# Proof about ideals that should not involve Zorn's Lemma

I am learning abstract algebra by myself using Dummit & Foote, and sometimes by working very hard I find very complicated non-proofs of things and I have no idea if they are the best or the worst ideas for those kind of things out there (obviously if they're not working they're probably wrong, but maybe not!). So I feel like talking about it and see what the audience thinks about it. Hence I am asking for a full solution only if you don't feel like giving a hint. If a hint has been given please don't try to come up with the answer and wait for me to accept.

So here's the exercise (page 259, Dummit & Foote's Abstract Algebra) :

Let $R = C[0,1]$ with addition and multiplication as the ring operations. For each $c \in [0,1]$ define $M_c = \{ f \in R \, | \, f(c) = 0 \}$.

It can be easily seen that $M_c$ is a maximal ideal of $R$ because $f(x) = (x-c) \in M_c$ and if $g \notin M_c$ then $g(c) \neq 0$, thus $f^2 > 0$ everywhere except at $c$ and $g^2(c)> 0$, hence $f^2 + g^2$ is a strictly positive continuous function, and since any ideal $I$ such that $M_c \subsetneq I \triangleleft R$ is such that $(M_c,g) \subseteq I$, $f^2 + g^2 \in (M_c,g)$, thus $f^2 + g^2$ is a unit in $(M_c,g)$ and that means that for every $g \notin M_c$, $R = (M_c, g) = I$.

Using this trick with $f^2 + g^2$ I wanted to show the following :

Prove that if $M$ is any maximal ideal of $R$ then there is a real number $c \in [0,1]$ such that $M = M_c$.

I tried considering $K(g) = \{ c \in [0,1] \, | \, g(c) = 0 \}$ and then defining decreasing sequences $K(g) \supsetneq K(g_1) \supsetneq K(g_2) \supseteq \dots$ (using the $f^2 + g^2$ trick) and then looking at the intersection of all those $K$'s but that got me to the weird need of using Zorn's Lemma and got me depressed after multiple tries. Also tried to use the fact that $R/M_c$ is a field but even if I had shown that it was isomorphic to $\mathbb R$ it wasn't clear what I could do in that direction so I gave up pretty fast on that one.

Any hints are welcome!

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You seem like you like intuition, so here's some. You are trying to prove something about a ring whose mere definition involves topology. Moreover, it's fairly obvious that this isn't true in general (i.e. you can find topological spaces $X$ for which $\text{MaxSpec}(C(X;\mathbb{R}))\ne{\mathfrak{m}_x:x\in X}$). Thus, you clearly need to use some nontrivial fact about the topology of $[0,1]$. Do any spring naturally to mind? They should, considering precisely what you noticed about the "$f^2+g^2$" problem. Namely, what you can note is that if $f$ doesn't vanish at $x$ on $[0,1]$ and $g$ doesn't vanish at $y$ on $[0,1]$ you can note by continuity that there exists neighborhoods $U,V$ of $x,y$ such that $f,g$ vanish nowhere on either. Aha! But then $f^2+g^2$ vanishes nowhere on $U\cup V$! It seems plausible then that if we assume that we have found an ideal for which there exists no such $c$ (as in the problem) then making an analogous jump we should be able to construct a function $f_x$ for each $x\in[0,1]$ for which $f_x$ doesn't vanish at $x$ and then consider $\displaystyle \sum_{x\in X}f_x^2$ to get a non-vanishing function on $[0,1]$. But, being non-vanishing we have that it's invertible (to a continuous function) and so our ideal contains a unit, and so is all of $C[0,1]$, a contradiction! Of course, this makes no sense since we can't take the infinite sum of all those squared functions. But, if we were somehow able to pick a finite subcollection which still "works" we'd be golden. Topology sounding a little nicer right about now?

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Very nice generalisation of the idea in this problem, I like your answer a LOT. I was looking for an answer but I am doing all the exercises in the Dummit & Foote not just for fun but to be a better at algebra in the future, so I just love to see those kind of answers, they're great for learning. –  Patrick Da Silva Dec 1 '11 at 5:27


• To show that we can employ completeness of $[0,1]$ instead of compactness to show this.
• Remark in Arturo's post sparked my interest whether AC is actually needed in the proof and if yes, then how much choice is needed.

I am aware that it can happen easily that use of AC goes unnoticed, H. Herrlich devoted the whole chapter named Hidden Choice in his book to this. So I hope that if I missed something, some of other users of MSE more experienced in working without choice will correct me.

Note 1: I've posted a question asking whether something more interesting can be said about role of AC in this result. I wasn't sure whether post this here on start a new question with giving my own answer - I hope the way I choose is not too of-topic. (In my post, the main topic was two proofs of the result OP is asking about, and then I only had a look where in these proofs AC was used and whether it can be eliminated. So I thought this answer is better suitable here.)

Note 2: This also appeared as Problem 6414 in Kostrikin's book Exercises in algebra: a collection of exercises in algebra, linear algebra and geometry. Hint at the end of this book suggests the approach using compactness.

Let me start by mentioning a different approach how to show that ideals of the form $M_c$ are maximal. It is easy to see that if we fix som $c\in[0,1]$ then $\varphi: C[0,1]\to\mathbb R$ defined by $\varphi(f)=f(c)$ is a ring homomorphism such that $\operatorname{Ker}(\varphi)=M_c$. Since it is clearly surjective, we see that $C[0,1]/M_c\cong \mathbb R$. Since the quotient ring is a field, the ideal $M_c$ is maximal.

Let us now turn to the proof of the fact that every maximal ideal has form $M_c$. Let us start with the proof using completeness of $[0,1]$. The proof I give here is basically the same (perhaps more detailed) as in the paper Stephan C. Carlson: Cauchy Sequences and Function Rings, The American Mathematical Monthly, Vol. 88, No. 9 (Nov., 1981), pp. 700-701, jstor.

Let us denote $C=C[0,1]$.

Theorem: An ideal $A\subseteq C$ is maximal if and only if $A=M_p$ for some $p\in\intrv 01$.

Proof. If $A$ is a proper ideal, then every element $f\in A$ must vanish at some point of $\intrv01$. (Otherwise $f.\frac1f = 1 \in A$.) We can also see that if a sum of two non-negative functions vanishes at some point $p$, the value of each of these functions at $p$ must 0.

Let $A$ be a maximal ideal in $C$. We will show that there is a $p$ such that $A=A_p$. Let $n\in\NN$ and for $1\leq i \leq n$ let $h_i$ be a non-negative function from $C$ such that $$h_i(x)=0\text{ if and only if }\frac{i-1}n \leq x \leq \frac in.$$ Clearly $$h_1(x)\ldots h_n(x)=0,$$ that is $h_1\cdots h_n\in A$, thus one of the functions $h_1,\dots,h_n$ belongs to $A$. (Since $A$ is a prime ideal.) Let us denote the function $h_i$ such that $h_i\in A$ by $f_n$.

Now we have for each $n\in\NN$ a non-negative function $f_n\in A$ such that $\inv{f_n}(0)=\intrv{a_n}{b_n}$ and $b_n=a_n+\frac1n$. If $m<n$, $f_m,f_n \in A$, then $f_m+f_n\in A$. Thus there exists $x_{m,n}$ such that $f_m+f_n(x_{m,n})=0$. This implies that $f_m(x_{m,n})=0$, $f_n(x_{m,n})=0$. Thus $x_{m,n}\in\intrv{a_n}{b_n}\cap\intrv{a_m}{b_m}$ and $\intrv{a_n}{b_n}\cap\intrv{a_m}{b_m} \neq \emptyset$, and thus $$\abs{a_n-a_m}\leq \frac1m, \abs{b_n-b_m}\leq \frac1m.$$

The sequences $(a_n)$, $(b_n)$ are Cauchy sequence and by completeness of $\intrv01$ they have a limit. Since the length of the interval $\intrv{a_n}{b_n}$ is $\frac1n$, both sequences have the same limit, let us denote it by $p$.

Now let $f\in A$. For any $n\in\NN$ we have $f^2+f_n \in A$. Thus there is a point $x_n$ such that $f^2(x_n)+f(x_n)=0$. Hence $x_n\in\intrv{a_n}{b_n}$ and $f_n(x_n)=0$. Since $x_n\to p$, we get $f(p)=0$ by the continuity of $f$. $\square$

Note that the above approach can be applied to a more general situation.

Theorem: Let $X$ be a complete and totally bounded metric space and let $C(X)$ be the ring of continuous real-valued functions on $X$. An ideal $A$ is a maximal ideal in $C(X)$ if and only if $A=M_p$ for some $p\in X$.

Proof. By total boundedness there is a finite $\frac1n$-net for every $n\in\NN$. I.e., there is a finite cover $B_{x_1,\frac1n},\ldots,B_{x_k,\frac1n}$ of the space $X$. We put $h_i(x)=\max\{0, d(x_i,x)-\frac 1n\}$ and we construct the sequence $f_n$ analogously as in the above proof. The zero set of $f_n$ will be a closed ball in $X$, the diameters of these balls converge to $0$ and their centers form a Cauchy sequences. The rest of the proof is the same as in the case $X=[0,1]$. $\square$

Now let us have a look at the approach which uses compactness. This is the approach from Arturo's post.

Theorem: Let $X$ be a compact Hausdorff space. Then $A$ is a maximal ideal in the ring $C(X)$ if and only if there is a point $p\in X$ such that $$A=M_p=\{f\in C(X); f(p)=0\}.$$

The name Gelfand-Kolmogorov theorem is sometimes used for this result. Perhaps more often this name is used for the result that two Hausdorff compact spaces $X$, $Y$ are homeomorphic if and only if the rings $C(X)$ and $C(Y)$ are isomorphic. (Or for the corresponding facts about the Stone-Čech compactification.) The above theorem is part of the proof of this fact.

See e.g. Gillman, Jerison: Rings of continuous functions, p.102 or Johnstone: Stone space p.145.

Proof. Suppose that $A$ is an ideal in $C(X)$ which is not contained in any $M_p$.

This means that for every $p\in X$ there is $f_p\in A$ such that $f_p(p)\neq0$. Then $$f_p^2(p)>0\text{ and }f_p\in A.$$ The continuity of $f_p$ implies that $f_p^2>0$ holds on some neighborhood $U_p$ of the point $p$. By compactness of the space $X$ we get that there is a finite subcover $\{U_{p_1},\ldots,U_{p_n}\}$ of the cover $\{U_p, p\in X\}$. Let us define $$g(x)=\sum_{j=1}^n f^2_{p_j}(x).$$ Clearly $g\in A$ and $g>0$ on the whole space $X$, since $\{U_{p_1},\ldots,U_{p_n}\}$ is a cover. Hence $g.\frac 1g=1\in A$ and $A=C(X)$, thus $A$ is not a proper ideal. $\square$

How much choice we need for $C[0,1]$?

Let us have a closer look at the above proof for the case $X=[0,1]$ and check where we made some choices.

But I still think that in the case of the unit interval, no choice is needed, since there we can write the function $h_1,\dots,h_n$ explicitly.

EDIT: Originally I thought I've shown much more without invoking ZF. But I did not notice the using of choice to choose the $\frac1n$-net for each $n$ in one of the proofs and to choose the function $f_p$ in another one.

Proof using completeness: We have chosen $f_n$ to be one of the functions $h_1,\ldots,h_n$, but we can simply define $f_n$ to be that function $h_i\in A$ with minimal $i\in\{1,\ldots,n\}$.

So we have shown without using AC that: If $X$ is a complete totally bounded metric space, then every maximal ideal of $C(X)$ is of the form $M_c$.

Proof using compactness: We have chosen a neighborhood $U_p$ on which $p$ is positive. But we can simply take $U_p:=f^{-1}(0,\infty)$.

So we have in ZF: If $X$ is a compact space, then every ideal of $C(X)$ is contained in ideal of the form $M_c$.

How much choice we need for $C(X)$?

I will refer to the book H. Herlich: Axiom of Choice, Lecture Notes in Mathematics 1876. Further references and original sources of these results can be found there. Here are the results related to this question I was able to find.

• Definition 3.23, p.35: A topological space is called Tychonoff-compact provided it is homeomorphic to a closed subspace of some power $[0,1]^I$ of the closed unit interval.

• Proposition 3.24, p.35: These two conditions are equivalent:
(i) $X$ is Tychonoff compact if and only if it is compact and completely regular;
(ii) Ultrafilter theorem

• Exercise E3, p.40: A completely regular space $X$ is compact if and only if every ideal in $C(X)$ is fixed.

• Exercise E4, p.41: A completely regular space $X$ is Tychonoff compact if and only if every maximal ideal in $C(X)$ is fixed.

• p.36: In ZF we have for any metric space: compact $\Rightarrow$ totally bounded and complete.

• Theorem 3.27, p.37: Axiom of countable choice (CC) is equivalent to: Compact = totally bounded and complete.

• Theorem 4.70, p.86: Equivalent are:

1. Products of compact Hausdorff spaces are compact.
2. Products of finite discrete spaces are compact.
3. Products of finite spaces are compact.
4. Hilbert cubes $[0, 1]^I$ are compact.
5. Cantor cubes $2^I$ are compact.
6. PIT.
7. UFT.
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Your answer is quite long, and I am pretty tired right now, so I'm not in the right mood to look at it. I promise I'll take a look later and see if I like it. =) –  Patrick Da Silva Jan 10 '12 at 3:15
No problem. It is quite possible that some other interesting proofs will appear in the question I've linked to: math.stackexchange.com/questions/97692/… (about the role of AC in tis claim) –  Martin Sleziak Jan 10 '12 at 6:32

Note that a function which is not zero at any point on $[0,1]$ is invertible, so if an ideal contains such a function, then the ideal is the whole ring.

Let $M$ be an ideal, and assume that $M\neq M_c$ for all $c$; we will prove that $M=(1)$.

Since $M\neq M_c$ for each $c\in [0,1]$, then for each $c$ there exists a function $f_c\in M$ such that $f_c(c)\neq 0$. Since $f$ is continuous, there exists $\delta_c\gt 0$ such that if $x\in[0,1]$ and $|x-c|\lt \delta_c$, then $f_c(x)=0$.

Now, $[0,1]$ is compact, and $\cup_{c\in [0,1]}(c-\delta_c,c+\delta_c)$ is an open cover, therefore...

(Of course, we are using the Axiom of Choice in selecting $\delta_c$ for each $c$, so in a sense we are using Zorn's Lemma...)

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Mannnnn, are you kidding me. I thought about all of those facts but never recalled the definition of compacity with open sets. I'm thinking about the fact that I couldn't sleep well for days because of such a simple argument... Thanks a lot! Great answer, very quick. –  Patrick Da Silva Dec 1 '11 at 5:23
Hahaha, nice adding between the parentheses. That doesn't hurt me much ; I don't mind using AC, I am not worried about using it or not, I just think that proofs involving explicit use of Zorn's Lemma (i.e. not something equivalent to it, but using it explicitly) makes things either very pretty for standards arguments in big theorems, either very ugly in small useless theorems. AC is often just hidden between the lines so he's not much of a messy guy. –  Patrick Da Silva Dec 1 '11 at 5:30