Is the ring of holomorphic functions on $S^1$ Noetherian?

Let $S^1={\{ z \in \Bbb{C} : |z|=1 \}}$ be the unit circle.

Let $R= \mathcal{H}(S^1)$ be the ring of holomorphic functions on $S^1$, i.e. the ring of functions $f: S^1 \longrightarrow \Bbb{C}$ which can be extended to an holomorphic function on an open neighbourhood of $S^1$. My question is:

Is $R$ is a Noetherian ring?

What I tried:

We can think $R$ as a direct limit $$R= \lim_{\longrightarrow} \mathcal{H}(\Omega_n)$$ where $\Omega_n = \{ z \in \Bbb{C} : 1-\frac{1}{n} < |z| < 1+\frac{1}{n}\}$ denotes the open annulus of amplitude $2/n$ for $n\ge3$, and $$\mathcal{H}(\Omega_3) \longrightarrow \mathcal{H}(\Omega_4) \longrightarrow \mathcal{H}(\Omega_5) \longrightarrow \dots$$ is a direct system of rings whose maps are restrictions. Since restrictions are injective ring morphisms, we can think $R= \bigcup_n \mathcal{H}(\Omega_n)$

In particular $R$ is a Bezout domain (every finitely generated ideal of $R$ is principal). Now, if $R$ were Noetherian, then $R$ would be a PID.

And here I got stuck:

1. it could be possible that this is non-Noetherian since it is a union of non-Noetherian rings

2. on the other hand $S^1$ is compact, so every function $f \in R$ can only have a finite number of zeroes: this gives me intuition that maybe $R$ is a UFD.

• This could be relevant. Note that the only non-finitely generated ideals they found are trivial in the holomorphic setting. It remains true that the maximal ideals consist of the functions vanishing at some point, and I think they're generated by a single function. Any idea what the prime ideals look like? – Ben Jul 29 '15 at 19:51

The ring $R= \mathcal{H}(S^1)$ is both noetherian and factorial.

a) Noetherianity follows from Theorème (I,9) page 123 of this Inventiones paper by J.Frisch.
He proves that given a complex analytic space $X$ and a compact subset $K\subset X$, the ring $\mathcal{H}(K)$ is noetherian as soon as $K$ is real semi-analytic and has a basis of open Stein neighbourhoods.
That last condition of course holds in your case $K=S^1$ since any open subset of $X=\mathbb C$ is Stein !

b) That $R$ is factorial follows from theorem 1 page 89 of this article by Dales.
He proves that given a complex smooth analytic space (aka holomorphic manifold!) and a compact subset $K\subset X$ as in part a) one has the equivalence $$\mathcal{H}(K) \operatorname {is a UFD} \iff H^2(K,\mathbb Z)=0$$ Since $H^2(S^1,\mathbb Z)=0$ we see, by taking $K=S^1\subset X=\mathbb C$, that indeed $\mathcal{H}(S^1)$ is a UFD.

• Theorem 1 is pretty interesting, but if the author already knows it is bezout, therefore a PID and factorial, that's a little more accessible. Can't have too much variety, I guess – rschwieb Jul 30 '15 at 2:45
• Wow. I don't know any of these tools you are using, however I am glad for the exhaustive answer. So, if I had another compact connected set $K \subset \Bbb{C}$, does the same argument apply? Can we say that $\mathcal{H}(K)$ is a PID? – Crostul Jul 30 '15 at 11:33
• @user26857 Actually I had this little question which remained unanswered... But you are right: I will accept this answer. – Crostul Sep 15 '15 at 19:27
• @Crostul: As in Georges' answer, you need $K$ to be real semi-analytic. The closed unit disc is fine; the topologist's sine curve is not. – user98602 Sep 15 '15 at 19:36