TITLE: Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?
QUESTION [12 upvotes]: More specifically, I just want to prove that in a Noetherian (even Dedekind) ring, every ideal in contained in a maximal ideal. Is the axiom oh choice needed here? The usual proofs (for general rings use Zorn's lemma; for Noetherian rings one can build an infinite ascending chain if there is no maximal element) seem to rely on it.
If the axiom of dependent choice can be avoided as well, even better.

REPLY [2 votes]: Fred Richman's webpage has some papers on mathematics without countable choice (and without the principle of excluded middle), including a choiceless development of the theory of Noetherian rings.
http://math.fau.edu/richman/HTML/DOCS.HTM
http://math.fau.edu/richman/Docs/new-acc.htm
Defining Noetherian by the ascending chain condition causes a reliance on countable dependent choice. To build a theory without nonconstructive principles of choice or excluded middle, one has to formulate an alternative definition of Noetherian that (in the presence of those principles) is equivalent to the usual chain condition, is inequivalent without the constructive principles, and is strong enough to allow the important parts of the nonconstructive theory to be constructivized.
I did not see whether Richman's paper discusses maximal ideals (in their classical formulation), since they are less useful constructively, or whether he proves the result you are looking for.  But I think his work goes some way toward showing that in principle the whole theory could be rebuilt without any use of AC, DC or CC.  This is maybe similar to the gradual elimination of Noetherian hypotheses in (some) theorems of algebraic geometry where it is habitual to assume finite dimensionality but the results could be stated in greater generality, after rephrasing in a form that is equivalent in the Noetherian case but also true in the general case.