A different proof of exercise 4.6 from Atiyah-Macdonald

Question

I know that this problem has already been solved a couple of times but I was here because while going through the exercises I had written a slightly different proof and since I have no one to check it, I wanted to ask the community.

My proof goes like this:

If, by contradiction, the zero ideal is decomposable then by proposition 4.7 of the book, the set of all zero divisors is the union of the prime ideals belonging to the zero ideal.

In particular, the number of prime ideals belonging to the zero ideal is finite and each of them is contained in at least one maximal ideal, therefore all the zero divisors are contained in a finite union of maximal ideals. Since all maximal ideals are of the form seen in exercise 1.26, we have that all of the zero divisors must be zero on at least one of the points $P=\{x_{1},...,x_{n}\}$ corresponding to these maximal ideals.

Since $X$ is an infinite topological space, we can choose a point $a\in X$ such that $a\notin P$. By the fact that $X$ is Hausdorff we can find a closed neighbourhood $V$ of $a$ which does not intersect the set $P$. Hence By Urysohn lemma we have a continuous function $f$ such that $f(V)=\{0\}$ and $f(P)=\{1\}$. Therefore $f$ is a zero divisor which is never zero on $P$, a contradiction.

As I said above, this proof is only slightly different to the other ones but I would like to be sure of its correctness and I have no other way but to ask you kindly.

Edit. Since I have been asked to cite the exercises and proposition I refer to, here is the list:

Exercise 4.6: Let $X$ be an infinite compact Hausdorff space and let $C(X)$ be the ring of real-valued continuous functions on $X$. Does $(0)$ have a primary decomposition in this ring? We want to prove the answer is no.

Proposition 4.7: [...]In particular, if the zero ideal is decomposable, the set $D$ of zero-divisors of $A$ is the union of the prime ideals belonging to $0$. This is the only part of the proposition I use.

Exercise 1.26: Of this exercise we only use the fact that all the maximal ideals of the above mentioned $C(X)$ are of the form $M_{x}=\{f\in C(X)| f(x)=0\}$ where $x$ is a point of $X$.

They are an excercise and a propositon from the above mentioned and quite famous Atiyah-Macdonald — Jabberwocky, Jan 25 '22 at 09:04
While I'm very aware of the book, I don't know every word in it by heart. What is the exercise and what is the proposition? — Arthur, Jan 25 '22 at 09:36
$(0)$ is just ${0}$ right? What is a primary decomposition ? What does proposition 1 in the book say? I don't know the book. — Henno Brandsma, Jan 25 '22 at 10:44
I had seen that answer, and a couple others, but I still asked my own question since I do not make use of the fact that in $C(X)$ a primary ideal is contained in only one maximal ideal and I wanted to know if the proof was still correct . — Jabberwocky, Jan 25 '22 at 10:52
@Jabberwocky you use that each maximal ideal contains a prime ideal. And as we have many of those (one for each point) we have at least as many prime ideals which is a contradiction. — Henno Brandsma, Jan 25 '22 at 11:32
Actually, I use the fact that every prime ideal is contained in a maximal ideal. I don't see where you get the contradiction from the fact that there are infinitely many prime ideals contains — Jabberwocky, Jan 25 '22 at 11:48

score 1 · Answer 1 · answered Jan 25 '22 at 13:15

I think the argument works, I'd write it more as follows:

If $P_1, \ldots, P_n$ is the list of prime ideals belonging to the $0$-ideal, we know that each of them is contained in a unique maximal ideal $M_i \supseteq P_i$. Each $M_i$ is of the form $M_i=\{f \in C(X)\mid f(p_i) = 0\}$ for some unique $p_i \in X, i=1,2,\ldots n$.

Then find $p \in X$ with $p \notin \{p_1,\ldots,p_n\}$ as $X$ is infinite.

There is some Urysohn function $f$ with $f(p)=0$ and $f(p_i)=1$ for $i=1,2,\ldots n$. It's easy to see that $f$ is a zero-divisor of $C(X)$ that is in none of the $P_i$ or $M_i$, which seemingly is a contradiction.

A different proof of exercise 4.6 from Atiyah-Macdonald

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