# In ZF, does the ring of continuous functions $C([0,1], \mathbb{R})$ have prime ideals which is not maximal?

In ZFC, it is known that the ring of continuous functions $$C([0,1], \mathbb{R})$$ have prime ideals which is not maximal. But all proofs of this which I saw uses the axiom of choice.

Then, in ZF, does the same statement hold?

If ZF cannot prove this, how strong is $$\mathrm{ZF}+(\text{C([0,1], \mathbb{R}) has prime ideals which is not maximal})$$ as an intermediate between ZF and ZFC?

• Can you give an example of a prime ideal which is not maximal in $C([0,1], \mathbb R)$ just for the sake of completeness? :) Jan 2, 2021 at 0:58
• @Patrick Da Silva Let $S$ be the set of all non-zero polynomials on $[0,1]$. It is multiplicatively closed. Using Zorn's lemma we can show there is an ideal $P$ which doesn't intersect $S$ and is maximal with that property. Such an ideal $P$ must be prime, this is a very standard argument in commutative algebra. (we use the fact that $S$ is multiplicative). However, it can't be maximal. It is well known that all maximal ideals in that ring have the form $N_{x_0}=\{f: f(x_0)=0\}$ for some $x_0\in [0,1]$. However, $P$ can't have that form, since then it would contain the polynomial $x-x_0\in S$.
– Mark
Jan 2, 2021 at 1:33
• @Mark : Very clear, thanks! Jan 2, 2021 at 1:41

No, the existence of such a nonmaximal prime ideal cannot be proved in ZF. In fact, the existence of such a nonmaximal prime ideal is equivalent to the existence of a nonprincipal ultrafilter on $$\mathbb{N}$$.

First, suppose $$F$$ is a nonprincipal ultrafilter on $$\mathbb{N}$$. Let $$I$$ be the set of continuous $$f:[0,1]\to \mathbb{R}$$ such that $$\{n\in\mathbb{N}:f(1/n)=0\}\in F$$. It is then easy to see that $$I$$ is an ideal in $$C([0,1],\mathbb{R})$$, and it is prime since $$F$$ is an ultrafilter. However, it is not maximal since it is strictly contained in the ideal of functions that vanish at $$0$$.

Conversely, suppose no nonprincipal ultrafilter on $$\mathbb{N}$$ exists and $$I\subset C([0,1],\mathbb{R})$$ is a prime ideal; we will prove $$I$$ is maximal. For each $$a\in[0,1]$$ and each $$n\in\mathbb{N}$$, let $$f_{n,a}:[0,1]\to\mathbb{R}$$ be the function that is $$0$$ on $$[a-1/n,a+1/n]$$, $$1$$ off of $$[a-2/n,a+2/n]$$, and interpolates linearly in between. I claim that there exists some $$a\in[0,1]$$ such that $$f_{n,a}\in I$$ for all $$n$$. To prove this, suppose no such $$a$$ exists; for each $$a$$, let $$n_a$$ be minimal such that $$f_{n_a,a}\not\in I$$. By compactness, finitely many of the intervals $$(a-1/n_a,a+1/n_a)$$ cover $$I$$. But then the product of the corresponding $$f_{n_a,a}$$s is $$0$$, contradicting primeness of $$I$$ since each $$f_{n_a,a}$$ is not in $$I$$.

So, we have a point $$a\in[0,1]$$ such that $$f_{n,a}\in I$$ for all $$n$$. It follows that every function that vanishes in a neighborhood of $$a$$ is in $$I$$, since every such function is divisible by some $$f_{n,a}$$. I now claim that in fact $$I$$ contains every function that vanishes at $$a$$, and thus is maximal. To prove this, suppose that $$f\in C([0,1],\mathbb{R})$$ is a function which vanishes at $$a$$. To show that $$f\in I$$, it suffices to show that $$f^2\in I$$ since $$I$$ is prime. Replacing $$f$$ by $$f^2$$, we may assume $$f\geq 0$$ and we wish to show $$f\in I$$.

Fix an increasing sequence $$(a_n)$$ converging to $$a$$ with $$a_0=0$$. Let $$F$$ be the set of all $$A\subseteq\mathbb{N}$$ such that there exists $$g\in I$$ such that $$g\geq 0$$ everywhere and $$g\geq f$$ on $$[a_{2n},a_{2n+1}]$$ for all $$n\not\in A$$. It is easy to see that $$F$$ is a filter on $$\mathbb{N}$$, and it contains all cofinite sets since $$I$$ contains all functions that vanish in a neighborhood of $$a$$. Note also that if $$A\subseteq\mathbb{N}$$, then we can construct a function $$g$$ which is $$f$$ on $$[a_{2n},a_{2n+1}]$$ for all $$n\not\in A$$ and $$0$$ outside of small neighborhoods of these intervals, and similarly we can construct a function $$h$$ with the same property with respect to $$\mathbb{N}\setminus A$$. Then $$gh=0$$ so either $$g\in I$$ or $$h\in I$$, so either $$A\in F$$ or $$\mathbb{N}\setminus A\in F$$.

Since by hypothesis, $$F$$ cannot be a nonprincipal ultrafilter on $$\mathbb{N}$$, the only remaining possibility is that $$F$$ is the improper filter, i.e. $$\emptyset\in F$$. So there is a nonnegative function $$g_1\in I$$ such that $$g_1\geq f$$ on $$[a_{2n},a_{2n+1}]$$ for all $$n$$. Similarly, there is a nonnegative function $$g_2\in I$$ such that $$g_2\geq f$$ on $$[a_{2n-1},a_{2n}]$$ for each $$n$$. Adding up $$g_1$$ and $$g_2$$, we get a nonnegative element of $$I$$ which is bounded below by $$f$$ on all of $$[0,a]$$. We can similarly get a nonnegative element of $$I$$ that is bounded below by $$f$$ on all of $$[a,1]$$. Adding these functions together, we get a function $$g\in I$$ such that $$g\geq f$$ on all of $$[0,1]$$. Now note that $$f^2/g$$ extends continuously to all the points where $$g$$ vanishes, since $$f$$ also vanishes at those points and $$f^2/g$$ is bounded above by $$f$$ outside them. Thus $$f^2$$ is a multiple of $$g$$ and hence is in $$I$$. Since $$I$$ is prime, this means $$f\in I$$, as desired.

• At the first time in the part "Then $gh=0$", I could not understand $\{x : f(x) = 0\}$ and $\{x : g(x) = 0\}$ cover entire $[0, 1]$. But I drawed a picture and I understood it. Thank you. Jan 2, 2021 at 11:49
• And I could not understand that $f^2/g$ extends continuously, but after a time I noticed that one can just put $h(x) = f^2(x)/g(x)$ if $g(x) \ne 0$, otherwise $h(x) = 0$. Jan 2, 2021 at 11:56

I haven't found a construction in ZF, but I found this paper which explores the spectrum of $$C(X, \mathbb R)$$ for $$X$$ an arbitrary topological space. It gets a bit heavy when the interaction with the Stone-Cech compactification shows up and partially ordered groups/rings appear, and I couldn't find a constructive approach to finding a prime ideal in $$C(X,\mathbb R)$$ either. But the relationship between $$\mathfrak m_x = \{ f \in C(X,\mathbb R) \, | \, f(x) = 0 \}$$ and $$\mathfrak n_x = \{ f \in C(X,\mathbb R) \, | \, \exists V \text{ neighborhood of } x \, \mathrm{s.t.} f|_V = 0 \}$$ seems to be extensively studied.

I'm curious to see if one could simplify the paper for the case of $$C([0,1],\mathbb R)$$ (so that the need for the Stone-Cech compactification goes away, simplifying many results) and shed some light on this question that has been triggering me (and probably a lot of other people) for many years. Let me know if you bump into anything interesting.

Hope that helps,