When can we prove constructively that a ring with unity has a maximal ideal?

Many commutative algebra textbooks establish that every ideal of a ring is contained in a maximal ideal by appealing to Zorn's lemma, which I dislike on grounds of non-constructivity. For Noetherian rings I'm told one can replace Zorn's lemma with countable choice, which is nice, but still not nice enough - I'd like to do without choice entirely.

So under what additional hypotheses on a ring $R$ can we exhibit one of its maximal ideals while staying in ZF? (I'd appreciate both hypotheses on the structure of $R$ and hypotheses on what we're given in addition to $R$ itself, e.g. if $R$ is a finitely generated algebra over a field, an explicit choice of generators.)

Edit: I guess it's also relevant to ask whether there are decidability issues here.

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My memory is that "every ring has a maximal ideal" is equivalent to AC so there are hence decidability issues. –  Kevin Buzzard Nov 28 '09 at 8:00

I suspect that the most general reasonable answer is a ring endowed with a constructive replacement for what the axiom of choice would have given you.

How do you show in practice that a ring is Noetherian? Either explicitly or implicitly, you find an ordinal height for its ideals. Once you do that, an ideal of least height is a maximal ideal. This suffices to show fairly directly that any number field ring has a maximal ideal: The norms of elements serve as a Noetherian height.

The Nullstellensatz implies that any finitely generated ring over a field is constructively Noetherian in this sense.

Any Euclidean domain is also constructively Noetherian, I think. A Euclidean norm is an ordinal height, but not at first glance one with the property that $a|b$ implies that $h(a) \le h(b)$ (with equality only when $a$ and $b$ are associates). However, you can make a new Euclidean height $h'(a)$ of $a$, defined as the minimum of $h(b)$ for all non-zero multiples $b$ of $a$. I think that this gives you a Noetherian height.

I'm not sure that a principal ideal domain is by itself a constructive structure, but again, usually there is an argument based on ordinals that it is a PID.

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I'm not sure I understand what "constructively Noetherian" means. It seems like it should mean: given a recursive set S of elements of R, give an algorithm to find a finite subset T of R such that <S> = <T>. But I worry that even Z is not constructively Noetherian in this sense. Take for instance the ideal generated by the set x_n = 2^n + e_n, where e_n = 0 if the nth Fourier coefficient of \Delta is nonzero and e_n = 1 otherwise. Then finding a finite generating set for <{x_n}> is equivalent to resolving Lehmer's Conjecture. –  Pete L. Clark Nov 29 '09 at 7:45
Of course it can mean more than one thing. What I mean by it is an order-preserving map from ideals (under reverse inclusion) to ordinals. So that the inductive structure of set of the ideals is constructive. As you argue, it does not really mean that their entire structure is constructive. –  Greg Kuperberg Nov 29 '09 at 16:09
OK, let's look at the condition on a commutative ring R that there exist an ordinal number (or equivalently, a well-ordered set) O and a map f from the set of nonzero ideals R to O with the property that I strictly contains J implies f(I) < f(J). Such rings are definitely Noetherian. Examples include Euclidean domains and also Samuel's generalization to "ordinally-normed" Euclidean domains. Does the coordinate ring of an affine variety over a field necessarily satisfy this property? –  Pete L. Clark Nov 29 '09 at 22:45
I'm a fake expert at best with the algebraic geometry and even worse with the logic. But my impression is that you can make an ordinal height from the Hilbert series of a graded algebra over a field. I referred to the Nullstellensatz above, but the Hilbert series is more directly to the point. –  Greg Kuperberg Nov 29 '09 at 22:57

Unlike full choice, you probably use countable choice all over the place without even recognizing it. Every time you do something iteratively and then take some sort of limit to your construction, you're using countable choice. In many cases, if you do very careful bookkeeping, you can eliminate it on a case by case basis. But you have to be very careful. Without countable choice, the countable union of countable sets isn't necessarily countable.

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It seems to me like there are two points here.

(1) A ring is called noetherian if any ascending chain of ideals terminates. I think all the standard facts about noetherian rings can be proved without choice: $\mathbb{Z}$ is noetherian; fields are noetherian; $A$ noetherian implies $A[x]$ noetherian; that quotients of noetherian rings are noetherian; localizations of noetherian rings are noetherian; completions of noetherian rings are noetherian.

That should take care of most the rings we need in algebraic geometry.

However, I am worried about another issue:

(2) The usual proof that noetherian rings have maximal ideal goes as follows. Let $A$ be a ring without maximal ideals. Take the ideal $I_0 = \{ 0 \}$. Since it is not maximal, choose an ideal $I_1$ which contains it. Choose an ideal $I_2$ which contains $I_1$. Continue in this manner to produce a chain $I_0 \subsetneq I_1 \subsetneq I_2 \subsetneq \cdots$. This doesn't terminate, so $A$ is not noetherian.

This proof, of course, uses countable choice. My gut feeling is that this can be eliminated for the same sort of rings I address in (1). But does anyone know a reference for this?

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This argument uses the principal of Dependent Choice, rather than merely countable choice, since the choices are made in succession. DC is known to be strictly stronger than countable choice, but strictly weaker than AC. –  Joel David Hamkins Jan 27 '11 at 20:58

There are certain kinds of rings that came to my mind when I saw this question. $K[v]:=$ Coordinate ring of an affine variety $V$ over a field $K$ , and $C(X, F)$:= the ring of continuous $F$-valued ($F$ a topological field) functions on a compact space $X$. In both examples one can construct maximal ideals as zero sets of minimal closed subsets of certain topological space (In one case $V$ and in other case $X$). So for $C(X, F)$ some max ideals correspond to points of $X$, and for $K[V]$ some max ideals correspond to points in $V$.

Another way to think about the question is the following analogy to it. Under what hypothesis can we exhibit a basis of a vector space without using any form of choice? here the clear answer should be finite dimensional spaces! This gives a clear picture about what rings $R$ we should consider, they are $R$'s that are Artinian rings.

About decidability I'd say think about boolean rings. The category of Boolean rings is equivalent to the category of Boolean algebras. Under that equivalence one have a correspondence between ideals and filters that takes maximal ideals to ultrafilters. If I'm not mistaken the existence of ultrafilters is equivalent to some form of choice (see http://www3.interscience.wiley.com/cgi-bin/fulltext/103520653/PDFSTART ) so the same form of choice should be equivalent to the existence of maximal ideals.

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The existence of ultrafilters is easy: take a principal one. It's the non-principal ones that need AC. So one can find maximal ideals of Boolean algebras constructively, just not interesting ones ;-) –  Kevin Buzzard Nov 28 '09 at 14:30
Of these Boolean algebras. Consider instead the quotient of $\Omega(\mathbb{N})$ by the ideal of finitely supported Boolean functions. Any maximal ideal of that is a non-principal ultrafilter. –  Greg Kuperberg Nov 28 '09 at 17:11
Yes good point Greg. So there's a very natural candidate for a ring that might well have no maximal ideals if AC fails sufficiently badly. My memory is that it's standard that there are models of ZF where the only ultrafilters on Z are the principal ones. –  Kevin Buzzard Nov 28 '09 at 17:26
According to this authority, that's correct. at.yorku.ca/cgi-bin/… The existence of ultrafilters is also supposedly weaker than the axiom of choice. –  Greg Kuperberg Nov 28 '09 at 17:52
More generally, in any atomless Boolean algebra, there are no principal ultrafilters. –  Joel David Hamkins Jan 27 '11 at 23:04

Here is a type of example coming from Analytic Geometry (in the sense of the second "GA" in Serre's "GAGA").

Consider a domain $D$ in $\mathbb C$. Then every $\mathbb C$-algebra morphism (aka character) $\chi : \mathcal O (D) \to \mathbb C$ is of the form $ev_d:f \to f(d)$ with $d=\chi (z) \in D$. This is completely elementary: just write $f(z)= f(d)+(z-d)g(z)$ and let $\chi$ act on both sides of the equality.[You have to convince yourself that $d$ is in $D$, not just in $\mathbb C$: else $1=(z-d).(z-d)^{-1}$ would lead to 1=0 by applying $\chi$].

[From this Lipman Bers proved in 1948 that, given two domains $D,D'\subset \mathbb C$, a purely algebraic isomorphism $u:\mathcal O (D) \to\mathcal O (D')$ necessarily comes from an analytic isomorphism $f=u(z): D' \to D$.]

A vast generalization is that for any Stein manifold (or even Stein space) X, all characters $\mathcal O (X) \to \mathbb C$ are evaluations at a point of $X$, and yield maximal ideals $ker \chi$ of $X$. This is proved in Grauert-Remmert's book "Theory of Stein spaces" and looking rather superficially at the proof I THINK the axiom of choice is not used. This is certainly not a satisfactory answer to Qiaochu's question (in particular I know nothing of other maximal ideals in $\mathcal O (X)$: do such exist?) but maybe these not so well-known results might interest some reader.

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You have to be careful about the basic structure of the complex numbers without countable choice. A Cauchy sequence may or may not be constructively Cauchy, and I read that the fundamental theorem of algebra only holds for polynomials whose coefficients are constructively Cauchy numbers. Possibly different definitions of Stein become inequivalent; and how do you know that O(X) even has a character? –  Greg Kuperberg Nov 28 '09 at 17:03
Evaluation at a point of X is a character, independantly of X being Stein or not, so I know that O(X) does have a character. I have no idea of what would happen to Stein theory if the fundamental theorem of algebra didn't hold, nor to be frank to the rest of mathematics. –  Georges Elencwajg Nov 28 '09 at 17:45

You should take a look at Coquand and Lombardi's "A Logical Approach to Abstract Algebra".

They observe that commutative rings have a purely equational description, and so there are very strong metatheorems that apply to this theory: Birkhoff's completeness theorem for equational logic, of course; and also Barr's theorem, which states that if a geometric sentences is a consequences of a geometric theories with classical logic plus choice, it's also intuitionistically valid. (And all equational theories are also geometric theories.)

They strengthen Barr's theorem a bit, by characterizing the relevant intuitionistic proofs, and then "de-Noetherian-ize" several basic theorems which are typically proved using maximal ideals.

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