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 $\mathbb C[S^1] = \mathbb C[X,Y]/(X^2+Y^2-1) \simeq \mathbb C[e^{\pm i \theta}]$).
Best Answer
Consider an ideal $\mathfrak I$ which is not contained in any $\mathfrak{m}_p$. Thus, for every $p\in S^1$, there is some function $f\in\mathfrak I$ with $f(p)\ne0$. Such an $f$ is nonzero in a neighbourhood of $p$. Since $S^1$ is compact, you can find $f_1$, …, $f_n\in\mathfrak I$ so that the $n$ sets $\{p\in S^1\colon f_k(p)\ne0\}$ cover $S^1$. Thus $f_1^2+\cdots+f_n^2\in I$ has no zeros, so it is invertible in $\mathfrak I$. (In the complex case, use $\bar f_1f_1+\cdots+\bar f_nf_n$ instead.)
Edited to add: Now consider any proper, nonzero ideal $\mathfrak I$. Let $\{p_1,\ldots,p_n\}$ be the common zeros of $\mathfrak I$ (it is finite because analytic, nonzero functions only have isolated zeros). Let $e_k$ be the minimal order of any zero at $p_k$. Clearly $\mathfrak I\subseteq \mathfrak m_1^{e_1}\cdots\mathfrak m_n^{e_n}$.
Edit the second: To prove the opposite inclusion in the even case (i.e., $e_1+\cdots+e_n$ even), note that any function with a zero of order $e_k$ at $p_k$ for each $k$, and no other zeros, is a generator of $\mathfrak m_1^{e_1}\cdots\mathfrak m_n^{e_n}$. I only need to find a member of $\mathfrak I$ with this property. First, find some member $f\in\mathfrak I$ with a zero of exactly order $e_k$ at $p_k$ for each $k$, but possibly other zeros as well. (Taking linear combinations of functions $f_k\in\mathfrak I$ with a zero of order $e_k$ at $p_k$ will do for this.) Assume $p_1$, …, $p_k$ are listed in clockwise order around $S^1$, and add indices modulo $n$ where necessary. Use obvious interval notation like $[p_k,p_{k+1}]$ for arcs of $S^1$. $f$ must suffer an even number of sign changes around the circle. Since we consider the even case, that means the arcs $[p_k,p_{k+1}]$ in which $f$ has an odd number of sign changes must itself be even in number. Multiply $f$ by some function which has a single zero in the interior of each such arc, and no other zeros. For economy of notation, call the result $f$ once more. So now $f$ has an even number of sign changes in each arc $[p_k,p_{k+1}]$. Moreover, the new $f$ has a sign change at an even number of the $p_k$. Pick a function $h\in C^\omega(S^1)$ with a single zero at each of those $p_k$ and no other zeros; negate it if necessary so that $f$ and $h$ have the same sign in a neighbourhood of each $p_k$.
Next, pick a function $g\in\mathfrak I$ which is positive in every open arc $(p_k,p_{k+1})$. (To do this, start by squaring any member of $\mathfrak I$. It has only a finite number of zeros outside the $p_k$, and we can eliminate those by adding squares of more members of $\mathfrak I$ which don't vanish at those points.) If $M$ is large enough, then $f+Mgh\in\mathfrak I$ will have no zero other than the $p_k$. (Sign considerations guarantee this in a neighbourhood of each $p_k$, and $gh$ is bounded away from zero elsewhere.) This is the function sought, and the proof of the even case is complete (modulo details that I might explain in the comments if asked).
For the odd case, pick any $f\in \mathfrak m_1^{e_1}\cdots\mathfrak m_n^{e_n}$. Since it must have an even number of zeros (couning multiplicity), it must have zero $q\notin\{p_1,\ldots,p_n\}$, or it has a zero of order $>e_k$ in some $p_k$ (in which case we put $q=p_k$). Let $\mathfrak I'=\{f\in\mathfrak I\colon f(q)=0\}$ (with the obvious modification if $q=p_k$, that $f$ have a zero of order $<e_k$ at $p_k$). Now $\mathfrak I'=\mathfrak m_1^{e_1}\cdots\mathfrak m_n^{e_n}\mathfrak m_q$ by the even case, and $f$ belongs to the right hand side. Thus $f\in\mathfrak I'\subset\mathfrak I$.
The above was for the real case; I see no difficulty in adapting it to the complex case.