arXiv:1708.02546v1 [] 8 Aug 2017

A PRINCIPAL IDEAL THEOREM FOR COMPACT SETS OF
RANK ONE VALUATION RINGS
BRUCE OLBERDING

Abstract. Let F be a field, and let Zar(F ) be the space of valuation rings of F
with respect to the Zariski topology. We prove that if X is a quasicompact set of
rank one valuation rings in Zar(F ) whose maximal ideals do not intersect to 0,
then the intersection of the rings in X is an integral domain with quotient field F
such that every finitely generated ideal is a principal ideal. To prove this result, we
develop a duality between (a) quasicompact sets of rank one valuation rings whose
maximal ideals do not intersect to 0, and (b) one-dimensional Prüfer domains with
nonzero Jacobson radical and quotient field F . The necessary restriction in all
these cases to collections of valuation rings whose maximal ideals do not intersect
to 0 is motivated by settings in which the valuation rings considered all dominate
a given local ring.

1. Introduction
Throughout this article, F denotes a field and Zar(F ) denotes the set of valuation
rings of F ; i.e., the subrings V of F such that for all 0 ≠ t ∈ F , t ∈ V or t−1 ∈ V . In
this article we are interested in subrings A of F which are an intersection of rank
one valuation rings in a quasicompact subset of Zar(F ). The rank of a valuation
ring, which coincides with its Krull dimension, is the real rank of its value group.
Thus the rank one valuation rings have valuations that take values in R ∪ {∞}.
The Zariski topology on Zar(F ) is the topology having as a basis of open sets the
subsets of Zar(F ) of the form {V ∈ Zar(F ) ∶ t1 , . . . , tn ∈ F } for t1 , . . . , tn ∈ F . With
this topology, Zar(F ) is the Zariski-Riemann space of F . The main purpose of this
article is to prove the following instance of what Roquette [38] calls a Principal Ideal
Theorem, that is, a theorem which guarantees a given class of integral domains has
the property that every finitely generated ideal is principal.
Main Theorem. If X is a quasicompact set of rank one valuation rings in Zar(F )
whose maximal ideals do not intersect to 0, then the intersection of the valuation
rings in X is an integral domain with quotient field F and Krull dimension one such
that every finitely generated ideal is a principal ideal.
The theorem, which most of the paper is devoted to proving, asserts that the
intersection of such valuation rings is a Bézout domain, a domain for which every
MSC: 13A18; 13F05; 14A15.
1

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finitely generated ideal is principal. Such rings belong to the extensively studied
class of Prüfer domains, those domains A for which AM is a valuation domain for
each maximal ideal M of A. The problem of when an intersection of valuation rings
is a Prüfer domain is a difficult but well-studied problem that has applications to real
algebraic geometry (e.g., [1, 2, 3, 42]), non-Noetherian commutative ring theory (e.g.,
[15, 27, 39]), formally p-adic holomorphy rings [38] and the study of integer-valued
polynomials [4, 24]. Using the equivalence of this problem to that of determining
when a subspace of Zar(F ) yields an affine scheme, a geometric criterion involving
morphisms of the Zariski-Riemann space into the projective line was given in [33].
In that approach, treating the Zariski-Riemann space as a locally ringed space, not
simply a topological space, is crucial. By contrast, the main theorem shows that
unlike in the general case, the geometry of the locally ringed space structure is not
needed to distinguish a Bézout intersection when the valuation rings satisfy the
hypotheses of the theorem. Instead, the question of whether the intersection of such
a collection of rank one valuation rings is a Prüfer domain is purely topological.
Moving beyond rank one valuation rings, quasicompactness of a subset X of
Zar(F ) is far from sufficient to guarantee that the intersection of valuation rings in
X is a Bézout domain. For example, if D is a Noetherian local domain with quotient
field F , then for any t1 , . . . , tn ∈ F ,
U = {V ∈ Zar(F /D) ∶ x1 , . . . , xn ∈ V }
is quasicompact, but the intersection of the valuation rings in U is a Bézout domain if and only if the integral closure of D[x1 , . . . , xn ] is a principal ideal domain,
something that occurs only for very special choices of D and x1 , . . . , xn . In fact,
the main theorem is optimal in the sense that if any one of the hypotheses “quasicompact”, “rank one”, or “the maximal ideals do not intersect to 0” is omitted,
the conclusion is false; see Example 5.7. Note that the condition that the maximal
ideals of the rank one valuation rings in X do not intersect to 0 occurs naturally
in settings where the valuation rings in X are assumed to dominate a local ring of
Krull dimension > 0. Examples of such settings of recent interest include Berkovich
spaces and tropical geometry (see for example [13]) and valuative trees of regular
local rings [9, 17].
Interest in compactness in the Zariski-Riemann space dates back to Zariski’s introduction of the topology on Zar(F ) in [44]. If D is a subring of F , then Zar(F /D),
the subspace of Zar(F ) consisting of the valuation rings in Zar(F ) that contain D
as a subring, is the Zariski-Riemann space of F /D. That Zar(F /D) is quasicompact was proved by Zariski [44] in 1944 as a step in his program for resolution of
singularities of surfaces and three-folds. In more recent treatments of the topology
of Zar(F /D) such as [10, 36], quasicompactness is viewed as part of a more refined
topological picture that treats Zar(F /D) as a spectral space or as a locally ringed

COMPACT SETS AND HOLOMORPHY RINGS

3

space that is a projective limit of projective schemes. The latter point of view also
has its origins in Zariski’s work [44].
However, in restricting to the subspace of rank one valuation rings of F , key
topological features of Zar(F ) are lost. For example, this subspace need not be
spectral, nor even quasicompact. Yet, in passing to the space of rank one valuation
rings, the main theorem shows that the topology becomes much more consequential
for the ring-theoretic structure of an intersection of valuation rings. One of the key
steps in proving this is first showing that in the setting of the main theorem, the
intersection of valuation rings in X is a Prüfer domain. We prove this in Theorem 5.6
by establishing the following lemma. For a subset X of Zar(F ), we let A(X) =
⋂V ∈X V be the holomorphy ring1 of X and J(X) = ⋂V ∈X MV be the ideal of A(X)
determined by the intersection of the maximal ideals MV of the valuation rings V .
For a ring A, Max(A) denotes its set of maximal ideals.
Main Lemma. The mappings

X ↦ A(X) and A ↦ {AM ∶ M ∈ Max(A)}

define a bijection between the quasicompact sets X of rank one valuation rings in
Zar(F ) with J(X) ≠ 0 and the one-dimensional Prüfer domains A with nonzero
Jacobson radical and quotient field F .
Corollary 5.9 gives another version of this result in which the spaces X need not
be assumed a priori to satisfy J(X) ≠ 0. In this case, “quasicompact” is replaced
with “compact” (= quasicompact and Hausdorff), and the holomorphy rings are
all assumed to have quotient field F . A consequence of the main lemma is that
if X is quasicompact, J(X) ≠ 0 and X consists of rank one valuation rings, then
X = {A(X)M ∶ M ∈ Max(A(X))}. As this suggests, the main lemma can be recast
in the language of schemes, and we do this in Corollary 5.14.
The applicability of the main theorem depends on whether a space of rank one
valuation rings can be determined to be quasicompact. The key technical observation behind our approach is that if a subset X of Zar(F ) consists of rank one
valuation rings and J(X) ≠ 0, then X is quasicompact if and only if X is closed in
the patch topology of Zar(F ). (See Section 2.) On the level of proofs and examples,
this reduces the issue of quasicompactness to calculation of patch limits point of X
in Zar(F ), and specifically whether such limit points have rank one. In a future
paper [37] we use the results of the present article along with additional methods
to show how to apply these ideas to divisorial valuation overrings of a Noetherian
local domain of Krull dimension two. This is discussed in Remark 6.4.
The present paper is motivated by recent work such as in [10, 11, 31, 32, 33,
34, 35, 36] on understanding how topological or geometric properties of a space of
1See Roquette [38] for an explanation of this terminology.

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valuation rings are reflected in the algebraic structure of the intersection of these
valuation rings.
2. Topological preliminaries
In this section we outline the topological point of view needed for the later sections. Recall that throughout the paper, F denotes an arbitrary field.
Notation 2.1. For each subset S of the field F , let

U (S) = {V ∈ Zar(F ) ∶ S ⊆ V } and V (S) = {V ∈ Zar(F ) ∶ S ⊆/ V }.

The Zariski topology on Zar(F ) has as a basis of nonempty open sets the sets of the
form U (x1 , . . . , xn ), where x1 , . . . , xn ∈ F . The set Zar(F ) with the Zariski topology
is the Zariski-Riemann space of F . Several authors have established independently
that the Zariski-Riemann space Zar(F ) is a spectral space2, meaning that (a) Zar(F )
is quasicompact and T0 , (b) Zar(F ) has a basis of quasicompact open sets, (c) the
intersection of finitely many quasicompact open sets in Zar(F ) is quasicompact, and
(d) every nonempty irreducible closed set in Zar(F ) has a unique generic point. See
[10] and [36] for more on the history of this central result. A spectral space admits
the specialization order ≤ given by x ≤ y if and only if y is in the closure of {x}.
In the Zariski topology on Zar(F ), we have for V, W ∈ Zar(F ) that V ≤ W if and
only if W ⊆ V . We use the specialization order in the results in this section but not
elsewhere in the paper.
As a spectral space, Zar(F ) admits two other useful topologies, the inverse and
patch topologies. The inverse topology on Zar(F ) is the topology that has, as a
basis of closed sets, the subsets of Zar(F ) that are quasicompact and open in the
Zariski topology; i.e., the nonempty closed sets are intersections of finite unions of
sets of the form U (x1 , . . . , xn ), x1 , . . . , xn ∈ F . The inverse topology is useful for
dealing with issues of irredundance and uniqueness of representations of integrally
closed subrings of F ; for example, see [35]. In the present article, we use the inverse
topology in a limited way.
The most important topology on Zar(F ) for the purposes of this article is the
patch topology on Zar(F ), which is given by the topology that has as a basis of open
sets the subsets of Zar(F ) of the form
U (x1 , . . . , xn ), V (y1 ) or U (x1 , . . . , xn ) ∩ V (y1 ) ∩ ⋯ ∩ V (ym ),

where x1 , . . . , xn , y1 , . . . , ym ∈ F . The complement in Zar(F ) of any set in this basis
is again open in the patch topology. Thus the patch topology has as a basis sets
that are both closed and open (i.e., the patch topology is zero-dimensional). The
patch topology is also spectral and hence quasicompact [23, p. 45]. Unlike the
2The terminology is motivated by a theorem of Hochster [23] that shows a space is spectral if

and only if it is homeomorphic to the prime spectrum of a ring

COMPACT SETS AND HOLOMORPHY RINGS

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Zariski and inverse topologies on Zar(F ), the patch topology is always Hausdorff
[23, Theorem 1].
Convention. In the article we work with all three topologies, inverse, patch
and Zariski, sometimes even in the same proof. To avoid confusion, we insert the
adjective “patch” before a topological property when working with it in the patch
topology. For example, a “patch open set” is a set that is open in the patch topology.
Similarly, we insert “inverse” as an adjective when working with the inverse topology.
If no adjective is present (e.g., “the set Z is quasicompact”), this is always to be
understood as indicating we are working in the Zariski topology. Thus the Zariski
topology is the default topology if no other topology is specified. Recall also from
the Introduction that by compact we mean both quasicompact and Hausdorff.
One of our main technical devices in the paper is that of a patch limit point.
Let X ⊆ Zar(F ). Then V ∈ Zar(F ) is a patch limit point of X if each patch open
neighborhood U of V in X contains a point in X distinct from V ; equivalently
(since the patch topology is Hausdorff), every patch open neighborhood U of V
contains infinitely many valuation rings in X. Applying the relevant definitions, it
follows that V is a patch limit point of X if and only if for all finite (possibly empty)
subsets S of V and T of MV there is a valuation ring U in X ∖ {V } such that S ⊆ U
and T ⊆ MU .3
Notation 2.2. Let ∅ ≠ X ⊆ Zar(F ). We use the following notation.
●
●
●
●

lim(X) = the set of patch limit points of X in Zar(F ).
patch(X) = X ∪ lim(X) = closure of X in the patch topology of Zar(F ).
A(X) = ⋂V ∈X V = holomorphy ring of X.
J(X) = ⋂V ∈X MV .

In the next lemma we collect some properties of patch closure that are needed in
later sections. More systematic treatments of Zariski, patch and inverse closure in
Zar(F ) can be found in [10] and [36] and their references.

Lemma 2.3. Let X be a nonempty subset of Zar(F ). Then

(1) A(X) = ⋂V ∈patch(X) V and J(X) = ⋂V ∈patch(X) MV .
(2) The set patch(X) is spectral in the subspace Zariski topology.

Suppose in addition that X is quasicompact and consists of rank one valuation rings.
(3) The set patch(X) is contained in X ∪ {F }.
(4) If S is a multiplicatively closed subset of A = A(X), then AS = ⋂AS ⊆V ∈X∪{F } V .

Proof. (1) The first assertion can be found in [10, Proposition 4.1] or [36, Proposition
5.6]. Since patch(X) = X ∪ lim(X), to see that the second assertion holds, it suffices
to show that J(X) ⊆ MU for each U ∈ lim(X). Let 0 ≠ a ∈ J(X), and let U ∈ lim(X).
3Throughout the paper, we denote the maximal ideal of a valuation ring V by M .
V

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If a ∈/ MU , then U ∈ U (a−1 ). Since U ∈ lim(X), there exists V ∈ X ∩ U (a−1 ), so that
a−1 ∈ V . However, a ∈ MV by the choice of a, a contradiction. Thus J(X) ⊆ MU ,
which verifies (1).
(2) A patch closed subspace of a spectral space is spectral in the subspace topology
[18, Proposition 9.5.29, p. 433].
(3) Suppose X is quasicompact. This implies that if U ∈ patch(X), then there
exists V ∈ X such that V ⊆ U [36, Proposition 2.2]. Since X consists of rank one
valuation rings, we have V = U or U = F . Thus patch(X) ⊆ X ∪ {F }.
(4) By [36, Corollary 5.7],
AS =

⋂

V,

AS ⊆V ∈Y

where Y is the set of all V ∈ Zar(F ) such that V ⊇ U for some U ∈ patch(X). Thus
(4) follows from (3) and the fact that every valuation ring in X has rank one.

The following proposition reinterprets for rank one valuation rings the property of
compactness in the Zariski topology of Zar(F ) in terms of the patch topology. This
enables us to work with the patch topology– and specifically, patch limit points– in
the algebraic arguments of the next sections. The proposition also shows that the
Hausdorff condition for a set X of rank one valuation rings is closely connected with
the algebraic property that J(X) ≠ 0.
Proposition 2.4. The following statements hold for every nonempty set X of rank
one valuation rings in Zar(F ).

(1) X is compact if and only if X is patch closed in Zar(F ).
(2) If J(X) ≠ 0, then the Zariski and patch topologies agree on X, and hence X is
Hausdorff and zero-dimensional.
(3) Suppose A(X) has quotient field F . Then J(X) ≠ 0 if and only if X is Hausdorff.

Proof. (1) Suppose that X is compact. By Lemma 2.3(3), patch(X) ⊆ X ∪{F }, so to
show that X is patch closed it suffices to show that F /∈ patch(X). If X consists of a
single valuation ring, then patch(X) = X and the claim is clear since this valuation
ring must have rank one. Suppose X contains at least two distinct valuation rings
V and W . Since X is Hausdorff, there exist x1 , . . . , xn , y1 , . . . , ym ∈ F such that
V ∈ U (x1 , . . . , xn ), W ∈ U (y1 , . . . , ym ) and
U (x1 , . . . , xn ) ∩ U (y1 , . . . , ym ) ∩ X = ∅.

Let

Y = V (x1 , . . . , xn ) ∪ V (y1 , . . . , ym ).

Then X ⊆ Y and Y is patch closed, so patch(X) ⊆ Y . Since F ∈/ Y , we have
F /∈ patch(X). Therefore, X is patch closed.

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7

Conversely, suppose that X is patch closed in Zar(F ). By Lemma 2.3(2), X is
a spectral space with respect to the Zariski topology and hence is quasicompact.
Since the valuation rings in X have rank one and F ∈/ X, the elements of X are
minimal in X with respect to the specialization order. This implies the patch and
Zariski topologies agree on X [41, Corollary 2.6]. Thus X is Hausdorff since the
patch topology is Hausdorff.
(2) Let 0 ≠ x ∈ J(X). Then X ⊆ V (x−1 ). Since the valuation rings in X have
rank one and F ∈/ V (x−1 ), the elements of X are minimal in V (x−1 ) with respect
to the specialization order induced by the Zariski topology. As a patch closed
subset of Zar(F ), the set V (x−1 ) is, with respect to the Zariski topology, a spectral
space (Lemma 2.3(2)). Thus the Zariski and patch topologies agree on the set of
elements of V (x−1 ) that are minimal with respect to the specialization order [41,
Corollary 2.6]. Since the patch topology is Hausdorff and zero-dimensional, the last
statement of (2) now follows.
(3) Suppose A = A(X) has quotient field F and X is Hausdorff. If X consists
of a single valuation ring, then it is clear that J(X) ≠ 0. Assume that X has more
than one valuation ring. Since X is Hausdorff and A has quotient field F , there
exist nonzero a1 , . . . , an , b1 , . . . , bm , c ∈ A such that
U (a1 /c, . . . , an /c) ∩ U (b1 /c, . . . , bm /c) ∩ X = ∅.

Therefore, U (1/c) ∩ X = ∅. Since each V ∈ X is a valuation ring, this implies
c ∈ MV , so that 0 ≠ c ∈ J(X). The converse follows from (2).

3. Residually transcendental limit points
The main result of this section, Theorem 3.2, is of a technical nature and involves
the existence of patch limit points that are residually transcendental over a local
subring of F . For the purpose of proving the results in this section, we recall the
notion of a projective model of a field; see [45, Chapter 6, §17] for more background
on this topic. Let D be a subring of the field F , and let t0 , t1 , . . . , tn be nonzero
elements of F . For each i, let Di = D[t0 /ti , . . . , tn /ti ], and let

M = ⋃ {(Di )P ∶ P ∈ Spec(Di )}.
n

i=0

Then M is the projective model of F /D defined by t0 , t1 , . . . , tn . Alternatively, a
projective model of F /D can be viewed as a projective integral scheme over Spec(D)
whose function field is a subfield of F . Motivated by this interpretation, it is often
convenient to view M as a locally ringed space; see for example [36, Section 3].
We do not need to do this explicitly in the present paper, but the notation remains
helpful here when we wish to view the local rings in M as points. In particular,

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viewing the set M as a topological space with respect to the Zariski topology4, the
local rings in M are points in M. For this reason, and in keeping with the locally
ringed space point of view, we denote a local ring x ∈ M by OM,x , despite the
redundancy in doing so. A subset Y of M is an affine submodel of M if there exists
a D-subalgebra R of F such that Y = {RP ∶ P ∈ Spec(R)}. (We differ here from [45]
in that we do not require R to be a finitely generated D-algebra.)
For each x ∈ M, there exists a valuation ring in Zar(F ) that dominates OM,x , and
for each V ∈ Zar(F /D), there exists a unique x ∈ M such that V dominates OM,x ;
see [45, pp. 119–120] or apply the valuative criterion for properness [20, Theorem 4.7,
p. 101]. For a nonempty subset X of M, we denote by M(X) the set of all x ∈ M
such that OM,x is dominated by some V ∈ X. More formally, M(X) is the image
of X under the continuous closed surjection Zar(F ) → M given by the domination
mapping [45, Lemma 5, p. 119].
Lemma 3.1. Let X be a nonempty subset of Zar(F ) such that A = A(X) is a
local ring. Let D be a subring of A, let t0 , . . . , tn be nonzero elements of F , and
let M be the projective model of F /D defined by t0 , . . . , tn . If M(X) is finite, then
t0 /ti , . . . , tn /ti ∈ A for some i = 0, 1, . . . , n.

Proof. Suppose M(X) is finite. Each finite subset of the projective model M is
contained in an affine open submodel of M. (This is a consequence of homogeneous
prime avoidance; see for example [43, Lemma 01ZY] or the proof of [33, Corollary 3.2]). Thus M(X) is contained in an affine submodel of M, and hence there
is a D-subalgebra R of F such that for each x ∈ M(X), OM,x is a localization of
R. For each V ∈ X, there is x ∈ M(X) such that V dominates OM,x . Since A is a
local ring, so is the subring
B = ⋂ OM,x
x∈M(X)

of A. (Indeed, if b ∈ B is a unit in A, then since, for each x ∈ M(X), OM,x is
dominated by a valuation ring in X, b is a unit in B. Thus if b, c ∈ B are nonunits
in B, then b + c is a nonunit in the local ring A, so that b + c is a nonunit in B.)
For each x ∈ M(X), since OM,x is a localization of R and R ⊆ B ⊆ OM,x with B
a local ring, we have that OM,x is a localization of B at a prime ideal of B. Since
M(X) is finite, B is a local ring that is a finite intersection of the local rings OM,x ,
x ∈ M(X), each of which is a localization of B at a prime ideal. It follows that
B is equal to one of these localizations; i.e., B = OM,x for some x ∈ M(X). Since
B ∈ M, there is i such that t0 /ti , . . . , tn /ti ∈ B ⊆ A.

The proof of the next lemma can be streamlined by using the language of schemes
and morphisms into the projective line (see [33] for more on this point of view in
4The basis for this topology is given by sets of the form {R ∈ M ∶ x , . . . , x ∈ R}, where
1
n

x1 , . . . , xn ∈ F .

COMPACT SETS AND HOLOMORPHY RINGS

9

our context), but in order to make the proof self-contained, we develop the needed
ideas in the course of the proof.
Theorem 3.2. Let X be a nonempty subset of Zar(F ) such that A = A(X) is a
local ring, and let 0 ≠ t ∈ F . Suppose D is a local subring of A, t ∈/ A, 1/t ∈/ A, and
all but at most finitely many V ∈ X dominate D. Then there exists U ∈ lim(X) such
that t, 1/t ∈ U , U dominates D and the image of t in U /MU is transcendental over
the residue field of D.
Proof. We first show that we can replace D with a local subring D ′ of A such that D ′
is integrally closed in D ′ [t] and all but finitely many valuation rings in X dominate
D ′ . Let D denote the integral closure of D in D[t]. Since A in integrally closed in F ,
D ⊆ A. Let m = M ∩ D, where M is the maximal ideal of A, and let D ′ = D m . Then
A dominates D ′ . Since D ⊆ D[t], we have D[t] = D[t], and so D is integrally closed
in D[t]. Thus D ′ = D m is integrally closed in D ′ [t] = D m [t] since D ′ and D ′ [t] are
localizations of D and D[t], respectively, at the same multiplicatively closed set.
Therefore, D ′ is a local ring that is integrally closed in D ′ [t] and is dominated by
A.
To see next that all but finitely many valuation rings in X dominate D ′ , suppose
that V ∈ X and V dominates D. We claim that V dominates D ′ . Let n denote the
maximal ideal of D, and let p = MV ∩ D. Since V dominates D, p lies over n. Thus,
since D is integral over D, p is a maximal ideal of D. If p ≠ m, then there exists
d ∈ p ∖ m so that 1/d ∈ D m = D ′ ⊆ A ⊆ V . Since d ∈ MV , this is a contradiction that
implies p = m. Thus V dominates D ′ . This shows that every valuation ring in X
that dominates D also dominates D ′ . Thus, if V ∈ X does not dominate D ′ , then
V does not dominate D. By assumption, there are at most finitely many valuation
rings in X that do not dominate D, so there are at most finitely many valuation
rings in X that do not dominate D ′ . With this in mind, we work for the rest of the
proof with D ′ instead of D and draw the desired conclusion for D in the last step
of the proof. The advantage of working with D ′ is that D ′ is a local ring that is
integrally closed in D ′ [t].
Let M be the projective model of F /D ′ defined by 1, t. Let f ∶ M → Spec(D ′ )
be the canonical mapping that sends a local ring R in M to D ′ ∩ N , where N is the
maximal ideal of R. Let C = f −1 (m′ ), where m′ is the unique maximal ideal of D ′ .
Since f is continuous in the Zariski topology [45, Lemma 5, p. 119], C is the closed
subset of M consisting of all the local rings in M that dominate D ′ . Since D ′ is
integrally closed in D ′ [t] and t, t−1 ∈/ D ′ (indeed, by assumption, t, t−1 ∈/ A), Seidenberg’s Lemma [29, Exercise 31, pp. 43–44] implies that the rings D ′ [t]/mD ′ [t] and
D ′ [t−1 ]/m′ D ′ [t−1 ] are each isomorphic to the polynomial ring (D ′ /m′ )[T ], where T
is an indeterminate. These isomorphisms are induced by the mappings t ↦ T and
t−1 ↦ T , respectively. Thus (t, m′ )D ′ [t] is a maximal ideal of D ′ [t], while m′ D ′ [t]
is a prime ideal in D ′ [t] and m′ D ′ [t−1 ] is a prime ideal in D ′ [t−1 ].

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Now, since m′ extends to a prime ideal in both D ′ [t] and D ′ [t−1 ], C is irreducible in the Zariski topology with generic point the local ring D ′ [t]m′ D′ [t] =
D ′ [t−1 ]m′ D′ [t−1 ] . Using the fact that the rings D ′ [t]/m′ D ′ [t] and D ′ [t−1 ]/m′ D ′ [t−1 ]
are each PIDs isomorphic to the polynomial ring (D ′ /m′ )[T ], it follows that the
closed points in C are the local rings in the set
{D ′ [t]P ∶ P is a maximal ideal in D ′ [t], m ⊆ P } ∪ {D ′ [t−1 ](m,t−1 ) }.

The only point in C that is not closed is D ′ [t]m′ D′ [t] , the unique generic point of C.
This accounts for all the local rings in C. The rest of the proof consists in showing
that there is a valuation ring U in lim(X) that dominates D ′ [t]m′ D′ [t] .
To this end, we next describe the local rings in M(X) ∩ C; i.e., we describe the
local rings in C that are dominated by valuation rings in X. In particular, we show
that there are infinitely many such local rings in C and that the Zariski closure of
M(X) ∩ C in M is C.
Let X ∗ be the set of valuation rings in X that dominate D ′ . We have established
that all but finitely many valuation rings in X dominate D ′ ; that is, X ∖X ∗ is finite.
The image M(X ∗ ) of X ∗ in M under the domination mapping is contained in C,
so that C(X ∗ ) = M(X ∗ ). By Lemma 3.1 the fact that A is a local ring and t and
1/t are not elements of D ′ implies M(X) is infinite. Since also X ∖ X ∗ is finite,
it follows that M(X ∗ ) = C(X ∗ ) is infinite. Thus, since C consists only of closed
points and a unique generic point for C, there are infinitely many closed points of
C in C(X ∗ ), which means there is a subset X ′ of X ∗ such that the image C(X ′ ) of
X ′ in C is infinite and consists of rings of the form D ′ [t]P , where P is a maximal
ideal of D ′ [t] that contains m′ . Therefore, the valuation rings in X ′ contain D ′ [t],
and the image of X ′ in Spec(D ′ [t]) under the map that sends a valuation ring to
its center in D ′ [t] consists of infinitely many maximal ideals of D ′ [t], all of which
contain the dimension one prime ideal m′ D ′ [t]. Since D ′ [t]/m′ D ′ [t] is a PID, the
intersection of these infinitely many maximal ideal is m′ D ′ [t].
Next, to see how this fact is reflected in C, let M1 be the affine submodel of M
given by
M1 = {D′ [t]P ∶ P ∈ Spec(D′ [t])}.

Let g ∶ M1 → Spec(D ′ [t]) denote the homeomorphism that sends a local ring D ′ [t]P
in M1 to its center in D ′ [t]. We have shown that the Zariski closure of C(X ′ ) in
Spec(D ′ [t]) is the set of all prime ideals of D ′ [t] containing the prime ideal m′ D ′ [t].
Since g is a homeomorphism, it follows that D ′ [t]m′ D′ [t] is in the Zariski closure of
C(X ′ ). Hence the Zariski closure of C(X ′ ) in M is C.
Now, since the generic point of the closed set C is the local ring D ′ [t]m′ D′ [t] and
C(X ′ ) is infinite and Zariski dense in C, we apply [36, Lemma 2.7(4)] to obtain that
there is a valuation ring U in the patch closure of X ′ that dominates
D ′ [t]m′ D′ [t] = D ′ [1/t]m′ D′ [1/t] .

COMPACT SETS AND HOLOMORPHY RINGS

Since

11

D ′ [t]/m′ D ′ [t] ≅ (D ′ /m′ )[T ],

the image of t in the residue field of U is transcendental over the residue field of D ′ .
Since all the valuation rings in X ′ are centered on maximal ideals of D ′ [t], U is not
a member of X ′ . Therefore, since U is in the patch closure of X ′ but not in X ′ , it
must be that U ∈ lim(X ′ ) ⊆ lim(X). Finally, since D ′ dominates D, we conclude
that the image of t in U /MU is transcendental over the residue field of D, which
completes the proof of the theorem.

4. Limit points of rank greater than one
We show in Theorem 4.3 that if X ⊆ Zar(F ) and A(X) is local but not a valuation
domain, then there is a patch limit valuation ring of X of rank > 1. This is the key
result needed in the next section to prove the main results of the paper. The proof of
Theorem 4.3 relies on the following lemma, which gives a criterion for the existence
of a valuation ring of rank > 1 to lie in the patch closure of X.
Lemma 4.1. Let A be a local integrally closed subring of F with maximal ideal M ,
and let X be a patch closed subset of Zar(F ) such that A = A(X). Suppose that
there exist 0 ≠ t ∈ F and m ∈ M such that mt /∈ A and t−1 /∈ A. If for each i > 0
there exists Vi ∈ X such that m ∈ MVi and mti is a unit in Vi , then X contains a
valuation ring of rank > 1.
Proof. Let i > 0. If mti ∈ A, then (mt)i ∈ A, so that since A is integrally closed
we have mt ∈ A, contrary to assumption. Thus mti ∈/ A for all i > 0. Moreover,
(mti )−1 ∈/ A, since otherwise t−i = m(m−1 t−i ) ∈ A, which since A is integrally closed
forces t−1 ∈ A, contrary to the choice of t. By assumption, there exists Vi ∈ X such
that mti is a unit in Vi and m ∈ MVi . If t ∈ Vi , then since mti is a unit in Vi it is
the case that m is a unit in Vi and an element of MVi , a contradiction. Thus t ∈/ Vi .
Using Notation 2.2, let
Ci = U (mti ) ∩ V (t) ∩ X.

Then Vi ∈ Ci , so that Ci is a nonempty patch closed subset of X.
We use compactness to show that ⋂i>0 Ci is nonempty. To this end, we claim
that the collection {Ci ∶ i > 0} has the finite intersection property. Let i > 0, and let
0 < j < i. Then
Vi ∈ Cj = U (mtj ) ∩ V (t) ∩ X

since mti ∈ Vi and t−1 ∈ Vi implies mtj = mti (t−1 )i−j ∈ Vi . For each i > 0, it
follows that C1 ∩ C2 ∩ ⋯ ∩ Ci contains the valuation ring Vi . Therefore, the collection
{Ci ∶ i > 0} of patch closed subsets of X has the finite intersection property. Since
X is a patch closed subset of the patch quasicompact space Zar(F ), X is patch

12

BRUCE OLBERDING

quasicompact. Thus the set
i
⋂ Ci = X ∩ V (t) ∩ ( ⋂ U (mt ))

i>0

i>0

is nonempty. Let U be a valuation ring in this intersection. Then U ∈ X, and, for
each i > 0, we have 0 ≠ m ∈ (t−1 )i U . Also, since t /∈ U , we have t−1 ∈ MU . Thus
m ∈ (t−1 )i U ⊆ MU . If U has rank 1, then the radical of mU in U is MU . In this
case there exists n > 0 such that
(t−1 )n ∈ mU ⊆ (t−1 )n+1 U,

a contradiction to the fact that t−1 is in U but is not a unit in U . We conclude that
U has rank > 1.

In the proof of Theorem 4.3 we pass to a subfield K of F . In doing so, we
need that features of the topology of X are preserved in the image of X in the
Zariski-Riemann space of K. This is given by the next lemma.
Lemma 4.2. Let K be a subfield of F . Then the mapping f ∶ Zar(F ) → Zar(K) ∶
V ↦ V ∩ K is a surjective map that is closed and continuous in the patch topology.
Proof. That f is surjective follows from the Chevalley Extension Theorem [8, Theorem 3.1.1, p. 57]. To see that f is continuous in the patch topology observe that
for x1 , . . . , xn , y ∈ K, the preimages under f of the subbasic patch open sets
{V ∈ Zar(K) ∶ x1 , . . . , xn ∈ V } and {V ∈ Zar(K) ∶ y ∈/ V }

are U (x1 , . . . , xn ) and V (y), respectively, and hence are patch open in Zar(F ).
Finally, with the patch topologies on Zar(F ) and Zar(K), f is a continuous map
between compact Hausdorff spaces, and so f is closed [7, Theorem 3.1.12, p. 125]. 
Theorem 4.3. Let X be a nonempty subset of Zar(F ) such that J(X) ≠ 0. If A(X)
is a local ring that is not a valuation domain, then patch(X) contains a valuation
ring of rank > 1.
Proof. Let A = A(X), J = J(X), and let M denote the unique maximal ideal of A.
We prove the lemma by establishing a series of claims. In the proof, for an ideal I
of A, we denote by End(I) the subring of F given by {t ∈ F ∶ tI ⊆ I}. Thus End(I)
is the largest ring in F in which I is an ideal.
Claim 1. If A is not completely integrally closed5, then patch(X) contains a
valuation ring of rank > 1.
5A domain R is completely integrally closed if for each nonzero ideal I of R and each element

t of the quotient field of R, tI ⊆ I if and only if t ∈ R; equivalently, R = End(I) for each nonzero
ideal I of R.

COMPACT SETS AND HOLOMORPHY RINGS

13

Proof of Claim 1. If every valuation ring in patch(X) has rank ≤ 1, then A,
as an intersection of completely integrally closed domains, is completely integrally
closed.

Claim 2. If M is a principal ideal of A, then patch(X) contains a valuation ring
of rank > 1.
Proof of Claim 2. Suppose M = mA for some m ∈ M . Since A is not a valuation
domain, A is not field, and hence M ≠ 0. Since M is principal, the ideal P = ⋂i>0 M i
is the unique largest nonmaximal prime ideal of A and P AP = P [29, Exercise 1.5,
p. 7]. If P = 0, then A is valuation ring (in fact, a DVR), contrary to assumption.
Thus P ≠ 0, and, since P = P AP , it follows that AP ⊆ End(P ). Since P is a
nonmaximal prime ideal, this implies A ⊊ End(P ), so that A is not completely
integrally closed. By Claim 1, patch(X) contains a valuation ring of rank > 1,
which proves Claim 2.

Claim 3. If all the valuation rings in X dominate A, then patch(X) contains a
valuation ring of rank > 1.
Proof of Claim 3. If M is a principal ideal of A, Claim 2 implies that patch(X)
contains a valuation ring of rank > 1, and the proof of the claim is complete. Thus
we assume M is not a principal ideal of A. Also, by Claim 1, we may assume
that A is completely integrally closed and hence End(M ) = A. It remains to prove
Claim 3 in the case in which End(M ) = A, M is not a principal ideal of A and all
the valuation rings in X dominate A.
Since A is not a valuation ring, there exists t ∈ F such that t /∈ A and t−1 ∈/ A.
Since M is not an invertible ideal of A, we have
M ⊆ M (A ∶F M ) ⊊ A,

which forces M = M (A ∶F M ). Thus

(A ∶F M ) = End(M ) = A,

so t ∈/ A = (A ∶F M ). Let m ∈ M such that mt ∈/ A. Since t−1 , mt ∈/ A, we have
mti , (mti )−1 ∈/ A for each i > 0, as noted at the beginning of the proof of Lemma 4.1.
By Theorem 3.2 (applied in the case where “D” is A), there exists, for each i > 0,
a valuation ring Vi ∈ patch(X) that dominates A and for which mti is a unit in Vi .
By assumption, every valuation ring in X dominates A, so Lemma 2.3(1) implies
that every valuation ring in patch(X) dominates A. Thus m ∈ MVi for all i. By
Lemma 4.1, patch(X) contains a valuation ring of rank > 1, which completes the
proof of Claim 3.

Claim 4. If there are valuation rings in X that do not dominate A, then there
exists a two-dimensional Noetherian local subring D of A such that D is dominated
by A and D contains nonzero elements a, b with a ∈ M ∖ J and b ∈ J.

14

BRUCE OLBERDING

Proof of Claim 4. Since there are valuation rings in X that do not dominate A,
we have 0 ≠ J ⊊ M . To prove the existence of the ring D, suppose first A contains
a field k. In this case, choose a ∈ M ∖ J and 0 ≠ b ∈ J. Let C = k[a, b] and let
D = CM ∩C . Then D is a Noetherian local subring of A that is dominated by A.
Since C is generated by two elements over a field, D has Krull dimension at most
two. To see that D has exactly Krull dimension two, observe that since a ∈ M ∖ J
there is V ∈ X such that a ∈/ MV . Thus 0 ≠ b ∈ MV ∩ D ⊊ M ∩ D, so that D has
Krull dimension two. This verifies Claim 4 if A contains a field.
Suppose next that A does not contain a field. Then the contraction of M to the
prime subring of A is a nonzero principal ideal, and hence there is a DVR W that
is dominated by A. Let p be the generator of the maximal ideal of W .
If p ∈ J, then let b = p and choose a ∈ M ∖ J.
If p /∈ J, then let a = p and choose 0 ≠ b ∈ J.
In either case, we have a ∈ M ∖ J and 0 ≠ b ∈ J. Let C = W [a, b], and let
D = CM ∩C . Since W is a DVR and either a ∈ W or b ∈ W , the ring D has Krull
dimension at most two. As in the case in which A contains a field, since a ∈ M ∖ J
and J ≠ 0, it follows that D has Krull dimension two. This verifies Claim 4.

Claim 5. The set patch(X) contains a valuation ring of rank > 1.
Proof of Claim 5. If every valuation ring in X dominates A, then, by Claim 3,
patch(X) contains a valuation ring of rank > 1. It remains to consider the case
where there are valuation rings in X that do not dominate A. By Claim 4, there is
a two-dimensional local Noetherian subring D of A such that D is dominated by A
and D contains nonzero elements a, b such that a ∈ M ∖ J and b ∈ J. The next step
of the proof involves a reduction that allows us to work in the quotient field of D.
Let K denote the quotient field of D. Let
A′ = A ∩ K,

M ′ = M ∩ K,

and

X ′ = {V ∩ K ∶ V ∈ patch(X)}.

Then A′ is a local integrally closed overring6 of D with maximal ideal M ′ and
A′ = ⋂V ∈X ′ V . Also, by Lemma 4.2, X ′ is patch closed in Zar(K).
Claim 5(a). All but finitely many valuation rings in X ′ dominate D.

Proof of Claim 5(a). Suppose that V ∈ X ′ and V does not dominate D. Then
MV ∩ D is a nonmaximal prime ideal of D that by Lemma 2.3(1) contains b. Since
D is a Noetherian ring of Krull dimension 2, there are only finitely many height
one prime ideals P1 , . . . , Pn of D containing b. The valuation ring V contains the
integral closure D ′ of the semilocal one-dimensional Noetherian ring DP1 ∩ ⋯ ∩ DPn .
Since the ring D ′ is a semilocal PID, there are only finitely many valuation rings
between D ′ and its quotient field K. Therefore, the set of valuation rings in X ′ that
do not dominate D is finite.

6By an overring of a domain, we mean a ring between the domain and its quotient field.

COMPACT SETS AND HOLOMORPHY RINGS

15

Returning to the proof of Claim 5, to show that patch(X) contains a valuation
ring of rank > 1, it is enough to prove that X ′ contains a valuation ring of rank > 1.
This is because if U ∈ X ′ has rank > 1, there is a valuation ring V ∈ patch(X) with
V ∩ K = U , and V is necessarily of rank > 1 since V extends U .
Thus we need only show that X ′ contains a valuation ring of rank > 1. We prove
this by verifying two claims.
Claim 5(b). If A′ [1/a] is a valuation ring, then X ′ contains a valuation ring of
rank > 1.
Proof of Claim 5(b). Since a ∈/ J, there exists V ∈ X ′ such that a ∈/ MV , and hence
1/a ∈ V . Since 0 ≠ b ∈ J ∩ A′ ⊆ MV , the valuation ring V does not have rank 0.
Therefore, A′ [1/a] ⊆ V ⊊ K. As an overring of the two-dimensional Noetherian ring
D, A′ has Krull dimension at most two. (This follows from the Dimension Inequality
[28, Theorem 15.5, p. 118].) Since a ∈ M ′ , the Krull dimension of A′ [1/a] is less
than that of A′ , and hence, since A′ [1/a] ⊊ K, it follows that A′ [1/a] has Krull
dimension one (and A′ has Krull dimension two). Thus the valuation ring A′ [1/a]
has rank one.
Now suppose by way of contradiction that every valuation ring in X ′ has rank
≤ 1. Then A′ , as an intersection of valuation rings in X ′ , is completely integrally
closed. Thus, since A′ has Krull dimension 2, A′ is not a valuation domain.
By [35, (4.2)] every patch closed representation of a domain contains a minimal
patch closed representation. Thus there is a patch closed subset Y of X ′ (recall that
X ′ is patch closed) such that A′ = ⋂V ∈Y V and no proper patch closed subset of Y
is a representation of A′ . By Lemma 2.3(4), the rank one valuation ring A′ [1/a], as
a proper subring of K, is an intersection of valuation rings in Y . Since A′ [1/a] has
the same quotient field as the valuation rings in X ′ , we conclude that A′ [1/a] is in
Y.
Since Y ⊆ X ′ and J ∩ A′ = ⋂V ∈X ′ MV , we have by Lemma 2.3(1) that 0 ≠ b ∈
J ∩ A′ ⊆ MV for all V ∈ Y . In particular, K ∈/ Y . Let W = A′ [1/a]. Since W has
rank one, we conclude that {W } = U (1/a) ∩ Y , so that W is a patch isolated point
in Y . We show this leads to a contradiction to the fact that Y is a minimal patch
closed representation of A′ .
Since W is a patch isolated point in Y , Y ∖ {W } is a patch closed set, and hence
by the minimality of Y we have have
A′ ⊊

⋂

U.

U ∈Y ∖{W }

Let
I =

⋂

U ∈Y ∖{W }

MU .

16

Then

BRUCE OLBERDING

⋂

U ⊆ End(I).

U ∈Y ∖{W }

Since A′ is completely integrally closed, it follows that I = 0 or I ⊆/ A′ . The former
case is impossible, since 0 ≠ b ∈ J ∩ A′ ⊆ I. Thus we conclude that I ⊆/ A′ . Let
t ∈ I ∖ A′ . Then t−1 ∈/ A′ since for any U ∈ Y ∖ {W }, we have t ∈ MU and A′ ⊆ U .
Thus t, t−1 /∈ A′ .
By Claim 5(a), all but finitely many valuation rings in X ′ , hence also in Y ,
dominate D. Applying Theorem 3.2 to D, A′ , Y and t, there exists a valuation ring
U ∈ lim(Y ) such that t, t−1 ∈ U . Since Y is patch closed, U ∈ Y , and, since t is a
unit in U , t /∈ MU . By the choice of t, t is in the maximal ideal of every valuation
ring in Y except W . This forces W = U ∈ lim(Y ), contradicting the fact that W is
a patch isolated point in Y . This contradiction shows that X ′ contains a valuation
ring of rank > 1. This completes the proof of Claim 5(b).

Claim 5(c). If A′ [1/a] is not a valuation ring, then X ′ contains a valuation ring
of rank > 1.
Proof of Claim 5(c). Since A′ [1/a] is not a valuation ring and A′ has quotient
field K, there exists 0 ≠ t ∈ K such that t ∈/ A′ [1/a] and t−1 ∈/ A′ [1/a]. Thus t, t−1 ∈/ A′
and at ∈/ A′ . With the aim of applying Lemma 4.1 to A′ and X ′ , we fix i > 0 and we
show that there is a valuation ring V ∈ X ′ such that ati is a unit in V and a ∈ MV .
Once this is proved, Lemma 4.1 implies that X ′ contains a valuation ring of rank
> 1, and the proof of Claim 5(c) is complete.
Let s = ati . Since t ∈/ A′ [1/a] and A′ [1/a] is integrally closed, it follows that
i
t ∈/ A′ [1/a], and hence s ∈/ A′ . If s−1 ∈ A′ , then t−i = a(a−1 t−i ) = as−1 ∈ A′ , so that,
since A′ is integrally closed in K, t−1 ∈ A′ , contrary to the choice of t. Therefore,
s, s−1 ∈/ A′ . By Claim 5(a), all but finitely many valuation rings in X ′ dominate D.
By Theorem 3.2, with D, A′ , X ′ and s playing the roles of “D”, “A”, “X” and “t” in
the theorem, we obtain U ∈ X ′ such that s, 1/s ∈ U and U dominates D. Therefore,
s = ati is a unit in U and a ∈ MU since U dominates D. By Lemma 4.1, X ′ contains
a valuation of rank > 1, which proves Claim 5(c).

Finally, to complete the proof of Claim 5, we note that Claim 5(b) and 5(c) show
that X ′ contains a valuation ring of rank > 1. As discussed after the proof of Claim
5(a), this implies that patch(X) contains a valuation of rank > 1. Therefore, with
the proof of Claim 5 complete, the proof of the theorem is complete also.

5. Compact sets and holomorphy rings
The main results of the paper involve one-dimensional Prüfer domains. We collect
in the next lemma some basic properties of such rings that are needed for the

COMPACT SETS AND HOLOMORPHY RINGS

17

theorems in this section. We denote by J(A) the Jacobson radical of a ring A, and
by Max(A) the space of maximal ideals of A endowed with the Zariski topology.
Lemma 5.1. Let A be a one-dimensional Prüfer domain with quotient field F , and
let X ⊆ Zar(F ) such that F ∈/ X and A = A(X). Then J(A) = J(X). If J(A) ≠ 0,
then the Zariski, patch and inverse topologies all coincide on X and
patch(X) = {AM ∶ M ∈ Max(A)}.
Proof. It is straightforward to check that J(X) ⊆ J(A); for example, see [19, Remark 1.3]. To see that the reverse inclusion holds, let V ∈ X. Since F /∈ X and A is
a one-dimensional Prüfer domain, V = AM , where M = MV ∩ A is a maximal ideal
of A. Since J(A) ⊆ M ⊆ MV , we have J(A) ⊆ J(X). This proves the first assertion
of the lemma.
Suppose now that J(A) ≠ 0. Since A has Krull dimension one, Max(A) is homeomorphic to the spectral space Spec(A/J(A)). In a spectral space for which every
point is both minimal and maximal with respect to the specialization order, the
Zariski, patch and inverse topologies all agree; cf. [41, Corollary 2.6] or use the fact
that A/J(A) is a von Neumann regular ring. Thus these three topologies all agree
on X since X is homeomorphic to a subspace of Max(A).
Finally, since A is a one-dimensional Prüfer domain represented by X, the set of
all valuation overrings of A (each of which must have rank ≤ 1) is patch(X)∪{F } [10,
Corollary 4.10]. Since J(A) ≠ 0, Lemma 2.3(1) implies F /∈ patch(X). Therefore,
patch(X) = {AM ∶ M ∈ Max(A)}.

Remark 5.2. Topological aspects and factorization theory of one-dimensional Prüfer
domains with nonzero Jacobson radical are studied in [22].
The first application of the results of the previous section is the following characterization of subsets of Zar(F ) whose holomorphy ring is a one-dimensional Prüfer
domain with nonzero Jacobson radical and quotient field F .
Theorem 5.3. The following are equivalent for a nonempty subset X of Zar(F )
with J(X) ≠ 0.
(1) A(X) is a one-dimensional Prüfer domain with quotient field F .
(2) X is contained in a quasicompact set of rank one valuation rings in Zar(F ).
(3) Every valuation ring in patch(X) has rank one.
Proof. Let A = A(X) and J = J(X).
(1) ⇒ (2) Since A is a one-dimensional Prüfer domain with quotient field F , the
subset of Zar(F ) given by Y = {AM ∶ M ∈ Max(A)} consists of rank one valuation
rings in Zar(F ). The only other valuation overring of A is F . Since J ≠ 0, we have
F ∈/ X, which forces X ⊆ Y . Moreover, Y is homeomorphic to Max(A) and the
maximal spectrum of a ring is quasicompact, so statement (2) follows.

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BRUCE OLBERDING

(2) ⇒ (3) Let Y be a quasicompact set of rank one valuation rings in Zar(F ) such
that X ⊆ Y . Since Y is quasicompact, Lemma 2.3(3) implies patch(X) ⊆ Y ∪ {F }. If
F ∈ patch(X), then from Lemma 2.3(1) it follows that J = 0, contrary to assumption.
Therefore, patch(X) ⊆ Y , so that patch(X) consists of rank one valuation rings.
(3) ⇒ (1) By Lemma 2.3(1), A = ⋂V ∈patch(X) V and 0 ≠ J = ⋂V ∈patch(X) MV .
Thus we can assume without loss of generality that X = patch(X). We claim first
that A has quotient field F . Let 0 ≠ a ∈ J. By Lemma 2.3(2), X is quasicompact,
so by Lemma 2.3(4) we have
A[1/a] =

⋂

V = F,

1/a∈V ∈X∪{F }

where the last equality follows from the fact that every valuation ring V in X has
rank one and satisfies a ∈ MV . Since A[1/a] = F , we conclude that A has quotient
field F .
To prove that A is a one-dimensional Prüfer domain, it suffices to show that AM
is a rank one valuation domain for each maximal ideal M of A. Let M be a maximal
ideal of A. Let Y = {V ∈ X ∶ AM ⊆ V }. Since
Y = X ∩ ( ⋂ U (t)),
t∈AM

Y is patch closed in X. By Lemma 2.3(2) and the fact that X is patch closed in
Zar(F ), Y is a quasicompact subset of Zar(F ). Since Y is quasicompact and consists
of rank one valuation rings, Lemma 2.3(4) implies that AM = ⋂V ∈Y V . Thus Y is
a patch closed representation of AM consisting of rank one valuation rings. Since
J ≠ 0, Theorem 4.3 implies that AM is a valuation domain. Since AM has quotient
field F and Y is a representation of AM consisting of rank one valuation domains,
it follows that AM ∈ Y . Hence AM is a rank one valuation domain, which proves
that A is a Prüfer domain with Krull dimension one.

A domain A is an almost Dedekind domain if for each maximal ideal M of A, AM
is a DVR. There exist many interesting examples of almost Dedekind domains; see
for example [22, 25, 30] and their references. For the factorization theory of such
rings, see [12, 22, 26].
Corollary 5.4. Let X be a nonempty set of Zar(F ) such that J(X) ≠ 0. Then X
is contained in a quasicompact set of DVRs in Zar(F ) if and only if A(X) is an
almost Dedekind domain with quotient field F .
Proof. Let A = A(X). Suppose X is contained in a quasicompact set Y of DVRs in
Zar(F ). By Theorem 5.3, A is a one-dimensional Prüfer domain with quotient field
F and nonzero Jacobson radical. By Lemma 2.3(3),
patch(X) ⊆ patch(Y ) = Y ∪ {F }.

COMPACT SETS AND HOLOMORPHY RINGS

19

Since J(X) ≠ 0, Lemma 2.3(1) implies F ∈/ patch(X), so patch(X) ⊆ Y . Thus
patch(X) consists of DVRs. By Lemma 5.1, we have that for each maximal ideal M
of A, AM is in patch(X) and hence AM is a DVR. Thus A is an almost Dedekind
domain. The converse follows from Lemma 5.1 and Theorem 5.3.

Remark 5.5. Let X be a nonempty quasicompact set of DVRs in Zar(F ) such
that J(X) ≠ 0. By Corollary 5.4, A = A(X) is an almost Dedekind domain, and
J(A) = J(X) by Lemma 5.1. If there is t ∈ J(A) such that J(A) = tA (equivalently,
MV = tV for each V ∈ X), then A has the property that every proper ideal is
a product of radical ideals. For this and related results on such rings, which are
known in the literature as SP-domains or domains with the radical factorization
property, see [12, 22, 30].
We prove next our main theorem of this section (the “main lemma” of the introduction) regarding the correspondence between quasicompact sets and holomorphy
rings in the space of rank one valuation rings.
Theorem 5.6. The mappings

X ↦ A(X) and A ↦ {AM ∶ M ∈ Max(A)}

define a bijection between the quasicompact sets X of rank one valuation rings in
Zar(F ) with J(X) ≠ 0 and the one-dimensional Prüfer domains A with quotient
field F and nonzero Jacobson radical.
Proof. Let X be a quasicompact set of rank one valuation rings in Zar(F ) with
J(X) ≠ 0. By Lemma 5.1 and Theorem 5.3, A = A(X) is a one-dimensional
Prüfer domain such that J(A) ≠ 0 and A has quotient field F . By Lemma 2.3(3),
patch(X) ⊆ X ∪ {F }. Since J(A) ≠ 0, Lemma 2.3(1) implies F /∈ patch(X). Thus
X = patch(X), and, by Lemma 5.1, X = {AM ∶ M ∈ Max(A)}. Conversely, suppose
A is one-dimensional Prüfer overring with quotient field F and 0 ≠ t ∈ J(A). Since
Max(A) is quasicompact, X = {AM ∶ M ∈ Max(A)} is a quasicompact set of rank
one valuation rings such that A = A(X) and 0 ≠ t ∈ J(X).

The next example shows the necessity of the hypotheses in Theorem 5.6.
Example 5.7. Let X be a nonempty subset of Zar(F ). Theorem 5.6 shows that if
(a) X consists of rank one valuation rings, (b) X is quasicompact and (c) J(X) ≠ 0,
then A(X) is a Prüfer domain. These hypotheses are necessary in the sense that
A(X) need not be a Prüfer domain if any one of (a), (b) or (c) is omitted. The
following classes of examples illustrate this.
(1) An example in which A(X) is not a Prüfer domain but in which (a) and (b)
hold. Let D be an integrally closed Noetherian domain of Krull dimension > 1.
Then the set X of localizations of D at height one prime ideals is a quasicompact
set of rank one valuation rings for which A(X) = D.

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BRUCE OLBERDING

(2) An example in which A(X) is not a Prüfer domain but in which (b) and (c)
hold. Let k be a field, and let V be the DVR k(S)[T ](T ) , where S and T are
indeterminates for k. Let R = k + MV . Then R is a one-dimensional integrally
closed local domain. With X the set of valuation overrings of R of rank > 0,
we have that R = A(X) and J(X) ≠ 0. Moreover, X is quasicompact since X
is the closed subset of the (quasicompact) space of valuation overrings. Indeed,
X is the set of valuation overrings of R contained in V . Thus (b) and (c) hold,
but A(X) is not a Prüfer domain since A(X) is a local domain contained in
more than one valuation ring of Krull dimension 2.
(3) An example in which A(X) is not a Prüfer domain but in which (a) and (c) hold.
Let D be an integrally closed Noetherian local domain of Krull dimension > 1
with quotient field F . To exhibit the desired example, it suffices to show that
D is the intersection of the DVRs in Zar(F /D) that dominate D. Let m denote
the maximal ideal of D, and let x ∈ F ∖ D. If x−1 ∈ D, then x−1 ∈ m. Since
every Noetherian local domain is birationally dominated by a DVR [6, p. 26],
there exists a DVR V in Zar(F /D) such that x−1 ∈ MV . Thus x ∈/ V . On the
other hand, if x−1 /∈ D, then the ring D[x−1 ] has a maximal ideal generated by
m and x−1 [29, Exercise 31, pp. 43–44], so, again by [6, p. 26], there is a DVR
V in Zar(F /D) that dominates D and for which x−1 ∈ MV and hence x ∈/ V . It
follows that D is the intersection of all the DVRs in Zar(F /D) that dominate
it.
Restricting to DVRs in Theorem 5.6 yields a correspondence with almost Dedekind
domains.
Corollary 5.8. The mappings

X ↦ A(X) and A ↦ {AM ∶ M ∈ Max(A)}

define a bijection between the quasicompact sets X of DVRs in Zar(F ) with J(X) ≠ 0
and the almost Dedekind domains A with quotient field F and nonzero Jacobson
radical.
Proof. If X is a quasicompact set of DVRs in Zar(F ) with J(X) ≠ 0, then, by
Corollary 5.4, A is an almost Dedekind domain. Conversely, if A is almost Dedekind
domain, then clearly X = {AM ∶ M ∈ Max(A)} is a quasicompact set of DVRs. Thus
the corollary follows from Theorem 5.6.

By Proposition 2.4(2) the quasicompact sets in Theorem 5.6 are compact, so the
theorem alternatively can be stated for compact sets instead. Along these lines, in
the case in which all the valuation rings under consideration occur as overrings of a
domain with quotient field F , the restriction that J(X) ≠ 0 in Theorem 5.6 can be
omitted (or, more correctly, hidden) if the quasicompact hypothesis is strengthened
to that of being compact.

COMPACT SETS AND HOLOMORPHY RINGS

21

Corollary 5.9. Let R be a proper subring of F with quotient field F . There is a
bijection given by
X ↦ A(X) and A ↦ {AM ∶ M ∈ Max(A)}

between the compact sets X of rank one valuation overrings of R and the onedimensional Prüfer overrings A of R with nonzero Jacobson radical.
Proof. If X is a compact set of rank one valuation overrings, then J(X) ≠ 0 by
Proposition 2.4(3). By Theorem 5.6, A = A(X) is a one-dimensional Prüfer domain
with J(A) ≠ 0 and X = {AM ∶ M ∈ Max(A)}. Conversely, if A is one-dimensional
Prüfer overring of R with J(A) ≠ 0, then X = {AM ∶ M ∈ Max(A)} is quasicompact
by Theorem 5.6, and X is Hausdorff by Proposition 2.4(2). Clearly, A = A(X). 
Remark 5.10. As in Corollary 5.8, the bijection in Corollary 5.9 restricts to a
bijection between compact sets of DVRs and almost Dedekind overrings A of R
with nonzero Jacobson radical.
In general it is difficult to determine when an intersection of two one-dimensional
Prüfer domains with quotient field F is a Prüfer domain. For example, it is easy
to see any Noetherian local UFD A of Krull dimension 2 can be written as an
intersection A = A1 ∩ A2 where A1 is a DVR overring and A2 is a PID overring.
In this example, A is not a Prüfer domain despite being an intersection of two
PIDs. Significantly, the ring A2 here has J(A2 ) = 0. In our context, the topological
characterization in Corollary 5.9 shows that the difficulty here is removed if J(Ai ) ≠
0, a fact we prove in the next corollary.
Corollary 5.11. Let A1 , . . . , An be one-dimensional Prüfer domains, each with quotient field F . Let J = J(A1 ) ∩ ⋯ ∩ J(An ), and let A = A1 ∩ ⋯ ∩ An . If J(Ai ) ∩ A ≠ 0
for all i, then A is a one-dimensional Prüfer domain with J(A) = J and quotient
field F . If also each Ai is an almost Dedekind domain, then so is A.
Proof. For each i = 1, 2, . . . , n, let

Xi = {(Ai )M ∶ M ∈ Max(Ai )}.

By Corollary 5.9, each Xi is compact. Thus X = X1 ∪ ⋯ ∪ Xn is quasicompact. Since
J(Ai ) ∩ A ≠ 0 for all i, we have
J(X) = J(A1 ) ∩ ⋯ ∩ J(An ) ≠ 0.

Thus Theorem 5.6 implies A = A(X) is a one-dimensional Prüfer domain, and, by
Lemma 5.1, J(A) = J(A1 ) ∩ ⋯ ∩ J(An ). If also Ai is an almost Dedekind domain,
then each Xi is a quasicompact set of DVRs, so that X is also a quasicompact set
of DVRs. In this case, A is an almost Dedekind domain by Corollary 5.8.

Remark 5.12. It is an open question as to whether the intersection A = A1 ∩ A2
of one-dimensional Prüfer domains A1 and A2 with quotient field F and nonzero

22

BRUCE OLBERDING

Jacobson radical has quotient field F . If the answer is affirmative, then A is a
one-dimensional Prüfer domain by Corollary 5.11. In any case, let D be the prime
subring of A, let Si = {1 + d ∶ d ∈ J(Ai )} and let Bi = DSi + J(Ai ). Then B1 and
B2 are local domains of Krull dimension one with quotient field F . A necessary
condition for A to have quotient field F is that B1 ∩ B2 has quotient field F . In
[16, Question 2.1] Gilmer and Heinzer ask whether the intersection of two onedimensional local domains, each with the same quotient field F , has quotient field
F?
Corollary 5.13. Let X be a nonempty quasicompact set of rank one valuation rings
in Zar(F ). If J(X) ≠ 0, then the patch, inverse and Zariski topologies agree on X.
Proof. By Theorem 5.6, A = A(X) is a one-dimensional Prüfer domain with J(A) ≠ 0
and quotient field F . By Lemma 5.1, the inverse, patch and Zariski topologies agree
on X.

As the last application of this section, we describe schemes in Zar(F ) consisting
of rank ≤ 1 valuation rings. Let X be a subspace of Zar(F ), and let X ∗ = X ∪ {F }.
Let OX ∗ be the sheaf on X ∗ defined for each nonempty open set U of X ∗ by
OX ∗ (U ) = ⋂V ∈U V . (The reason for appending F to X is to guarantee that OX ∗
is a sheaf.) We say that X ∗ is a scheme in Zar(F ) if the locally ringed space
(X ∗ , OX ∗ ) is a scheme, and that X ∗ is an affine scheme in Zar(F ) if (X ∗ , OX ∗ ) is
an affine scheme. Thus X ∗ is an affine scheme in Zar(F ) if and only if the set of
all localizations A(X)P , P a nonzero prime ideal of A(X), is X. The question of
whether a subset of Zar(F ) is an affine scheme is closely connected to the question
of whether the intersection of valuation rings in the set is a Prüfer domain with
quotient field F . For more on this, see [36].
A necessarily condition for X to be an affine scheme in Zar(F ) is that X is
quasicompact; similarly, for X to be a scheme, X must be locally quasicompact
(i.e., every point has a quasicompact neighborhood). The corollary shows that
these conditions are also sufficient for sets X of valuation rings of rank ≤ 1 with
J(X) ≠ 0.
Corollary 5.14. Let X be a nonempty set of rank one valuation rings in Zar(R)
with J(X) ≠ 0. Then X ∗ is an affine scheme in Zar(F ) if and only if X is quasicompact; X ∗ is a scheme in Zar(F ) if and only X is locally quasicompact.

Proof. An affine scheme in Zar(F ) is quasicompact since the prime spectrum of
a ring is quasicompact. Conversely, if X is quasicompact, then, with A = A(X),
Theorem 5.6 implies that X ∗ = {AP ∶ P ∈ Spec(A)}, so that X ∗ is an affine scheme
in Zar(F ).
Now suppose X is locally quasicompact. Let V ∈ X, and let Z be a quasicompact
neighborhood of V in X. Then there is an open subset Y of X of the form Y =

COMPACT SETS AND HOLOMORPHY RINGS

23

U (x1 , . . . , xn ) ∩ X, where x1 , . . . , xn ∈ V and Y ⊆ Z. Since Z is quasicompact with
J(Z) ≠ 0, Corollary 5.13 implies that the Zariski and inverse topologies agree on
Z. Thus Y is a Zariski closed subset of Z. Since Z is quasicompact, Y is also
quasicompact, and hence Y is an affine scheme that is open in X. This shows that
X is a union of affine open schemes, so that X is a scheme in Zar(F ). Conversely, if
X is a scheme in Zar(F ), then X is a union of open sets that are affine schemes in
Zar(F ). Thus X is a union of quasicompact open subsets, proving that X is locally
quasicompact.

6. Proof of Main Theorem
In this section we prove the main theorem of the introduction. In light of Theorem 5.6, what remains to be shown is that the one-dimensional Prüfer domains with
nonzero Jacobson radical are Bézout domains. In fact, we prove more generally
that any one-dimensional domain with nonzero Jacobson radical has trivial Picard
group.
Theorem 6.1. If A is a one-dimensional domain with J(A) ≠ 0, then every invertible ideal of A is a principal ideal.
Proof. Let I be an invertible ideal of A. We show that I is a principal ideal of A.
After multiplying I by a nonzero element of J(A), we can assume I ⊆ J(A). Since
A has Krull dimension one and I is invertible, there exist 0 ≠ a, b ∈ I such that
I = (a, b)A; see [40, Corollary 4.3], or [21, Theorem 3.1] for a more general result.
Let J = J(A), and let
X = {M ∈ Max(A) ∶ (aA ∶A b) ⊆/ M } and Y = {M ∈ Max(A) ∶ (aA ∶A b) ⊆ M }.
First we show that there is e ∈ A such that e2 − e ∈ J and
X = {M ∈ Max(A) ∶ e ∈ M } and Y = {M ∈ Max(A) ∶ 1 − e ∈ M }.
Since A/J is a reduced ring of Krull dimension 0, A/J is a von Neumann regular
ring, and hence every finitely generated ideal of A/J is generated by an idempotent.
In order to apply this observation to the image of (aA ∶A b) in A/J, we claim that
(aA ∶A b) is a finitely generated ideal of A. Since I = (a, b)A, we have (aA ∶A b) =
(aA ∶F I). Since I is invertible, it follows that
(aA ∶A b)I = (aA ∶F I)I = aA.

With I −1 = (A ∶F I), we have (aA ∶A b) = aI −1 . Since I −1 is invertible, I −1 is
a finitely generated A-submodule of F , and it follows that (aA ∶A b) is a finitely
generated ideal of A. Therefore, since A/J is a von Neumann regular ring, there is
f ∈ (aA ∶A b) such that f A + J = (aA ∶A b) + J and f 2 − f ∈ J. Set e = 1 − f . Then
e2 − e ∈ J, and, since J is contained in every maximal ideal of A, it follows that

24

BRUCE OLBERDING

X = {M ∈ Max(A) ∶ e ∈ M } and Y = {M ∈ Max(A) ∶ f ∈ M }.
√
√
Next, since ab ∈ J and A/ abA has Krull dimension 0, it follows that J = abA.
(Recall we have assumed that I ⊆ J.) Since
√
√
ef = e − e2 ∈ J = abA = abJ,
there is k > 0 such that ek f k ∈ abJ. Let c = (a − ek )(b − f k ). We claim that I = cA.
It suffices to check that this equality holds locally.
Let M be a maximal ideal of A. Suppose first that M ∈ X. Then (aA ∶A b) ⊆/ M ,
so there exists d ∈ A ∖ M such that db ∈ aA. It follows that bAM ⊆ aAM , and hence
IAM = aAM . Thus to show that IAM = cAM , it suffices to show that aAM = cAM .
If b ∈/ M , then since d ∈/ M we have db ∈/ M . However, with b ∈/ M , the fact that
ab ∈ M implies db ∈ aA ⊆ M , a contradiction. Thus b ∈ M . Now, since e ∈ M , we
have f = 1 − e ∈/ M and hence (because b ∈ M ) we conclude that b − f k ∈/ M . This
implies that cAM = (a − ek )AM . Since
ek AM = ek f k AM ⊆ aJAM ,

there is j ∈ J and h ∈ A ∖ M such that hek = aj. Thus
h(a − ek ) = ha − hek = ha − aj = (h − j)a.

Since j ∈ M and h ∈/ M , we have h − j /∈ M . From the fact that h(a − ek ) = (h − j)a
we conclude that
cAM = (a − ek )AM = aAM ,

which proves the claim that for each M ∈ X, IAM = cAM .
Now suppose that M ∈ Y , so that f ∈ M . We show that IAM = cAM in this case
also. Since (aA ∶A b) ⊆ M , we have bAM ⊆/ aAM . Since I is invertible, IAM is a
principal ideal of AM . The following standard argument shows that this implies that
IAM = bAM . Let z ∈ I such that IAM = zAM . Then there exist x, y, s, t ∈ AM such
that a = zx, b = zy and z = as + bt. If x is a unit in AM , then bAM ⊆ zAM = aAM ,
a contradiction. Thus x is not a unit in AM . Since a = zx = (as + bt)x, we have
a(1 − sx) = btx, with 1 − sx a unit in AM since x ∈ M AM . Therefore, aAM ⊆ bAM ,
which proves that IAM = bAM .
We have shown that for M ∈ Y , we have IAM = bAM . To complete the proof of
the lemma, it suffices to show that bAM = cAM . The proof proceeds as in the case
where M ∈ X. Since bAM ⊆/ aAM , it follows that a ∈ M , and hence a − ek ∈/ M . Thus
cAM = (b − f k )AM . Also,
f k AM = ek f k AM ⊆ bJAM ,

so that there is h ∈ A ∖ M such that hf k = bj for some j ∈ J. Thus
h(b − f k ) = hb − bj = b(h − j),

COMPACT SETS AND HOLOMORPHY RINGS

25

with h, h − j units in AM . Therefore, (b − f k )AM = bAM , which shows that
cAM = (b − f k )AM = bAM = IAM .

This proves that IAM = cAM for all maximal ideals M of Y . Since Max(A) = X ∪Y ,
we conclude that I = cA.

Corollary 6.2. If A is a one-dimensional Prüfer domain with J(A) ≠ 0, then A is
a Bézout domain.
Proof. This follows from Theorem 6.1 and the fact that every finitely generated ideal
of a Prüfer domain is invertible [14, Theorem 22.1].

A special case of the lemma in which it is assumed in addition that A is an almost
Dedekind domain for which every maximal ideal of A has finite sharp degree was
proved by Loper and Lucas [26, Theorem 2.9] using different methods.
With Corollary 6.2 and the results of the preceding sections, we can now prove
the main theorem from the introduction.
Theorem 6.3. If X is a quasicompact set of rank one valuation rings in Zar(F )
such that J(X) ≠ 0, then A(X) is a Bézout domain of Krull dimension one with
quotient field F .
Proof. By Theorem 5.6, A(X) is a one-dimensional Prüfer domain with quotient
field F and nonzero Jacobson radical. By Corollary 6.2, A(X) is a Bézout domain,
which proves the theorem.

Remark 6.4. In [37] we apply the results of this article to rank one valuation
overrings of a two-dimensional Noetherian local domain D with quotient field F .
We focus on the divisorial valuation overrings of D, i.e., the DVRs that birationally
dominate D and are residually transcendental over D. It is shown, for example,
that if n is a positive integer, then the subspace X of Zar(F /D) consisting of all
divisorial valuation rings that can be reached through an iterated sequence of at
most n normalized quadratic transforms of D is quasicompact. By Corollary 5.8,
A(X) is an almost Dedekind domain with nonzero Jacobson radical.
Acknowledgment. I thank the referee for helpful comments that improved the
presentation of the article and for suggesting the version of Example 5.7(2) that is
included here.
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Department of Mathematical Sciences, New Mexico State University, Las Cruces,
NM 88003-8001

