0
$\begingroup$

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?

$\endgroup$
  • 6
    $\begingroup$ What continuous functions? Between what and what?... $\endgroup$ – DonAntonio Nov 17 '13 at 5:31
  • 1
    $\begingroup$ ay ay ay, what a mess. $\endgroup$ – Nick Nov 17 '13 at 9:37
  • $\begingroup$ It's certainly clearer, but is it a legitimate edit? @JonathanY., how do you know this is what OP meant? $\endgroup$ – tomasz Nov 24 '13 at 3:14
  • $\begingroup$ @tomasz because OP posted a duplicate. I flagged that for moderation and also explained what happened in a request to reopen on meta. $\endgroup$ – Jonathan Y. Nov 24 '13 at 6:31
3
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ What about when we do not assume $X$ is compact? The ring is well-defined in any case. $\endgroup$ – Henno Brandsma Nov 17 '13 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.