This question was inspired by

How to prove that the subrings of the rational numbers are noetherian?

which some people found too routine to be of interest. So I have decided to liven things up a bit with the following questions. In the interest of full disclosure, I have not thought seriously about these questions, and I think that I probably could answer at least some of them myself, but I do think they are interesting and, if I may say so, educational.

Find all (commutative!) fields $K$ such that every (unital!) subring $R$ of $K$ is:

a) a principal ideal domain.
b) a Dedekind domain.
c) a Noetherian domain.

I mean here to be asking three different questions, one for each condition. Evidently the classes of such fields are nondecreasing from a) to b) and from b) to c).

If you would like to answer the question with a), b) or c) replaced by some other standard property of commutative rings -- especially if it yields a different class of fields than in the first three questions -- please feel free.

Addendum: How about

d) a Dedekind domain if it is integrally closed?
e) a PID if it is integrally closed?

  • $\begingroup$ @Prof. Clark: The problem I see with this question is that there are going to be five correct answers, and you will only be able to accept one of them. Would you be averse to splitting it up into several questions? $\endgroup$ – Harry Gindi Mar 27 '10 at 21:26
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    $\begingroup$ @fpqc: I don't think anything is gained by splitting up five very closely related questions into separate posts. As for choosing an answer, I'll do what people always do when there are multiple correct answers: I'll pick the one that seems best to me. (The point of my CW answer is to perform the service of compiling different contributions into a single answer. But I will not choose this one as correct.) $\endgroup$ – Pete L. Clark Mar 27 '10 at 22:18
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    $\begingroup$ On the whole, I don't think it's really a matter for "moderator attention". To the extent that my personal opinion counts, I think it's a difficult question to decide how to split up questions, but it is better addressed by asking "Which choice best serves someone interested in this subject?" than by any consideration involving the mechanics of the site (i.e. reputation, accepted answers, etc.), and it seems that this is what Pete has tried to do. $\endgroup$ – Scott Morrison Mar 27 '10 at 23:32
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    $\begingroup$ @fpqc: The appropriate thing to do is to start a thread on meta, not to flag for moderator attention. The ♦ comes with some super powers, but it doesn't mean we automatically know what community standards should be. Such things are best answered by discussing the pros and cons, not by appealing to some oracle. $\endgroup$ – Anton Geraschenko Mar 28 '10 at 0:39
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    $\begingroup$ My (first look) opinion: this is best as a single question, especially since the implicit goal is to compare the answers to the different questions. If this were split into five different questions, each one would seem kinda random without the context of the others. In the same way that it would be silly to ask two separate questions: "what are necessary conditions for X?" and "what are sufficient conditions for X?" $\endgroup$ – Anton Geraschenko Mar 28 '10 at 0:39

Regarding question (c), I can tell you exactly which integral domains have only Noetherian subrings by quoting the aptly titled Integral domains with Noetherian subrings by Robert Gilmer:

If $K$ is the field of fractions and $char(K)=0$, we just need $[K:\mathbb{Q}]<\infty$.

If $char(K)=p$ with prime subfield $k$, we need $K$ to be either finite or a finite algebraic extension of a $k[X]$ for some transcendental $X$.

I guess this pretty much restricts the answers to questions (a) and (b)...

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  • $\begingroup$ Sounds good. With those fields as an upper bound, the other two questions should be easier to answer. Would you consider sketching the proof? $\endgroup$ – Pete L. Clark Mar 27 '10 at 17:37
  • $\begingroup$ Also, when the characteristic is positive, I believe the correct condition is that either $K$ is algebraic (not necessarily finite) over $\mathbb{F}_p$ or is a finite extension of $\mathbb{F}_p(t)$. $\endgroup$ – Pete L. Clark Mar 27 '10 at 17:56
  • $\begingroup$ Ha yes, of course, I used the wrong brackets and inadvertently missed infinitely many examples. I need to work on my copying skills... $\endgroup$ – dke Mar 28 '10 at 15:23

Let me put together the previous two answers (plus epsilon) to give an answer to all three questions.

Step 1: By Gilmer's theorem, a field $K$ has all its subrings Noetherian iff:

(i) It is a finite extension of $\mathbb{Q}$, or
(ii) It is an algebraic extension of $\mathbb{F}_p$ or a finite extension of $\mathbb{F}_p(t)$.

Step 2: Suppose $K$ is a number field which is not $\mathbb{Q}$. We may write $K = \mathbb{Q}[\alpha]$ for some algebraic integer $\alpha$. Then $R = \mathbb{Z}[2\alpha]$ is a non-integrally closed subring of $K$ so is not a Dedekind domain. So the only field of characteristic $0$ which has every subring a Dedekind domain is $\mathbb{Q}$, in which case (by the previous question) every subring is a PID.

Step 3: Suppose $K$ has characteristic $p > 0$. If $K$ is algebraic over $\mathbb{F}_p$, then every subring is a field, hence also Dedekind and a PID. If $K$ is a finite extension of $\mathbb{F}_p(t)$ then it admits a subring of the form $\mathbb{F}_p[t^2,t^3]$, which is not integrally closed.

So the fields for which every subring is a Dedekind ring are $\mathbb{Q}$ and the algebraic extensions of $\mathbb{F}_p$. For all such fields, every subring is in fact a PID.

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For (a) and (b), when the characteristic is positive and K is not algebraic over the prime field, then there is a subring of the form k[t^2,t^3] which is not a PID and is not Dedekind.

When the characteristic is zero, for (b), since a Dedekind domain is required to be integrally closed by defition, once K is a number field different from ℚ, one can find proper subrings of the ring of integers, and these rings will not be integrally closed.

For (d), note that if R is Dedekind with field of fractions K, and if R' is any ring between R and K, then the local rings of R' (localisation with respect to a maximal ideal) will be a subset of the local rings of R. Thus they will all be DVRs. Since the ring of integers in an algebraic number field is Dedekind, this shows that for any number field K, we have that every integrally closed subring is a Dedekind domain.

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  • $\begingroup$ oh dearie me, some poor soul decided to increase the load time by latexifying unnecessarily. i've fixed this. $\endgroup$ – Peter McNamara Sep 12 '14 at 13:25

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