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Assume $S$ to be 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.

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I have serious doubts that this works without the continuity assumption on those functions. – user26857 Apr 28 '13 at 18:17
@YACP: You are right. For example, let $I:=\{f\in S: f$ vanishes on $[0,1]$ except for finitely many points $\}$. Then $I$ is an ideal of $S$, but it is not contained in any $M_{x_0}$. – 23rd Apr 28 '13 at 18:25
I guess that you mean continuous functions. – Sangchul Lee May 3 '13 at 11:24
@YACP:S be all continuous functions. my teacher give us this question with out continuity assumption but after 3 days i ask him he said S is all continuous functions – Maisam Hedyelloo May 3 '13 at 14:31
See a similar question posted here. – Martin Sleziak May 3 '13 at 14:48
up vote 14 down vote accepted

New Answer.

Let $I$ be a proper ideal of the ring $C([0,1])$ of continuous functions on $[0, 1]$. We prove that there exists $x \in [0, 1]$ such that $f(x) = 0$ for all $f \in I$.

Assume that for each point of $x \in [0, 1]$ there exists a function $f \in I$ such that $f(x) \neq 0$. Then there exists a neighborhood $N_{x}$ of $x$ in $[0, 1]$ such that $f \neq 0$ on $N_{x}$. Since $[0, 1]$ is compact and $\{ N_x \}$ is an open cover, there exists a finite cover. That is, there are functions $f_1, \cdots, f_n \in I$ such that for each $x \in [0, 1]$, we have $f_i(x) \neq 0$ for some $i$ (depending on $x$). Then the function $f = f_{1}^2 + \cdots + f_{n}^2$ is also a member of $I$ such that $f > 0$ on $I$. Therefore $1 = f/f \in I$ and $I = C([0, 1])$, a contradiction.

Thus if $M$ is maximal, then the set $Z = \{ z \in [0, 1] : f(z) = 0 \text{ for all } f \in M \}$ is non-empty. But for each $z \in Z$, we have $M \subset M_{z}$. Thus by maximality, $Z$ cannot have two distinct element and we must have $M = M_{z}$ for some $z$.

Old Answer.

Let $X$ be a set and for each subset $Y \subset X$, define

$$S(Y) = \{ f : X \to \Bbb{R} : f|_{Y} \equiv 0 \}$$

and $S = S(\varnothing)$. In OP's case, we have $X = [0, 1]$. Also, for each $f \in S$ we define

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

Then for each ideal $I$ of $S$ and $f \in I$, it is easy to observe that $S(Z_{f}) \leq I$. Indeed, for each $g \in S(Z_{f})$, define $\tilde{g} \in S(Z_{f})$ 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 $X$ we can retrieve a proper ideal $I$ of $S$ that yields $\mathcal{F} = \mathcal{F}(I)$. Thus 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).

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Why is $\bar{g}$ continuous? – D_S Mar 1 '15 at 23:11

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.

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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.

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There's another answer by sos440 that shows that it is essentially equivalent to asking whether there exists a non-principal ultrafilter on $[0,1]$ – xyzzyz Apr 28 '13 at 19:01
True. But doesn't Lanscape's example above show, via Krull, that there exists a maximal ideal which is not of the form $M_{x_0}$?. There is no direct reference to any ultrafilter here. Just Zorn. I know that's close, but maybe some people feel more comfortable invoking Zorn rather than manipulating ultrafilters. – 1015 Apr 28 '13 at 19:16

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