Question

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?

Answer 1

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 $m_U$.