maximal spectrum

Definition

Let $X$ be a compact, Hausdorff, topological space and $R=C^0(X,\mathbb{R})$ the ring of $\mathbb{R}$-valued continuous functions on $X$. The maximal spectrum of $R$, denoted by $\text{MSpec }R$, is the set of maximal ideals of $R$.

Geometric interpretation

There is a bijection $X\simeq \text{MSpec }R$ using the following steps.

Given $x\in X$, let $I_x$ be the subset of $R$ consisting of all functions which vanish at $x\in X$.

Claim: $I_x$ is a maximal ideal of $R$ and hence construct a map $X\to \text{MSpec }R$, $x\mapsto I_x$.

Proof:

We know that $I_x$ is maximal if and only if $C^0(X,\mathbb{R})$ is a field.

For $x\in X$, consider the ring homomorphism

$$\begin{array}{rl}e{v}_x:C^0(X,\mathbb{R})& ⟶\mathbb{R}\\ f& ⟼f(x)\end{array}$$

We can note that $I_x=\ker (e{v}_x)$, therefore $C^0/I_x\cong \mathbb{R}$ which is a field so $I_x$ is a maximal ideal.

Thus we have a map

$$\begin{array}{rl}X& ⟶\text{MSpec }\\ x& ⟼I_x\end{array}$$

Claim: Every maximal ideal of $R$ is of the form $I_x$ for some $x$. (This step uses compactness of $X$.) Moreover, $x$ is unique. (This step uses that $X$ is Hausdorff.)

Topology on \text{MSpec }R

There is a natural topology on $\text{MSpec }R$, which I do not define here, for which the bijection $X\simeq \text{MSpec }R$ becomes a homeomorphism. This is the starting of non-commutative geometry, whereby topological spaces are studied via their rings of functions. This is also the key perspective of algebraic geometry, whereby algebraic varieties are built by gluing together spectra of certain commutative rings.

Proof:

First, I will prove that every maximal ideal of $R$ is of the form $I_x$.

Let $J$ be a maximal ideal. Note if for some $x\in X$, $f(x)=0,\forall f\in J$ then $J\subset I_x$ which is maximal so $J=I_x$ and there is nothing to do.

Therefore, assume that there the vanishing locus of $J$ is empty. For each $x\in X$ there exists a function $f_x$ such that $f_x(x)\ne 0$.

Since each function is continuous, there is a neighborhood $U_x$ of $x$ such that $f$ is non-zero on this neighborhood. Collecting all these $U_x$ gives an open cover of $X$, so by compactness there is some finite sub-cover $U_{x_1},\dots U_{x_n}$. So we now can take each of the functions $f_{x_i}$ and build the new function

$$g=f_{x_1}^2+\cdots +f_{x_n}^2\in J$$

This function is always non-zero so we can look at the continuous function

$$\frac{1}{f_{x_1}^2+\cdots +f_{x_n}^2}$$

multiplying this by $g$ above gives $1_R\in J$ so $J=R$ which is a contradiction. Therefore, there is some element $x\in X$ such that $f(x)=0$ for all $f\in J$, so by the above $J=I_x$.

Next, I will show the uniqueness of $x$. Consider a maximal ideal $I_x$. It suffices to show that $\exists f\in I_x$ such that $f(y)\ne 0$ for each $y\ne x$.

By Urysohn’s lemma, since $X$ is Hausdorff for $x\ne y$ there exists some function $g\in C^0(X,\mathbb{R})$ such that $g(x)\ne g(y)$. Then take the function $h=g-C_{g(x)}$ where $C_{g(x)}$ is the constant function with value $g(x)$.

Thus,

$$h(x)=g(x)-g(x)=0\text{and}h(y)=g(y)-g(x)\ne 0$$

Therefore, for each $y\ne x$ we can construct $h\in I_x$, such that $h\notin I_y$. Hence, each maximal ideal is of the form $I_x$ for a unique $x$.