TITLE: Maximal ideal and Zorn's lemma
QUESTION [12 upvotes]: 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.)

REPLY [6 votes]: 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:

Let $I_0 = J$.
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.