 2-dimensional torus
I decided to have a go at finding a parametrisation by following the instructions of Coray and Tsfasman (reference in my comment above), using Magma. Amazingly enough, it works, even working generically with $a,b$ variables.


Here's what I did. Working over the field $K(\omega, \sqrt[3]{a^2b})$, find the three singular points of $X$. Writing $c = \sqrt[3]{a^2b}$, they are $(0:c:a/c:1)$, $(0:\omega^2 c:\omega a/c:1)$, $(0:\omega c:\omega^2 a/c:1)$. There's an obvious rational point $(1:1:0:0)$. The Cremona transformation $f \colon \mathbb{P}^3 \to \mathbb{P}^3$ associated to these four points can be found by linearly mapping them to the standard basis points, applying the standard Cremona transformation $(1/X\_0: 1/X\_1: 1/X\_2: 1/X\_3)$, and reversing the linear map. It turns out that $f^{-1}(X)$ is (modulo some rubbish supported on the four planes where $f$ is not defined) a quadric surface $Y$, defined by
$$
X\_0^2 + \frac{1}{3a} X\_0 X\_2 + \frac{1}{27a^2} X\_2^2 + \frac{1}{27a^2b}(X\_0-X\_1)X\_3. 
$$


The surface $Y$ has an obvious rational point $(0:1:0:0)$, so projecting away from that gives an isomorphism $Y \to \mathbb{P}^2$. The inverse of that isomorphism, composed with $f$, gives a rational map from $\mathbb{P}^2$ to $X$. The equations are (cut & pasted from Magma):



```
729*c^6*x^6 + 729*c^6/a*x^5*y + 324*c^6/a^2*x^4*y^2 + 81*c^6/a^3*x^3*y^3 + 
    12*c^6/a^4*x^2*y^4 + c^6/a^5*x*y^5 + 1/27*c^6/a^6*y^6 + -3*c^3/a*x^2*y*z^3 +
    -c^3/a^2*x*y^2*z^3 + -2/27*c^3/a^3*y^3*z^3 + 1/27*z^6
729*c^6*x^6 + 729*c^6/a*x^5*y + 324*c^6/a^2*x^4*y^2 + 81*c^6/a^3*x^3*y^3 + 
    12*c^6/a^4*x^2*y^4 + c^6/a^5*x*y^5 + 1/27*c^6/a^6*y^6 + -9*c^3*x^3*z^3 + 
    -6*c^3/a*x^2*y*z^3 + -c^3/a^2*x*y^2*z^3 + -2/27*c^3/a^3*y^3*z^3 + 1/27*z^6
243*a*c^3*x^5*z + 162*c^3*x^4*y*z + 45*c^3/a*x^3*y^2*z + 6*c^3/a^2*x^2*y^3*z + 
    1/3*c^3/a^3*x*y^4*z + -1/3*x*y*z^4
-9*c^3/a*x^3*y*z^2 + -3*c^3/a^2*x^2*y^2*z^2 + -1/3*c^3/a^3*x*y^3*z^2 + 1/3*x*z^5

```

I'm sure these can be tidied up a lot, but notice that they are at least defined over the original base field, since $c$ only ever appears as $c^3$.


I'll post the Magma code (only 20 lines) if anybody's interested.


