# Commutative Algebra without the axiom of choice

It is well known that in a commutative ring with unit, every proper ideal is contained in a maximal ideal. The proof uses the axiom of choice. This fact, and others that are proved using essentially the same argument, anchor a large part of commutative algebra.

Suppose now that we disallow the use of the axiom of choice. My feeling is that this fact should still hold except for very pathological rings. I would find it odd if commutative algebra was entirely dependent on the axiom of choice. I also recall hearing that there was a workaround argument that did not use the axiom of choice for sufficiently nice rings, so this question is not entirely speculative.

So, assuming we work without the axiom of choice, for which rings can we prove that every proper ideal is contained in a maximal ideal? How is this done? And what characterizes the rings where we can't?

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See also mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma . It is tempting to say "things are good for Noetherian rings" but the problem is that in bare ZF different definitions of Noetherianness which are equivalent in the presence of some form of choice are no longer equivalent, so one has to pick one (and then show that the rings one cares about satisfy it!). – Qiaochu Yuan Jun 17 '12 at 4:47
Thanks for the link! I'm not sure how I feel about Dependence Choice (it requires assuming more than just ZF, right?). The in-equivalence of the definitions of Noetherian is also going to be problematic for finding a good way to work without AC. – Potato Jun 17 '12 at 4:55
Yes, DC is independent of ZF. Again that comes from using the "wrong" definition of Noetherianness. In that thread the correct definition of Noetherianness in ZF is suggested to be "every nonempty collection of ideals has a maximal element." With this definition of course every Noetherian commutative ring has a maximal ideal, and then from here the main issue is to verify that the proofs of standard properties of Noetherianness go through with this definition (e.g. Hilbert's basis theorem), which they should (Google "constructive Hilbert basis theorem"); I haven't checked the details though. – Qiaochu Yuan Jun 17 '12 at 5:01

It should not be surprising at all that the axiom of choice is so involved in most parts of modern mathematics. The reason is simple, too: finitely generated things are by nature very well-behaved, however as time progressed we began exploring things which are not finitely generated (e.g. measure spaces, function rings, etc.) and the axiom of choice is a great tool to control these things.

True, if you are only interested in, say, rings of cardinality less than $2^{2^{\aleph_0}}$ then you probably don't need the entire axiom of choice, small fragments would suffice for everything to be well-behaved. However why limit yourself by cardinality when the argument would hold for all similar structures? So we just assume the axiom of choice and run with it.

What sort of things could break? Here are a few quick examples:

1. The existence of maximal ideals in unital rings;
2. Not every vector space has a basis, and a similar argument would show that not every projective module over $\mathbb Z$ is free;
3. For abelian groups injectivity is no longer equivalent to divisibility;
4. If things break real bad then there are not enough projective objects in $\mathbf{Ab}$, and not enough injectives either;
5. Not every set is an underlying set of a group.

If you get even further to the place where topology begins to take an important part of the work (e.g. topological groups/vector spaces/etc.) then the axiom of choice becomes an even more important tool.

However not all is lost. If your ring can be well-ordered then you can still apply most of the standard arguments to it. If its power set can also be well-ordered then you have even more.

What sort of common sets are guaranteed to be well-ordered in ZF? Well... countable set. That's it. If you assume more, then more. However there is no guarantee to that happening.

So to answer your later question: I am not aware to the existence of such class, furthermore I do not believe there is a "nice" description of rings you can work with in ZF. If anything, I would expect most things to fail at size continuum or less.

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The size of the ring doesn't strike me as a good measure of its complexity from a commutative-algebraic perspective even without choice. Arbitrarily large fields still behave nicely from the perspective of commutative algebra (no nontrivial ideals). – Qiaochu Yuan Jun 17 '12 at 7:35
@Qiaochu: This would be additional structure to the one of an arbitrary ring. Furthermore fields are not that interesting, if anything we care about polynomials over a field. If a field contains an amorphous set then its polynomial ring begins to behave very badly. We can construct examples such that $F$ contains an amorphous set and has no algebraic closure. You need to remember that without the axiom of choice "every cardinality" is not just "up" but also "to the sides" (where up means well-orderability, and sideways means large non well-orderability). – Asaf Karagila Jun 17 '12 at 7:50
I guess what I'm saying is that I believe it is still true that finitely generated rings over fields have maximal ideals in ZF and that you're being too pessimistic considering that the OP does not care (the way I read the question anyway) about arbitrary rings. – Qiaochu Yuan Jun 17 '12 at 12:44
@Qiaochu: I cannot really answer that because I don't know too much about local rings, for example I cannot tell you for certainty that localization does not require any choice at all in the general case. You can always talk about a very specific localization, etc. but from what I recall in the courses I took, there is no essential drive to do so, because assuming AC we can prove things so much more general. I read in this question (and others like it) somewhat of a philosophical problem with AC present in mathematics, and this is the key point to my answer: infinitary objects are strange. – Asaf Karagila Jun 17 '12 at 12:53
(cont.) However assuming the axiom of choice we can really control their behavior and tame them to be nice. So if the problem is mostly philosophical and motivational, I think that I gave a good reason why we should use the axiom of choice and why it appears there in the first place. I suppose this is why my answer was accepted. – Asaf Karagila Jun 17 '12 at 12:54

As you can see in this related question, the existence of maximal ideals in rings with unity is equivalent to the axiom of choice.

If you are interested in consequence of the axiom of choice, read this.

For rings that have no maximal ideals, read this. Also this paper provides a characterization of commutative rings with no maximal ideals.

EDIT: After seeing the OP's rephrased question in the comments, see this related question.

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Yes, in all rings. The question is whether one can do without this assumption for rings of "practical interest" (e.g. finitely generated algebras over a field). – Qiaochu Yuan Jun 17 '12 at 4:48
Thank you for the references. But an Qiaochu says, I am more interested in what we can do for rings of "practical interest." – Potato Jun 17 '12 at 4:53
@Potato I saw that comment and included a list of rings that have no maximal ideals. – Eugene Jun 17 '12 at 4:55
Perhaps it would be better if I rephrased my question to be "Which rings can we still prove to have maximal ideals without using AC?" – Potato Jun 17 '12 at 4:58
In fact, the statement "every unique factorization domain has a maximal ideal" is equivalent to AC in ZF. – Cameron Buie Jun 17 '12 at 5:31

[So, assuming we work without the axiom of choice, for which rings can we prove that every proper ideal is contained in a maximal ideal? How is this done?]

Most of rings you encounter in algebraic geometry. For example, rings which are finitely generated over a field and their localizations. For the proof, see this thread.

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