# Maximal ideal and Zorn's lemma

It is known that any ring A (say commutative with 1) has a maximal ideal. The proof uses Zorn's lemma.

Now I heard some people saying that if we assume A to be noetherian, then we don't need to use Zorn's lemma. The argument would basically be as follows:

"Suppose it doesn't have a maximal ideal. Then we can build an ascending chain of distinct ideals."

But, as far as I know it, we have to use Zorn's lemma in order to construct such an ascending chain. Am I right?

If I am right, is it still true (via some other argument) that we don't need to use Zorn's lemma to prove the result?

(EDIT: My definition of noetherian ring is that any ascending chain of ideals stabilizes.)

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The equivalence between different definitions of Noetherian requires the axiom of choice, so first of all you have to pick one. –  Qiaochu Yuan Jan 27 '11 at 19:16
I've often been bothered by arguments like this in which people try to construct something infinite one step at a time, without explicitly appealing to Zorn's lemma. –  Dan Ramras Jan 27 '11 at 19:25
Potential duplicate: mathoverflow.net/questions/7025/… –  David Speyer Jan 27 '11 at 19:44
"Suppose it doesn't have a maximal ideal. Then we can build an ascending chain of distinct ideals." I don't think that's the "correct" bogus argument. What you want to say is "In ZF, you can extract cross-sections from finite families of non-empty sets (or finite sequences from infinite sets). But that's all you'll ever need for a Noetherian ring before you reach a maximal ideal." –  David Feldman Jan 28 '11 at 0:35

With your definition of Noetherian, then you don't need full AC to carry out the argument, but only the weaker principle known as Dependent Choices (DC), which asserts that one can make countably many choices in succession. In your argument, if there is no maximal ideal, then by DC you could successively pick larger and larger ones, violating the Noetherian property.

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Thanks, I didn't know about DC but I am still curious as to my final question. Is there a proof of this result that does not use anything beyond ZF axioms? –  expmat Jan 27 '11 at 19:41
To rephrase expmat's question, is the statement "All rings without infinite, proper, ascending chains have maximal ideals" equivalent to DC? –  Dan Ramras Jan 27 '11 at 19:43
These kinds of questions come up so much I am beginning to wonder if consequences.emich.edu/CONSEQ.HTM should not be in the FAQ for MathOverflow. (DC) is form 43, btw, for those interesting in looking. –  Willie Wong Jan 27 '11 at 19:50
I think the answer to expmat's question is that the ACC is simply not the right definition of Noetherian in plain ZF; the right definition that every nonempty set of ideals has a maximal element. With the right definition, the existence of maximal ideals is trivial. In fact, this is not the only standard theorem about Noetherian rings which is easy to prove in plain ZF with the right definition. –  François G. Dorais Jan 27 '11 at 23:41
See related question mathoverflow.net/questions/39872/… –  Joel David Hamkins Jan 28 '11 at 22:22

This is an elaboration of Joel's answer. Suppose $R$ is a ring in which every ascending chain $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots$ of ideals stabilizes.

We show that every (non-trivial) ideal can be extended to a maximal ideal. Given an ideal $J$, define a sequence of ideals $I_n$ as follows:

1. Let $I_0 = J$.
2. For $n \geq 0$, if $I_n$ is a maximal ideal then let $I_{n+1} = I_n$, otherwise choose an ideal $I_{n+1}$ which is strictly larger than $I_n$.

We used Dependent Choice to construct the sequence of ideals $I_n$. Because $I_n$ is an ascending chain of ideals it stabilizes. When it does, it reaches a maximal ideal that extends $J$.

So we used Dependent Choice to construct the ascending chain of ideals and Excluded Middle to decide whether an ideal is maximal or not.

In a particular case it may be possible to extend a non-maximal ideal to a larger one in a canonical way (for example, if we know how to well-order $R$ then we generate $I_{n+1}$ by adjoining to $I_n$ the first element which is not in $I_n$ yet). In this case we do not even need Dependent Choice.

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Thanks, Andrej! –  Joel David Hamkins Jan 27 '11 at 23:19