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## Spectra of rings of continuous functions

An analyst thinks of the ring $\mathbb{C}[ t]$ of polynomials as a useful tool because, on intervals, it is dense in the continuous functions $\mathbb{R} \to \mathbb{C}$ in the uniform topology. If we want to understand the relationship between $\mathbb{Z}$ and polynomial rings in a more general context, it might pay off to expand our scope from polynomial rings to more general types of well-behaved rings.

The rings we’ll be considering today are the commutative rings $C( X) = \text{Hom}_{\text{Top}}(X, \mathbb{R})$ of real-valued continuous functions $X \to \mathbb{R}$ on a topological space $X$ with pointwise addition and multiplication. It turns out that one can fruitfully interpret ring-theoretic properties of this ring in terms of topological properties of $X$, and in certain particularly nice cases one can completely recover the space $X$. Although the relevance of these rings to number theory seems questionable, the goal here is to build geometric intuition. You can consider this post an extended solution to Exercise 26 in Chapter 1 of Atiyah-Macdonald.

Basic properties

$C( X)$ cannot have nonzero nilpotents: if a function has a nonzero value somewhere so does all of its powers. So these rings cannot be too badly behaved; their nilradical is zero. This will be true of all the rings we are concerned with.

$C( X)$ has nontrivial idempotents if and only if $X$ is disconnected: if a nonconstant function satisfies $f^2 = f$, then $X = X_0 \cup X_1$ where $f( X) = 0$ on $X_0$ and $f( X) = 1$ on $X_1$. The preimage of neighborhoods of both of these points is open, so $X_0, X_1$ are disjoint open sets. Conversely if $X$ is disconnected such a function always exists. Ring-theoretically this means $C( X) = C(X_0) \oplus C(X_1)$. (Recall that a continuous function on a space is determined by a tuple of continuous functions on each connected component.)

The units of $C( X)$ are precisely those functions which are nonzero everywhere. This is because the inverse function $x^{-1} : \mathbb{R} - \{ 0 \} \to \mathbb{R} - \{ 0 \}$ is continuous, so the inverse of a continuous function which is nonzero everywhere is a continuous function which is nonzero anywhere. This unit group is huge, which is something to keep in mind if we want a sensible notion of unique factorization for $C( X)$.

Given a subset $S$ of $X$, the set of all functions such that $f(s) = 0 \forall s \in S$ is an ideal of $C( X)$. By continuity it suffices to only consider $S$ closed. When $S$ is a single element we get ideals $m_x$ corresponding to $\mathbb{R}$-algebra homomorphisms $e_x : f \mapsto f( X)$ into $\mathbb{R}$, which is an integral domain (in fact a field). It follows that the ideals $m_x$ are prime. But now that we’ve seen so many (surjective) homomorphisms into fields we might as well name them separately: they are the maximal ideals. Among prime ideals, the maximal ideals are distinguished by the fact that no other proper ideal (i.e. one which is not the whole ring) contains them. In rings like $\mathbb{Z}$ and $\mathbb{C}[ t]$ every nonzero prime ideal is maximal, but this is not always true. For example, if $R = \mathbb{C}[x, y]$ then the ideal generated by $x$ is prime but not maximal, since $R/( X) \simeq \mathbb{C}[y]$ is an integral domain but not a field.

Maximal spectrum

The upshot of all this is that we can recover information about the points of $X$ by looking at some of the maximal ideals of $C( X)$. This is very important, and raises some questions:

• Are these the only maximal ideals of $C( X)$?
• When are two maximal ideals $m_x, m_y, x \neq y$ the same?

We’d like the second condition to hold so that we have some hope of identifying $X$ with (possibly a subset of) the maximal ideals. For this to be true there must be some function $f$ which vanishes at $x$ but not at $y$. The preimages of disjoint neighborhoods of $f( X), f(y)$ are open and disjoint, so it follows that $X$ must be Hausdorff. But we don’t know yet if this is sufficient.

We’d like the first condition to hold so that we can identify $X$ with the maximal ideals (once we know when the second condition holds). If $I$ is some other ideal which isn’t contained in $m_x$ for any $x$ (hence which is contained in some other maximal ideal), then for every $x$ there is some function $f_x \in I$ which is nonzero at $x$. What we’d like to do is to use these functions to build a function $f \in I$ which is nonzero everywhere, i.e. a unit, since that will show that $m$ must be the whole ring. If finitely many of them sufficed, we could take the sum of their squares.

Now, notice that the preimages $f_x^{-1}(\mathbb{R} - \{ 0 \})$ form an open cover of $X$. If $X$ were, say, compact, then we could take finitely many $f_x$ and find a unit in $m$. And for now this is exactly what we’re going to do. With this additional hypothesis we have now ruled out the possibility of any weird maximal ideals running around, so to every point $x \in X$ we can associate a maximal ideal $m_x$, and this map is surjective. It remains to determine when this map is bijective. The condition that there exists a continuous function to $\mathbb{R}$ taking different values at two points is called “separation by a continuous function,” and a sufficient condition for this to be possible is the subject of a standard fact in topology.

Urysohn’s Lemma: Let $X$ be a normal space, i.e. disjoint closed sets in $X$ can be separated by neighborhoods. Then disjoint closed sets can be separated by a continuous function $X \to \mathbb{R}$.

Since closed subsets of a compact space are compact and disjoint compact sets can be separated by neighborhoods in a Hausdorff space, it follows that compact Hausdorff spaces are normal.

Corollary: If $X$ is compact Hausdorff, then $x \mapsto m_x$ is a bijection between $X$ and the set $\text{MaxSpec } C( X)$ of maximal ideals of $C( X)$.

$\text{MaxSpec}$ stands for maximal spectrum, a terminology hailing from functional analysis.

The obvious question from here is whether we can do better than recover just the points of $X$: can we recover its topology? Well, there are open sets we know we have to have. Our goal is to find a topology on $\text{MaxSpec } X$ such that $C( X)$ is precisely the set of all continuous functions on $\text{MaxSpec } X$. But given a set of functions on a space there is a unique coarsest topology making all of those functions continuous; this is just the initial topology obtained by taking the inverse image of all open sets in sight.

Proposition: If $X$ is compact Hausdorff and $\text{MaxSpec } X$ is given the initial topology coming from $C( X)$, then $x \mapsto m_x$ is a homeomorphism.

Proof. This is not really a statement about $\text{MaxSpec}$; it suffices to show that the topology on $X$ is initial for the continuous functions $X \to \mathbb{R}$. To do this we’ll show that the zero sets $\{ f^{-1}(0) | f \in C( X) \}$ form a basis for the closed sets of $X$, as follows: given a closed set $V \subset X$ and a point $x \not \in V$ we can find, by Urysohn’s lemma, a function $f$ such that $f( X) \neq 0$ but such that $f^{-1}(0) \supset V$. Taking the intersection over all $x \not \in V$ gives the desired result.

(Note that we used a weaker condition than the conclusion of Urysohn’s lemma, which was that points and closed sets could be separated. This condition is called complete regularity, and it turns out to be much nicer than normality.)

Primes and ideals

Again let’s summarize the situation.

1. Ideals in $C( X)$ arise as sets of the form “functions satisfying $f( X) = 0$ for all $x$ in some subset of $X$,” which is without loss of generality closed by continuity.
2. Maximal ideals (nonzero prime ideals) occur by choosing a single point. When $X$ is compact Hausdorff these are the only maximal ideals, and one can recover the topology on $X$ as the initial topology on $\text{MaxSpec}$. This description requires only knowledge about $C( X)$.
3. More generally, closed sets in $\text{MaxSpec}$ arise as sets of the form “points satisfying $f( X) = 0$ for all $f$ in some subset of $C( X)$,” which is without loss of generality an ideal. Again, this description requires only knowledge about $C( X)$. This is the Zariski topology on $\text{MaxSpec}$.

This suggests a very interesting way to study more general commutative rings: think of them as continuous functions on their maximal spectrum in the Zariski topology!

But we have to be careful. The property of $C( X)$ that makes it a true ring of functions is that its residue fields are all the same – they’re all $\mathbb{R}$. This is also true for the polynomial ring $\mathbb{C}[ t]$ – the residue fields are all $\mathbb{C}$. But for $\mathbb{Z}$ the residue fields are all different: the residue field at $(p)$ is $\mathbb{F}_p$. So integers do form functions on the set of primes, but not into the same set for each prime…

In the next post we’ll see how this works for polynomial rings $\mathbb{C}[x_1, ... x_n]$.

### 4 Responses

1. […] keep in mind as examples in this post. Keep in mind the following, which was discussed previously here and […]

2. […] with the argument from this previous blog post about recovering the topology of a compact Hausdorff space from the ring . There we argued that […]

3. Is it sufficient only to consider the maximal spectrum in the general case?

The weak* topology on the space of continuous homomorphisms $C(X) \to \mathbb{C}$ (which are uniquely determined) is also the same as the $MaxSpec$ topology in this case. I think this is how the theorem that a compact Hausdorff space is determined by its ring of continuous functions is usually proved (i.e. using Banach algebras) even though it is entirely the same thing.

• The general case is odd because points being separated by a continuous real-valued function is a strictly stronger condition than Hausdorff, so $\text{MaxSpec}$ ends up identifying certain points.

I rather like the Banach algebra approach but it leads in sort of a different direction, and I want to get to varieties at some point.