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What is the Krull dimension of the ring of continuous real-valued functions on an infinite compact Hausdorff space?

If the Krull dimension is not finite, we will say that it is infinite, and not try to define it as a cardinal or an ordinal.

Of course the Krull dimension of $C(X)$ depends a priori on $X$, but I'd be very happy if it could be computed even in the most particular cases.

Pierre-Yves Gaillard
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    There are infinitely many prime ideals even for $X$ a closed interval: you can find a lovely proof here: http://pages.uoregon.edu/adding/notes/cont_ag.pdf - it may take some extra arguments to extend this fact to a proof that the Krull dimension is infinite. – Joppy Jun 15 '17 at 14:02
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    Near-duplicate: https://math.stackexchange.com/questions/2078755/finding-a-space-x-such-that-dim-cx-n – Eric Wofsey Jun 15 '17 at 19:05

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The Krull dimension of $C(X)$ is infinite for any infinite compact Hausdorff space $X$. By my answer here, it suffices to show there exists a continuous function $f:X\to\mathbb{R}$ which is not locally constant. To show this, let $A=\{x_n\}$ be a countably infinite discrete subset of $X$. For each $n$, by Urysohn's lemma there exists a continuous function $f_n:X\to [0,1/2^n]$ such that $f_n(x_m)=0$ for $m\geq n$ and $f_n(x_m)=1/2^n$ for $m<n$. The sum $\sum f_n$ then converges uniformly on $X$ to a continuous function $f:X\to[0,2]$, and $f$ is injective when restricted to $A$. Now let $y\in X$ be any accumulation point of $A$. Since $f$ is injective on $A$ and every neighborhood of $y$ contains infinitely many points of $A$, $f$ is not constant in any neighborhood of $y$.

Pierre-Yves Gaillard
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Eric Wofsey
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  • Thank you very much for this awesome answer! I find the dichotomy in your other answer quite striking! I had never seen this condition of admitting a non locally constant real valued function before. I wonder if it is related to other properties (beside the Krull dimension of $C(X)$)? I agree that most of the job is done in the other answer, but I still find your argument above very ingenious (I wouldn't have found it by myself). – Pierre-Yves Gaillard Jun 16 '17 at 11:44
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    Yeah, I haven't seen that exact condition anywhere else either--I just encountered it as exactly what I needed when writing my other answer. – Eric Wofsey Jun 16 '17 at 16:07