1 Introduction
A (commutative integral) domain is factorial if and only if it is Krull with trivial class group. Every Krull monoid (in particular, every Krull and every Dedekind domain) has a transfer homomorphism to the monoid of zero-sum sequences over the subset of classes of its class group that contain prime divisors. Transfer homomorphisms allow to pull back arithmetic invariants (such as sets of lengths) from the target monoid back to the original object of interest. Thus, the existence of transfer homomorphisms justifies the classic philosophy that the arithmetic of rings of integers in algebraic number fields and of Dedekind domains depends only on their class group. Monoids of zero-sum sequences are Krull monoids, which are usually studied with combinatorial methods [Reference Geroldinger and Ruzsa44, Reference Grynkiewicz48]. If the class group is finite, then additive combinatorics plays a central role. It provides precise results on sets of lengths (and on invariants controlling their structure, such as elasticities and sets of distances) in terms of the group invariants.
Factorization theory has its origin in algebraic number theory. Carlitz and Narkiewicz are considered as pioneers in this area (see Carlitz’s characterization of half-factorial principal orders in algebraic number fields [Reference Carlitz18] and Narkiewicz’s first edition of his monograph on algebraic number theory [Reference Narkiewicz61]). Pushed forward by seminal papers of Anderson et al. [Reference Anderson, Anderson and Zafrullah2, Reference Anderson, Anderson and Zafrullah3], factorization theory expanded into commutative ring theory. From that time on, factorization theory in commutative rings developed hand in hand with multiplicative ideal theory (see the classic monographs by [Reference Gilmer47, Reference Halter-Koch50, Reference Jaffard55, Reference Larsen and McCarthy59] and the more recent ones [Reference Elliott27, Reference Halter-Koch53, Reference Wang and Kim72]). This development is documented in periodic conference proceedings, including [Reference Anderson1, Reference Anderson and Papick4, Reference Badawi8, Reference Badawi and Coykendall9, Reference Brešar, Geroldinger, Olberding and Smertnig16, Reference Brewer, Glaz, Heinzer and Olberding17, Reference Chabert, Fontana, Frisch, Glaz and Johnson19, Reference Chapman, Fontana, Geroldinger and Olberding21, Reference Chapman and Glaz22, Reference Dobbs, Fontana and Kabbaj26, Reference Facchini, Fontana, Geroldinger and Olberding29, Reference Facchini, Houston and Salce30, Reference Fontana, Frisch and Glaz33–Reference Fontana, Kabbaj and Wiegand37]. Apart from commutative ring theory, factorization theory has expanded its scope into non-commutative ring theory, module theory, abstract semigroup theory, and additive combinatorics (we can refer only to some monographs and surveys [Reference Baeth and Wiegand12, Reference Facchini28, Reference Fan and Tringali32, Reference Geroldinger, Kim and Loper43, Reference Geroldinger and Ruzsa44, Reference Geroldinger and Zhong46, Reference Smertnig71]).
Factorization theory studies the arithmetic of a wide range of domains, including weakly Krull domains, monoid algebras, and rings of integer-valued polynomials. Orders in Dedekind domains are the simplest class of non-Krull domains in this scenario. They are weakly Krull, whence they admit transfer homomorphisms to T-block monoids (constructed from zero-sum sequences and localizations at primes containing the conductor). These transfer homomorphisms allow to obtain arithmetic finiteness results, but in general, these results are less precise than those obtained for Krull domains (see, e.g., [Reference Geroldinger and Halter-Koch41, Theorem 3.7.1]). But not only are the methods weaker but the arithmetic of general non-principal orders shows different features compared with the arithmetic of the corresponding principal orders (which are Dedekind and hence Krull). To give an example, the set of distances of a ring of integers (whose class group has at least three elements) is an interval with minimum one. This property does not even hold for arbitrary orders in quadratic number fields [Reference Reinhart67].
In his seminal paper [Reference Smertnig70], Smertnig characterized the maximal orders in central simple algebras over algebraic number fields which allow a transfer homomorphism to a monoid of zero-sum sequences. The characterization shows that, apart from an exceptional case, these maximal orders (which are non-commutative Dedekind domains) allow such a transfer homomorphism. His work initiated the search for rings and monoids, that are not Krull, but nevertheless allow a transfer homomorphism to a monoid of zero-sum sequences (or equivalently, to a commutative Krull monoid). They are called transfer Krull and include non-commutative rings, Noetherian domains that are not integrally closed, and others (e.g., [Reference Baeth and Smertnig11, Reference Bashir and Reinhart13, Reference Geroldinger, Schmid and Zhong45, Reference Smertnig71]). At the same time, other classes of rings and monoids were revealed that do not allow such transfer homomorphisms. They include power monoids, rings of integer-valued polynomials, and further classes of commutative and non-commutative rings (e.g., [Reference Bell, Brown, Nazemian and Smertnig14, Reference Fan and Tringali32, Reference Frisch, Nakato and Rissner38]). For a list of transfer Krull and non-transfer Krull monoids and domains, we refer to the survey [Reference Geroldinger and Zhong46].
The arithmetic of orders in Dedekind domains (with a focus on orders in algebraic number fields) has been studied for decades and found a renewed interest in recent years (e.g., [Reference Brantner, Geroldinger and Reinhart15, Reference Coykendall and Moles25, Reference Gao, Liu, Tringali and Zhong39, Reference Geroldinger, Kainrath and Reinhart42, Reference Kettinger and Moles58, Reference Philipp62–Reference Rago66]). In the present article, we establish an algebraic characterization of transfer Krull orders in Dedekind domains. The mild conditions, which we impose, are satisfied by rings of integers in algebraic number fields [Reference Halter-Koch51, Corollary 4.4.4], by holomorphy domains in global fields and associated S-class groups [Reference Halter-Koch52, Corollary 7.9.3], and by other, more abstract classes of Dedekind domains (see, e.g., [Reference Chang and Geroldinger20] and the references given there). Our main result can be stated as follows.
Theorem 1.1 Let R be a Dedekind domain with torsion class group and let
$\mathcal O \subseteq R$
be an order with conductor
$\mathfrak f$
such that
$R/\mathfrak f$
is finite.
-
(1) Suppose that
$|\operatorname {\mathrm {Cl}} (R)| \ne 2$
and that every class of
$\operatorname {\mathrm {Cl}} (R)$
contains a prime ideal comaximal to
$\mathfrak f$
. Then
$\mathcal O$
is transfer Krull if and only if the following two conditions hold: -
(a)
$\mathcal O\cdot R^{\times } = R$
. -
(b)
$\mathsf v_{\mathfrak p} (u)=1$
for all atoms
$u \in \mathcal O_{\mathfrak p}$
and all
$\mathfrak p \in \operatorname {\mathrm {Spec}} ( \mathcal O)$
.
Moreover, if
$\mathcal O$
is transfer Krull, then the inclusion
$\mathcal O^{\bullet } \hookrightarrow R^{\bullet }$
is a transfer homomorphism and there is a transfer homomorphism
$\theta \colon \mathcal O^{\bullet } \to \mathcal B (\operatorname {\mathrm {Cl}} (R) )$
, where
$\mathcal B (\operatorname {\mathrm {Cl}} (R) )$
is the monoid of zero-sum sequences over the class group
$\operatorname {\mathrm {Cl}} (R)$
. -
-
(2) Suppose that
$|\operatorname {\mathrm {Cl}} (R)| = 2$
and that every class of
$\operatorname {\mathrm {Pic}} (\mathcal O)$
contains a prime ideal comaximal to
$\mathfrak f$
. Then
$\mathcal O$
is transfer Krull if and only if the following two conditions hold: -
(a)
$\mathcal O \cdot R^{\times } = R$
. -
(b)
$\mathsf v_{\mathfrak p} (u) \in \{1,2\}$
for all atoms
$u \in \mathcal O_{\mathfrak p}$
and all
$\mathfrak p \in \operatorname {\mathrm {Spec}} ( \mathcal O)$
and if the prime ideal lying over
${\mathfrak {p}}$
is principal, then
$\mathsf v_{\mathfrak p} (u) =1$
for all atoms
$u\in {\mathcal {O}}_{\mathfrak {p}}$
.
-
Theorem 1.1 implies that the arithmetic results derived for R hold true for a transfer Krull order
$\mathcal O \subseteq R$
as well. If
$\operatorname {\mathrm {Cl}} (R)$
is finite, then a survey of Schmid [Reference Schmid69] provides a good overview of results achieved with methods from additive combinatorics. If
$\operatorname {\mathrm {Cl}} (R)$
is infinite, then every finite nonempty subset of
${\mathbb N}_{\ge 2}$
occurs as a set of lengths of R [Reference Geroldinger and Halter-Koch41, Theorem 7.4.1], whence also as a set of lengths of
$\mathcal O$
.
The characterization in the case
$|\operatorname {\mathrm {Cl}}(R)|=2$
is a generalization of the result in [Reference Rago66], where half-factorial orders in algebraic number fields are studied. We discuss the particularities of this case in a final remark at the end of Section 4.
We will proceed as follows. In Section 2, we recall several definitions and concepts from factorization theory, in particular, the notion of transfer Krull. Moreover, we present some algebraic properties of Dedekind domains and orders.
In Section 3, we study the relationship of regular elements of transfer Krull orders with the monoid
$\operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
, and we prove that
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\simeq \operatorname {\mathrm {Cl}}(R)$
, except possibly in the special case
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
.
In the final section, we show that the Spec-map is bijective and then turn our attention to the irregular localizations
${\mathcal {O}}_{\mathfrak {p}}$
, before proving Theorem 1.1.
2 Notation and preliminaries
2.1 Monoids and factorization theory
Let H be a commutative semigroup with identity. H is called cancellative if for all
$a,b,c\in H$
with
$ac=bc$
, it follows that
$a=b$
.
-
Throughout this article, a monoid is always a commutative, cancellative semigroup with identity.
We denote by
$\mathbb {N}$
the set of positive integers and by
$\mathbb {N}_0$
the set of nonnegative integers. For
$n,m\in \mathbb {N}_0$
, we write
$[n,m]=\{x\in \mathbb {N}_0 \ | \ n\leq x\leq m\}$
. Let H be a multiplicatively written monoid with identity
$1_H$
. We let
$H^\times $
denote the group of units (invertible elements) of H and by
$H_{\text {red}}=H/H^\times =\{aH^\times \ |\ a\in H\}$
the associated reduced monoid. In general, we call a monoid H reduced if
$H^\times =\{1\}$
. A subset
$T\subseteq H$
is called a submonoid of H if
$1_H\in T$
and
$ab\in T$
for all
$a,b\in T$
. We let
$\mathsf {q}(H)=\{ab^{-1}\ | \ a,b\in H\}$
denote the quotient group of H. This group is uniquely determined up to canonical isomorphism. Any monoid T between H and
$\mathsf {q}(H)$
is called an overmonoid of H. For
$a,b\in H$
, we set
$a\sim _H b$
if
$a=\varepsilon b$
for some
$\varepsilon \in H^\times $
and we say that a and b are associated in H.
Let a be a non-unit of H. Then a is called an atom if a cannot be written as a product of two non-units of H, and a is said to be prime if, whenever a divides
$bc$
for elements
$b,c\in H$
, then a divides b or c. Moreover, we say that an atom a is absolutely irreducible (also the terms completely irreducible and strong atom can be used) if, for every
$n\in \mathbb {N}$
,
$a^n$
has precisely one factorization into atoms (up to associates). Absolutely irreducible elements are frequently studied in factorization theory (the oldest reference we are aware of is a paper by Kaczorowski [Reference Kaczorowski56]), and they also found attention in quite recent literature (e.g., [Reference Angermüller5, Reference Angermüller6, Reference Fadinger, Frisch, Nakato, Smertnig and Windisch31, Reference Hiebler, Nakato and Rissner54, Reference Rissner and Windisch68]).
We denote by
$\mathcal {A}(H)$
the set of atoms of H and say that H is atomic if every non-unit can be written as a finite product of atoms. For each non-unit
$a\in H$
, we let
$$\begin{align*}\mathsf{L}_H(a)=\{k\in\mathbb{N}\ | \ a \text{ is a product of } k \text{ atoms of } H \}\end{align*}$$
be the set of lengths of a, and we set
$\mathsf {L}_H(a)=\{0\}$
if
$a\in H^\times $
. An atomic monoid is said to be factorial if every element of H has precisely one factorization up to associates, or equivalently, if every atom of H is prime. Moreover, H is called half-factorial if
$|\mathsf {L}_H(a)|=1$
for every
$a\in H$
. Clearly, every factorial monoid is half-factorial. Let R be an integral domain. We will denote the multiplicative monoid of nonzero elements of R by
$R^\bullet $
. All arithmetic terms defined for monoids carry over to domains and we set
$\mathcal {A}(R)=\mathcal {A}(R^\bullet )$
and so on.
Let G be an additively written, abelian group and let
$g\in G$
. We denote by
$\operatorname {\mathrm {ord}}(g)$
the order of the element g in G. Let now
$\mathcal {F}(G)$
be the (multiplicative) free abelian monoid over G. For not necessarily distinct elements
$g_1,\ldots ,g_n\in G$
, we denote by
${\sigma (g_1\cdot \dots \cdot g_n)=g_1+\cdots +g_n\in G}$
the sum of the sequence
$g_1\cdot \dots \cdot g_n\in \mathcal {F}(G)$
. Moreover, we denote the monoid of all sequences
$x\in \mathcal {F}(G)$
with
$\sigma (x)=0$
by
$\mathcal {B}(G)$
, and we call
$\mathcal {B}(G)$
the monoid of zero-sum sequences over G.
We denote by
$$\begin{align*}\widehat{H}=\{x\in\mathsf{q}(H) \ | \ \text{there exists some } c\in H, \text{ such that } cx^n\in H \text{ for all } n\in\mathbb{N} \}\end{align*}$$
the complete integral closure of the monoid H, and we say that H is completely integrally closed if
$H=\widehat {H}$
. A monoid H is called Krull if H is completely integrally closed and satisfies the ascending chain condition for v-ideals. For the definition of v-ideals and a detailed treatment of Krull monoids, we refer to [Reference Geroldinger and Halter-Koch41]. If R is an integral domain, then
$R^\bullet $
is a Krull monoid if and only if R is a Krull domain. Every Krull monoid H has a class group
$\mathcal {C}(H)$
associated with it.
Let H and B be monoids. A monoid homomorphism
$\theta :H\to B$
is said to be a transfer homomorphism if the following two properties are satisfied:
-
(T1)
$B=\theta (H)\cdot B^\times $
and
$\theta ^{-1}(B^\times )=H^\times $
. -
(T2) If
$u\in H$
,
$b,c\in B$
, and
$\theta (u)=bc$
, then there exist
$v,w\in H$
such that
$u=vw$
,
$\theta (v)\sim _B b$
, and
$\theta (w)\sim _B c$
.
Henceforth, we will refer to property (T2) as the transfer property. Transfer homomorphisms preserve many arithmetic invariants, such as the set of distances, the elasticity (for definitions, see [Reference Geroldinger and Halter-Koch41, Chapter 1]) and, most importantly, sets of lengths. We have
$\mathsf {L}_H(u)=\mathsf {L}_B(\theta (u))$
for all
$u\in H$
. If H is a Krull monoid with the additional property that every class in
$\mathcal {C}(H)$
contains a prime divisor, then H admits a transfer homomorphism
$\boldsymbol {\beta }:H\to \mathcal {B}(\mathcal {C}(H))$
.
Transfer homomorphisms were introduced by Halter-Koch [Reference Halter-Koch49] and their theory was developed in [Reference Geroldinger and Halter-Koch40]. They got generalized to the non-commutative setting by Baeth, Smertnig, and others (e.g., [Reference Bachman, Baeth and Gossell7, Reference Baeth and Smertnig10]; if the involved objects are commutative, then every weak transfer homomorphism is a transfer homomorphism). Apart from transfer homomorphisms, there is a variety of homomorphisms relating the arithmetic of the involved objects. We would like to mention the Dedekind–Hasse norm (as studied by Coykendall, see [Reference Coykendall23] for a first paper and [Reference Coykendall and Hasenauer24] for more recent work in this direction) and equimorphisms (as studied by Fan and Tringali, e.g., [Reference Fan and Tringali32]). Moreover, if
$R \subseteq S$
is an extension of integral domains, and the embedding
$R^{\bullet } \hookrightarrow S^{\bullet }$
is a transfer homomorphism, then
$R \subseteq S$
is an inert extension and a quasi-divisor-closed extension (see [Reference Mondal60]).
A monoid H is said to be transfer Krull if there exists a Krull monoid B and a transfer homomorphism
$\theta :H\to B$
. Since the identity is a transfer homomorphism, every Krull monoid is transfer Krull, but the converse does not necessarily hold. For example, every half-factorial monoid H is transfer Krull via the transfer homomorphism
$\theta :H\to (\mathbb {N}_0,+)$
, which maps every element to its unique factorization length, but not every half-factorial monoid is Krull. Thus, transfer homomorphisms are a useful tool to pull back arithmetic properties of a Krull monoid, whose arithmetic is very well understood, to H. The following proposition summarizes some properties of transfer Krull monoids.
Proposition 2.1 Let S be a monoid with quotient group
$\mathsf {q}(S)$
. Then S is transfer Krull if and only if there is a Krull monoid H with
$S\subseteq H\subseteq \mathsf {q}(S)$
, such that the inclusion
$S\hookrightarrow H$
is a transfer homomorphism. If this holds, then we have
$H=SH^\times $
,
$H^\times \cap S=S^\times $
, and
$\mathcal {A}(H)=\{u\varepsilon \ | \ u\in \mathcal {A}(S),\varepsilon \in H^\times \}$
.
Proof See [Reference Geroldinger and Zhong46, Proposition 5.3] and [Reference Bashir and Reinhart13, Proposition 2.7].
A monoid H is called finitely primary if there exist positive integers
$\alpha ,s$
with the following properties:
-
(1) H is a submonoid of a factorial monoid F, such that F is generated by its unit group
$F^\times $
along with s pairwise non-associated prime elements
$p_1,\ldots , p_s$
. -
(2)
$H\setminus H^\times \subseteq p_1\ldots p_sF$
. -
(3)
$(p_1\ldots p_s)^\alpha F \subseteq H$
.
If this is the case, then we say that H is finitely primary of rank s and exponent
$\alpha $
.
Let
$\varepsilon \in F^\times $
and let
$x=\varepsilon p_1^{k_1}\ldots p_s^{k_s}\in H$
for nonnegative integers
$k_i$
. For each
${i\in [1,s]}$
, we define the function
$\mathsf {v}_{p_i}:H\to \mathbb {N}_0$
, where
$\mathsf {v}_{p_i}(x)=k_i$
.
Lemma 2.2 If
$H\subseteq F$
is finitely primary of rank
$s\geq 2$
and exponent
$\alpha $
, then
$|\mathsf {v}_{p_i}(\mathcal {A}(H))|=\infty $
for each prime element
$p_i\in F$
.
Proof Let k be a positive integer and consider the element
$x=p_i^{\alpha k}(p_1\ldots p_s)^\alpha $
. By definition, we have
$x\in H$
. For every
$y\in \mathcal {A}(H)$
and
$j\in [1,s]$
, we have
$\mathsf {v}_{p_j}(y)\geq 1$
and thus
$\max \mathsf {L}(x)\leq \alpha $
. Hence, in every factorization of x, there is an atom y satisfying
$\mathsf {v}_{p_i}(y)\geq k+1$
. Since k was chosen arbitrarily, we obtain
$|\mathsf {v}_{p_i}(\mathcal {A}(H))|=\infty $
.
2.2 Dedekind domains and orders
Let R be a Dedekind domain with quotient field K. Since R is a Krull domain,
$R^\bullet $
is a Krull monoid and the class group
$\mathcal {C}(R^\bullet )$
coincides with the usual ideal class group
$\operatorname {\mathrm {Cl}}(R)$
. We denote the class of an ideal
$I\subseteq R$
by
$[I]$
and we write the class group additively, whence
$[IJ]=[I]+[J]$
for ideals I and J. If every class in
$\operatorname {\mathrm {Cl}}(R)$
contains a prime ideal, then there is a transfer homomorphism
$\boldsymbol {\beta }:R^\bullet \to \mathcal {B}(\operatorname {\mathrm {Cl}}(R))$
, defined in the following way. If
$a\in R^\bullet $
with
$aR=P_1^{n_1}\ldots P_k^{n_k}$
for prime ideals
$P_1,\ldots , P_k\in \operatorname {\mathrm {Spec}}(R)$
, then
$\boldsymbol {\beta }(a)=[P_1]^{n_1}\cdot \dots \cdot [P_k]^{n_k}$
. It is easy to see that an element
$a\in R$
is an atom (equivalently, that
$\boldsymbol {\beta }(a)$
is an atom) if and only if
$aR=P_1^{n_1}\ldots P_k^{n_k}$
does not contain any proper non-empty subproduct that is principal. Moreover, if
$\operatorname {\mathrm {Cl}}(R)$
is a torsion group, then a is absolutely irreducible if and only if
$aR=P^{\operatorname {\mathrm {ord}}([P])}$
for a prime ideal P (for details, see [Reference Geroldinger and Halter-Koch41, Proposition 7.1.5]) and for every element
$b\in R$
, there is
$n\in \mathbb {N}$
such that
$b^n$
can be written as a product of absolutely irreducible elements.
We will now gather several facts and properties about orders in Dedekind domains. For references, see, for example, [Reference Geroldinger and Halter-Koch41, Chapter 2.10] and [Reference Halter-Koch51, Chapter 2.11].
An order in a Dedekind domain R is a subring
${\mathcal {O}}\subseteq R$
with quotient field
$\mathsf {q}({\mathcal {O}})=\mathsf {q}(R)=K$
such that R is a finitely generated
${\mathcal {O}}$
-module. Every order is one-dimensional and Noetherian but is integrally closed if and only if it is Dedekind. From now on, we will always assume that
${\mathcal {O}}$
is a proper subring of R. In particular, this means that
${\mathcal {O}}^\bullet $
cannot be Krull. The conductor of
${\mathcal {O}}$
is the unique R-ideal
${\mathfrak {f}}$
that is maximal with respect to the property of also being an ideal of
${\mathcal {O}}$
. The conductor is always a nonzero ideal and we have
$$\begin{align*}{\mathfrak{f}}=\{x\in R \ | \ xR\subseteq{\mathcal{O}}\}.\end{align*}$$
We will denote the set of R-prime ideals comaximal to
${\mathfrak {f}}$
by
$\operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
.
Let
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
. Then
${\mathcal {O}}_{\mathfrak {p}}$
is a DVR of rank 1 if and only if
${\mathfrak {p}}\not \supseteq {\mathfrak {f}}$
. The integral closure
$\overline {{\mathcal {O}}_{\mathfrak {p}}}$
is a semilocal PID with n maximal ideals, where n is the number of R-prime ideals lying over
${\mathfrak {p}}$
. Moreover,
${\mathcal {O}}_{\mathfrak {p}}^\bullet $
is a finitely primary monoid of rank n. Let p be a prime element of
$\overline {{\mathcal {O}}_{\mathfrak {p}}}$
and let
$a\in \overline {{\mathcal {O}}_{\mathfrak {p}}}$
. The element p naturally corresponds to a prime ideal
$P\in \operatorname {\mathrm {Spec}}(R)$
. We denote by
$\mathsf {v}_p(a)$
the highest power of p that divides a and if
$a\in R$
, then
$\mathsf {v}_p(a)=\mathsf {v}_P(a)$
, where
$\mathsf {v}_P(a)$
denotes the highest power of P that divides
$aR$
. If
$a\in {\mathcal {O}}_{\mathfrak {p}}$
and
$x\in ({\mathcal {O}}_{\mathfrak {p}}^\bullet )_{\text {red}}$
such that
$x=a{\mathcal {O}}_{\mathfrak {p}}^\times $
, then we write
$\mathsf {v}_p(a)=\mathsf {v}_p(x)$
. If there is only one R-prime ideal lying over
${\mathfrak {p}}$
, then
$\overline {{\mathcal {O}}_{\mathfrak {p}}}$
is a DVR of rank 1 with precisely one prime element up to associates. In this case, we simply write
$\mathsf {v}_{\mathfrak {p}}(a)$
.
An element
$a\in {\mathcal {O}}$
is said to be regular if
$a+{\mathfrak {f}}\in ({\mathcal {O}}/{\mathfrak {f}})^\times $
. We denote the (multiplicative) monoid of regular elements of
${\mathcal {O}}$
by
$\operatorname {\mathrm {Reg}}({\mathcal {O}})$
. Similarly, we define the monoid
$$\begin{align*}\operatorname{\mathrm{Reg}}_{\mathfrak{f}}(R):=\{a\in R \ | \ a+{\mathfrak{f}}\in (R/{\mathfrak{f}})^\times\}.\end{align*}$$
Note that, if
$a\in \operatorname {\mathrm {Reg}}({\mathcal {O}})$
and
$b\in R$
, then
$ab\in {\mathcal {O}}$
implies that
$b\in {\mathcal {O}}$
. Moreover,
$\operatorname {\mathrm {Reg}}({\mathcal {O}})$
is a divisor-closed submonoid of
${\mathcal {O}}^\bullet $
, meaning that whenever
$ab\in \operatorname {\mathrm {Reg}}({\mathcal {O}})$
for elements
$a,b\in {\mathcal {O}}^\bullet $
, then
$a,b\in \operatorname {\mathrm {Reg}}({\mathcal {O}})$
. Similarly,
$\operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
is a divisor-closed submonoid of
$R^\bullet $
. We call the extension
${\mathcal {O}}\subseteq R$
a root extension if for every
$a\in R,$
there is
$n\in \mathbb {N}$
such that
$a^n\in {\mathcal {O}}$
. If
$R/{\mathfrak {f}}$
is a finite ring, then
${\mathcal {O}}\subseteq R$
is a root extension if and only if the map
$$\begin{align*}\operatorname{\mathrm{Spec}}(R)\to\operatorname{\mathrm{Spec}}({\mathcal{O}})\end{align*}$$
$$\begin{align*}P\mapsto P\cap{\mathcal{O}}\end{align*}$$
is bijective, for a proof, see [Reference Kainrath57, Theorem 2]. We will call this map the Spec-map. More precisely, if
$R/{\mathfrak {f}}$
is finite and
$a\in R$
, then
$a^n\not \in {\mathcal {O}}$
for all
$n\in \mathbb {N}$
if and only if there are
$P,Q\in \operatorname {\mathrm {Spec}}(R)$
, lying over the same prime ideal
${\mathfrak {p}}$
, such that
$\mathsf {v}_P(a)\geq 1$
and
$\mathsf {v}_Q(a)=0$
. In particular,
$\operatorname {\mathrm {Reg}}({\mathcal {O}})\subseteq \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
is a root extension of monoids.
Let
$\operatorname {\mathrm {Pic}}({\mathcal {O}})$
denote the Picard group of
${\mathcal {O}}$
. There is a natural exact sequence
$$\begin{align*}1\to R^\times/{\mathcal{O}}^\times\to(R/{\mathfrak{f}})^\times/({\mathcal{O}}/{\mathfrak{f}})^\times\to\operatorname{\mathrm{Pic}}({\mathcal{O}})\to\operatorname{\mathrm{Cl}}(R)\to 0,\end{align*}$$
see, for example, [Reference Halter-Koch51, Theorem 2.11.5], which implies that the canonical surjection
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\to \operatorname {\mathrm {Cl}}(R)$
is an isomorphism if and only if
$ R^\times /{\mathcal {O}}^\times \to (R/{\mathfrak {f}})^\times /({\mathcal {O}}/{\mathfrak {f}})^\times $
is an isomorphism. Note that this is also equivalent to
$\operatorname {\mathrm {Reg}}({\mathcal {O}})\cdot R^\times =\operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
. Moreover, we have
${\mathcal {O}}\cdot R^\times =R$
if and only if
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\simeq \operatorname {\mathrm {Cl}}(R)$
, the Spec-map is bijective and
$1\in \mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))$
for all
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
. Note also that the canonical surjection
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\to \operatorname {\mathrm {Cl}}(R)$
implies that if every class of
$\operatorname {\mathrm {Pic}}({\mathcal {O}})$
contains a prime ideal, then the same holds true for
$\operatorname {\mathrm {Cl}}(R)$
. In particular, our assumption in Theorem 1.1(2) is stronger than the one made in (1).
Let
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
, let
$P_1,\ldots ,P_k$
be the R-prime ideals lying over
${\mathfrak {p}},$
and let
$p_1,\ldots ,p_k$
be prime elements of
$\overline {{\mathcal {O}}_{\mathfrak {p}}}$
such that
$p_i$
corresponds to
$P_i$
for all
$i\in [1,k]$
. Moreover, let
$a\in R$
and
$x\in {\mathcal {O}}_{\mathfrak {p}}$
such that
$aR^\times \cap {\mathcal {O}}\neq \emptyset $
and
$\mathsf {v}_{P_i}(a)=\mathsf {v}_{p_i}(x)$
for all
$i\in [1,k]$
. Then there is
$b\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$ab\sim _{{\mathcal {O}}_{\mathfrak {p}}}x$
and the choice of b only depends on its class in
$(R/{\mathfrak {f}})^\times /({\mathcal {O}}/{\mathfrak {f}})^\times $
. If we now suppose in addition that
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\simeq \operatorname {\mathrm {Cl}}(R)$
, then we can choose b to be a unit and hence we find an element
$c\in {\mathcal {O}}$
such that
$$\begin{align*}a\sim_R c\sim_{{\mathcal{O}}_{\mathfrak{p}}} x.\end{align*}$$
We will use this fact several times in the final section.
Let H be an overmonoid of
$R^\bullet $
,
$a\in R$
and let
$I,I_1,I_2$
be principal ideals of R. We write
$a\sim {\mathcal {O}}$
if there is
$b\in {\mathcal {O}}$
such that
$a\sim _R b$
, and
$I\sim {\mathcal {O}}$
if there is
$b\in {\mathcal {O}}$
with
$bR=I$
. Moreover, we write
$I_1\sim _H I_2$
if for all
$a_1,a_2\in R$
with
$(a_1)=I_1,(a_2)=I_2$
, we have
$a_1\sim _H a_2$
.
-
Throughout the rest of this article, R is always a Dedekind domain with torsion class group and
${\mathcal {O}}\subseteq R$
is an order with conductor
${\mathfrak {f}}$
such that
$R/{\mathfrak {f}}$
is finite and such that every class in
$\operatorname {\mathrm {Cl}}(R)$
contains a prime ideal, comaximal to
${\mathfrak {f}}$
.
Using Proposition 2.1, we can deduce the following basic facts about transfer Krull orders, which we will use freely from now on.
Lemma 2.3 Suppose that
${\mathcal {O}}$
is transfer Krull with Krull monoid
$H\supseteq {\mathcal {O}}^\bullet $
as in Proposition 2.1. Then
(i)
$R^\bullet \subseteq H$
.
(ii)
$\mathcal {A}(\operatorname {\mathrm {Reg}}({\mathcal {O}}))\subseteq \mathcal {A}(R)$
.
(iii) If
${\mathcal {O}}\subseteq R$
is a root extension, then
$\mathcal {A}({\mathcal {O}})\subseteq \mathcal {A}(R)$
.
Proof (i) Since H is a completely integrally closed overmonoid of H, we obtain
$\widehat {{\mathcal {O}}^\bullet }=R^\bullet \subseteq H$
.
(ii) Let
$a\in \mathcal {A}(\operatorname {\mathrm {Reg}}({\mathcal {O}}))$
and suppose that we can write
$a=bc$
for non-units
$b, c\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
. Since a is an
${\mathcal {O}}$
-atom, it is an H-atom by Proposition 2.1. By (i), we have
$b,c\in H$
and thus
$b\in H^\times $
or
$c\in H^\times $
. Suppose without loss of generality that
$b\in H^\times $
and let
$n\in \mathbb {N}$
such that
$b^n\in {\mathcal {O}}$
. Then
$b^n\in H^\times \cap {\mathcal {O}}={\mathcal {O}}^\times $
, a contradiction.
(iii) See the proof of (ii).
3 Regular elements of transfer Krull orders
The aim of this section is to prove that
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\simeq \operatorname {\mathrm {Cl}}(R)$
for a transfer Krull order
${\mathcal {O}}$
, provided that
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
. This establishes a strong connection between the regular elements of
${\mathcal {O}}$
and the monoid
$\operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
.
-
Throughout this section, we suppose in addition to our initial assumptions that
${\mathcal {O}}$
is transfer Krull with Krull monoid
$H\supseteq {\mathcal {O}}^\bullet $
as in Proposition 2.1.
Lemma 3.1 Let
$a\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
be absolutely irreducible in R. Then
$a\sim {\mathcal {O}}$
.
Proof Let
$n\geq 2$
such that
$a^n\in {\mathcal {O}}$
. By the transfer property, there is
$b\in {\mathcal {O}}$
such that b divides
$a^n$
in
${\mathcal {O}}$
and
$a\sim _H b$
. Since a is absolutely irreducible in R, we have
$b\sim _R a^k$
for some
$k\leq n$
and consequently
$a\sim _H a^k$
. If
$k\neq 1$
, then
$a\in H^\times $
and consequently
${a^n\in H^\times \cap {\mathcal {O}}={\mathcal {O}}^\times }$
, a contradiction. Hence,
$k=1$
and
$a\sim {\mathcal {O}}$
.
Before continuing, we recall a simple group-theoretical fact. If G is an (additive) abelian torsion group,
$|G|\neq 2$
and
$g\in G$
with
$\operatorname {\mathrm {ord}}(g)=2$
, then there is
$h\neq g$
such that
$\operatorname {\mathrm {ord}}(h)=2$
or
$\operatorname {\mathrm {ord}}(h)=2k$
for a positive integer
$k\geq 2$
and
$kh=g$
.
Lemma 3.2 Suppose that
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
and let
$P_1,P_2,P_3\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
.
(i) If
$[P_1]=-[P_2]$
, then
$P_1P_2\sim {\mathcal {O}}$
.
(ii) If
$\operatorname {\mathrm {ord}}([P_1])=\operatorname {\mathrm {ord}}([P_2])=2$
,
$[P_1]\neq [P_2],$
and
$[P_3]=[P_1P_2]$
, then
$P_1P_2P_3\sim {\mathcal {O}}$
.
(iii) If
$\operatorname {\mathrm {ord}}([P_1])=2 and \operatorname {\mathrm {ord}}([P_2])=2k$
for some
$k\geq 2$
such that
$k[P_2]=[P_1]$
, then
$P_1P_2^k\sim {\mathcal {O}}$
.
Proof (i) We will first prove the assertion for all
$P_1\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
for which
${n:=\operatorname {\mathrm {ord}}([P_1])\neq 2}$
. If
$n=1$
, the claim follows immediately from Lemma 3.1. Hence, we can assume that
$n\geq 3$
. Let
$a\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$aR=P_1P_2$
. Since
$a^n$
is a product of absolutely irreducibles in R, we obtain
$a^n\sim {\mathcal {O}}$
by Lemma 3.1, and we find
$b\in {\mathcal {O}}$
with
$a^n\sim _R b$
. Consequently, the transfer property yields
$\mathsf {L}_H(a^n)=\mathsf {L}_{\mathcal {O}}(b)$
. Our claim is now proved by using the fact that
$n\in \mathsf {L}_H(a^n)$
and by observing that, since neither
$P_1^2$
nor
$P_2^2$
are principal, b has an
${\mathcal {O}}$
-factorization of length n only if
$P_1P_2\sim {\mathcal {O}}$
.
Before continuing with the case
$n=2$
, we will first prove the other two statements.
(ii) Let
$a\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$aR=P_1P_2P_3$
. Since
$a^2$
is a product of absolutely irreducibles, Lemma 3.1 yields
$a^2\sim {\mathcal {O}}$
and we find
$b\in {\mathcal {O}}$
such that
$b\sim _R a^2$
. By the transfer property, we obtain
$c,d\in {\mathcal {O}}\setminus {\mathcal {O}}^\times $
such that
$cd=b$
and
$c\sim _H a\sim _H d$
. If
${P_1P_2P_3 \not \sim {\mathcal {O}}}$
, then without loss of generality,
$cR=P_{\sigma (1)}^2$
and
$dR=P_{\sigma (2)}^2P_{\sigma (3)}^2$
, where
$\sigma :[1,3]\to [1,3]$
is a permutation. But then
$c\in \mathcal {A}({\mathcal {O}})\subseteq \mathcal {A}(H)$
and
$d\not \in \mathcal {A}(H)$
, a contradiction.
(iii) Let
$P_3\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
with the property that
$[P_2]=-[P_3]$
, let
$a,b\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$aR=P_1P_2^k$
and
$bR=P_1P_3^k$
and suppose that
$a\not \sim {\mathcal {O}}$
. We claim that also
$b\not \sim {\mathcal {O}}$
. Otherwise, we may assume that
$b\in {\mathcal {O}}$
and observe that
$P_1^2\sim {\mathcal {O}}$
by Lemma 3.1 and, since
$\operatorname {\mathrm {ord}}([P_2])>2$
, that
$P_2P_3\sim {\mathcal {O}}$
by (i). Hence,
$$\begin{align*}abR=P_1^2P_2^kP_3^k\sim{\mathcal{O}}\end{align*}$$
and, after possibly replacing a with an R-associate, we obtain
$ab\in {\mathcal {O}}$
. However, since
$b\in \operatorname {\mathrm {Reg}}({\mathcal {O}})$
, this yields
$a\in {\mathcal {O}}$
, a contradiction. Hence,
$P_1P_2^k\not \sim {\mathcal {O}}$
and
$P_1P_3^k\not \sim {\mathcal {O}}$
.
Since
$a^2$
is a product of absolutely irreducibles in R, Lemma 3.1 yields an element
$c\in {\mathcal {O}}$
such that
$a^2\sim _R c$
. By the transfer property, there are
$d,e\in {\mathcal {O}}$
such that
$de=c$
and
$d\sim _H a\sim _H e$
. Clearly, either d or e must generate the principal ideal
$P_1^2$
and we obtain
$$\begin{align*}P_1^2\sim_H P_1P_2^k.\end{align*}$$
Repeating the same argument for
$P_3$
in place of
$P_2$
, we obtain
$$\begin{align*}P_1^2\sim_H P_1P_3^k\end{align*}$$
and consequently
$$\begin{align*}P_1^4\sim_H P_1^2P_2^k P_3^k,\end{align*}$$
which implies that
$$\begin{align*}P_1^2\sim_H P_2^kP_3^k.\end{align*}$$
As we already proved, we have
$P_1^2\sim {\mathcal {O}}$
and
$P_2^kP_3^k\sim {\mathcal {O}}$
. However, elements in
${\mathcal {O}}$
generating
$P_1^2$
are atoms, while those that generate
$P_2^kP_3^k$
are not atoms in H, a contradiction.
We will now prove the remaining case in (i). Suppose that
${\operatorname {\mathrm {ord}}([P_1])\kern-1pt =\kern-1pt \operatorname {\mathrm {ord}}([P_2])\kern-1pt =\kern-1pt 2}$
and let
$a\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$aR=P_1P_2$
. Since
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
, there is
$P_3\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
with the property that
$\operatorname {\mathrm {ord}}([P_3])=2$
and
$[P_1]\neq [P_3]$
or such that
$\operatorname {\mathrm {ord}}([P_3])=2k$
for some
$k\geq 2$
and
$k[P_3]=[P_1]$
.
In the first case, let
$P_4 \in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[P_4]=[P_1P_3]$
. By (ii), we have
$$\begin{align*}P_1P_3P_4\sim{\mathcal{O}},\end{align*}$$
$$\begin{align*}P_2P_3P_4\sim{\mathcal{O}},\end{align*}$$
and consequently
$$\begin{align*}P_1P_2P_3^2P_4^2\sim{\mathcal{O}}.\end{align*}$$
By Lemma 3.1, we find
$b,c\in \operatorname {\mathrm {Reg}}({\mathcal {O}})$
such that
$bR=P_3^2$
and
$cR=P_4^2$
. After possibly replacing a with an R-associate, we have
$abc\in {\mathcal {O}}$
and since
$bc\in \operatorname {\mathrm {Reg}}({\mathcal {O}})$
, we obtain
$a\in {\mathcal {O}}$
and consequently
$P_1P_2\sim {\mathcal {O}}$
.
In the second case, we have
$P_1P_3^k\sim {\mathcal {O}}$
and
$P_2P_3^k\sim {\mathcal {O}}$
by (iii) and consequently
$$\begin{align*}P_1P_2P_3^{2k}\sim{\mathcal{O}}.\end{align*}$$
By noting that
$P_3^{2k}\sim {\mathcal {O}}$
, we can proceed in the same manner as in the previous case to obtain
$P_1P_2\sim {\mathcal {O}}$
.
Lemma 3.3 Suppose that
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
and let
$a,b\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$a\sim _H b$
. If a is absolutely irreducible in R, then
$\boldsymbol {\beta }(a)=\boldsymbol {\beta }(b)$
.
Proof First, we will assume that b is absolutely irreducible as well. Let
$aR=P_1^n, bR=P_2^m$
, let
$Q_1\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[P_1]=-[Q_1]$
and suppose that
$n>m$
. By Lemmas 3.1 and 3.2, there are
$c,d\in {\mathcal {O}}$
such that
$cR=P_1Q_1$
and
$dR=P_2^mQ_1^n$
. By assumption, we have
$c^n\sim _H d$
and thus
$\mathsf {L}_{\mathcal {O}}(c^n)=\mathsf {L}_{\mathcal {O}}(d)$
. Since
$n>m\geq 1$
, we have
$c\in \mathcal {A}({\mathcal {O}})$
and thus
$n\in \mathsf {L}_{\mathcal {O}}(c^n)$
. It is then easy to see that
$n\in \mathsf {L}_{\mathcal {O}}(d)$
is only possible if
$P_2$
is principal and
$n=2$
. Thus, we will suppose that
$n=2$
and
$m=1$
.
Since
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
, there is
$P_3\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
with the property that
$\operatorname {\mathrm {ord}}([P_3])=2$
and
$[P_1]\neq [P_3]$
or such that
$\operatorname {\mathrm {ord}}([P_3])=2k$
for some
$k\geq 2$
and
$k[P_3]=[P_1]$
.
In the first case, let
$P_4 \in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[P_4]=[P_1P_3]$
. By Lemmas 3.1 and 3.2, there are
$c,d\in {\mathcal {O}}$
such that
$cR=P_1P_3P_4$
and
$dR=P_2P_3^2P_4^2$
. By assumption, we have
$c^2\sim _H d$
, since
$P_1^2\sim _H P_2$
. Moreover, we have
$c\in \mathcal {A}({\mathcal {O}})$
and thus
$2\in \mathsf {L}_{\mathcal {O}}(c^2)$
. However, any
${\mathcal {O}}$
-factorization of d of length two yields an
${\mathcal {O}}$
-atom, that is not an R-atom, contradicting Lemma 2.3.
In the second case, let
$P_4 \in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[P_3]=-[P_4]$
. By Lemmas 3.1 and 3.2, we find
$c,d,e\in {\mathcal {O}}$
such that
$cR=P_1P_3^k$
,
$dR= P_1P_4^k$
, and
$eR=P_2^2P_3^{k}P_4^k$
. We then have
$cd\sim _H e$
and
$2\in \mathsf {L}_{\mathcal {O}}(cd)$
, since
$c,d\in \mathcal {A}({\mathcal {O}})$
. Recalling that
$k\geq 2$
, it is easy to see that
$2\in \mathsf {L}_{\mathcal {O}}(e)$
implies the existence of an
${\mathcal {O}}$
-atom that is not an R-atom, contradicting Lemma 2.3 again.
Hence, we have shown that
$n=m$
. We will now prove that
$[P_1]=[P_2]$
. This is clearly true if
$n=1$
. Suppose now that
$n\geq 2$
and let
$Q_1\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
${[P_1]=-[Q_1]}$
. Again, by Lemmas 3.1 and 3.2, we find
$c,d\in {\mathcal {O}}$
such that
$cR=P_1Q_1$
and
$dR=P_2^nQ_1^n$
. Then
$c^n\sim _H d$
and
$n\in \mathsf {L}_{\mathcal {O}}(c^n)$
. If
$[P_1]\neq [P_2]$
, then
$P_2Q_1$
is not principal, which clearly implies that
$n=2$
.
Let
$P_3\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[P_3]=[P_1P_2]$
. Again, by Lemmas 3.1 and 3.2, we find elements
$e,f\in {\mathcal {O}}$
such that
$eR=P_1P_2P_3$
and
$fR=P_1^4P_3^2$
. Then we have
$e^2\sim _H f$
and
$2\in \mathsf {L}_{\mathcal {O}}(e^2)=\mathsf {L}_{\mathcal {O}}(f)$
, which contradicts Lemma 2.3. Hence, our claim is proved, when b is absolutely irreducible in R.
Let now
$aR=P^n$
and
$bR=P_1^{n_1}\ldots P_k^{n_k}$
for prime ideals
$P,P_1,\ldots , P_k\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
and positive integers
$n, n_1,\ldots ,n_k$
, where
$n=\operatorname {\mathrm {ord}}([P])$
and suppose that
$a\sim _H b$
. There is a positive integer m such that
$a^m,b^m\in {\mathcal {O}}$
and such that
$b^m$
is a product of absolutely irreducible elements in R. By the transfer property, there is
$c\in {\mathcal {O}}$
, dividing
$a^m$
, such that
$c\sim _H d$
, where
$dR=P_i^{\operatorname {\mathrm {ord}}([P_i])}$
and
$i\in [1,k]$
. By Lemma 3.1, we can assume that
$d\in \mathcal {A}({\mathcal {O}})$
. Hence,
$c\in \mathcal {A}({\mathcal {O}})$
and
$a\sim _R c$
. By the first part of the proof, we obtain
$\operatorname {\mathrm {ord}}([P_i])=n$
and
$[P_i]=[P]$
. The only thing left to show is that
$n=\sum _{i=1}^{k}n_i$
. This follows from the fact that a is an R-associate of an
${\mathcal {O}}$
-atom, whence b is an R-atom by Lemma 2.3.
We can now extend this result for arbitrary elements of
$\operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
.
Lemma 3.4 Suppose that
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
and let
$a,b\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$a\sim _H b$
. Then
$\boldsymbol {\beta }(a)=\boldsymbol {\beta }(b)$
.
Proof There is a positive integer m such that
$a^m\in {\mathcal {O}}$
and such that
$b^m$
can be written as a product of absolutely irreducible elements in R. Clearly, we have
${\boldsymbol {\beta }(a)=\boldsymbol {\beta }(b)}$
if and only if
$\boldsymbol {\beta }(a^m)=\boldsymbol {\beta }(b^m)$
and thus we can assume that
$m=1$
. By Lemma 3.1, there are
$b_1,\ldots ,b_k\in {\mathcal {O}}$
, absolutely irreducible elements in R such that
$\prod _{i=1}^{k}b_i\sim _R b$
. We proceed by induction on the positive integer k. If
$k=1$
, we are done by Lemma 3.3. Otherwise, since we have
$a\sim _H \prod _{i=1}^{k}b_i$
, the transfer property implies the existence of elements
$c,d\in {\mathcal {O}}$
such that
$cd=a$
,
$c\sim _H b_k$
, and
$d\sim _H \prod _{i=1}^{k-1}b_i$
. By the induction hypothesis and Lemma 3.3, the claim is then proved.
Proposition 3.5 Suppose that
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
. Then
$\operatorname {\mathrm {Cl}}(R)\simeq \operatorname {\mathrm {Pic}}({\mathcal {O}})$
.
Proof It is enough to show that
$a\sim {\mathcal {O}}$
for every element
$a\in \mathcal {A}(\operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R))$
. Let
$aR=P_1^{n_1}\ldots P_k^{n_k}$
for pairwise distinct prime ideals
$P_1,\ldots ,P_k\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
and positive integers
$n_1,\ldots ,n_k$
. By Lemmas 3.1 and 3.2, we can assume that
$\sum _{i=1}^{k}n_i\geq 3$
. Let now
$Q_1,\ldots , Q_k\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[Q_i]=-[P_i]$
and
$Q_i=P_i$
if
$\operatorname {\mathrm {ord}}([P_i])=2$
for every
$i\in [1,k]$
. Moreover, let
$b\in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
such that
$bR=Q_1^{n_1}\ldots Q_k^{n_k}$
. By Lemma 3.2, we have
$P_iQ_i\sim {\mathcal {O}}$
for every
$i\in [1,k]$
and thus we can find
$c\in {\mathcal {O}}$
with the property
$c\sim _R ab$
. By the transfer property, there is
$d\in {\mathcal {O}}$
, dividing c such that
$d\sim _H a$
. By Lemma 3.4, we then have
$\boldsymbol {\beta }(a)=\boldsymbol {\beta }(d)$
.
We will show that for every
$i\in [1,k]$
,
$P_i^{n_i}$
divides
$dR$
, which, together with the fact that
$\boldsymbol {\beta }(a)=\boldsymbol {\beta }(d)$
, implies that
$a\sim _R d$
and proves our claim. Suppose that
${\operatorname {\mathrm {ord}}([P_i])=2}$
. Then
$n_i=1$
and
$[P_i]\neq [P_j]$
for all
$j\neq i$
, since we are assuming that
$\sum _{i=1}^{k}n_i\geq 3$
. Similarly, we have
$[P_i]\neq [Q_j]$
for all
$j\neq i$
. Then
$\boldsymbol {\beta }(a)=\boldsymbol {\beta }(d)$
implies that
$P_i$
or
$Q_i$
divides
$dR$
. However, we have
$P_i=Q_i$
by construction. Suppose now that
$\operatorname {\mathrm {ord}}([P_i])\geq 3$
(note that
$\operatorname {\mathrm {ord}}([P_i])=1$
is not possible since a is an R-atom and
$\sum _{i=1}^{k}n_i\geq 3$
) and that
$P_i^{n_i}$
does not divide
$dR$
. Then there is a
$Q_j$
, dividing
$dR$
, such that
$[Q_j]=[P_i]$
. We clearly have
$i\neq j$
, whence
$P_iP_j$
is principal, again contradicting the fact that
$\sum _{i=1}^{k}n_i\geq 3$
.
4 Localizations of transfer Krull orders
In this section, we use Proposition 3.5 to study the irregular prime ideals and localizations of a transfer Krull order
${\mathcal {O}}$
before proving our main theorem.
Proposition 4.1 Suppose that
${\mathcal {O}}$
is transfer Krull with Krull monoid
$H\supseteq {\mathcal {O}}^\bullet $
as in Proposition 2.1 and that
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\simeq \operatorname {\mathrm {Cl}}(R)$
. Then
${\mathcal {O}}\subseteq R$
is a root extension.
Proof By [Reference Kainrath57, Theorem 2], we need to show that the Spec-map is bijective. If this is not the case, then there is
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
with
${\mathfrak {f}}\subseteq {\mathfrak {p}}$
with at least two R-prime ideals lying over
${\mathfrak {p}}$
. Let
$P_1,\ldots , P_s$
denote these prime ideals, where
$s\geq 2$
and let
$n_i=\text {ord}([P_i])$
for each
$i\in [1,s]$
. Then
${\mathcal {O}}_{\mathfrak {p}}^\bullet $
is a finitely primary monoid of rank s. Let
$p_1,\ldots ,p_s$
be pairwise non-associate prime elements of
$\overline {{\mathcal {O}}_{\mathfrak {p}}}$
such that
$p_i\in P_i$
for all
$i\in [1,s]$
. Moreover, for each
$i\in [1,s]$
, let
$k_i\in \mathbb {N}$
such that
$k_i>n_i$
.
By Lemma 2.2, there is
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
with
$\mathsf {v}_{p_i}(x)\geq k_i$
for every
$i\in [1,s]$
. If
$$\begin{align*}P_1^{\mathsf{v}_{p_1}(x)}\ldots P_s^{\mathsf{v}_{p_s}(x)}\end{align*}$$
is not principal, take
$Q\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$$\begin{align*}QP_1^{\mathsf{v}_{p_1}(x)}\ldots P_s^{\mathsf{v}_{p_s}(x)}\end{align*}$$
is principal. Otherwise, set
$Q=R$
. Moreover, let
$a\in R$
be such that
$$\begin{align*}aR=QP_1^{\mathsf{v}_{p_1}(x)}\ldots P_s^{\mathsf{v}_{p_s}(x)}.\end{align*}$$
Since
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\simeq \operatorname {\mathrm {Cl}}(R)$
, we can assume that
$a\in {\mathcal {O}}$
and
$a\sim _{{\mathcal {O}}_{\mathfrak {p}}}x$
. Since x is an atom of
${\mathcal {O}}_{\mathfrak {p}}$
, we obtain that
$a\in \mathcal {A}({\mathcal {O}})\subseteq \mathcal {A}(H)$
. Let
$b_i\in R$
such that
$b_iR=P_i^{n_i}$
. Since we have
$\mathsf {v}_{p_i}(x)> n_i$
by assumption, we can write
$a=b_ic$
, where
$c\in R$
is contained in every prime ideal
$P_1,\ldots , P_s$
. Then either
$b_i$
or c is a unit in H. However,
$c\not \in H^\times $
, since
$c^n\in {\mathcal {O}}$
for some
$n\in \mathbb {N}$
, which implies that
$b_i\in H^\times $
.
Hence, we find
$b_1,\ldots ,b_s\in R\cap H^\times $
such that
$b_iR=P_i^{n_i}$
for all
$i\in [1,s]$
. However, there exists
$n\in \mathbb {N}$
such that
${(b_1\ldots b_s)}^n \in {\mathcal {O}} \cap H^\times ,$
which is a contradiction.
Proposition 4.2 Suppose that
${\mathcal {O}}$
is transfer Krull with Krull monoid
$H\supseteq {\mathcal {O}}^\bullet $
as in Proposition 2.1 and that
$\operatorname {\mathrm {Cl}}(R)\simeq \operatorname {\mathrm {Pic}}({\mathcal {O}})$
. Let
$P\in \operatorname {\mathrm {Spec}}(R)$
with
$n=\operatorname {\mathrm {ord}}([P])$
such that P divides
${\mathfrak {f}}$
and let
${\mathfrak {p}}=P\cap {\mathcal {O}}$
. Then
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))\subseteq \{1,2\}$
, if
$n=2$
and
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))= \{1\}$
otherwise. In particular, we have
${\mathcal {O}}\cdot R^\times =R$
.
Proof Suppose first that there is
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
with
$\mathsf {v}_{\mathfrak {p}}(x)>n$
. If
$P^{\mathsf {v}_{\mathfrak {p}}(x)}$
is not principal, let
$Q\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$QP^{\mathsf {v}_{\mathfrak {p}}(x)}$
is principal. Otherwise set
$Q=R$
. Since
$\operatorname {\mathrm {Cl}}(R)\simeq \operatorname {\mathrm {Pic}}({\mathcal {O}})$
, we find
$a\in {\mathcal {O}}$
with
$aR=QP^{\mathsf {v}_{\mathfrak {p}}(x)}$
such that
$a\sim _{{\mathcal {O}}_{\mathfrak {p}}}x$
. Then, since
$\mathsf {v}_{\mathfrak {p}}(x)>n$
, we see that a is an
${\mathcal {O}}$
-atom but not an R-atom, a contradiction. This proves the claim for
$n\leq 2$
.
Suppose now that
$n\geq 3$
and that there is
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
with
$\mathsf {v}_{\mathfrak {p}}(x)=m$
, where
${1<m<n}$
. Similar to above, we find
$a\in {\mathcal {O}}$
and
$Q_1,Q_2\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
satisfying
$aR=Q_1Q_2P^m$
,
$[Q_1]=-[P]$
,
$[Q_2]=-[P^{m-1}]$
, and
$a\sim _{{\mathcal {O}}_{\mathfrak {p}}}x$
. Since a is not an R-atom, it cannot be an
${\mathcal {O}}$
-atom, whence there are non-units
$b,c\in {\mathcal {O}}$
such that
$a=bc$
. However, both b and c are contained in P and thus both b and c are proper divisors of x in
${\mathcal {O}}_{\mathfrak {p}}$
, a contradiction.
Suppose now that there is
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
with
$\mathsf {v}_{\mathfrak {p}}(x)=n$
. Take
$Q\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
with
$[Q]=-[P]$
and
$a\in {\mathcal {O}}$
such that
$aR=Q^nP^n$
and
$a\sim _{{\mathcal {O}}_{\mathfrak {p}}}x$
. Since a is not an R-atom, it is not an
${\mathcal {O}}$
-atom, whence there is
$b\in {\mathcal {O}}$
, properly dividing a. Note that, if b is contained in P, then
$bR=P^n$
, since
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
. Otherwise, we have
$bR=Q^n$
. Let
$d,e\in R$
, satisfying
$dR=QP$
and
$eR=Q^{n-1}P^{n-1}$
. Then, by the transfer property, there are
$b,c\in {\mathcal {O}}$
such that
$a=bc, b\sim _H d$
, and
$c\sim _H e$
. However, c is an
${\mathcal {O}}$
-atom, but e is not an R-atom, a contradiction.
Hence,
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))\subseteq \{1,2\}$
, if
$n=2$
and
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))= \{1\}$
otherwise. Moreover, we have
$1\in \mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))$
for all
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
(note that
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))=\{2\}$
is impossible since
${\mathcal {O}}_{\mathfrak {p}}^\bullet $
is a finitely primary monoid of finite exponent), whence it follows that
${\mathcal {O}}\cdot R^\times =R$
.
Proof of Theorem 1.1
(1) Suppose that
$|\operatorname {\mathrm {Cl}}(R)|\neq 2$
. Let
$P_1,\ldots ,P_k\in \operatorname {\mathrm {Spec}}(R)$
be the prime ideals that divide
${\mathfrak {f}}$
, let
${\mathfrak {p}}_i=P_i\cap {\mathcal {O}}$
and let
$p_i$
be a prime element of
$\overline {{\mathcal {O}}_{{\mathfrak {p}}_i}}$
for every
$i\in [1,k]$
. If
${\mathcal {O}}$
is transfer Krull, we obtain
$\operatorname {\mathrm {Cl}}(R)\simeq \operatorname {\mathrm {Pic}}({\mathcal {O}})$
by Proposition 3.5. Then, by Proposition 4.2, we have
${\mathcal {O}}\cdot R^\times =R$
and
$\mathsf {v}_{{\mathfrak {p}}_i}(\mathcal {A}({\mathcal {O}}_{{\mathfrak {p}}_i}))=\{1\}$
for every
$i\in [1,k]$
such that
$\operatorname {\mathrm {ord}}([P_i])\neq 2$
.
Suppose now that
$\operatorname {\mathrm {ord}}([P_i])=2$
for some
$i\in [1,k]$
. Then
$\mathsf {v}_{{\mathfrak {p}}_i}(\mathcal {A}({\mathcal {O}}_{{\mathfrak {p}}_i}))\subseteq \{1,2\}$
by Proposition 4.2. Write
$P=P_i$
and
${\mathfrak {p}}={\mathfrak {p}}_i$
and suppose that there is
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
with
$\mathsf {v}_{\mathfrak {p}}(x)=2$
. Then we find non-principal prime ideals
$Q_1,Q_2,Q_3,Q_4\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
such that
$[Q_1]\neq [P]$
,
$[Q_1]=-[Q_2],[Q_3]=[Q_1P]$
, and
$[Q_3]=-[Q_4]$
and an element
${a\in {\mathcal {O}}}$
with
$aR=Q_1Q_2Q_3Q_4P^2$
such that
$a\sim _{{\mathcal {O}}_{\mathfrak {p}}}x$
. By Proposition 4.2, we have
${\mathcal {O}}\cdot R^\times =R$
, whence there are
$b,c\in {\mathcal {O}}$
with
$bR=Q_2Q_3P$
and
$cR=Q_1Q_4P$
. We clearly have
$b,c\in \mathcal {A}({\mathcal {O}})\subseteq \mathcal {A}(H)$
and consequently
$2\in \mathsf {L}_H(bc)=\mathsf {L}_H(a)$
. However, if
$d,e\in \mathcal {A}({\mathcal {O}})\subseteq \mathcal {A}(R)$
such that
$a=de$
, then d and e are both contained in P, which contradicts the fact that
$x\in \mathcal {A}({\mathcal {O}}_{\mathfrak {p}})$
.
Hence, we have shown that
$\mathsf {v}_{{\mathfrak {p}}_i}(\mathcal {A}({\mathcal {O}}_{{\mathfrak {p}}_i}))=\{1\}$
for all
$i\in [1,k]$
.
To prove the converse, we will show that the inclusion
${\mathcal {O}}^\bullet \hookrightarrow R^\bullet $
is a transfer homomorphism. By assumption, we have
${\mathcal {O}}\cdot R^\times =R$
and also
${\mathcal {O}}\cap R^\times ={\mathcal {O}}^\times $
. Thus, the only thing left to show is that if
$a,b\in R$
with
$ab\in {\mathcal {O}}$
, then there is
$\varepsilon \in R^\times $
such that
$\varepsilon a,\varepsilon ^{-1}b \in {\mathcal {O}}$
.
For
$i\in [1,k]$
, let
$a_i=\mathsf {v}_{P_i}(a)$
,
$b_i=\mathsf {v}_{P_i}(b)$
and let
$\Phi :{\mathcal {O}}^\bullet \rightarrow \prod _{i=1}^{k}{{({\mathcal {O}}_{{\mathfrak {p}}_i}^\bullet )}_{\text {red}} }$
denote the canonical homomorphism. Since
$\mathsf {v}_{{\mathfrak {p}}_i}(\mathcal {A}({\mathcal {O}}_{{\mathfrak {p}}_i}))=\{1\},$
we find elements
$x,y\in \prod _{i=1}^{k}{{({\mathcal {O}}_{{\mathfrak {p}}_i}^\bullet )}_{\text {red}} }$
such that
$\mathsf {v}_{p_i}(x)=a_i$
and
$\mathsf {v}_{p_i}(y)=b_i$
for all
$i\in [1,k]$
and
$\Phi (ab)=xy$
. Moreover, we find
$\varepsilon _1,\varepsilon _2\in R^\times $
such that
$\varepsilon _1 a,\varepsilon _2 b\in {\mathcal {O}}$
and
$\Phi (\varepsilon _1 a)=x,\Phi (\varepsilon _2 b)=y$
. Then, since
$ab\sim _R \varepsilon _1 a \cdot \varepsilon _2 b$
and
$\Phi (ab)=\Phi (\varepsilon _1 a \cdot \varepsilon _2 b)$
, we also have
$ab\sim _{\mathcal {O}} \varepsilon _1 a \cdot \varepsilon _2 b$
. Hence, there is
$\varepsilon \in R^\times $
such that
$\varepsilon a\in {\mathcal {O}}$
and
$ab= \varepsilon a\cdot \varepsilon _2 b$
. This implies that
$\varepsilon _2=\varepsilon ^{-1}$
and proves the claim.
(2) Suppose that
$|\operatorname {\mathrm {Cl}}(R)|=2$
, which implies that R is half-factorial. If
${\mathcal {O}}$
is transfer Krull, the claim follows from Proposition 4.2, once we prove that
$\operatorname {\mathrm {Cl}}(R)\simeq \operatorname {\mathrm {Pic}}({\mathcal {O}})$
. By Lemma 3.1, we have
$\pi \sim {\mathcal {O}}$
for every prime element
$\pi \in \operatorname {\mathrm {Reg}}_{\mathfrak {f}}(R)$
, which, by assumption implies that the kernel of the canonical surjection
$\operatorname {\mathrm {Pic}}({\mathcal {O}})\to \operatorname {\mathrm {Cl}}(R)$
is trivial.
For the converse, we claim that it is enough to show that
$\mathcal {A}({\mathcal {O}})\subseteq \mathcal {A}(R)$
. Indeed, for every
$a\in {\mathcal {O}}$
, we have
$\mathsf {L}_{\mathcal {O}}(a)\subseteq \mathsf {L}_R(a)$
and since R is half-factorial, we have
$|\mathsf {L}_{\mathcal {O}}(a)|\leq |\mathsf {L}_R(a)|=1$
. Hence,
${\mathcal {O}}$
is half-factorial and thus transfer Krull.
Let
$a\in \mathcal {A}({\mathcal {O}})$
and suppose that we can write
$a=bc$
for non-units
$b,c\in R$
. Then for all
$\varepsilon _1,\varepsilon _2\in R^\times $
such that
$\varepsilon _1 b,\varepsilon _2 c\in {\mathcal {O}}$
, we have
$a\not \sim _{\mathcal {O}} \varepsilon _1b\cdot \varepsilon _2c$
. This implies that there is
$P\in \operatorname {\mathrm {Spec}}(R)$
such that
$a\not \sim _{{\mathcal {O}}_{\mathfrak {p}}}\varepsilon _1b\cdot \varepsilon _2c$
for every such
$\varepsilon _1,\varepsilon _2\in R^\times $
, where
${\mathfrak {p}}=P\cap {\mathcal {O}}$
. Since we have
${\mathcal {O}}\cdot R^\times =R$
, this implies that for every
$x,y\in {\mathcal {O}}_{{\mathfrak {p}}}$
with
$\mathsf {v}_{\mathfrak {p}}(x)=\mathsf {v}_P(b)$
and
$\mathsf {v}_{\mathfrak {p}}(y)=\mathsf {v}_P(c)$
, we have
$a\not \sim _{{\mathcal {O}}_{\mathfrak {p}}}xy$
. This is clearly only possible if
$2\in \mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))$
(and thus
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))=\{1,2\}$
since the finitely primary monoid
${\mathcal {O}}_{\mathfrak {p}}^\bullet $
has finite exponent),
$\mathsf {v}_P(a)$
is even and both
$\mathsf {v}_P(b)$
and
$\mathsf {v}_P(c)$
are odd. In particular, P is not principal by assumption and we have
$\mathsf {v}_P(a)\geq 2$
. Hence, we can write
$a=de$
for elements
$d,e\in R$
, where
$dR=P^2$
. Since a is not an R-atom, e is a non-unit and for all
$\varepsilon _1,\varepsilon _2\in R^\times $
such that
$\varepsilon _1 d,\varepsilon _2 e\in {\mathcal {O}}$
, we have
$a\not \sim _{\mathcal {O}} \varepsilon _1d\cdot \varepsilon _2e$
. However, as we have shown, this implies that
$\mathsf {v}_Q(d)$
is odd for some
$Q\in \operatorname {\mathrm {Spec}}(R)$
, a contradiction.
Remark (1) Although it is unknown whether the additional assumption about the Picard group
$\operatorname {\mathrm {Pic}}({\mathcal {O}})$
is necessary for the second part of Theorem 1.1, there is evidence that the case
$|\operatorname {\mathrm {Cl}}(R)|$
=2 is indeed a special one. In [Reference Rago66], an example of a half-factorial order
${\mathcal {O}}$
in a cubic number field and a prime ideal
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
, for which
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))=\{1,2\}$
can be found.
(2) If
${\mathcal {O}}$
is transfer Krull with
$|\operatorname {\mathrm {Cl}}(R)|=2$
and there is
${\mathfrak {p}}\in \operatorname {\mathrm {Spec}}({\mathcal {O}})$
with
$\mathsf {v}_{\mathfrak {p}}(\mathcal {A}({\mathcal {O}}_{\mathfrak {p}}))=\{1,2\}$
, then both R and
${\mathcal {O}}$
are half-factorial, however,
${\mathcal {O}}^\bullet \hookrightarrow R^\bullet $
is not a transfer homomorphism. Indeed, let P be the non-principal prime ideal lying over
${\mathfrak {p}}$
. Then we find a non-principal
$Q\in \operatorname {\mathrm {Spec}}_{\mathfrak {f}}(R)$
and an element
$a\in {\mathcal {O}}$
with
$aR=P^2Q^2$
such that a is an atom in
${\mathcal {O}}_{\mathfrak {p}}$
. We can write
$a=bc$
, where
$b,c\in R$
with
$bR=cR=PQ$
. Then, by the transfer property, we find
$d,e\in {\mathcal {O}}$
such that
$d\sim _R b, e\sim _R c$
, and
$a=de$
. However, we have
$\mathsf {v}_{\mathfrak {p}}(d)=\mathsf {v}_{\mathfrak {p}}(e)=1$
, contradicting the fact that a is an atom in
${\mathcal {O}}_{\mathfrak {p}}$
.
