Let $k$ be a field with $\operatorname{char} k \neq 2$. I think it's true that: $$\def\Proj{\operatorname{Proj}}\Proj k[x,y,z] / (x^2 + y^2 -z^2) \cong \Proj k[\lambda, \mu] = \mathbb{P}_k^1$$
The left hand side is the projective version of a "circle". My guess that these schemes are isomorphic comes from the following "isomorphism", which is constructed using the well known "line-intersection-trick" with the point "$(-1 : 0 : 1)$" on the circle: $$ \begin{align} (x : y : z) &\mapsto \begin{cases} (y : x + z) & \text{for }(x : y : z) \neq (-1 : 0 : 1) \\ (x - z : y) & \text{for }(x : y : z) \neq (1 : 0 : 1)\end{cases} \\ (\mu^2 - \lambda^2 : 2 \lambda \mu : \mu^2 + \lambda^2) &\leftarrow (\lambda : \mu) \end{align} $$ This is how one would give a morphisms of classical varieties. Applying those two morphisms after each other (and cancelling!) yields the identity, respectively. It makes me pretty confident that this yields an isomorphism of schemes. However, I don't know how to show this. Or at least, I don't know how to do it effectively. There should be a pretty elegant+fast way to do it from this prototype, no? But difficulties I came across are:
- We can not just show that these induce isomorphisms of graded rings, because they don't. This is for one thing because we need two charts (see second point), but also because we need cancellation. For example, applying our maps $\mathbb{P}_k^1 \to \Proj k[x,y,z]/I \to \mathbb{P}_k^1$ with the first chart yields $(\lambda : \mu) \mapsto (2\lambda \mu : 2 \mu^2)$. This is the identity, but to see it we need to cancel by $2\mu$, and in particular it doesn't define an isomorphism of rings (also not localized ones, no?).
- To define the morphism to $\mathbb{P}^1$, we need two charts. (I think it's not possible to do it with only one chart, no?) So the proof would need to use at least some localizations.
If anybody could show me how to elegantly+efficiently turn my sketch of the morphisms into a proof that the schemes are isomorphic, I would be very grateful!
