Here's another question in the same spirit as my previous one:

Are there any integrally closed BFDs which are not Krull domains?

Some background information:

A BFD (bounded factorization domain) is defined as an atomic domain where the number of factors is bounded for each element. (More precisely: every nonzero nonunit $x$ can be factorized into irreducibles, but not necessarily uniquely, and there is a number $N(x)$ such that whenever $x=a_1 \dots a_r$ is such a factorization, then $r \le N(x)$.)

Unless I've misread something, the following relations hold (with strict inclusions): $$ \{ \text{Noetherian domains} \} \subset \{ \text{BFDs} \} \quad \Bigl( \subset \{ \text{ACCP domains} \} \Bigr), $$ $$ \{ \text{Noetherian domains} \} \cap \{ \text{integrally closed domains} \} = \{ \text{Noetherian domains} \} \cap \{ \text{Krull domains} \}, $$ $$ \{ \text{Krull domains} \} \subset \{ \text{integrally closed domains} \}, $$ $$ \{ \text{Krull domains} \} \subset \{ \text{BFDs} \}. $$ (Disclaimer: Trying to navigate in this jungle of classes of integral domains has got my head spinning, so I'm not 100% sure about anything at the moment...)

For the purposes of drawing a correct Venn diagram involving these relations (and many more!), I would need to know whether there is equality or strict inclusion in the relation $$ \{ \text{Krull domains} \} \subseteq \{ \text{BFDs} \} \cap \{ \text{integrally closed domains} \}. $$


1 Answer 1


Let $D$ be a domain with quotient field $K$. The ring of integer valued polynomials is defined as $$ \operatorname{Int}(D) = \{\, f \in K[X] \mid f(D) \subset D \,\}. $$ It is known that if $D$ is a BF-domain, then so is $\operatorname{Int}(D)$. If $D$ is a Dedekind domain with finite residue fields, then $\operatorname{Int}(D)$ is a Prüfer domain (in particular integrally closed), but does not satisfy the ascending chain condition on divisorial ideals. Hence it is not a Krull domain.

Thus, for instance, the rather beautiful ring $\operatorname{Int}(\mathbb Z)$ is a non-Krull integrally closed BF-domain.

See the book Integer Valued Polynomials by Cahen and Chabert, in particular Chapter VI, for background on integer valued polynomials. (But note that some of the results on factorization properties are outdated.)

An example which is moreover not Prüfer:

This example can be adapted to obtain a ring which is (completely) integrally closed, not Krull, not Prüfer, and BF. Let me first recall how a bunch of ring-theoretic properties behave for $\operatorname{Int}(D)$.

If $\operatorname{Int}(D)$ is Prüfer, then $D$ is an almost Dedekind domain (all localizations at maximal, equivalently prime, ideals are discrete valuation rings) with all residue fields finite. In particular, almost Dedekind domains are $1$-dimensional Prüfer domains.

Moreover, $\operatorname{Int}(D)$ is integrally closed if and only if $D$ is. Similarly, $\operatorname{Int}(D)$ is completely integrally closed (c.i.c.) if and only if $D$ is. (For Noetherian domains, being c.i.c. and being integrally closed are equivalent. In the non-Noetherian case, being c.i.c is a stronger property.)

Recall a convenient ring-theoretic characterization of Krull domains: A domain $D$ is Krull if and only if it is completely integrally closed and a Mori domain (satisfies the ascending chain condition on divisorial ideals).

It is also known (see P.-J. Cahen, S. Gabelli, E. Houston, Mori domains of integer-valued polynomials, J. Pure. Appl. Algebra, 153(1), 2000) that $\operatorname{Int}(D)$ Mori implies $D$ Mori.

So, to get $\operatorname{Int}(D)$ to be BF, integrally closed, non-Prüfer, non-Krull we want that $D$ is

  • BF,
  • integrally closed,
  • not an almost Dedekind domain (to ensure non-Prüfer),
  • not c.i.c. or not Mori (to ensure non-Krull).

Actually, it would be better if $D$ is even c.i.c. but not Mori. Here are two possibilities:

  1. Take $D=K[X,XY,XY^2,XY^3,\ldots] \subset K[X,Y]$. By exercise I.3.14 in L. Fuchs, L. Salce, Modules Over Non-Noetherian Domains, $D$ is integrally closed but not c.i.c. It is easy to see that it is 2-dimensional and BF.

  2. Incidentally, $D=\operatorname{Int}(\mathbb Z)$ is c.i.c., non-Krull, 2-dimensional and BF. So $$ \operatorname{Int}(\operatorname{Int}(\mathbb Z)) $$ is non-Krull, non-Prüfer, c.i.c., BF. I haven't seen this ring before; it's rather amusing that this works! You can in fact replace $\mathbb Z$ by any Dedekind domain with finite residue fields and get the same result.

Edit: The following may be an easier example. If $D$ is integrally closed, non-Krull, BF then so is $D[X]$. ($D[X]$ is Krull iff $D$ is Krull.) $D[X]$ can moreover not be Prüfer ($D[X]$ is Prüfer iff $D$ is a field). So, for instance $\operatorname{Int}(\mathbb Z)[X]$ also works. Unfortunately, I don't know enough about integer valued polynomials to see quickly whether actually $\operatorname{Int}(\mathbb Z)[X]$ is a proper subring of $\operatorname{Int}(\operatorname{Int}(\mathbb Z))$ or whether they are equal. (If a domain $D$ is an intersection of localizations at prime ideals with infinite residue fields, then $\operatorname{Int}(D)=D[X]$ is trivial.)

  • $\begingroup$ Great, thank you! This was a somewhat unexpected answer, since apparently (when I look back at my old Venn diagram) I had gotten the impression somewhere that a BFD which is Prüfer must automatically be Dedekind (hence Krull). But this example shows that that's wrong. So a follow-up question would be if there are any BFDs which are integrally closed but not Krull and not Prüfer either... $\endgroup$ Commented Sep 3, 2015 at 7:52
  • $\begingroup$ Exactly. In fact $\operatorname{Int}(\mathbb Z)$ is a somewhat classical example of a non-Dedekind Prüfer domain. It is $2$-dimensional. (Unlike $\overline{\mathbb Z}$, the other classical example). $\endgroup$
    – moonlight
    Commented Sep 3, 2015 at 10:04
  • $\begingroup$ Wow, that was a quick update! I'll have to digest it for a while (and learn some more algebra too...), but this should be very helpful! $\endgroup$ Commented Sep 3, 2015 at 11:01
  • $\begingroup$ Sure. Btw, being a Mori domain is a sufficient property for a ring to be a BF-domain. The c.i.c. property of a Krull domain is somewhat orthogonal to that. It ensures that all nonzero divisorial fractional ideals are invertible with respect to the divisorial product; so the nonzero div. frac. ideals form a (lattice-ordered) abelian group. The Mori property then ensures that the group is free abelian. Compare Krull/c.i.c. to Dedekind/Prüfer: In a Prüfer domain the nonzero frac. ideals form a group, the ACC of the Dedekind domain ensures it's free abelian. $\endgroup$
    – moonlight
    Commented Sep 3, 2015 at 11:15

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