# What is the ideal class group of the ring $\mathbb{R}[x,y]/(x^2+y^2-1)$?

I would like to find the ideal class group of $$\mathbb{R}[x,y]/(x^2+y^2-1)$$. The motivation of this question comes from totally outside of algebraic number theory -- I was playing around with Lissajous curves, i.e. curves parametrized by $$x=A\sin(t+t_0),y=B\sin(\alpha t)$$. In the book Mathematical Methods of Classical Mechanics, Arnold claims that when $$\alpha$$ is rational, such curves are actually algebraic, and left the proof of that claim as an exercise. My main idea to prove this was just to analyze the associated ring $$\mathbb R[\cos t,\sin t]\cong\mathbb R[x,y]/(x^2+y^2-1)=:A$$. As a finite integral extension of $$\mathbb R[x]$$, it must be a Dedekind domain, but I strongly suspect that it is not a PID. Is there any clear way to calculate the ideal class group here?

$$A=\Bbb{R}[x,y]/(x^2+y^2-1) = \Bbb{R}[\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}], \qquad Frac(A) = \Bbb{R}(\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}-1})=\Bbb{R}(t)$$

For $$f(t) \in \Bbb{R}(t)$$ if its only pole is at $$t= \pm i$$ of order $$k$$ then $$f(t) = (a\pm ib) (t\pm i)^{-k}+O( (t\pm i)^{1-k}) \implies f(t) - \frac{a}2\frac{1-t^2}{1+t^2}+\frac{b}2\frac{2t}{1+t^2}=O( (t \pm i)^{1-k})$$

thus by induction on $$k$$ there is $$g(t) \in A$$ such that $$f(t)-g(t)\in \Bbb{R}(t)$$ has no poles which means $$f(t)-g(t) \in \Bbb{R}, f(t) \in A$$. Whence $$A$$ is the subring of $$\Bbb{R}(t)$$ of rational functions with poles only at $$\pm i$$.

Its maximal ideals are the $$m_p= \{ f(t) \in \Bbb{R}(t), f(p) = 0\} \qquad \text{ for each } \ p \in (\Bbb{R}\cup \infty - (\pm i)) / Gal(\Bbb{C/R})$$ Moreover $$m_p^2= (h_p(t))$$ is principal: for $$p \in \Bbb{R}, h_p(t)= \frac{(t-p)^2}{t^2+1}$$, for $$p \in \Bbb{C}-(\pm i), h_p(t)= \frac{(t-p)^2(t-\overline{p})^2}{(t^2+1)^2}$$, for $$p = \infty$$, $$h_p(t) = \frac1{1+t^2}$$.

Thus every maximal ideal is inversible and $$A$$ is a Dedekind domain.

For two maximal ideals $$m_p,m_q$$ there exists $$u(t),v(t)\in A$$ such that $$u(t) m_p = v(t)m_q$$ iff $$p,q$$ are both real or both complex. If $$p$$ is real and $$q$$ is complex then $$um_p^2 = vm_q$$.

Thus the ideal class group is $$Cl(A)=\{ m_q,m_p\}\cong \Bbb{Z}/2\Bbb{Z}$$ Every non-zero ideal is invertible thus the fractional ideals form a group $$\mathcal{I}(A)$$ which is isomorphic to $$Div(\Bbb{P^1_R}) / <\pm i>$$ where $$\Bbb{P^1_R}=(\Bbb{R}\cup \infty)/ Gal(\Bbb{C/R})$$ and $$Div(\Bbb{P^1_R})=Div(\Bbb{P^1_C})^{Gal(\Bbb{C/R})}$$ and $$Cl(A)=\mathcal{I}(A)/\mathcal{P}(A)$$ is isomorphic to $$Pic(\Bbb{P^1_R}) / <\pm i>$$

To see that the class group is nontrivial is pretty easy: I claim that $$\langle x-1, y \rangle$$ is not principal. If $$\langle x-1,y \rangle = \langle f \rangle$$ for some polynomial $$f(x,y)$$, then $$f(\cos \theta, \sin \theta)$$ would vanish with multiplicity $$1$$ at $$\theta =0$$ and not at any $$0 < \theta < 2 \pi$$. But a periodic smooth function always has an even number of zeroes (counted with multiplicity).

Working a little harder, it is easy to see that there is a surjection from the class group to $$\mathbb{Z}/(2 \mathbb{Z})$$, sending ideals of the form $$\langle x-\cos \theta, y - \sin \theta \rangle$$ to $$1$$ and all other maximal ideals to $$0$$. Again, this map vanishes on principal ideals because a periodic smooth function always has an even number of zeroes.

I don't know how to check, without getting your hands dirty as in reuns answer, that this surjection is an isomorphism. I believe that all maximal ideals of $$A$$ are either of the form $$\langle x-\cos \theta, y - \sin \theta \rangle$$ or of the form $$\langle (\cos \theta) x + (\sin \theta) y - r \rangle$$ with $$r>1$$ (in which case the ideal is principal, and $$A/\mathfrak{m} \cong \mathbb{C}$$), but I don't know a slick proof of this.