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Let $A = C^\omega(S^1)$ (resp. $C^\omega_{\mathbb C}(S^1)$) the ring of real-analytic real-valued (resp. complex valued) functions on the circle.

These rings have maximal ideals $\mathfrak m_p = \left \{ f \in A \, | \, f(p) = 0\right \}$ (for $p \in S^1$) and ideals $\mathfrak m_{p_1}^{e_1} \mathfrak m_{p_2}^{e_2} \cdots \mathfrak m_{p_n}^{e_n}$ (the ideal of functions having prescribed zeroes).

What I would like to prove is that there are no other ideals.

That would give a nice example of Dedekind rings: $C^\omega_{\mathbb C}(S^1)$ would be a PID (because it is not hard to give functions generating the aforementioned ideals) but $C^\omega(S^1)$ would be an example of Dedekind ring $A$ with $\mathrm{Cl}(A) = \mathbb Z/2$ (essentially because of the intermediate value theorem: only ideals $\mathfrak m_{p_1}^{e_1} \mathfrak m_{p_2}^{e_2} \cdots \mathfrak m_{p_n}^{e_n}$ with $e_1 + \cdots + e_n$ even are principal).

I feel like such a result, if true, must be classical, but I was unable to find references on those rings (unlike their algebraic counterpart: trigonometric polynomial rings $\mathbb R[S^1] = \mathbb R[X,Y]/(X^2+Y^2-1) \simeq \mathbb R[\cos \vartheta, \sin \vartheta]$ and $\mathbb C[S^1] = \mathbb C[X,Y]/(X^2+Y^2-1) \simeq \mathbb C[e^{\pm i \theta}]$).

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    While thinking about expanding my answer, I had a thought: I don't think that $\mathfrak{m}_p$ has a generator in the real case (no problem in the complex case). After all, the generator could only have a simple zero at $p$, so it changes sign there. But then it needs another zero for another sign change. (I noted this in a comment to the first answer, which has since been deleted by its author.)2012-06-06
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    $A$ is a Dedekind domain iff it is Noetherian and the localization at each maximal ideal is a discrete valuation ring. The latter condition is easy to show at this point, I think. Can we easily show it to be Noetherian?2012-06-06
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    @Harald: "I don't think that mp has a generator in the real case". That's what I alluded to in penultimate paragraph: I think the class group of this Dedekind-to-be ring is of order two, precisely because a periodic function has to change sign an even number of times.2012-06-06
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    Indeed; another point I had missed due to haste. Sorry not to have noticed it.2012-06-06

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