Maximal ideals in the ring of real functions on $[0,1]$

Question

Let $S$ be the ring of all continuous functions from $[0,1]$ to $\mathbb R$. How to prove that all maximal ideals of $S$ have the form $M_{x_0}=\{f\in S \mid f(x_0)=0\}$?

Thanks in advance.

Answer 1

New Answer.

Let $I$ be a proper ideal of the ring $C([0,1])$ of continuous functions on $[0, 1]$. We prove that

$$ Z(I) := \bigcap_{f\in I} f^{-1}(\{0\}) $$

is non-empty. This proves the existence of a point $x$ satisfying $f(x) = 0$ for all $f \in I$.

Assume otherwise that $Z(I) = \varnothing$. Then $\{ f^{-1}(\mathbb{R}\setminus\{0\}) : f \in I\}$ is an open cover of $[0, 1]$, and by the compactness, there exists a finite set $J \subset I$ such that $\{ f^{-1}(\mathbb{R}\setminus\{0\}) : f \in J \}$ is also an open cover of $[0, 1]$. Now define $g : [0, 1] \to \mathbb{R}$ by

$$ g(x) = \sum_{f \in J} f(x)^2. $$

Then $g \in I$ and $g$ is non-zero everywhere. Since $I$ is ideal and $1/g \in C([0, 1])$, this implies that $1 = g\cdot(1/g) \in I$ as well, which in turn implies that $I = C([0, 1])$, a contradiction.

Finally, if $M$ is a maximal ideal, then for each $z \in Z(M)$ we have $M \subset M_{z}$, where $M_z$ is as in OP's notation. Thus by the maximality, $M = M_z$ (and, in particular, $Z(M)$ is a singleton).

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Old Answer. (In the early version of OP's question, $S$ was assumed to be the set of all functions $[0, 1] \to \Bbb{R}$. That is, $S = \Bbb{R}^{[0, 1]}$. This is an answer for that version.)

We consider a more general situation. Fix a non-empty set $X$ and let $S = \Bbb{R}^X$ be the space of all real-valued functions on $X$. For each subset $Y \subset X$, we define

$$ \mathsf{S}(Y) = \{ f \in S : f \text{ vanishes on } Y \}$$

and $\mathsf{S}(\varnothing) = S$. In OP's case, we have $X = [0, 1]$. Also, for each $f \in S$ we define the zero set of $f$ by

$$Z_{f} = \{ x \in X : f(x) = 0 \}. $$

Then for each ideal $I$ of $S$ and $f \in I$, it is easy to check that $\mathsf{S}(Z_{f}) \leq I$. Indeed, for each $g \in \mathsf{S}(Z_{f})$, define $\tilde{g} \in \mathsf{S}(Z_{f})$ by

$$ \tilde{g}(x) = \begin{cases} g(x) / f(x) & x \notin Z_{f} \\ 0 & x \in Z_{f} \end{cases}. $$

Then $g = \tilde{g}f \in I$ and the claim follows. Now for each proper ideal $I$ of $S$, we associate a family

$$ \mathcal{F}(I) = \{ Z_f : f \in I \}. $$

Then it is clear that

1. $\varnothing \notin \mathcal{F}(I)$, since $I$ is proper.
2. If $A \in \mathcal{F}(I)$ and $A \subset B \subset X$, then $B \in \mathcal{F}(I)$.
3. If $A, B \in \mathcal{F}(I)$, then $A \cap B \in \mathcal{F}(I)$.

That is, $\mathcal{F}(I)$ is a filter on $X$. Conversely, for each filter $\mathcal{F}$ on $X$, the union

$$ I = \bigcup_{A \in \mathcal{F}} \mathsf{S}(A) $$

is a proper ideal of $S$ for which $\mathcal{F} = \mathcal{F}(I)$ holds. In this way, the family of proper ideals of $S$ corresponds to the family of filters on $X$.

It is clear that $I \leq I'$ if and only if $\mathcal{F}(I) \leq \mathcal{F}(I')$. Thus $I$ is a maximal ideal if and only if $\mathcal{F}(I)$ is an ultrafilter. Then OP's claim that every maximal ideal of $S$ is of the form $M_{x_0}$ corresponds to the claim that every ultrafilter on $X$ is principal, which is refuted by the existence of a non-principal ultrafilter on $X$ (if $X$ is infinite).

Answer 2

Here is another, slightly silly, way to see this. One can use the Stone-Weierstrass Theorem.

Let $R$ denote the ring of continuous functions on $[0,1]$, and let $M$ be a proper maximal ideal. For the sake of contradiction, suppose there is no $x\in [0,1]$ such that $f(x) = 0$ for all $f\in M$.

Observe that $M$ is an algebra of real continuous functions on $[0,1]$. By assumption, it vanishes at no point of $[0,1]$. Suppose there are $x_1$, $x_2 \in [0,1]$ such that $f(x_1) = f(x_2) \neq 0$ for all $f\in M$. Then the property of ideals tells us $x\cdot f(x) \in M$, so $$x_1 f(x_1) = x_2 f(x_2) \implies x_1 = x_2$$ Thus, $M$ separates points on $[0,1]$. The Stone-Weierstrass Theorem implies $M$ is dense in $R$. In particular, there is some $g\in M$ such that $g>0$ on $[0,1]$. Then $1/g \in R$, so $g \cdot (1/g) = 1 \in M$. Hence $M = R$, contradiction.

Answer 3

Without considering only continuous functions, you're out of luck.

Pick a non-principal ultrafilter $\mathcal{F}$ on $[0,1]$, and put $I = \{f: [0, 1] \to \mathbb{R}: f^{-1}(0) \in \mathcal{F}$}. I say that $I$ is maximal.

Indeed, if $g \not \in I$, then $g^{-1}(0) \not \in \mathcal{F}$. Since $\mathcal{F}$ is an ultrafilter, $\mathcal{F} \ni [0, 1] - g^{-1}(0) = g^{-1}(\mathbb{R}-\{0\})$. Thus, if we consider a function $h: [0, 1] \to \mathbb{R}$, $h(x) = 0$ if $x \in g^{-1}(\mathbb{R} - \{0\})$ and $h(x) = 1$ otherwise, the function $g+h$ is nowhere zero, and thus invertible, but $h \in I$, and thus $I + (g)$ is not a proper ideal.