# 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 $\mathbb{P}_k^1 = Proj \ k[u,v]$.

• The intersection of this with $(z \neq 1) \simeq \mathbb{A}^2$ is the circle $C$ that he talks about just before this. He's just built a morphism $C \to \mathbb{P}^1_k$. Maybe this extends?
– Hoot
Jul 31, 2015 at 1:24
• ive just realized that Takumi's map actually goes in the other direction. so probably it is the inverse to the one i define above. so it remains to check this as painlessly as possible. Also I wrote P^2 a couple of times when i meant P^1. Jul 4, 2019 at 17:20

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.

• I was wondering if you could possibly explain how does one come up with these maps? Thanks! Aug 1, 2015 at 0:20
• @JohnnyT. Which maps? Do you mean how one can find the map I wrote down? I'm pretty sure this is the map you'd get as the inverse of Mohan's projection map. If you mean the scheme-theoretic maps, the idea is to cover the conic with open affines, define the maps there according to the formula I gave, and then making sure this glues together. Aug 1, 2015 at 0:58
• @JohnnyT. Another way to construct it is to use Exercise 6.4.A in Vakil's notes (ver. 4/29/15), where the graded ring homomorphism in question would be $\frac{k[x,y,z]}{x^2+y^2-z^2} \to k[u,v]$ given by $x \mapsto u^2 - v^2$, $y \mapsto 2uv$, $z \mapsto u^2 + v^2$. Now the image of the irrelevant ideal is $(u^2-v^2,2uv,u^2+v^2) = (u^2,v^2,uv)$, for which $V(u^2,v^2,uv) = \emptyset$, and so you get a well-defined map $\mathbf{P}^1 \to \operatorname{Proj} \frac{k[x,y,z]}{x^2+y^2-z^2}$. Aug 1, 2015 at 1:00

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.

• Could you possibly explain a little more what "project from $p$ to $\mathbb{P}^1$" by any chance? Thank you! Aug 1, 2015 at 16:12
• If $p\in\mathbb{P}^2_k$ is a $k$-rational point and $L=\mathbb{P}^1_k$ is a line in the plane, we can project from $p$ to $L$ to get a morphism $\mathbb{P}^2-\{p\}\to L$ as follows. If $q\neq p$ is a point in the plane, take the line joining $p,q$ and it intersects $L$ in a point $r$. The map taking $q\mapsto r$ is called the projection from $p$. Aug 1, 2015 at 16:45

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.

$$k[m]_{m^2+1} \cong \big(k[x,y]/(x^2+y^2-1)\big)_{x-1},$$

$$m\mapsto y/(x-1),~~~~~~~~ (x,y)\mapsto (\frac{m^2-1}{m^2+1}, \frac{-2m}{m^2+1}).$$

 $$k[n]_{n^2+1}\cong \big(k[x,y]/(x^2+y^2-1)\big)_{x+1},$$

$$n\mapsto -y/(x+1) ,~~~~~~~~~(x,y)\mapsto (\frac{1-n^2}{n^2+1}, \frac{-2n}{n^2+1}) .$$

 $$k[m]_{m^2-1}\cong \big(k[y,z]/(y^2-z^2+1)\big)_{z-1},$$

$$m\mapsto y/(1-z),~~~~~~~~~ (z,y)\mapsto (\frac{m^2+1}{m^2-1}, \frac{-2m}{m^2-1}).$$



$$k[n]_{1-n^2}\cong \big(k[y,z]/(y^2-z^2+1)\big)_{z+1},$$

$$n\mapsto -y/(1+z), ~~~~~~~~~~(z,y)\mapsto (\frac{n^2+1}{1-n^2}, \frac{-2n}{1-n^2})$$



$$k[m]_m\cong \big(k[x,z]/(x^2-z^2+1)\big)_{x-z},$$

$$m\mapsto 1/(x-z),~~~~~~~~~ (z,x)\mapsto (\frac{-m^2-1}{2m}, \frac{1-m^2}{2m}).$$

 $$k[n]_n\cong \big(k[x,z]/(x^2-z^2+1)\big)_{x+z},$$

$$n\mapsto -1/(x+z), ~~~~~~~~~~~ (z,x)\mapsto (\frac{-n^2-1}{2n}, \frac{1-n^2}{2n}).$$