If $(X,d)$ is a finite metric space , then is every prime ideal of $C(X, \mathbb R)$ maximal?

Question

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 .

Answer 1

No. These two questions discuss the characterization of zero dimensional rings:

If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).

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.

Prime ideals in $C[0,1]$