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

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

$Z(f)\cup Z(g)=Z(f\cdot g)$. 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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