# Spectrum of the commutative ring of $R$-valued functions on a compact Hausdorff space

In Atiyah and Macdonald chapter 1, it is an exercise to prove the following:

Let $X$ be a compact Hausdorff space. Let $C(X)$ be the commutative ring of $\mathbb{R}$-valued continuous functions on $X$. Then $\operatorname{Max}(C(X))$, the subspace of $\operatorname{Spec}(C(X))$ consisting of maximal ideals, is homeomorphic to $X$.

An important step in this proof is demonstrating that every maximal ideal of $C(X)$ is of the form $\left\{f \in C(X) : f(x) = 0\right\}$ for some $x\in X$. I understand the proof, but I am struggling to assign geometric intuition to $C(X)$. (Ideally I would prefer to visualize the associated topological spaces in the context of a familiar manifold like $S_n$.)

What does $\operatorname{Spec}(C(X))$ look like? I'm not sure how to approach this question, but my conjecture is that it's not that much different from $\operatorname{Max}(C(X))$. This is my reasoning:

1. A prime ideal of $C(X)$ should probably be identified by its zeros on $X$. I'm not sure how to prove this exactly, but it seems to me that requiring zeros is the only way to ensure that the a subset of $C(X)$ is closed as a $C(X)$-module.
2. If you cut out an ideal $\mathfrak{a} \subsetneq C(X)$ by $\left\{f\in C(X) : f(x) = 0 \ \forall x\in Z\right\}$, $Z \subseteq X$; $\mathfrak{a}$ cannot possibly be prime if $Z$ is finite. (The product of two functions with one zero can have two zeros.)
3. As above, if $Z$ is infinite, it seems that $\mathfrak{a}$ is in fact prime.

So I suspect that $\operatorname{Spec}(C(X))$ is, as a poset, isomorphic to $\left\{A \in \mathscr{P}(X) : A \ \text{infinite}\right\}$. Am I on the right track?

Any geometric/topological insights?

• Paper reference, see also Gilman and Jerrison, Rings of Continuous functions, a classic in this area. – Henno Brandsma Jun 18 '18 at 22:09
• Counterexample: if $X = Z = [-1, 1]$, then the ideal is $\{ 0 \}$. But $(x - |x|) (x + |x|) = 0$ so that ideal is not prime. – Daniel Schepler Jun 19 '18 at 0:31
• It is rather hard to give a geometric insight on something that shouldn't be thought of geometrically (the algebraic structure has more benefit imo). But one possibility to think how this would look like if $X$ were a finite discrete space (then $C(X)$ reduces to the ring $\mathbb R^n$) – quantum Jun 19 '18 at 7:05
• What is $\mathcal P(X)$ ? the power set of $X$? You are almost right, but I think you have to think subsets of the powerset (for instance if $X$ is 0-dimensional then these are the ultrafilters of $X$). – quantum Jun 19 '18 at 7:13
• @quantum yes I mean the power set. Can you explain how the subsets of the power set are relevant? – Simon Kuang Jun 19 '18 at 7:18

The prime spectrum is very similar to the maximal spectrum, but not quite in the way you imagine. For any $x \in X$, let $M_x$ denote the (maximal) ideal of all functions that vanish at $x$ and $O_x$ the ideal of all functions that vanish on a neighbourhood of $x$. Since $X$ is compact, there is for any prime ideal $P$ an $x$ such that $P \subset M_x$. It is easy to show that $P$ in turn must contain $O_x$. (Use Urysohn's lemma to show that if $f$ vanishes on a neighbourhood of $x$, there is a function $g$ such that $fg = 0$ and $g \notin M_x$)
It is also not hard to see that $O_x \not\subset M_y$ unless $x = y$, so any prime ideal $P$ is contained in a unique maximal ideal $\mu(P)$. It can be shown that the mapping $\mu: \operatorname{Spec} C(X) \to \operatorname{Max} C(X)$ is in fact a strong deformation retraction.