Let $A=C(0,1)$ be the ring of continuous real valued functions on the **open** interval $(0,1)$. It is not too difficult to show that if $\mathfrak{m}\subseteq A$ is a maximal ideal with residue field $A/\mathfrak{m}\simeq \mathbb{R}$ then $\mathfrak{m}=\mathfrak{m}\_a:=ker(ev\_a)$ where for $a\in(0,1)$,
$ev\_a:A\rightarrow\mathbb{R}$ is the evaluation map at $a$. It is easy to show that there are
maximal ideals not of the for $\mathfrak{m}\_a$. For instance one may look at the
ideal 


$$
I=\{f\in A:\exists n\in\mathbf{Z}\_{\geq 1}, f\equiv 0\;\;\mbox{on $(0,1/n]$}\}
$$
Then $I$ is not contained in any $\mathfrak{m}\_a$ but by Zorn's lemma it is contained in
some maximal ideal $\mathfrak{M}$. So here are two natural questions on the ring $A$ for which I don't have an answer: 


Q1: Do we have a structure theorem for the possible residue fields of maximal
ideals of $A$.


Q2: How does one show the existence (or construct) of a prime ideal of $A$ which is not maximal? 



non-maximal prime ideal in the ring of continuous functions
Take any free ultrafilter $U$ on $(0,1)$, and let $m\_U$ be the set of functions which are $0$ "almost everywhere", i.e., on a set in the ultrafilter. It seems to me that is a prime ideal, which is in general not maximal. If all ultrafilter sets have, say, the number 1/2 as a limit point, then $m\_U$ is properly contained in the ideal of functions vanishing at 1/2.


Why is it an ideal? That looks easy. E.g., if you have two functions vanishing on $A\_1$, $A\_2$ respectively, then their sum vanishes on $A\_1\cap A\_2$, which is again in the ultrafilter. 


Why is it prime? Let $f\_3:=f\_1\cdot f\_2$. Let $A\_i:= \{x:f\_i(x)=0\}$.
Then $f\_3\in m\_U$ iff $A\_3 \in U$ iff $A\_1\cup A\_2\in U$ iff $A\_1\in U $ or $A\_2\in U$ iff one of $f\_1$, $f\_2$ is in $m\_U$. 


