Conic in projective plane isomorphic to projective line

Question

I've got to show that any irreducible conic in $\mathbb{P^2(\mathbb{K})}$ where $\mathbb{K}$ is algebraically closed is isomorphic as a quasiprojective variety to the projective line. I wrote the conic as equivalent to the zero locus of $ x^2 +y^2+z^2$ in $\mathbb{P^2(\mathbb{K})}$. I tried to use the standard affine charts to cover this simpler variety and parametrize it but i didn't manage to do that. What am i missing?

Answer 1

For $\mathbb K=\mathbb R$ your specific conic $x^2+y^2+z^2=0$ seems like a poor choice, since it does not contain any points at all.

The general approach I'd suggest is the following: pick three distinct points $A,B,C$ on the conic to establish a projective basis, just as you would pick three points on a projective line to establish a basis there. Also pick a fourth point $P$ on the conic; it does not matter where as long as it is distinct from $A,B,C$. Then you can characterize every point $D$ on the conic via the cross ratio seen from $P$:

$$f(D)=(A,B;C,D)_P=\frac{[A,C,P][B,D,P]}{[A,D,P][B,C,P]}$$

The value of this fraction does not depend on $P$, except that for $D\sim P$ you would divide zero by zero. But choosing a different $P$ can remove this singularity. You could treat numerator and denominator as two homogeneous coordinates describing a point $D$ on a projective line.

For the special case of the conic $x^2+y^2-z^2=0$ i.e. the unit circle, with $A=[-1:0:1],B=[1:0:1],C=[0:1:1]$ this corresponds to the tangent half-angle substitution:

$$f(D)=\frac{u}{v} \quad\Leftrightarrow\quad D \sim [v^2-u^2:2uv:v^2+u^2]$$

If you really want a map for $x^2+y^2+z^2=0$ over a field where this has some points to begin with, using those points as $A,B,C,P$ should give you a suitable parametrization, too. Over $\mathbb C$ you could multiply all $z$ coordinates by $i$, as Jan-Magnus Økland suggested in a comment.

See also this answer of mine for some more background, based on the related question of how to parametrize a conic. Also see this question for how to map one conic to another, to see why having an isomorphism for one non-degenerate conic will yield an isomorphism for others as well.

Answer 2

I cannot comment so I might as well just write a simple answer for the case when characteristic of K is 2. In this case, write $f = x^2+y^2+z^2 = (x+y+z)^2$ which gives a map into $\mathbb{P}^1$ directly, $x\to t,\ y\to s, \ z\to -t-s$. In this case $f$ is not irreducible and the conic can be identified as the zero locus of $g = x+y+z$. The map above gives an isomorphism of this zero locus with the projective line.

Edited:

As pointed out by Captain Lama, the conic would rather look like $x^2+yz$. I would like to rectify the error by introducing a parametrization for this conic in the field of characteristic $2$. Let $i$ be the root of the polynomial $x^2+1=0$ (using the fact that $K$ is an algebraically closed field). Then the parametrization below would work just fine:

$x \to i\cdot t+s$

$y \to t-s$

$z \to t+s$