# When is the ring of continuous functions Noetherian?

Let $X$ be a topological space. Is there any topological property on $X$ that be equivalent to $C(X,\mathbb R)$ being a noetherian ring?

• What continuous functions? Between what and what?... – DonAntonio Nov 17 '13 at 5:31
• ay ay ay, what a mess. – Nick Nov 17 '13 at 9:37
• It's certainly clearer, but is it a legitimate edit? @JonathanY., how do you know this is what OP meant? – tomasz Nov 24 '13 at 3:14
• @tomasz because OP posted a duplicate. I flagged that for moderation and also explained what happened in a request to reopen on meta. – Jonathan Y. Nov 24 '13 at 6:31

I assume you are referring to the space $C(X)$ of continuous complex/real valued functions on some compact Hausdorff space $X$. Check that:
1. If $X$ is finite, then $C(X)$ is Noetherian.
2. If $R$ is a commutative ring, let $Y := \text{mspec}(R)$, the collection of maximal ideals of $R$ with the Zariski topology. If $R$ is Noetherian, then every subset of $Y$ is compact. In particular, if $Y$ is Hausdorff, then $Y$ is finite.
3. If $R = C(X)$, then $\text{mspec}(R) \cong X$.
Conclude that, $C(X)$ is Noetherian iff $X$ is finite.
• What about when we do not assume $X$ is compact? The ring is well-defined in any case. – Henno Brandsma Nov 17 '13 at 12:37