Bijection between compact space $K$ and maximal ideals of real-valued functions on $K$

Question

Let $K$ be a compact topological space, and denote by $R$ the ring of continuous functions $K \to \mathbb{R}$, with addition and multiplication defined pointwise. We prove that there is a bijection between $K$ and the "maximal spectrum" of $R$ as follows:

For $p \in K$, let $M_p = \{f \in R | f(p) = 0\}$. Prove that $M_p$ is a maximal ideal in $R$.
Prove that if $f_1, \dots, f_r \in R$ have no common zeros, then the ideal generated by $f_1, \dots, f_r$ (let's call it $(f_1, \dots, f_r)$) is equal to $R$. (Hint: Consider $f_1^2 + \cdots + f_r^2$)
Prove that ever maximal ideal $M$ in $R$ is of the form $M_p$ for some $p \in K$. (Hint: you will use compactness of $K$ and part 2)

(This is problem III.4.17 in Paolo Aluffi's Algebra Chapter 0) Part 1 is not difficult, but for part 2 I thought I had a proof, that I now realize is horribly faulty, and didn't use the hint. Am I missing something fairly obvious? Thanks for the help.

@Ravi so because $f_1^2 + \cdots + f_r^2$ is strictly positive on all $p \in K$, it makes sense to divide by it (pointwise, that is). So we have $(f_1^2 + \cdots + f_r^2)/(f_1^2 + \cdots + f_r^2) = 1$, and thus $(f_1, \cdots, f_r) = (1)$, yes? Thanks for the help! — Reuben Stern, Jun 11 '16 at 20:51
Okay, awesome :) And for part 3, where does compactness come into play? Is it that all $f \in R$ are bounded? — Reuben Stern, Jun 11 '16 at 21:01
Think of the other definition of compactness. The open cover - finite subcover one. — Arkady, Jun 11 '16 at 21:05
@Ravi we suppose $M \not\subseteq M_p$ for any $p \in K$. In this case, for every $p$ there is an $f_p \in R$ such that $f_p(p) \neq 0$. By continuity of $f_p$, there is then some open neighborhood of $p$ (let us call it $N_p$) such that $f_p(q) \neq 0$ for any $q \in N_p$. The set of all $N_p$ forms an open cover of $K$, so by compactness, there is a finite subset ${N_{p_1}, \dots, N_{p_n}}$ that covers $K$. The corresponding $f_{p_i}$ share no common zeros, for if they did, the $N_{p_i}$ would not cover $K$. So by 2, the ideal generated by the $f_{p_i} = R$. The rest follows simply. — Reuben Stern, Jun 12 '16 at 13:19

score 1 · Answer 1 · answered Jun 12 '16 at 06:27

1

You already saw part 2 in the comments: if $f_1,\ldots,f_r$ have no common zeroes, then $f_1^2 + \ldots f_r^2$ is never $0$ on $K$ (it's only $0$ iff all $f_i$ are), and so is a unit (because $\frac{1}{f_1^2 + \ldots f_r^2}$ is well-defined continuous and its multiplicative inverse). And so the ideal $(f_1,\ldots,f_r)$ contains a unit and so equals $R = C(K)$.

Now, suppose $M$ is a maximal ideal in $C(X)$. Suppose $M \nsubseteq M_p$ for every $p \in K$. So for every $p \in K$, there is some $f_p \in M$, such that $f_p(p) \neq 0$, and so by continuity we have an open neighbourhood $U_p$ of $K$ such that $f_p(x) \neq 0$ for all $x \in U_p$ (why?). Finitely many $U_{p_i}, i= 1,\ldots,n$ cover $K$ by compactness and so now note that $f_{p_1},\ldots,f_{p_n} \in M$ have no common zero (a common zero would not be covered by any of the $U_{p_i}$), so apply 2, and we have a contradiction ($1 \in M$). So $M \subseteq M_p$ for some $p$, and by maximality we have equality.

answered Jun 12 '16 at 06:27

Henno Brandsma

242,131

In regards to your first paragraph: How do we know $\frac{1}{f_1^2 + \ldots f_r^2}$ is continuous? – user5826 Aug 14 '19 at 19:03
@AlJebr squaring and adding are continuous on the reals. – Henno Brandsma Aug 14 '19 at 19:04
@AlJebr and inversion too on the reals minus zero. – Henno Brandsma Aug 14 '19 at 19:10
Thanks @HennoBrandsma. So, if $f: X \to \mathbb R$ is continuous and $f(x) \ne 0$ for all $x \in X$, then $1/f(x)$ is also continuous? – user5826 Aug 14 '19 at 19:13
@AlJebr indeed. Composition of continuous maps etc. – Henno Brandsma Aug 14 '19 at 19:14
What would be the composition here? We can't consider $1/x$ since $x \in X$ is just an arbitrary top space. – user5826 Aug 14 '19 at 19:21
@AlJebr call the inversion map on the reals $i$. Then $\frac{1}{f(x)}$ equals $i \circ f$. – Henno Brandsma Aug 14 '19 at 19:23

Bijection between compact space $K$ and maximal ideals of real-valued functions on $K$

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