Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Please can someone tell me the history of the simple argument that any maximal ideal of a commutative ring or distributive lattice is prime? (It is understood that we have found the maximal one using Zorn's Lemma.)

How were prime ideals found before the enactment of the Axiom of Choice? How do constructive algebraists find them now? What similar arguments (directly referring to polynomials or algebraic numbers) were used before ideals were invented?

My own interest is really in distributive lattices and locales, from a constructive point of view, but I am aware that these notions appeared much earlier for commutative rings.

Incorporating my comments:

For locales/frames, (the relevant analogue of) prime ideals are (formal) points. To be precise, these are completely coprime filters.

In the draft paper [http://www.paultaylor.eu/ASD/loccbv] on which I am working (and a propos of which I asked this question) I have a partly constructive argument that I also use to find points of inhabited overt subspaces.

I am really more interested in the history of the arguments than the definitions (or even concepts). Maybe factorisation is the relevant idea to pursue in order to answer my history question. As Mamuka says, it all comes from prime numbers and therefore probably from the argument by infinite descent in Elements VII 31.

share|improve this question
    
First thing that comes to mind is that prime ideals correspond to quotients which are integral domains. While in the lattice context this means something relatively exotic (every element is dense), in ring theory domains are everywhere (btw this is an essentially infinite notion as finite domains are fields). In any case it must not be impossible to construct homomorphisms to fields; kernels of such are precisely prime ideals. However I do not really know history well enough, hence - just a comment. –  მამუკა ჯიბლაძე Jul 28 at 16:13
    
Well then you probably need "very" primes (and filters rather than ideals) - those prime with respect to arbitrary rather than just binary families. I doubt they have analogs in the ring-theoretic origins of the notion of prime ideal. –  მამუკა ჯიბლაძე Jul 28 at 16:26
1  
In fact let me dare to mention some history. I believe the first appearance of prime ideals was Kummer's attempt at restoring unique factorization into primes for the needs of Fermat's last theorem. Rings of integers in cyclotomic fields fail to have uniqueness of factorization into prime elements but factorization into product of prime ideals is still unique. So seemingly prime ideals first appeared as "idealizations" of prime numbers. –  მამუკა ჯიბლაძე Jul 28 at 16:29
1  
I apologize for telling the obvious thing, but maybe a way to understand this kind of question consists to think geometrically: the main source of examples of (locally) integral domains are regular rings. If we restrict to algebras of finite type over algebraically closed fields, these correspond to (ideals generated by) finite families of polynomials satisfying the jacobian criterium. There are plenty of such things, and they historically are the main subject of study of classical algebraic geometry. In these examples, maximal ideals correspond to (closed) points of the corresponding variety. –  Denis-Charles Cisinski Jul 28 at 20:58
2  
On the question of how one deals with the lack of maximal ideals in constructive mathematics, I think Coquand and Lombardi have their "dynamic method", which replaces a maximal ideal by an ideal which "grows" as needed during the proof. This works nicely in simple situations such as showing that if two polynomials have content $1$ then so does their product (see also Reiner/Messing arXiv:1209.6307 for this particular proof). I am not sure if this is still functional when higher-order concepts such as Noetherianness are involved. –  darij grinberg Jul 29 at 11:22

2 Answers 2

In general, one cannot expect commutative rings (with unit) to have prime ideals if the axiom of choice fails. For a specific example, consider the ring obtained from the direct product $(\mathbb Z/2)^{\mathbb N}$ (i.e., infinite sequences of $0$'s and $1$'s with addition and multiplication componentwise mod $2$) by dividing by the ideal of sequences that have only finitely many $1$ entries. A prime ideal in this quotient ring amounts to a nonprincipal ultrafilter on $\mathbb N$, and it is known that the existence of such things is not provable in ZF set theory without choice (not even if one adds certain weak choice principles like, for example, the axiom of dependent choice).

share|improve this answer
    
Thanks, Andreas, but I wanted to know the origin of the argument when it is valid, rather than discussing constructivity. On the other hand, I would also like to know what constructive arguments there are, with a view to adapting them to my situation. –  Paul Taylor Jul 28 at 16:57
1  
Sorry, Paul. I wrote my answer without noticing who asked the question. Had I noticed, I wouldn't have bothered, because I know that you already know what I wrote here. Well, maybe my answer will be useful for some other readers. –  Andreas Blass Jul 28 at 17:23
2  
Of course, Andreas, you are completely right in providing information that students may not know. I am completely against the narrow-minded "rules" of this site in this respect. –  Paul Taylor Jul 28 at 17:41

It is perhaps worth mentioning that you only need a choice function for nonempty subsets of the ring that you are working with. Indeed, given such a function $c$ and an ideal $I$ you can define $d(I)=\{x\not\in I : I + Rx \neq R\}$ and then $$ e(I) = \begin{cases} I + R c(d(I)) & \text{ if } d(I)\neq\emptyset \\ I & \text{ if } d(I) = \emptyset \end{cases} $$

You can then iterate transfinitely by the rule $$ e^\alpha(I) = \begin{cases} e(e^\beta(I)) & \text{ if } \alpha = \beta + 1 \\ \bigcup_{\beta<\alpha}e^\beta(I) & \text{ if } \alpha \text{ is a limit ordinal}. \end{cases} $$ Now $e^\alpha(I)$ will be independent of $\alpha$ when $\alpha$ is large enough, equal to $I^*$ say; and $I^*$ will be maximal provided that $I$ is proper.

If you prefer a description without ordinals, we can say that $a\in I^*$ if for each family $\mathcal{J}$ of ideals such that

  • $I\in\mathcal{J}$
  • $e(J)\in\mathcal{J}$ whenever $J\in\mathcal{J}$
  • The union of any nonempty chain in $\mathcal{J}$ is also in $\mathcal{J}$

there exists $J\in\mathcal{J}$ such that $a\in J$.

Moreover, the class of rings for which one can write down a choice function is quite large. It certainly contains all rings for which we have an explicit enumeration $\mathbb{N}\to R$, and is closed under taking quotients, products, tensor products, finitely generated algebras and so on.

Of course we do not have choice functions for nontrivial algebras over $\mathbb{R}$ or $\mathbb{Q}_p$. However, in some cases like this the ring will have a topology and it will be sufficient to have a choice function for nonempty open subsets of nonempty closed subsets, which can be arranged explicitly.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.