# Algebra without Zorn's lemma

One can't get too far in abstract algebra before encountering Zorn's Lemma. For example, it is used in the proof that every nonzero ring has a maximal ideal. However, it seems that if we restrict our focus to Noetherian rings, we can often avoid Zorn's lemma. How far could a development of the theory for just Noetherian rings go? When do non-Noetherian rings come up in an essential way for which there is no Noetherian analog? For example, Artin's proof that every field has an algebraic closure uses Zorn's lemma. Is there a proof of this theorem (or some Zorn-less version of this theorem) that avoids it?

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What is your definition of Noetherian ring? Note that the equivalence of the usual three definitions (every increasing chain stabilizes; every ideal is finitely generated; every nonempty collection of maximal ideals has a maximal element) requires AC/Zorn's Lemma for some of the implications, so which one you pick may be important if you drop AC. – Arturo Magidin Jan 7 '11 at 21:31
@Arturo This is an interesting point. I guess I am interested in the answer for any of the three definitions you mentioned. – Vitaly Lorman Jan 7 '11 at 21:36
I've added the tag [axiom-of-choice], I think it's fitting. – Asaf Karagila Jan 7 '11 at 21:46
Of course, one can do all of finite group theory (which seems to me to go very far) without Zorn's Lemma. – Arturo Magidin Jan 7 '11 at 21:50

The proof of existence and uniqueness of algebraic closures goes through assuming only the ultrafilter lemma, which is strictly weaker than AC; see this MO question. Exactly how strong this assumption is relative to other well-known forms of AC appears to be unknown. I don't know what "Noetherian version of this theorem" means, since every field is Noetherian.

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Thanks! I didn't have anything specific in mind when I wrote that, but I was thinking that a version of this theorem for only a certain subclass of all fields might be easier to prove without Zorn. For example, the last part of the answer to the question on MO mentions that there is something special about proving the uniqueness of the algebraic closure for countable fields (such as the rationals). – Vitaly Lorman Jan 7 '11 at 21:50
@Vitaly: existence one can often prove in other ways. For example, you can prove that C is algebraically closed in ZF, so you can prove the existence of the algebraic closure of any subfield of C without choice. There are also a few other fields for which it is possible to explicitly write down an algebraic closure. But uniqueness seems very different to me. – Qiaochu Yuan Jan 7 '11 at 21:58
@Vitaly: anyway, for many "practical purposes" you don't have to work in algebraically closed fields: just adjoin roots of specific polynomials as necessary. This is constructive, although tedious. For example, I think you can phrase a constructive form of the Nullstellensatz in this language. – Qiaochu Yuan Jan 7 '11 at 22:03