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?
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6$\begingroup$ What continuous functions? Between what and what?... $\endgroup$ – DonAntonio Nov 17 '13 at 5:31
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1$\begingroup$ ay ay ay, what a mess. $\endgroup$ – Nick Nov 17 '13 at 9:37
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$\begingroup$ 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
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$\begingroup$ @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
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I assume you are referring to the space $C(X)$ of continuous complex/real valued functions on some compact Hausdorff space $X$. Check that:
- If $X$ is finite, then $C(X)$ is Noetherian.
- 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.
- If $R = C(X)$, then $\text{mspec}(R) \cong X$.
Conclude that, $C(X)$ is Noetherian iff $X$ is finite.
answered Nov 17 '13 at 6:11
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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
