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It is clear that each maximal ideal in ring of continuous functions over $[0,1]\subset \mathbb R$ corresponds to a point and vice-versa.

So, for each ideal $I$ define $Z(I) =\{x\in [0,1]\,|\,f(x)=0, \forall f \in I\}$. But map $I\mapsto Z(I)$ from ideals to closed sets is not an injection! (Consider the ideal $J(x_0)=\{f\,|\,f(x)=0, \forall x\in\hbox{ some closed interval which contains }x_0\}$)

How can we describe ideals in $C([0,1])$ ? Is it true that prime ideals are maximal for this ring?

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    $\begingroup$ Former colleagues, more knowledgeable than I am about these things, usually recommend looking in Gillman and Jerison's book "Rings of continuous functions". (Try Googling the two names in conjunction with "prime ideal".) On the other hand, since you are dealing with functions on [0,1] and not some more general Hausdorff space, it should be possible to resolve your questions more directly. $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2010 at 20:44
  • $\begingroup$ Maybe I am just naive, but why wouldn't the space of ideals just be the set of all subsets of [0,1] ? I don't see how having the functions be continuous would get you out of it, since you could always take a limit of continuous functions until it converges to whichever subset you wanted. (This is not a rigorous argument and I am not trying to fully justify this assertion, rather it is a related subquestion.) $\endgroup$
    – Mikola
    Commented Aug 16, 2010 at 20:53
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    $\begingroup$ @Mikola: take a dense proper subset S of [0,1]. Any continuous function which vanishes on that set is identically zero, and so there is no ideal in C[0,1]) - closed or otherwise - whose zero set is S. $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2010 at 21:05
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    $\begingroup$ Any filter of closed subsets of the interval gives a proper ideal in $C(0,1)$ consisting of functions whose zero set lies in this filter. (Indeed, the zero set of $f+g$ contains the intersection of the zero sets of $f$ and $g$, so this is actually an ideal.) This handles the example you gave. Gilman and Jerison call ideals defined in this manner $z$-ideals. I suspect that there are conditions where every ideal is a $z$-ideal, but I'm not aware of them. $\endgroup$
    – Akhil Mathew
    Commented Aug 16, 2010 at 21:24
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    $\begingroup$ Since I was recently thinking about your last question and I don't see this argument in the answers below, I thought I'd add it: Suppose every prime ideal in this ring is maximal. Then $C([0,1])/N$ is absolutely flat, where $N$ is the nilradical, which is trivial, so $C([0,1])$ is absolutely flat. Then for every $f$ there is a $g$ such that $f^2g=f$. But this is clearly impossible, take $f \doteq x$. There is no continuous $g$ which satisfies this. $\endgroup$
    – M10687
    Commented May 29, 2017 at 5:58

6 Answers 6

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Here is a way to construct a non-maximal prime ideal: consider the multiplicative set $S$ of all non-zero polynomials in $C[0,1]$. Use Zorn lemma to get an ideal $P$ that is disjoint from $S$ and is maximal with this property. $P$ is clearly prime (for this you only need $S$ to be multiplicative.) On the other hand $P$ cannot be any one of the maximal ideals, since it does not contain $x-c$ for every $c \in [0,1]$.

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  • $\begingroup$ Hm. Let us assume that $ab\in P$ but $a,b \ne \in P$. Add $a$ to $P$, so $<P,a>$ intersects with $S$. And what should I see for the next? $\endgroup$
    – Nikita Kalinin
    Commented Aug 20, 2010 at 10:28
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    $\begingroup$ You are on the right track: $p+au \in S$ for some $p \in P$ and $u \in C[0,1]$. Similarly $p'+bv \in S$. Multiplying the two you see an element of $P$ in $S$. $\endgroup$
    – Keivan Karai
    Commented Aug 20, 2010 at 12:41
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There are many prime ideals that are not maximal. You can find some things by Googling "prime ideals in $C(X)$" (e.g. that every maximal ideal is the sum of two proper prime ideals).

The problem is that prime ideals are not closed unless they are maximal. The closed ideals in $C(X)$ for $X$ compact Hausdorff are in 1-1 correspondence with the closed subsets of $X$. This fact is in many books; e.g., it is an exercise in chapter 11 of Rudin's Functional Analysis.

EDIT 8/17: There is a lot of information about prime ideals in $C(X)$ in Chapters 7 and 14 of the book Rings of Continuous Functions by Gillman and Jerison, mentioned already by Yemon.

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In answer to your first question, you probably want to think about closed ideals (i.e. closed in the sup norm). (Bars and overlines seem to render poorly, so I'll use a prime to denote closure.) For $A \subset [0,1]$ we let $I(A) = \lbrace f : f = 0 \text{ on } A \rbrace$, which is clearly an ideal.

The following are all true and mostly easy to show:

  1. $Z(I)$ is closed in $[0,1]$ for any ideal $I$
  2. $Z(I) = Z(I')$
  3. $Z(I(A))=A'$
  4. $I(A)$ is closed in $C([0,1])$ for any $A \subset [0,1]$
  5. $I(A) = I(A')$
  6. $I(Z(I))=I'$ for any ideal $I$.
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I found this paper on Project Euclid, Have a look.

PRIME IDEALS IN RINGS OF CONTINUOUS FUNCTIONS by CARL W.KOHLS

http://projecteuclid.org/download/pdf_1/euclid.ijm/1255454113

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Let $ \mathbf U\ $ be a non-trivial ultrafilter of sets in an infinite Hausdorff compact space $\ X\ $ (e.g. $\ X:=[0;1]).\ $ Then

$$ J_\mathbf U:=\{f\in C(X): f^{-1}(0)\in\mathbf U\} $$

is a prime ideal; this ideal is contained in exactly one maximal ideal, namely $\ J_\mathbf U\subseteq J_{(x)},\ $ where $\ x\ $ is the (unique) limit point of $\ \mathbf U,\ $ and $\ (x)\ $ is a filter (maximal) with the base $\ \{\{x\}\}$. When $\ \{x\}\ $ is a $\ G_\delta$-set (as for instance for $\ X:=[0;1])\ $ then the inclusion for the said ideals is sharp, $\ J_\mathbf U\subset J_{(x)}$.

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    $\begingroup$ In my opinion, this answer is much more precise and illuminating than all others, for me this is THE answer to this question. Kudos! $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 10, 2023 at 7:09
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I think this paper may be of some help. Although this may not be the earliest source, it has a proposition there which says that given a point $p\in [0,1]$, one can find non-maximal prime ideals $\mathfrak{p}_1,\mathfrak{p}_2$ such that $\mathfrak{m}_p$, the maximal ideal at $p$, is the sum of $\mathfrak{p}_1$ and $\mathfrak{p}_2$. They take a sequence of points $x_n$ converging to $p$ and then use two different ultrafilters on $D= \{ x_n \}_{n\geq 1}$ to define the two prime ideals.

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