Let's take a smooth algebraic curve $C$ over a field $k$. We know that its coordinate ring $\mathcal{O}$ is a Dedekind domain.
Now, isomorphic curves should have isomorphic Dedekind domains attached to them, right? But I'm confused by the following example:
$$\begin{align}
C_1 &:= \operatorname{Spec}\mathbb{R}[x,y]/(x^2 + y^2 – 1) \\
C_2 &:= \mathbb{P}_\mathbb{R}^1
\end{align}$$
As far as I can see:
- $C_1$ is isomorphic to $C_2$ via the usual line intersection trick
- $C_1$ has coordinate ring $\mathcal{O}_1 := \mathbb{R}[x,y]/(x^2 + y^2 – 1)$
- $C_2$ has coordinate ring $\mathcal{O}_2 := \mathbb{R}[t]$
- $\mathcal{O}_1$ is not isomorphic to $\mathcal{O}_2$, because $\mathcal{O}_2$ has class number 1 whereas $\mathcal{O}_1$ has class number 2
(That $\mathcal{O}_1$ has class number 2 isn't obvious, but I'm pretty sure it's true because it was claimed on a couple of MathOverflow posts; the non-trivial ideal class corresponds to the Moebius strip line bundle, apparently.)
I can't quite see where my mistake is. I think it might be connected to some "affine vs. projective" issues. If one of you could shine some light on it, that would be great!
Best Answer
It doesn't make sense to talk about coordinate rings of non-affine varieties. The ring of global section of any projective variety over a field is a finite dimensional algebra over the base field. Your description of $C_2$ can't be correct.
Moreover, you are conflating schemes and their points over a field. The $\mathbb{R}$ -points of your $C_1$ and $C_2$ might be in bijection, even homeomorphic as topological spaces, but these two curves are not isomorphic as schemes. $\operatorname{Spec} A$ and $\operatorname{Spec} B$ may easily have the same points over some field while not being isomorphic as affine schemes. This boils down to $\mathbb{R}$ not being algebraically closed.
Also, $\mathbb{R}[t]$ is a PID and hence has class number 1.