Amazed by how ring theory can be linked up to topology

I am reviewing some fundamental algebra, and I just learnt something beautifully suggesting the connection of ring theory and geometry, which made me eager to learn some algebraic geometry (which I was not very interested when I was in undergrad). It is an exercise in Aluffi’s book¹.

Lets’ jump into it.

The problem

Let $K$ be a compact topological space, and let $R$ be the ring of continuous real-valued functions on $K$, with addition and multiplication defined pointwise.
(i) For $p \in K$, let $M_p = \{f \in R | f(p) = 0\}$. Prove that $M_p$ is a maximal ideal in $R$.
(ii) Prove that if $f_1, \dots , f_r \in R$ have no common zeros, then $(f1, \dots , fr)$ = $(1)$.
(Hint: Consider $f^2_1 + \dots + f^2_r$ .)
(iii) Prove that every maximal ideal $M$ in $R$ is of the form $M_p$ for some $p \in K$.
(Hint: You will use the compactness of $K$ and (ii).)

Conclude that $p \mapsto M_p$ defines a bijection from $K$ to the set of maximal ideals of $R$.

…and for a complete picture, here is a related problem ² (also in the same book) to show that compactness of $K$ is necessary:

Let $R$ be the ring of continuous from $\mathbb{R}$ to $\mathbb{R}$. Show that there is a maximal ideal of $R$ such that it is not in the form of $M_p$.

Thought process (in a separate section so that you could skip if you are not interested)

(i) is trivial by constructing a ring homomorphism $\phi$ from $R$ to $\mathbb{R}$ such that $\phi (f) = f(p)$.

(ii) is also straight forward:
First notice that $1$ in $R$ is the function $Id$ on $K$ such that $Id(x) = 1 \forall x \in K$. So for any $f \in R$ if $f$ has no zeros it will have an inverse ($\dfrac{1}{f}$). So what happens if $f_1, \dots , f_r \in R$ have no common zeros? They can “cover” each other. Though it is not simply $f_1 + \dots + f_r$ since the “positive plus negative” cases will cause pathologies. But $f^2_1 + \dots + f^2_r$ obviously solve this problem: it is always positive. And hence has an inverse. Since the ideal $(f_1, \dots , f_r)$ contains $f_1, \dots , f_r$, it must contains $f^2_1 + \dots + f^2_r$ and hence the ideal equals to the whole $R$.

(iii) is a little bit tricky, but is the beautiful part of the problem. Below is a rough recap of my thought process:

First, from (ii) one should expect that $M$ cannot contains functions with no common zeros, otherwise $M = R$. So maybe let denote $Z$ as the set of common zeros of all the functions in $M$? What does this $Z$ should look like? It is natural to expect that it is non-empty. We want it to be a singleton because that will match the definition of $M_p$. We hope $M = \{f \in R | f(Z) = 0\}$ and $Z$ is a singleton, that seems natural and it will prove the statement straight away.

But…what if it is not? What will happen if there are more than one common zeros…? Oh actually it does not matter, since obviously $\{f \in R | f(Z) = 0\} \subseteq \{f \in R | f(p) = 0\}$ for any $p \in Z$, and since $M$ has to be maximal so as long as we can show $M \subseteq \{f \in R | f(Z) = 0\}$ it immediately implies that $M \subseteq \{f \in R | f(p) = 0\}$ for some $p \in Z$.

Ok, is it even possible for $M$ to be not contained in $\{f \in R | f(Z) = 0\}$…? It seems impossible, cuz the functions in $M$ must have common zeros or $M$ will eqaul to $R$. Hence any function $f \in M$ must also be in $\{f \in R | f(Z) = 0\}$ right…? Huh? Isn’t that complete the proof? But I didn’t use the compactness of $K$. I should expect it fails when $K$ is not compact, so let’s think about a counterexample…

[After trying to construct a few functions from $\mathbb{R}$ to $\mathbb{R}$ and let them generate an ideal so that the resulting ideal is not contained in $\{f \in R | f(p) = 0\}$ for some $p \in \mathbb{R}$, but with no luck, I turn into the following thoughts…]

It seems hard to imagine the statement is not true for general $K$, I must have done something wrong… Let’s look at (ii) again since it causes all the pathologies. Oh wait, it says $f_1, \dots , f_r \in R$ should have common zeros, there are only finitely many of them! So in order to fail one must construct a collection of continuous functions such that any finite subcollection of them have common zeros, but the whole collection of them have no common zeros. Looking at the word “compact”, what can I think of? Finite intersection property!

and the rest of proving (iii) was straight forward by considering the collection of closed set $f^{-1}_{\alpha}(0) \text{ where } M = \bigcup_{\alpha} f_{\alpha}$

Now I can come up with a counterexample for when $K$ is not compact, consider the the set $C = \bigcup_{i} \{f : \; f(x) = 0 \; \forall x \in (-\infty , i)\}$ for $i = -1, -2, -3, \dots$. It is obviously an ideal, it satisfies the condition that any finite subcollection $f_1, \dots , f_r$ have common zeros, and yet they are not contained in $M_p = \{f \in R | f(p) = 0\}$ for any $p$.

Afterthoughts

What we have shown is that a compact topological space can be viewed as a collection of maximal ideals of a ring (namely the ring of continuous functions to ANY field). (The problem used $\mathbb{R}$, but it should work for any field with a topology, since (ii) requires a field to construct the inverse but there is nothing specific to $\mathbb{R}$).

So that makes an interesting connection between the structure of the maximal ideals of a ring and the topology of a compact space.

What information about other topological properties can we deduce from studying those maximal ideals?

These are probably questions that people already answered, but I came up with some quick thoughts here:

What if we take $K = \{0, 1\} \subset \mathbb{R}$? Obviously any function from $K$ to \mathbb{R} is continuous, so the ring of continuous function $C(\mathbb{R})$ is like $\mathbb{R} \times \mathbb{R}$ with pointwise addition and multiplication. It is easy to see that the only maximal ideals are $ I = 0 \times \mathbb{R}$ and $J = \mathbb{R} \times 0$. Note that there are many nonzero continuous functions belong to $I$ but not to $J$ and vice versa. In general this is true for any finite discrete topological space.

1. But even when $K$ is not discrete the similar is also true.
   When $K = [0, 1] \cup [2, 3]$, the ring of continuous functions from $K$ to $\mathbb{R}$ looks like $C([0, 1]) \times C([2, 3])$. This time by continuity if a function $f$ is not in $M_{p_0}$ for some $p_0$, there must be a family of $p$ such that $f$ is also not in $M_p$.
   However, since there are 2 components in $K$, one can divide the set of maximal ideals into 2 disjoint subset: $I = \{M_p$ where $p \in [0,1] \}$ and $J = \{M_p$ where $p \in [2, 3] \}$. Now it is possible to have many nonzero $f$ belongs to ALL ideals in $I$ but not in ALL ideals in $J$, and vice versa.
2. So… it looks like if we can give a suitable topology to the set of maximal ideals, one can reflects the topology of $K$ 😕 for example for the case above when $K = [0, 1] \cup [2, 3]$ we hope $I$ and $J$ are open and closed, in order to reflect the corr. components of $K$.