Degree 2 Fermat curve I'm trying to solve the following exercise:
Prove that the variety $V\subset \mathbb{CP}^2$ defined by $x^2+y^2+z^2=0$ is isomorphic to $\mathbb{CP}^1$.
What I've done: I tried to define an explicit isomorphism $\mathbb{CP}^1\rightarrow V$ as follows: $[x,y]\mapsto [x+y,x-y,\sqrt2 \ i\sqrt{x^2+y^2}]$. The inverse is given by $[r,s,t]\mapsto [\frac{r+s}{2},\frac{r-s}{2}]$. The problem is that - because of the square root in the first map, this will not map regular functions to regular functions.
How do I find the correct isomorphism? Is finding it explicitly even the "good" way, or is there some more elegent argument that I'm missing?
Thanks for your help,
Paul
 A: My teacher in high school taught us the amazing equality $(u^2-v^2)^2+(2uv)^2=(u^2+v^2)^2$ .
 By  infinitesimal tinkering it yields the parametrization $\mathbb P^1\to V$ you want: $$[u:v]\mapsto [x:y:z]=[u^2-v^2:2uv: i(u^2+v^2)]$$ Try to show that this parametrization is an isomorphism  
I write this answer as an homage to that dear teacher of so long ago, Monsieur Devroegh. 
Edit
For those who couldn't find the inverse of the parametrization above , here it is (recall that $x^2+y^2+z^2=0$): $$ V\ni[x:y:z]\mapsto [u:v]=[-y: x+iz ]\in \mathbb P^1$$ and $[1:0:i]\mapsto [1:0], [1:0:-i]\mapsto [0:1]$
A: Here's one old trick. Up to isomorphism (replacing $z$ by $iz$), we may consider the curve $z^2 = x^2 + y^2$ and the point $P = [1:0:1]$ on it. On the affine piece with $z\neq0$, the equation becomes that of a circle $1=X^2+Y^2$, where $X=x/z$ and $Y=y/z$, and we may define a rational map to the projective line by sending a point $Q$ on the circle to the slope of the line through $P = (1,0)$ and $Q$. This map extends to a map on the whole curve. 
