I am new to the study of algebraic geometry and have decided to study it for a personal project of mine. As such I have been trying to understand what differentiates a rational and an irrational ruled surface.
What I know:
A ruled surface takes the parametric form of $x(u,v)=b(u)+vd(u)$
$b(u)$ is the directrix curve and $d(u)$ is the director curve.
What I think I know:
A ruled surface is rational if the genus of the directrix is 0
The genus of the directrix (base curve) can be calculated from its degree as such $g=0.5(d-1)(d-2)$
If the ruled surface can be parameterized such that $b(u)$ takes the form of a line then $b(u)$ is the line of striction
My question: is a ruled surface with the directrix defined by the parametric equation of a line a rational ruled surface even if the parameterization of the director is irrational?
For example. Let:
$b(u)=<X_0+x*u, Y_0+y*u, Z_0+z*u,>$
$d(u)=<\sqrt{ X_0+x*u }, \sqrt{ Y_0+y*u }, \sqrt{ Z_0+z*u },>$
Given b(u) is a line of degree 1 its genus is 0 thus the surface is a rational ruled surface. Is my logic correct or am I misunderstanding?
Can't you reparametrize so that the "director" is actually the generators (which are actually the lines, hence "ruled")? In any case, the answer is "Yes". If the directrix is paramerizable (i.e. genus 0) then the ruled surface is birational to $\mathbb P^1 \times \mathbb P^1$ so it is parametrizable if the ground field is algebraically closed.