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Let $A$ be a domain. Recall that $A$ is Cohen-Kaplansky (or CK) if

(CK) any nonzero nonunit of $A$ is a product of irreducible elements, and there are only finitely irreducible elements up to multiplication by units.

Consider the following condition on $A$:

(RFFF) the field of fractions $K$ of $A$ is ring-finite over $A$, that is, $K$ is a finitely generated $A$-algebra.

[RFFF stands for "ring-finite field of fractions".]

Clearly, (CK) implies (RFFF).

Naive Conjecture: (RFFF) implies (CK).

Question: Is the Naive Conjecture true?

One easily sees that (RFFF) is equivalent to

(RFFF') there is a nonzero element $a$ in $A$ such that, for any $b$ in $A$, there is a positive integer $n(b)$ such that $b$ divides $a^{n(b)}$.

In particular, the Naive Conjecture holds for unique factorization domains, and also for Dedekind domains.

EDIT A. Thank you very much to Hagen and Bill Dubuque for their answers! Here is an update. From now on, following Kaplansky, I'll call G-domain what I called RFFF-domain above.

$(1)$ Here is how I understand a part of Hagen's answer: Bourbaki, Algèbre Commutative, VI.$6.3$, provides the following example of a G-domain which is not a CK-domain. There exist a field $K$ and a surjective multiplicative monoid morphism $x\mapsto|x|$ from $K$ onto $\mathbb Q_{\ge0}$ satisfying $$ |x|=0\iff x=0,\quad|x+y|\le\max(|x|,|y|), $$ such that the closed ball of radius one is a local domain $A$ with maximal ideal $\mathfrak m$ equal to the open ball of radius one, the ideals of $A$ being the closed balls of radius $r$ with $0\le r\le1$ and the open balls of radius $r$ with $0 < r\le1$, and the group $A^\times$ of units of $A$ being the sphere of radius one. In particular $0$ and $\mathfrak m$ are the only prime ideals. Thus any nonzero $a$ in $\mathfrak m$ satisfies $A[a^{-1}]=K$, so $A$ is a G-domain. The group $K^\times/A^\times$, being isomorphic to $\mathbb Q_{ > 0}$, is not finitely generated, that is $A$ is not a CK-domain.

$(2)$ In the article

Rings with a finite number of primes, I. S. Cohen and Irving Kaplansky, Trans. Amer. Math. Soc. $60$ ($1946$), $468$-$477$,

it is proved that CK-domains are noetherian (Theorem $6$ p. $471$).

$(3)$ Recall from Bill Dubuque's answer that a domain $A$ is a noetherian G-domains if and only if $A$ has only a finite number of non-zero prime ideals, all of which are maximal.

$(4)$ The main question which remains open (at least for me) is this:

Is there a noetherian G-domain which is not a CK-domain?

That is:

Is there a noetherian domain $A$ such that

$\bullet\ A$ has only a finite number of non-zero prime ideals, all of which are maximal,

$\bullet\ A$ has infinitely many (association classes of) irreducibles?

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3 Answers

up vote 4 down vote accepted

The naive conjecture is not true: for every 1-dimensional valuation ring $A$ and every element $x\in M\setminus 0$, $M$ the maximal ideal of $A$, we have $A[1/x]=K$ because $A[1/x]$ must be a valuation ring of lower dimension than $A$ itself. A non-noetherian 1-dimensional valuation ring on the other hand has no irreducible elements at all.

If one assumes $A$ to be noetherian and integrally closed, then by the Krull-Zariski-Goldmann theorem (http://cr.yp.to/zgk.html) we have $K=A[1/x]$ if $A$ satisfies (RFFF). Consequently $A$ has only finitely many prime ideals of height $1$. By Krull's Principal Ideal Theorem there are infinitely many prime ideals between two prime ideal $p\subset q$, if there are any. Thus $A$ must be $1$-dimensional, that is a semi-local principal ideal domain.

Bill Dubuque has drawn my attention to something that I wasn't aware off: namely that if a valuation domain $A$ has a minimal non-zero prime ideal, then $K=A[1/x]$, $K$ the fraction field, for every $x\neq 0$ in this minimal non-zero prime. Proof: $A[1/x]$ is a localization of $A$ and $x$ is a unit in it. Thus all prime ideals of $A$ except $0$ are killed in $A[1/x]$.

Is there a noetherian G-domain, which is not CK ?

Answer: Yes.

In the article D. D. ANDERSON, J. L. MOTT, Cohen-Kaplansky Domains: Integral Domains with a Finite Number of Irreducible Elements, JOURNAL OF ALGEBRA 148, 17-41 (1992) it is shown (Theorem 2.4) that the following properties are equivalent:

  1. $A$ is a CK-domain.
  2. The integral closure $\overline{A}$ is $1$-dimensional semi-local principal ideal domain, for every non-principal maximal ideal $M$ of $A$ the field $A/M$ is finite, $\overline{A}$ is a finite $A$-module and the number of maximal ideals of $A$ and $\overline{A}$ coincide.

Example: Let $K$ be a finite field with $1+1\neq 0$. Let $K[x,y]:=K[X,Y]/(Y^2-X^2(X+1))$ and $A:=K[x,y]_{(x,y)}$. Then $A$ is $1$-dimensional, local, noetherian, hence $\overline{A}$ is a $1$-dimensional semi-local principal ideal domain.

The integral closure of $K[x,y]$ equals the polynomial ring $K[y/x]$, because $(\frac{y}{x})^2=x+1$ and $x,y\in K[y/x]$. We get $\overline{A}=(A\setminus (x,y))^{-1})K[y/x]$; in particular $\overline{A}$ is a finite $A$-module. Moreover $A/M=K$ is finite.

The prime ideals generated by $\frac{y}{x}-1$ and $\frac{y}{x}+1$ lie over $(x,y)$. Note that the two elements are not associated in $K[y/x]$, therefore they generate different prime ideals. Thus $\overline{A}$ must have at least $2$ maximal ideals (in fact it has exactly $2$). Consequently $A$ is not a CK-domain, but of course a noetherian G-domain.

Remark: I wasn't able to explicitely write down infinitely many non-associated irreducibles. Maybe someone else can do that. Candidates are the elements $x+y^k$, which are at least irreducibles.

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Maybe I'm missing something, but why is "no irreducible elements at all" a counterexample to "only finitely many irreducible elements"? –  Pete L. Clark Aug 17 '12 at 14:15
    
Dear @Pete: I think "no irreducible elements at all" is a counterexample to "any nonzero nonunit is a product of irreducible elements" (for a domain which is not a field). –  Pierre-Yves Gaillard Aug 17 '12 at 14:43
    
@Pierre: sorry, I missed the part about nonzero nonunits factoring into irreducibles. –  Pete L. Clark Aug 17 '12 at 18:23
    
Dear Hagen: Thanks for your answer, and especially for the example of a noetherian G-domain which is not CK!!! [QiL already gave such an example: math.stackexchange.com/questions/184487/… .] I don't understand the argument showing that $\frac{y}{x}-1$ and $\frac{y}{x}-1$ are not units. –  Pierre-Yves Gaillard Aug 23 '12 at 13:06
    
Well, it seems that my "argument" wasn't even one. I improved my post in that respect. Sorry. –  Hagen Aug 24 '12 at 6:54
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RFFF domains are studied in Kaplansky's Commutative Rings, where they are called G-domains (honoring Oscar Goldman), in his abstract treatment of Hilbert's Nullstellensatz. Their structure is easily determined in the Noetherian case (1-dim semi-local), but much more complex otherwise. Indeed, Kap writes on p. 13:

Some remarks will precede the development of the theory of G-domains. Of course any field is a G-domain. To get more examples, we examine principal ideal domains. We see immediately that a principal ideal domain is a G-domain if and only if it has only a finite number of primes (up to units).

Later we shall determine exactly which Noetherian domains are G-domains, the precise condition being that there are only a finite number of non-zero prime ideals, all of which are maximal. For non-Noetherian domains the facts are more complex, and we seem to lack even a reasonable conjecture concerning the structure of general G-domains. At any rate, examples show that the Noetherian facts do not at all generalize. It is easy to exhibit a valuation domain that is a G-domain but nevertheless possesses comparable non-zero prime ideals. Also (this is more difficult) there exist G-domains in which all nonzero prime ideals are maximal and there are an infinite number of them.

If memory serves correct, there has been significant work done on structure theory in the non-Noetherian case since Kap's book was published, but I don't recall the authors off the top of my head. They should be easy to locate by chasing links to the original papers of Artin and Tate, Goldman, and Krull (cf. Kap's book). See also subsequent work on CK-domains by D. Anderson and J. Mott.

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A less involved example of a Noetherian G-domain that is not a CK-domain:

Let $\mathbb R$ and $\mathbb C$ denote the fields of real and complex numbers respectively and consider $S = R + X\mathbb C[[X]]$ the ring of power series over $\mathbb C$ with real constant terms. Since the only nonunits of $S$ are powers of $X$ (with coefficients from $\mathbb C$) it is easy to see that

(1) $X\mathbb C[[X]]$ is the maximal ideal of $S$,

(2) $S$ is a G-domain and is $1$-dimensional.

Next note that $X\mathbb C[[X]] = XS+iXS$, where $i$ is the imaginary number. So $S$ is Noetherian. Now a CK-domain is an atomic domain that contains only a finite number of irreducible elements up to associates. But $S$ has infinitely many non-associated atoms: $(r + i)X$ where $r$ ranges over real numbers. (This is a simplified version of what follows Corollary 3.6 of [AM]. )

It has already been established, in a simple enough fashion, here that there are non-Noetherian G-domains and so (RFF) does not imply CK. But an example attributed to Bill Dubuque sort of attracted my attention. The example is: If a valuation domain $V$ has a nonzero minimal prime ideal $P$ then $V[1/x] =qf(V)$ for all nonzero $x\in P$. I have the following observation. (Incidentally, where is Bill Dubuque? Is he Gone?)

Observation: Let $D$ be a domain with a minimal nonzero prime ideal $P$ which is comparable with each principal ideal of $D$, then $D$ is a G-domain.

While it gives Bill’s example it delivers examples that may not be easy to see.

Example B: The ring $T= \mathbb Z+X\mathbb Q[[X]]$ is a G-domain with infinitely many maximal ideals.

Example C: The ring $\mathbb Z+X\mathbb R[[X]]$ is a G-domain with infinitely many maximal ideals.

Next the statement in (3) of Edit A seems a little I have a feeling that “a domain $A$ is a noetherian G-domain if and only if $A$ has only a finite number of non-zero prime ideals, all of which are maximal” needs to be changed to “a noetherian domain $A$ is a G-domain if and only if $A$ has only a finite number of non-zero prime ideals, all of which are maximal”.

Now a bit about the history tidbits by Gone (er Bill, how the heck am I to know?). If you really look up [AM] you would know that the logical next step was [AAZ], the study of atomic domains with almost all atoms prime. Unfortunately there was a massive mistake introduced in the paper, in an effort to get more bang without putting in any bucks. The mistake was caught by Martine Picavet-L’Hermite in [MP]. After a massive ignoring campaign there appeared [CA] where, the mistake was owned. My sympathies go to Martine, she is one of the folks who come up with something remarkable and have to face blank faces of the confirmed geniuses.

[AAZ] D.D. Anderson, D.F. Anderson and M. Zafrullah, (1992). Atomic domains in which almost all atoms are prime. Comm. Algebra 20:1447–1462.

[AM] D.D. Anderson, J. Mott, (1992). Cohen–Kaplansky domains: Integral domains with a finite number of irreducible elements. J. Algebra 148:17–41.

[CA] S. Chun and D.D. Anderson, A class of atomic rings, Comm. Algebra 40(2012) 1086-1095.

[MP] Picavet-L’Hermitte, M. (2000). Factorization in Some Orders with a PID as Integral Closure, Algebraic Number Theory and Diophantine Analysis. Berlin: de Gruyter, pp. 365–390.

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Thanks YACP. I have completed the [AAZ] paper. –  mzafrullah Jul 5 '13 at 2:18
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