Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
One might try to construct a multiplicatively closed set with certain bad properties. I don't think the set of functions with finitely many zeroes works, but it might be a start. I think one might have to classify the maximal ideals first. This might be impossible. –  Will Sawin Dec 9 '11 at 15:44
    
@Will: sharing the fact that your first thoughts have led nowhere seems not very helpful to me. –  user2035 Dec 9 '11 at 16:39
3  
Q1 is the same question as mathoverflow.net/questions/3871/…, where the answerer noted that the residue fields of maximal ideals not arising from points are necessarily more exotic than $\mathbf{C}$ (or $\mathbf{R}$ in the context of the current post). –  Jared Weinstein Dec 9 '11 at 18:34
    
Hi Hugo - you might want to have a look at work of Hung Le Pham, or at least at some of the prior work that he builds on, see homepages.ecs.vuw.ac.nz/~hlpham –  Yemon Choi Dec 9 '11 at 20:47

1 Answer 1

up vote 4 down vote accepted

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 $m_U$.

share|improve this answer
    
Could you please indicate how to construct an ultrafilter belonging to a non-maximal prime ideal (or show its existence)? –  user2035 Dec 9 '11 at 17:28
3  
@a-fortiori: Extend the filter generated by the sets $((1-1/n)/2,0)\cup(0,(1+1/n)/2)$ to an ultrafilter $U$. Note that that if $f \in m_U$ then $f(1/2) = 0$ by continuity, but $m_U$ is not the ideal of all functions that vanish at $1/2$ since $f(x) = x-1/2$ is not in $m_U$. –  François G. Dorais Dec 9 '11 at 18:45
    
(Of course, I meant the sets $((1-1/n)/2,1/2)\cup(1/2,(1+1/n)/2)$.) –  François G. Dorais Dec 9 '11 at 18:50
    
Thank you. I should have thought about it harder. –  user2035 Dec 9 '11 at 22:39
    
Thanks Goldstem for the construction. –  Hugo Chapdelaine Dec 10 '11 at 12:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.