If $(X,d)$ is a finite metric space , then is every prime ideal of $C(X, \mathbb R)$ maximal ? The thing is , since $X$ is finite , so it is compact , so ideal $M$ is maximal iff it is of the form $M_a:=\{f \in C(X, \mathbb R) : f(a)=0 \}$ , but I don't know the structure of prime ideals , please help . Thanks in advance .
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In your definition of $M_a$, do you mean $x=a$? – Clément Guérin Mar 23 '15 at 15:45
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@ClémentGuérin : Yes , sorry , I edited – Mar 23 '15 at 15:49
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23 questions within 20 minutes, all of which seem homework-like... I think you should consider slowing down with asking questions and also showing some more of your own significant attempts at answering these questions. – Dan Rust Mar 23 '15 at 15:52
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No. These two questions discuss the characterization of zero dimensional rings:
and
If every prime ideal is maximal, what can we say about the ring?
So in response to your question: it is easy to show your ring has no nonzero nilpotent elements. Thus maximal ideals will be prime iff the ring is von Neumann regular, but you can show this is not the case.
There are proofs somewhere on the site, but you can easily reconstruct one. The main thing to observe is what the idempotent elements of this ring must look like.
Finding nonmaximal primes is not trivial has been discussed before.
