From: brnstnd@nyu.edu (D. J. Bernstein)
Newsgroups: sci.math
Subject: Re: consequences of the Axiom of Choice
Date: 11 Oct 92 02:08:53 GMT
Organization: IR

In article <1992Oct5.090904.12816@news.Hawaii.Edu> lady@uhunix.uhcc.Hawaii.Edu (Lee Lady) writes:
> It seems to me that  uniform [noetherian] => strong => weak  and it seems
> plausible that none of these implications is reversible without AC.  
> Can you actually show that?  

Here's what's known. Uniform implies strong: Every nonempty collection
of submodules has a maximal element. We have a submodule N. The set of
finitely generated submodules F of N has a maximal element, which has to
be N.

Strong implies weak: Every submodule is finitely generated. We have an
ascending chain of submodules. The union of the chain must be finitely
generated and hence must stabilize past the submodules containing the
generators.

Weak, plus a well-ordering of the submodules, implies uniform: We have a
nonempty collection of submodules with no maximal element. Construct an
ascending chain of submodules inside the collection, by taking the
(well-)least submodule strictly containing the current submodule at each
step.

Weak, plus a well-ordering of the elements of our module, implies
strong: We have a submodule which is not finitely generated. Construct
an ascending chain of finitely generated submodules, by taking the
(well-)least element outside the current submodule at each step.

Of course uniform implies weak, and strong plus a well-ordering of the
elements implies uniform.

I don't know to what extent the upward implications depend on
well-ordering. I'll give you a set which can't be well ordered. Can you
show me a ring and an ideal in that ring which isn't contained in any
maximal ideal? I don't know if this falls out of Hodges's proof on the
subject. Producing a weakly Noetherian module which isn't uniformly
Noetherian shouldn't be much more difficult.

> And what would be some examples of standard theorems about noetherian
> rings (or modules) which require uniform (a.k.a. the maximum principle)
> and can't be proved with merely strong or weak?

A domain D is Dedekind if (1) it is _uniformly_ Noetherian; (2) it is of
dimension at most one, meaning that every nonzero prime ideal is
maximal; (3) it is integrally closed, meaning that I:I = D for every
finitely generated module I inside the field of fractions. (By previous
comments, D is strongly Noetherian, so any ideal I is finitely generated
and hence satisfies I:I = D.)

A Dedekind domain has unique factorization into prime ideals. The
fastest proof I've seen of this (from Frohlich & Taylor's new book) is
as follows. In any domain D we define a nonzero ideal I as _having
nontrivial inverse_ if D:I is not equal to D. Any ideal I which is
maximal among the ideals having nontrivial inverse must be prime (the
proof of this is short but clever). In this case, if D is of dimension
at most one, then I must be maximal (in the usual sense). If further D
is integrally closed, then I must be invertible. (Proof: (D:I)I is
contained in D but contains I, hence must equal either I or D. If (D:I)I
equals I then D:I is contained in I:I = D, but we assumed that I had
nontrivial inverse. Hence (D:I)I = D, i.e., I is invertible.)

Any ideal which is a product of invertible ideals is invertible. In
Dedekind domains a strong converse is true: every proper invertible
ideal J is a product of invertible maximal ideals. Indeed, suppose not.
By _uniform_ Noetherianness find a counterexample J maximal among the
counterexamples. Consider the collection of ideals having nontrivial
inverse and containing J. By _uniform_ Noetherianness this collection
has a maximal element I, which by the previous paragraph is an
invertible maximal ideal. Now J(D:I) is larger than J, contained in D,
and invertible, so it is a product of invertible maximal ideals. Hence
J = J(D:I)I is also a product of invertible maximal ideals.

Next let P be a nonzero prime ideal. Pick a nonzero x in P. By the
previous paragraph, xD is a product of invertible maximal ideals. Hence
P contains that product. By primality P must equal one of the invertible
maximal ideals. Hence any nonzero prime ideal is invertible.

We finally show that every ideal J is invertible. If not, pick a
counterexample maximal among the counterexamples. Pick a prime ideal P
containing J. Now P is invertible by the previous paragraph, so J(D:P)
is larger than J and contained in D, hence invertible. So J is
invertible as desired. Q.E.D.

Uniform Noetherianness is crucial for this proof. I don't mean to imply
that the books which develop number theory without distinguishing
between degrees of Noetherian modules are leaving something out---all
Noetherian conditions appearing in number theory are clearly uniform.
Nevertheless I believe that adding the word ``uniform'' (or ``strong''
or ``weak'' depending on the situation) helps the reader by giving him a
hint as to which properties are really at work.

---Dan