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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.) –  Harald Hanche-Olsen Jun 6 '12 at 11:25
    
$A$ 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? –  Hurkyl Jun 6 '12 at 12:44
    
@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. –  PseudoNeo Jun 6 '12 at 13:39
    
Indeed; another point I had missed due to haste. Sorry not to have noticed it. –  Harald Hanche-Olsen Jun 6 '12 at 13:55
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1 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.

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Great. That shows that only the $\mathfrak m_p$ are maximal. But that doesn't show that the products $\mathfrak m_{p_1}^{e_1} \cdots \mathfrak m_{p_n}^{e_n}$ are the only ideals, does it? –  PseudoNeo Jun 6 '12 at 10:45
    
You're right; I had overlooked that part (I was replying in a hurry because of a meeting I had to go to). I'll give it some thought. –  Harald Hanche-Olsen Jun 6 '12 at 11:12
    
Added a bit more detail. I noted that I had forgotten to upvote your question. How rude! –  Harald Hanche-Olsen Jun 6 '12 at 11:40
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The part about $\sum_k \overline{f_k} f_k$ in the complex case could use some more explanation to show that it is analytic on $S^1$, for example use that $\overline{z} = z^{-1}$ for $z \in S^1$. –  WimC Jun 6 '12 at 12:04
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@WimC: Maybe. I tend to think of these functions as periodic, real analytic functions on the real line, and then this becomes rather trivial (a complex function on the real line is real analytic iff the real and imaginary parts are both real analytic). –  Harald Hanche-Olsen Jun 6 '12 at 13:15
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