Recover an ultrafilter from a maximal ideal Let $\mathcal{F}(I,\mathbb{R})$ denote the algebra of functions from an infinite set $I$ to $\mathbb{R}$. We know that if $U$ is an ultrafilter on $I$, then $\{f: I \rightarrow \mathbb{R} \ \vert \ \exists Q \in U: \ \forall x \in Q, \ f(x) = 0\} \subseteq \mathcal{F}(I,\mathbb{R})$ is a maximal ideal. Is the converse true? More precisely, let $J \subseteq \mathcal{F}(I, \mathbb{R})$ be a maximal ideal: is the set $\{f^{-1}(0): f \in J\}$ an ultrafilter over $I$? If not, are there any sufficient conditions on $I$ that would make it an ultrafilter?
 A: Yes, this is true.  Let $R=\mathcal{F}(I,\mathbb{R})$ and let $U=\{f^{-1}(0):f\in J\}$.  First, note that if $f\in J$, then we can multiply $f$ by a function to assume $f$ only takes the values $0$ and $1$ without changing its zero set (just multiply by a function that is $1/f$ when $f$ doesn't vanish).  So, $A\in U$ iff the function $1_{I\setminus A}$ which is $1$ on $I\setminus A$ and $0$ on $A$ is in $J$.
We now see easily that $U$ is a filter: it is closed under supersets since $J$ is closed under multiplication by elements of $R$, and it is closed under intersections since $1_{I\setminus A}+1_{I\setminus B}$ vanishes only on $A\cap B$.  To see that it is an ultrafilter, it suffices to show that for any $A\subseteq I$, exactly one of $e=1_A$ and $f=1_{I\setminus A}$ is in $J$.  This follows from the fact that $e+f=1$ and $ef=0$: since $e+f=1$, they cannot both be in $J$, and since $ef=0$, at least one must be in $J$ since $J$ is prime.

Much more generally, let $R$ be any commutative ring.  Then the set $B$ of idempotents of $R$ (elements $e\in R$ such that $e^2=e$) forms a Boolean algebra under the operations $e\wedge f=ef$, $e\vee f=e+f-ef$, and $\neg e=1-e$.  If $J\subset A$ is any prime ideal, then $J\cap B$ is always an maximal ideal in $B$: that it is an ideal follows immediately from the fact that $J$ is an ideal and the definitions of $\wedge$ and $\vee$, and it is maximal since for any $e\in B$, the elements $e$ and $f=\neg e$ satisfy $e+f=1$ and $ef=0$ so $J$ contains exactly one of them.  (When you identify idempotent elements with clopen subsets of $\operatorname{Spec} R$, the Boolean operations on $B$ are just the usual set operations, and the maximal ideals/ultrafilters on $B$ correspond to the connected components of $\operatorname{Spec} R$, so the fact that $J\cap B$ is a maximal ideal just corresponds to the fact that a point of $\operatorname{Spec} R$ is in exactly one component.)
In the case $R=\mathcal{F}(I,\mathbb{R})$, the idempotent elements are exactly the elements of the form $1_{A}$, and their Boolean operations are the usual set operations, and so $B\cong \mathcal{P}(I)$ and an maximal ideal on $B$ corresponds to a maximal ideal (or taking complements, an ultrafilter) on the set $I$.
