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By inspection, I could find a parameterization for a surface $f(x_1,x_2,x_3)$ in $\mathbb{A}_{C}^3$. Such parameterization is of the form $x_i=\frac{\phi_i(s,t)}{\psi_i(s,t)}$ where $\phi_i(s,t)$ and $\psi_i(s,t)$ are polynomials in the variables $s$ and $t$. My question is: is there any criterion to prove that this parameterization is birational or not? A jacobian, a determinant, something like that?

I tried use MAGMA, via the command IsInvertible. The answer was no: that is not a birational parameterization. However, MAGMA works on $\mathbb{A}_{Q}^3$, and I don't know how to extend the field to complex numbers.

The explicit equation of the surface is,

$f(x_1,x_2,x_3)=2x_1^2x_2 - 2x_3 - 5x_1x_2x_3 + 2x_2x_3^2=0$

One possible parameterization is obtained by solving $f(x_1,x_2,x_3)=0$ for $x_2$, since the variable $x_2$ is linear , and by setting $x_1=s$ and $x_3=t$. Such parameterization is clearly birational, with inverse map $s=x_1$ and $t=x_3$. Explicitly,

$x_1 = s$

$x_2 = \frac{2t}{2s^2-5st+2t^2}$

$x_3 = t$

Another possible parameterization is given by,

$x_1 = \frac{s(-1 + 4t^2)}{3t}$

$x_2 = \frac{2(-1 + t^2)}{3st}$

$x_3 = \frac{2s(-1 + t^2)}{3t}$

In this case, I was not able to find the inverse map, and so I believe that this is not a birational parameterization. But how to prove it?

Thanks in advance,

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    $\begingroup$ Do you have a specific surface in mind? If so, is it possible to see the equation? Information on the degree of the surface and whether or not it is singular will help. $\endgroup$
    – Daniel Loughran
    Commented Dec 2, 2011 at 12:56
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    $\begingroup$ This is not a real question, since we need to see the surface. $\endgroup$
    – Igor Rivin
    Commented Dec 2, 2011 at 15:49
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    $\begingroup$ Map is 2-1 as $(s,t),(-s,-t)$ get mapped to the same point. I'll vote to close as too localized. $\endgroup$
    – Felipe Voloch
    Commented Dec 2, 2011 at 18:00
  • $\begingroup$ Thanks, Felipe. But is there a general criterion? $\endgroup$
    – pepper
    Commented Dec 2, 2011 at 23:00
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    $\begingroup$ Your surface has a few singularities at infinity. The usual way to construct a parametrisation for a singular cubic surface is to fix a singularity $P$ and consider the set of lines passing through each sufficiently general point and $P$. Such a line intersects the surface at only these two points (as long as the surface is not a cone) and yields a parametrisation. $\endgroup$
    – Daniel Loughran
    Commented Dec 3, 2011 at 0:11

1 Answer 1

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Your second "parametrization" is not birational.

Since you have a parametrization, it is easy to check. Indeed, the first parametrization is a birational map $\tau\colon \mathbb{A}^2\to S$, where $S$ is your surface. The second one is a rational map $\mu\colon \mathbb{A}^2\to S$. Because $\tau$ is birational, we can find that $\mu$ is birational if and only if $\tau^{-1}\circ \mu$ is birational.

This map is equal to

$(s,t)\mapsto (\frac{s(-1+4t^2)}{3t},\frac{2s(-1+t^2)}{3t})$

composing with the birational map $(s,t)\mapsto (3st,t)$ we obtain

$(s,t)\mapsto (s(-1+4t^2),2s(-1+t^2))$

which is clearly not birational ($(s,t)$ and $(s,-t)$ have the same image).

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