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It is a well-known fact that for any compact topological space $\rm S$, the set its points is in bijection with the set of maximal ideals of $\mathcal C(\rm S)$, the algebra of continuous functions $\rm S \to \mathbb R$.

Now consider the circle $\rm S^1$ with its usual $\sigma$-algebra. Note $\mathcal M$ the algebra of measurable functions $\rm S^1 \to \mathbb R$ and $\mathcal N$ the ideal of functions vanishing almost everywhere.

By Zorn's lemma, the algebra $\mathcal M / \mathcal N$ has at least one maximal ideal. However in this example, it is not possible to get a maximal ideal by considering functions vanishing at one point.

What is an example of a maximal ideal of this algebra?

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The existence of such a maximal ideal in this particular case requires some form of the axiom of choice, so one cannot hope to obtain an example that does not involve nonconstructive tools like Zorn's lemma. –  Dmitri Pavlov Feb 18 '13 at 23:19
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How do you prove that? –  Allen Knutson Feb 18 '13 at 23:39
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nice question to begin with. –  Koushik Feb 19 '13 at 0:24
    
In the continuous case, it holds for arbitrary compact Hausdorff space (not just for compact groups). And in the measure-theoretical case the topological nature of S is unimportant. It's really about what sometimes is called a Lebesgue space(?), i.e. about non-atomic space, etc. –  Wlodzimierz Holsztynski Feb 19 '13 at 1:50
    
@Damien, you may describe your algebra more completely. What multiplication do you have in mind? Should integrals be finite or you truly want all measurable functions? –  Wlodzimierz Holsztynski Feb 19 '13 at 1:57
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2 Answers

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We shall give a complete characterization of all the maximal ideals in $\mathcal{M}/\mathcal{N}$ in terms of ultrafilters. For simplicity, we shall classify all maximal ideals in $\mathcal{M}$ since the maximal ideals in $\mathcal{M}/\mathcal{N}$ are in a one-to-one correspondence with all maximal ideals in $\mathcal{M}$ extending $\mathcal{N}$. Let $\mathcal{B}$ denote the Boolean algebra of Lebesgue measurable sets in $S$. Let $S(\mathcal{B})$ be the set of all ultrafilters on the Boolean algebra $\mathcal{B}$, and let $Max(\mathcal{M})$ denote the collection of all maximal ideals in $\mathcal{M}$. We shall give a one-to-one correspondence between $Max(\mathcal{M})$ and $S(\mathcal{B})$ as follows. For this problem, let $Z(f)=\{x\in S|f(x)=0\}$ denote the zero set of $f$.

If $\mathcal{U}\in S(\mathcal{B})$, then let $V(\mathcal{U})=\{f\in\mathcal{M}|Z(f)\in\mathcal{U}\}$. Then clearly $V(\mathcal{U})$ is an ideal on $\mathcal{M}$. Furthermore, if $f+V(\mathcal{U})\neq 0$, then $Z(f)^{c}\in\mathcal{U}$, so let $g$ be a function where $g=\frac{1}{f}$ on $Z(f)^{c}$. Then the elements $f+V(\mathcal{U})$ and $g+V(\mathcal{U})$ are inverses, so $\mathcal{M}/V(\mathcal{U})$ is a field. Hence $V(\mathcal{U})$ is a maximal ideal. Therefore we may consider $V$ as a function from $S(\mathcal{B})$ to $Max(\mathcal{M})$.

Going the other direction, assume $I$ is a maximal ideal in $\mathcal{M}$. Consider the set $\{Z(f)|f\in I\}$. Then

  1. $Z(f)\cap Z(g)=Z(f^{2}+g^{2})$ (here we need the real field instead of the complex field) and

  2. $Z(f)\cup Z(g)=Z(f\cdot g)$. In particular, if $Z(f)\subseteq A$, then $Z(f\cdot\chi_{A^{c}})=Z(f)\cup Z(\chi_{A^{c}})=Z(f)\cup A=A$.

Therefore, the set $\{Z(f)|f\in I\}$ is a filter on $\mathcal{B}$. If $A\in\mathcal{B}$, then since $\mathcal{M}/I$ is a field, and $(\chi_{A}+I)^{2}=\chi_{A}+I$, we have either $\chi_{A}+I=0$ or $\chi_{A}+I=1$. If $\chi_{A}+I=0$, then $\chi_{A}\in I$, so $A^{c}=Z(\chi_{A})\in\{Z(f)|f\in I\}$. On the other hand, if $\chi_{A}+I=1$, then $\chi_{A^{c}}=1-\chi_{A}\in I$, so $A=Z(\chi_{A^{c}})\in\{Z(f)|f\in I\}$. Therefore, the set $\{Z(f)|f\in I\}$ is an ultrafilter. Therefore let $\mathbf{Z}:Max(\mathcal{M})\rightarrow S(\mathcal{B})$ be the mapping where $\mathbf{Z}(I)=\{Z(f)|f\in I\}$ for maximal ideals $I$.

We claim that the mappings $\mathbf{Z}$ and $V$ are inverses. Assume that $I$ is a maximal ideal in $\mathcal{M}$ and assume $f\in I$. Then $Z(f)\in \mathbf{Z}(I)=\{Z(f)|f\in I\}$ , so $f\in V(\mathbf{Z}(I))$. Therefore $I\subseteq V(\mathbf{Z}(I))$, so $V(\mathbf{Z}(I))=I$ by maximality.

Now assume that $\mathcal{U}\in S(\mathcal{B})$ is an ultrafilter. Let $A\in\mathcal{U}$, and assume that $Z(f)=A$. Then $Z(f)=A\in\mathcal{U}$, so $f\in V(\mathcal{U})$. Therefore $A=Z(f)=\{Z(f)|f\in V(\mathcal{U})\}=\mathbf{Z}(V(\mathcal{U}))$. We conclude that $\mathcal{U}\subseteq\mathbf{Z}(V(\mathcal{U}))$, so clearly $\mathcal{U}=\mathbf{Z}(V(\mathcal{U}))$. Therefore the functions $\mathbf{Z}:Max(\mathcal{M})\rightarrow S(\mathcal{B})$ and $V:S(\mathcal{B})\rightarrow Max(\mathcal{M})$ are inverses. Therefore the maximal ideals in $\mathcal{M}$ are in a one-to-one correspondence with the ultrafilters on $S(\mathcal{B})$. Furthermore, the maximal ideals extending $\mathcal{N}$ are clearly in a one-to-one correspondence with the ultrafilters that contain all sets of full measure. In particular, if $M\subseteq\mathcal{B}$ is the ideal of sets of measure zero, then the maximal ideals in $\mathcal{M}/\mathcal{N}$ are in a one-to-one correspondence with the ultrafilters on the Boolean algebra $\mathcal{B}/M$.

If we let $\mathcal{M}^{\sharp}$ denote the ring of bounded measurable functions in $\mathcal{M}$, then the maximal ideals in $\mathcal{M}^{\sharp}$ are also in a one-to-one correspondence with the ultrafilters in the Boolean algebra $\mathcal{B}$. To prove this correspondence, let's give the set $S$ the proximity where two sets are separated if and only if they are separated by a measurable set. In other words, $(S,\delta)$ becomes a proximity space where $A\not\delta B$ if and only if there is a measurable set $C$ such that $A\subseteq C,B\subseteq C^{c}$. It turns out that $\mathcal{M}^{\sharp}$ is the set of all bounded proximity maps from $(S,\delta)$ to $\mathbb{R}$. Therefore, if $X$ is the Smirnov compactification of the proximity space $(S,\delta)$, then the ring $\mathcal{M}^{\sharp}$ is isomorphic with the ring $C(X)$ of continuous real-valued functions on $X$. However, it is well known and it is easy to prove that the maximal ideals on $C(X)$ are simply the ideals of the form $\{f\in C(X)|f(x_{0})=0\}$ for some $x_{0}\in X$. Therefore the maximal ideals in $\mathcal{M}^{\sharp}\simeq C(X)$ are in a one-to-one correspondence with the set $X$. However, in the paper Zero-Dimensional Proximities and Zero-Dimensional Compactifications, it is remarked that the Smirnov compactification $X$ is simply the space $S(\mathcal{B})$ of ultrafilters on $\mathcal{B}$. Therefore, the maximal ideals in $\mathcal{M}^{\sharp}$ are also in a one-to-one correspondence with $S(\mathcal{B})$. In my paper A Generalization of the Notion of a $P$-Space to Proximity Spaces I outlined this same exact proof of the correspondence between the maximal ideals in $\mathcal{M}^{\sharp}$ and $S(\mathcal{B})$. In other words, the sets $S(\mathcal{B}),Max(\mathcal{M}),Max(\mathcal{M}^{\sharp})$ are all in a one-to-one correspondence.

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If, instead of all measurable functions, you use the bounded measurable functions, then you probably want to consult the literature known as "lifting" ... for example:

Topics in the Theory of Lifting (Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge)

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