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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?

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

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 $Pic(\Bbb{P^1_R}) / <\pm i>$

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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.

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