Question

This question was inspired by

http://mathoverflow.net/questions/19480/how-to-prove-that-the-subrings-of-the-rational-numbers-are-noetherian/19481#19481

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?

Answer 1

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)...

Answer 2

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.

Answer 3

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.