Exercise 6.5.F in Ravi Vakil's notes: Showing conic $x^2 + y^2=z^2$ in $\mathbb{P}_k^2$ is isomorphic to $\mathbb{P}_k^1$ I have been stuck on Exercise 6.5.F in Ravi Vakil's notes for a little while now, and I would greatly appreciate any hints/comments/solutions!
Let $k$ be a field that is not of characteristic $2$. I want to show that conic $x^2 + y^2=z^2$ in $\mathbb{P}_k^2 = Proj \ k[x,y,z]$ is isomorphic to $\mathbb{P}_k^1 = Proj \ k[u,v]$. 
 A: Here's a hint: if $k$ is algebraically closed, you can write down the map explicitly:
$$[u:v] \mapsto \left[u^2 - v^2 : 2uv : u^2 + v^2 \right]$$
You can make this rigorous scheme-theoretically by patching together maps on affine open sets.
A: The point $p=(1,0,1)$ is on the conic. Project from $p$ to $\mathbb{P}^1$, a line not passing through $p$, say $x=0$ and show that this gives an isomorphism.
A: Here are explicit local isomorphisms which you can easily check patch together.  Notice that the first inverse is given in vakil's notes above the exercise, and the rest are gotten the same way.  Also notice that you only need the first four of these to define the isomorphism since $\mathbb{P}^1$ (resp the conic $\mathcal{C}$) is covered by the domains of those isomorphisms.  
$$k[m]_{m^2+1} \cong \big(k[x,y]/(x^2+y^2-1)\big)_{x-1},$$
$$ m\mapsto y/(x-1),~~~~~~~~  (x,y)\mapsto (\frac{m^2-1}{m^2+1}, \frac{-2m}{m^2+1}). $$
$$$$
$$k[n]_{n^2+1}\cong \big(k[x,y]/(x^2+y^2-1)\big)_{x+1},$$
$$ n\mapsto -y/(x+1) ,~~~~~~~~~(x,y)\mapsto (\frac{1-n^2}{n^2+1}, \frac{-2n}{n^2+1}) .$$
$$$$
$$k[m]_{m^2-1}\cong \big(k[y,z]/(y^2-z^2+1)\big)_{z-1},$$
$$ m\mapsto y/(1-z),~~~~~~~~~ (z,y)\mapsto (\frac{m^2+1}{m^2-1}, \frac{-2m}{m^2-1}). $$
$$$$
$$k[n]_{1-n^2}\cong \big(k[y,z]/(y^2-z^2+1)\big)_{z+1},$$
$$  n\mapsto -y/(1+z), ~~~~~~~~~~(z,y)\mapsto (\frac{n^2+1}{1-n^2}, \frac{-2n}{1-n^2}) $$ 
$$$$
$$k[m]_m\cong \big(k[x,z]/(x^2-z^2+1)\big)_{x-z},$$
$$ m\mapsto 1/(x-z),~~~~~~~~~ (z,x)\mapsto (\frac{-m^2-1}{2m}, \frac{1-m^2}{2m}). $$
$$$$
$$k[n]_n\cong \big(k[x,z]/(x^2-z^2+1)\big)_{x+z},$$
$$ n\mapsto -1/(x+z), ~~~~~~~~~~~ (z,x)\mapsto (\frac{-n^2-1}{2n}, \frac{1-n^2}{2n}).$$
