quotient ring), the orders $E$ are not even atomic. However, these orders are
not Goldie, as they are not of finite left or right uniform dimension, and do
not satisfy the ACC on left or right annihilator ideals.

Before going to maximal orders in central simple algebras, we discuss
principal ideal rings.

######  Example  (Principal ideal rings).

Let $R$ be a bounded order in a quotient ring $Q$. Assume that every left
$R$-ideal and every right $R$-ideal is principal. By the characterization in
Section 5, $R$ is then already a maximal order, and it satisfies A1–A3. Thus
$\mathcal{H}_{R}=\mathcal{I}_{v}(\alpha)$, and facts about the rigid
factorizations in $\mathcal{I}_{v}(\alpha)$ trivially descend to facts about
rigid factorizations of $R^{\bullet}$. Examples we have in mind include
bounded skew polynomial rings $D[X,\sigma]$, where $D$ is a division ring and
$\sigma\colon D\to D$ is an automorphism, and the Hurwitz quaternions
$\mathbb{Z}[1,i,j,\frac{1+i+j+ij}{2}]$ with $i^{2}=-1$, $j^{2}=-1$ and
$ij=-ji$. Both of these examples are left- and right-euclidean domains, and
hence principal ideal rings. In this way we can for example rediscover Theorem
2 in [16, §5].

Let $Q$ be a quaternion algebra over a field $K$ with
$\operatorname{char}(K)\neq 2$, and $a\in Q^{\bullet}\setminus K^{\times}$.
Then $\operatorname{nr}(a)=a\overline{a}\in K^{\times}$ and
$\operatorname{tr}(a)=a+\overline{a}\in K$. For the polynomial ring $Q[X]$ in
the central variable $X$, therefore

$f=X^{2}-\operatorname{tr}(a)X+\operatorname{nr}(a)=(X-cac^{-1})(X-c\overline{a}c^{-1})\quad\text{for
all $c\in Q^{\bullet}$},$

and thus $\lvert\mathsf{Z}^{*}_{Q[X]}(f)\rvert=\infty$ if $K$ is infinite.
(But these rigid factorizations are usually considered to be identical
factorizations, and $Q[X]$, being left- and right-euclidean, is even a UFD
with suitable definitions, see for example [8, Chapter 3.2] and [14, Chapter
3].) In terms of ideal theory, every element $X-cac^{-1}$ with $c\in
Q^{\bullet}$ generates a maximal left $Q[X]$-ideal lying above the maximal
two-sided $Q[X]$-ideal $Q[X]f$. If also $d\in Q^{\bullet}$, then
$Q[X](X-cac^{-1})=Q[X](X-dad^{-1})$ if and only if $cac^{-1}=dad^{-1}$, i.e.,
$d^{-1}c\in K(a)$.

###  5.2. Classical maximal orders over Dedekind domains in CSAs

Let $\mathcal{O}$ be a commutative domain with quotient field $K$. By a
central simple algebra $A$ over $K$, we mean a $K$-algebra with
$\dim_{K}(A)<\infty$, which is simple as a ring, and has center $K$. Then $A$
is artinian because it is a finite-dimensional $K$-algebra, and hence it is a
quotient ring (in an artinian ring, every non-zero-divisor is invertible [42,
§3.1.1], hence it is a quotient ring and an element is left-cancellative if
and only if it is right-cancellative if and only if it is cancellative). By
Posner’s Theorem ([42, §13.6.6]), a ring $R$ is a prime PI ring if and only if
it is an order in a central simple algebra, and hence in particular, prime PI
rings are bounded Goldie rings. Furthermore, PI Krull rings are characterized
as those maximal orders in central simple algebras whose center is a
commutative Krull domain ([35, Theorem 2.4]). We start with a simple corollary
of Section 5.1.

######  Corollary .

If $R$ is a PI Krull ring, then $\mathsf{L}_{R^{\bullet}}(a)$ is finite and
non-empty for all $a\in R^{\bullet}$.

###### Proof.

We only have to verify the conditions of Section 5.1. By [35, Theorem 2.4] the
various notions of Krull rings coincide for prime PI rings. Thus $R$ is a
bounded Chamarie-Krull ring. The ACC on divisorial left $R$-ideals and
divisorial right $R$-ideals follows from [11] (or [31, Corollary 3.11]).
Moreover, for every divisorial prime $R$-ideal $P$, the set of regular
elements modulo $P$, denoted $\mathcal{C}(P)$, is cancellative, satisfies the
left and right Ore condition, and for the localization
$R_{\mathcal{C}(P)}={}_{\mathcal{C}(P)}R\subset Q$ every left (right)
$R_{\mathcal{C}(P)}$-ideal is principal ([11, Proposition 2.5]). The lattice
of divisorial fractional left (right) $R_{\mathcal{C}(P)}$-ideals is hence
modular. Using the ACC on divisorial left and right $R$-ideals, one checks as
in the commutative case that
$I_{v}R_{\mathcal{C}(P)}=(IR_{\mathcal{C}(P)})_{v}$ for a fractional right
$R$-ideal $I$.

Suppose now $I,J,K$ are divisorial fractional right $R$-ideals, and $K\subset
I$. We have to check $I\cap(J+K)_{v}=((I\cap J)+K)_{v}$. But
$(I\cap(J+K)_{v})R_{\mathcal{C}(P)}=IR_{\mathcal{C}(P)}\cap(JR_{\mathcal{C}(P)}+KR_{\mathcal{C}(P)})_{v}$
and $((I\cap J)+K)_{v}R_{\mathcal{C}(P)}=((IR_{\mathcal{C}(P)}\cap
JR_{\mathcal{C}(P)})+KR_{\mathcal{C}(P)})_{v}$, and thus, by modularity in the
localizations, they are equal for every divisorial prime $R$-ideal $P$. The
claim now follows from [11, Lemme 2.7], by which the global divisorial
fractional right $R$-ideals can be recovered as intersections from the local
ones. ∎

Using Remark 5.2, we get the above result even for more general classes of
rings, namely for Dedekind prime rings and bounded Chamarie-Krull rings (cf.
[11]).

But the aim of this subsection is to restrict to the situation where the base
ring $\mathcal{O}$ is a Dedekind domain, as a preparation for the structural
results on sets of lengths in the setting of holomorphy rings. Suppose that
$\mathcal{O}$ is a Dedekind domain. A ring $R$ is a _classical
$\mathcal{O}$-order_ of $A$ if $\mathcal{O}\subset R$, $R$ is finitely
generated as $\mathcal{O}$-module and $KR=A$. $R$ is a _classical maximal
$\mathcal{O}$-order_ if it is maximal with respect to set inclusion within the
set of all classical $\mathcal{O}$-orders. Such classical maximal
$\mathcal{O}$-orders as well as their ideal theory are well-studied, in
particular Reiner’s book [47] provides a thorough description of them. If $R$
is a classical $\mathcal{O}$-order, then it is a ring-theoretic order in $A$
in the sense we discussed, and it is a maximal order if and only if it is a
classical maximal $\mathcal{O}$-order (for this see [42, §5.3]). The set of
all classical maximal $\mathcal{O}$-orders forms an equivalence class of
(ring-theoretic) maximal orders, call it $\beta$ for a moment. If we write
$\alpha$ for the same semigroup-theoretic equivalence class of maximal orders
(i.e., $\alpha=\beta$ as sets, but we view the elements of $\beta$ as rings
and those of $\alpha$ just as semigroups), then
$\mathcal{F}_{v}(\alpha)=\mathcal{F}_{v}(\beta)$ by Section 5.1. Next, we
recall that our notion of ideals coincides with that of [47] and [53] in the
case of maximal orders, thereby seeing how the one-sided ideal theory of
classical maximal $\mathcal{O}$-orders is a special case of the semigroup-
theoretic divisorial one-sided ideal theory developed in this section. We also
recognize the abstract norm homomorphism $\eta$ of Section 4 as a
generalization of the reduced norm of ideals (in the sense of [47, §24]).

######  Lemma .

Let $I\subset A$ and let $T$ be a classical $\mathcal{O}$-order in $A$. The
following are equivalent:

  1. (a)

$I$ is a fractional left $T$-ideal in the ring-theoretic sense (i.e., as in
[42, §3.1.11]).

  2. (b)

$I$ is a finitely generated $\mathcal{O}$-module with $KI=A$ and $TI\subset I$
. 333Here $KI=\\{\,\lambda a\mid\lambda\in K,a\in
I\,\\}=\\{\,\sum_{i=1}^{n}\lambda_{i}a_{i}\mid\text{$\lambda_{i}\in K,a_{i}\in
I$}\,\\}=K\otimes_{\mathcal{O}}I$.

If $T$ is maximal, then in addition the following statements are equivalent to
the previous ones:

  1. (c)

$I$ is a divisorial fractional left $T$-ideal in the semigroup-theoretic sense
(Definitions 5.2 and 5.3).

  2. (d)

$I$ is a divisorial fractional left $T$-ideal in the ring-theoretic sense
(i.e., a reflexive fractional left $T$-ideal as in [42, §5.1]).

###### Proof.

(a) $\Rightarrow$ (b): Recall that $I$ is a fractional left $T$-ideal in the
ring-theoretic sense if $TI\subset I$, $I+I\subset I$ and there exist $x,y\in
A^{\times}$ with $x\in I$ and $Iy\subset T$. $\mathcal{O}$ is the center of
$T$, and $T$ is finitely generated over the noetherian ring $\mathcal{O}$.
Since $Iy\subset T$, therefore also $I$ is a finitely generated
$\mathcal{O}$-module. Writing $x^{-1}=rc^{-1}$ with $r\in T^{\bullet}$ and
$c\in\mathcal{O}^{\bullet}$ we see that $c=rx\in I\cap\mathcal{O}^{\bullet}$.
If $a\in A$ is arbitrary, then $a=r^{\prime}d^{-1}$ with $r^{\prime}\in T$,
$d\in\mathcal{O}^{\bullet}$ and therefore $a=(r^{\prime}c)(c^{-1}d^{-1})\in
KI$.

(b) $\Rightarrow$ (a): Certainly $TI\subset I$ and $I+I\subset I$. We have to
find $x,y\in A^{\times}$ with $x\in I$ and $Iy\subset T$. Since $KI=A$, there
exist $\lambda\in K^{\times}$ and $x\in I\cap A^{\times}$ with $1=\lambda x$
(in fact even $x\in K^{\times}$). If $I={}_{\mathcal{O}}\langle
y_{1},\ldots,y_{l}\rangle$ with $y_{1},\ldots,y_{l}\in I$, then due to $KT=A$
there exists a common denominator $y\in\mathcal{O}^{\bullet}$ with $y_{i}y\in
T$, hence $Iy\subset T$.

Let now $T$ be maximal. (d) $\Rightarrow$ (a) is trivial, and (a)
$\Rightarrow$ (d) follows because $T$ is a Dedekind prime ring, and hence
every fractional left $T$-ideal (in the ring-theoretic sense) is invertible
(see [42, §5.2.14] or [47, §22,§23] for the more specific case where $R$ is a
maximal order in a CSA), and therefore divisorial.

(c) $\Leftrightarrow$ (d) follows from Section 5.1. ∎

A subset $I\subset A$ satisfying the second condition of the previous lemma
and additionally $\mathcal{O}_{l}(I)=T$ is considered to be a left $T$-ideal
in [47] and [53]. Thus, a left $T$-ideal in the sense of [47, 53] is (in our
terms) a fractional left $T$-ideal in the ring-theoretic sense with
$\mathcal{O}_{l}(I)=T$. If $T$ is maximal, then the extra condition
$\mathcal{O}_{l}(I)=T$ is trivially satisfied, and the definitions are
equivalent, but for a non-maximal order the definitions do not entirely agree
(we will only need to work with ideals of maximal orders).

Since all $I\in\mathcal{F}_{v}(\alpha)$ are invertible (i.e.,
$II^{-1}=\mathcal{O}_{l}(I)$ and $I^{-1}I=\mathcal{O}_{r}(I)$ for the ring-
theoretic products), the $v$-product coincides with the usual proper product
of ideals: $I\cdot_{v}J=I\cdot J$ whenever $I,J\in\mathcal{F}_{v}(\alpha)$
with $\mathcal{O}_{r}(I)=\mathcal{O}_{l}(J)$. Therefore,
$\mathcal{F}_{v}(\alpha)$ is the groupoid of all normal ideals of $A$ in
Reiner’s terminology ($\mathcal{O}$ is fixed implicitly).

To be able to apply our abstract results we still have to check that A1
through A3 are true for $\alpha$: A1 follows because every $R\in\alpha$ is
noetherian, while A2 is true because every fractional left $R$-ideal with
$R\in\alpha$ in fact even contains a non-zero element of the center (cf. [42,
Prop. 5.3.8(i) and (ii)] or see “(b) $\Rightarrow$ (a)” of the last proof).
Since every fractional left (right) $R$-ideal is divisorial, A3 follows from
the modularity of the lattice of left (right) $R$-modules.

Writing $\mathcal{F}^{\times}(\mathcal{O})$ for the non-zero fractional ideals
of the commutative Dedekind domain $\mathcal{O}$, and $\mathbb{G}$ for the
universal vertex group of $\mathcal{F}_{v}(\alpha)$, we have the following.

######  Lemma .

If $R,R^{\prime}\in\alpha$ and $\mathcal{P}\in\mathbb{G}$, then
$\mathcal{P}_{R}\cap\mathcal{O}=\mathcal{P}_{R^{\prime}}\cap\mathcal{O}\in\max(\mathcal{O})$
and there is a canonical bijection

$\\{\,\mathcal{P}\mid\mathcal{P}\in\mathbb{G}\text{ maximal integral
\\}}\to\max(\mathcal{O}),$

inducing an isomorphism of free abelian groups
$r\colon\mathbb{G}\overset{\sim}{\rightarrow}\mathcal{F}^{\times}(\mathcal{O})$.
The inverse map is given by $\mathfrak{p}\mapsto(\mathfrak{P})$ where
$\mathfrak{P}$ is the unique maximal (two-sided) $R$-ideal lying over
$\mathfrak{p}$. If $R$ is unramified at $\mathfrak{p}$, then
$\mathfrak{P}=R\mathfrak{p}$.

If all residue fields of $\mathcal{O}$ are finite, and
$\eta\colon\mathcal{F}_{v}(\alpha)\to\mathbb{G}$ is the abstract norm
homomorphism, then $r\circ\eta=\operatorname{nr}_{A/K}$.

###### Proof.

All but the last statement follow from [47, Theorem 22.4]. Since $r\circ\eta$
and $\operatorname{nr}_{A/K}$ are both homomorphisms
$\mathcal{F}_{v}(\alpha)\to\mathcal{F}^{\times}(\mathcal{O})$, it suffices to
verify equality for $M$ a maximal integral left $R^{\prime}$-ideal with
$R^{\prime}\in\alpha$, where it holds due to [47, Theorem 24.13]. ∎

##  6\. Proof of Section 1

_Throughout this section, let $K$ be a global field and $\mathcal{O}$ be a
holomorphy ring in $K$. 444For us, $\mathcal{O}$ is a holomorphy ring if it is
an intersection of all but finitely many of the valuation domains associated
to valuations of $K$. Furthermore, let $A$ be a central simple algebra over
$K$, and $R$ a classical maximal $\mathcal{O}$-order. _

Setting $P_{A}=\\{\,aO\mid a\in K^{\times},\text{ $a_{v}>0$ for all
archimedean places $v$ of $K$ where $A$ is ramified}\,\\}$, and denoting by
$\operatorname{\mathcal{C}}_{A}(\mathcal{O})=\mathcal{F}^{\times}(\mathcal{O})/P_{A}$
the corresponding ray class group, we have the following.

######  Lemma .

Let $r$ be as in Section 5.2. Then $r$ induces an isomorphism

$\mathbb{G}/P_{R^{\bullet}}\,\cong\,\operatorname{\mathcal{C}}_{A}(\mathcal{O}),$

where $P_{R^{\bullet}}=\\{\,\eta(Rx)\mid x\in
A^{\times}\,\\}\subset\mathbb{G}$.

###### Proof.

By Section 5.2, $r\circ\eta=\operatorname{nr}_{A/K}$. The isomorphism follows
because $\operatorname{nr}(Rx)=\mathcal{O}\operatorname{nr}(x)$ for all $x\in
A^{\times}$, and $\operatorname{nr}(A^{\times})=\\{\,a\in K^{\times}\mid
a_{v}>0\text{ for all archimedean places $v$ of $K$ where $A$ is ramified
}\,\\}$ by the Hasse-Schilling-Mass theorem on norms ([47, Theorem 33.15]). ∎

######  Lemma .

For all classical maximal $\mathcal{O}$-orders $R^{\prime}$, and all
$g\in\operatorname{\mathcal{C}}_{A}(\mathcal{O})$, there exist infinitely many
maximal left $R^{\prime}$-ideals $I$ with $[\operatorname{nr}(I)]=g$.

###### Proof.

Let $g\in\operatorname{\mathcal{C}}_{A}(\mathcal{O})$. Then there exist
infinitely many distinct maximal ideals $\mathfrak{p}$ of $\mathcal{O}$ with
$[\mathfrak{p}]=g$: The number field case for $\mathcal{O}=\mathcal{O}_{K}$,
the ring of algebraic integers, can be found in [27, Corollary 2.11.16] or
[43, Corollary 7 to Proposition 7.9]. The general case then follows because
$\mathcal{O}$ is obtained from $\mathcal{O}_{K}$ by localizing at finitely
many maximal ideals, hence the induced epimorphism
$\operatorname{\mathcal{C}}_{A}(\mathcal{O}_{K})\to\operatorname{\mathcal{C}}_{A}(\mathcal{O})$
yields the statement. For the function field case see [27, Proposition 8.9.7].

For each $\mathfrak{p}\in\max(\mathcal{O})$ with $[\mathfrak{p}]=g$ and every
maximal left $R^{\prime}$-ideal $M$ with $\mathfrak{p}\subset M$, we have
$[\operatorname{nr}(M)]=[\mathfrak{p}]=g$ ([47, Theorem 24.13], or use Section
5.2). ∎

In the following equivalent characterizations of the first condition of
Section 1, “left” may be replaced by “right” in each statement; this follows
easily from the first statement. We write $\operatorname{\mathcal{LC}}(R)$ for
the finite set of isomorphism classes of fractional left $R$-ideals, i.e.,
$[I]=[J]$ in $\operatorname{\mathcal{LC}}(R)$ if and only if $J=Ix$ with $x\in
A^{\times}$. The reduced norm induces a surjective map of finite sets
$\mu_{R}\colon\operatorname{\mathcal{LC}}(R)\to\operatorname{\mathcal{C}}_{A}(\mathcal{O})$,
given by $[I]\mapsto[\operatorname{nr}(I)]$.

######  Lemma .

The following are equivalent.

  1. (a)

A fractional left $R^{\prime}$-ideal with $R^{\prime}$ conjugate to $R$ is
principal if and only if $\operatorname{nr}(I)\in P_{A}$.

  2. (b)

A fractional left $R$-ideal is principal if and only if
$\operatorname{nr}(I)\in P_{A}$.

  3. (c)

Every fractional left $R$-ideal $I$ with $[\operatorname{nr}(I)]=\mathbf{0}$
is principal.

  4. (d)

For the map of finite sets
$\mu_{R}\colon\operatorname{\mathcal{LC}}(R)\to\operatorname{\mathcal{C}}_{A}(\mathcal{O})$
it holds that $\lvert\mu_{R}^{-1}(\mathbf{0})\rvert=1$.

  5. (e)

Every stably free left $R$-ideal is free.

  6. (f)

Every finitely generated projective $R$-module that is stably free is free.

###### Proof.

The equivalence of (a), (b), (c) and (d) is trivial. The remaining
equivalences follow from standard literature: (f) $\Rightarrow$ (e) is true
because $R$ is hereditary noetherian.

(e) $\Rightarrow$ (f): Let $M\neq\mathbf{0}$ be a stably free finitely
generated projective $R$-module. Then $M\cong R^{n}\oplus I$ for some left
$R$-ideal $I$ and $n\in\mathbb{N}_{0}$ ([47, Theorem 27.8] or [42, §5.7.8]).
$I$ is stably free and hence free by (e), but then so is $M$.

To see (d) $\Leftrightarrow$ (e) it suffices to recall that
$\operatorname{\mathcal{C}}_{A}(\mathcal{O})$ is isomorphic to the projective
class group $\operatorname{\mathcal{C}}(R)$ (see e.g. [52, Corollary 9.5]) and
that $\operatorname{\mathcal{LC}}(R)$ is just the set of isomorphism classes
of locally free $R$-modules of rank one, i.e., the map $\mu_{R}$ corresponds
to $\mathsf{LF}_{1}\to\operatorname{\mathcal{C}}(R),[I]\mapsto[I]-[R]$ in the
notation of [52]. (See also [47, Theorem 35.14] or [22] for the number field
case.) ∎

###### Proof of Section 1.

By Section 5.1, $R$ is an arithmetical maximal order in $Q$. We verify
conditions (i) and (ii) of Section 5. Let $I$ be a fractional left $R$-ideal.
By Section 6, $\eta(I)\in P_{R^{\bullet}}$ if and only if
$\operatorname{nr}(I)\in P_{A}$. By Section 6, and the fact that every stably
free left $R$-ideal is free, this is the case if and only if $I$ is principal,
thus condition (i) holds. Condition (ii) holds due to Section 6. By Section 6,
$C\cong\operatorname{\mathcal{C}}_{A}(\mathcal{O})$, and by Section 6,
therefore $C=C_{M}$. Hence there exists a transfer homomorphism $\theta\colon
R^{\bullet}\to\mathcal{B}(\operatorname{\mathcal{C}}_{A}(\mathcal{O}))$. The
remaining claims in the theorem follow from this by Section 3. ∎

######  Remark .

If more generally $\mathcal{O}^{\prime}$ is an arbitrary Dedekind domain with
quotient field the global field $K$, then there is a transfer homomorphism to
either $\mathcal{B}(\operatorname{\mathcal{C}}_{A}(\mathcal{O}^{\prime}))$ or
$\mathcal{B}(\operatorname{\mathcal{C}}_{A}(\mathcal{O}^{\prime})\setminus\\{\mathbf{0}\\})$,
depending on whether or not $\mathcal{O}^{\prime}$ contains prime elements.
Only Section 6 has to be adapted: $\mathcal{O}^{\prime}$ is a localization of
a holomorphy ring $\mathcal{O}$, and hence there is an epimorphism
$\operatorname{\mathcal{C}}_{A}(\mathcal{O})\to\operatorname{\mathcal{C}}_{A}(\mathcal{O}^{\prime})$.
This implies that every class
$g\in\operatorname{\mathcal{C}}_{A}(\mathcal{O}^{\prime})\setminus\\{\mathbf{0}\\}$
contains a maximal ideal (see [13] for details). Therefore, for all classical
maximal $\mathcal{O}^{\prime}$-orders $R^{\prime}$ and all
$g\in\operatorname{\mathcal{C}}_{A}(\mathcal{O}^{\prime})\setminus\\{\mathbf{0}\\}$,
there exists a maximal left $R^{\prime}$-ideal $I$ with
$[\operatorname{nr}(I)]=g$. The trivial class however may or may not contain a
maximal ideal. In either case, the statements 1–3 of Section 1 hold true.
Thanks to Kainrath for pointing this out.

##  7\. Proof of Section 1

_Throughout this section, let $\mathcal{O}_{K}$ be the ring of algebraic
integers in a number field $K$, $A$ a central simple algebra over $K$, and $R$
a classical maximal $\mathcal{O}_{K}$-order in $A$ having a stably free left
$R$-ideal that is not free. Furthermore, the discriminant of $A$ is denoted
by_

$\mathfrak{D}=\prod_{\begin{subarray}{c}\mathfrak{p}\in\max(\mathcal{O}_{K})\\\
\textnormal{$A$ is ramified at
$\mathfrak{p}$}\end{subarray}}\mathfrak{p}\,\triangleleft\,\mathcal{O}_{K}.$

The aim of this section is to prove Theorem 1. The existence of a stably free
left $R$-ideal that is not free implies that $A$ is a totally definite
quaternion algebra and that $K$ is totally real (Note, that conversely, for
all but finitely many isomorphism classes of such classical maximal
$\mathcal{O}_{K}$-orders in totally definite quaternion algebras there exist
stably free left $R$-ideals that are non-free). We proceed in three
subsections.

###  7.1. Reduction.

We state two propositions and show how they imply Theorem 1. The proofs of
these two propositions will then be given in Section 7.3.

######  Proposition .

There exists a totally positive prime element $p\in\mathcal{O}_{K}$, a non-
empty subset $E\subset\\{\,2,3,4\,\\}$ and for every $l\in\mathbb{N}_{0}$ an
atom $y_{l}\in\mathcal{A}(R^{\bullet})$ such that

$\mathsf{L}_{R^{\bullet}}(y_{l}p)=\\{\,3\,\\}\,\cup\,(l+E).$

(We emphasize that $E$ does not depend on $l$.)

######  Proposition .

If $L\in\mathcal{L}(R^{\bullet})$ and $n\in\mathbb{N}$, then
$n+L=\\{\,n+l\,\mid\,l\in L\,\\}\in\mathcal{L}(R^{\bullet})$.

###### Proof of Section 1 (based on Section 7.1 and Section 7.1).

We first show that there is no transfer homomorphism
$R^{\bullet}\to\mathcal{B}(G_{P})$ for any subset $G_{P}$ of an abelian group.
Assume to the contrary that $\theta\colon R^{\bullet}\to\mathcal{B}(G_{P})$ is
such a transfer homomorphism.

  1. Claim A.

If $S\in\mathcal{B}(G_{P})$ and $U\in\mathcal{A}(\mathcal{B}(G_{P}))$, then
$\max\mathsf{L}_{\mathcal{B}(G_{P})}(SU)\leq\lvert S\rvert+1.$

###### Proof of A.

Let $S=g_{1}\cdot\ldots\cdot g_{l}$, with $l=\lvert S\rvert$ and
$g_{1},\ldots,g_{l}\in G_{P}$, and suppose that $SU=T_{1}\cdot\ldots\cdot
T_{k}$ with $k\in\mathbb{N}$ and
$T_{1},\ldots,T_{k}\in\mathcal{A}(\mathcal{B}(G_{P}))$. Then for every
$i\in[1,k]$ either $T_{i}\mid U$, but then already $T_{i}=U$, or $g_{j}\mid
T_{i}$ for some $j\in[1,l]$. This shows $k\leq\lvert S\rvert+1$. ∎

By Section 7.1, there exists a totally positive prime element
$p\in\mathcal{O}_{K}$, and for every $l\in\mathbb{N}_{0}$ an atom
$y_{l}\in\mathcal{A}(R^{\bullet})$ with
$\max\mathsf{L}_{R^{\bullet}}(y_{l}p)\geq l+2$. But, if
$l\geq\lvert\theta(p)\rvert$, then

$l+2\,\leq\,\max\mathsf{L}_{R^{\bullet}}(y_{l}p)=\max\mathsf{L}_{\mathcal{B}(G_{P})}(\theta(y_{l})\theta(p))\,\leq\,\lvert\theta(p)\rvert+1\,\leq\,l+1,$

a contradiction.

In order to show $\Delta(R^{\bullet})=\mathbb{N}$, we choose $d\in\mathbb{N}$.
Let $p$ and $E$ be as in Section 7.1 and set $\epsilon=\min E$. If
$l=d+3-\epsilon$ and $y_{l}$ as in Section 7.1, then we find
$d=(l+\epsilon)-3\in\Delta_{R^{\bullet}}(y_{l}p)$.

Let $k\in\mathbb{N}_{\geq 3}$. By definition, we have
$\mathcal{U}_{k}(R^{\bullet})\subset\mathbb{N}_{\geq 2}$. Thus it remains to
show that for every $k^{\prime}\geq 3$ there exists an element $a\in
R^{\bullet}$ with $\\{\,k,k^{\prime}\,\\}\subset\mathsf{L}(a)$. Assume without
restriction that $k\leq k^{\prime}$ and let $k=3+n$ with $n\in\mathbb{N}_{0}$.
Using Section 7.1, we find an element $a^{\prime}\in R^{\bullet}$ with
$\\{\,3=k-n,\,k^{\prime}-n\,\\}\in\mathsf{L}(a^{\prime})$, and hence by
Section 7.1 there exists an element $a\in R^{\bullet}$ with
$\\{\,k,k^{\prime}\,\\}\in\mathsf{L}(a)$. ∎

###  7.2. Preliminaries.

_Algebraic number theory:_ Our notation mainly follows Narkiewicz [43]. Let
$L/K$ be an extension of number fields. Then $D_{L/K}$ is the relative
different, $\operatorname{N}_{L/K}$ the relative field norm,
$d_{L/K}=\operatorname{N}_{L/K}(D_{L/K})$ is the relative discriminant and
$d_{K}=d_{K/\mathbb{Q}}$ the absolute discriminant (we tacitly identify ideals
of $\mathbb{Z}$ with their positive generators for the absolute discriminant
and norm). If $\mathcal{O}\subset\mathcal{O}_{K}$ is an order, then
$\mathfrak{f}_{\mathcal{O}}$ is the conductor of $\mathcal{O}$ in
$\mathcal{O}_{K}$ and
$h(\mathcal{O})=\lvert\operatorname{Pic}(\mathcal{O})\rvert$ is the class
number of $\mathcal{O}$. Given $a\in L$ with minimal polynomial $f\in K[X]$
over $K$, $\delta_{L/K}(a)=f^{\prime}(a)$ is the different of $a$. Completion
at a prime $\mathfrak{p}\in\max(\mathcal{O}_{K})$ is denoted by a subscript
$\mathfrak{p}$, e.g., $\mathcal{O}_{K,\mathfrak{p}}$, $K_{\mathfrak{p}}$, and
so on. If $\mathfrak{m}\,\triangleleft\,\mathcal{O}_{K}$ is a squarefree
ideal, then

$\operatorname{\mathcal{C}}^{+}_{\mathfrak{m}}(\mathcal{O}_{K})=\\{\,\mathfrak{a}\in\mathcal{F}^{\times}(\mathcal{O}_{K})\mid(\mathfrak{a},\mathfrak{m})=\mathcal{O}_{K}\,\\}\,\big{/}\,\\{\,a\mathcal{O}_{K}\mid\text{$a\in
K^{\times}$ is totally positive, $a\equiv 1\mod\mathfrak{m}$}\,\\}$

denotes the corresponding ray class group. We will repeatedly make use of the
fact that every class in
$\operatorname{\mathcal{C}}^{+}_{\mathfrak{m}}(\mathcal{O}_{K})$ contains
infinitely many maximal ideals of $\mathcal{O}_{K}$ ([43, Corollary 7 to
Proposition 7.9]).

_Quaternion algebras:_ We follow [53, 41], and [39, 40] for computational
aspects. Denote by
$\overline{\,\cdot\,}:A\overset{\sim}{\rightarrow}A^{\text{op}}$ the anti-
involution given by conjugation of elements. Then

$\operatorname{nr}_{A/K}(x)=\operatorname{nr}(x)=x\overline{x}=\overline{x}x\qquad\text{and}\qquad\operatorname{tr}_{A/K}(x)=\operatorname{tr}(x)=x+\overline{x}\qquad\text{for
all $x\in A$.}$

Every element $x\in A$ satisfies an equation of the form

$x^{2}-\operatorname{tr}(x)x+\operatorname{nr}(x)=0,$

and if $x\in A\setminus K$, then $K(x)/K$ is a quadratic field extension. From
the equation above we see that
$\operatorname{N}_{K(x)/K}=\operatorname{nr}_{A/K}|K(x)$ and
$\operatorname{Tr}_{K(x)/K}=\operatorname{tr}_{A/K}|K(x)$.

A classical $\mathcal{O}_{K}$-order $T$ of $A$ is called a _classical Eichler
( $\mathcal{O}_{K}$)-order_ if it is the intersection of two classical maximal
$\mathcal{O}_{K}$-orders. 555Though unconventional, we keep the qualifier
“classical” for consistency with the earlier sections. The reduced
discriminant of a classical $\mathcal{O}_{K}$-order $T$ takes the form
$\mathfrak{D}\mathfrak{N}$ where
$\mathfrak{N}\,\triangleleft\,\mathcal{O}_{K}$ is the level of $T$.
Furthermore, because $A$ is totally definite, we have
$[T^{\times}:\mathcal{O}_{K}^{\times}]<\infty$ for the unit group, and
$\operatorname{\mathcal{C}}_{A}(\mathcal{O}_{K})=\operatorname{\mathcal{C}}^{+}(\mathcal{O}_{K})$
is the narrow class group of $\mathcal{O}_{K}$.

As in the previous section, $\operatorname{\mathcal{LC}}(R)$ is the set of
isomorphism classes of left $R$-ideals, and
$\mu_{R}\colon\operatorname{\mathcal{LC}}(R)\to\operatorname{\mathcal{C}}^{+}(\mathcal{O}_{K}),[I]\mapsto[\operatorname{nr}(I)]$.

######  Proposition .

Let $C\in\mathbb{N}$. Let $\mathfrak{p}\in\max(\mathcal{O}_{K})$ with
$\mathfrak{p}\nmid d_{K}\mathfrak{D}$, and such that

$\frac{h^{+}}{Mw^{2}}(\operatorname{N}_{K/\mathbb{Q}}(\mathfrak{p})+1)-\frac{2}{w}\sqrt{\operatorname{N}_{K/\mathbb{Q}}(\mathfrak{p})}\geq
C.$

Then, for every $c\in\operatorname{\mathcal{LC}}(R)$ with
$\mu_{R}(c)=[\mathfrak{p}]$, there exist at least $C$ maximal left (right)
$R$-ideals of reduced norm $\mathfrak{p}$ and class $c$. Here
$h^{+}=\lvert\operatorname{\mathcal{C}}^{+}(\mathcal{O}_{K})\rvert$ is the
narrow class number, and $w$ and $M$ are constants depending on $\mathfrak{D}$
(see [39, 40]).

###### Proof.

Although not explicitly stated in this way, this is proved by Kirschmer and
Voight in [39, 40]: In the proof of [39, Proposition 7.7], a lower bound on
the entries of a matrix $T^{\prime}$ (in their notation) is derived, which
immediately gives a lower bound on the entries of a matrix $T(\mathfrak{p})$
(in their notation). This is exactly what we need, as is clear from their
definition of $T(\mathfrak{p})$. ∎

_Optimal embeddings:_ Let $L/K$ be a quadratic field extension, and $T$ a
classical Eichler $\mathcal{O}_{K}$-order in $A$ of squarefree level
$\mathfrak{N}\,\triangleleft\,\mathcal{O}_{K}$. If $\mathcal{O}$ is an order
in $L$, every embedding $\iota\colon\mathcal{O}\to T$ gives rise to a unique
embedding $\iota\colon L\to A$, and $\iota$ is an _optimal embedding_ if
$\iota(L)\cap T=\mathcal{O}$. For $a\in T^{\times}$, $\mathcal{O}\to
T,x\mapsto a\iota(x)a^{-1}$ is then another such embedding. The number of
optimal embeddings up to conjugation by units is bounded above by a constant
(depending only on $\mathfrak{D}$ and $\mathfrak{N}$) times $h(\mathcal{O})$
(see [53, Corollaire III.5.12]). Since
$[T^{\times}\colon\mathcal{O}_{K}^{\times}]$ is finite, the total number of
optimal embeddings of $\mathcal{O}$ into $T$ is still bounded by a constant
times $h(\mathcal{O})$.

_Quadratic forms:_ We use a theorem about representation numbers of totally
positive definite quadratic forms over totally real fields. Let $V$ be an
$n$-dimensional $K$-vector space. An $\mathcal{O}_{K}$-lattice $L$ of rank $n$
is a finitely generated $\mathcal{O}_{K}$-submodule of $V$ that generates $V$
(over $K$). Together with a quadratic form $q\colon V\to K$ with
$q(L)\subset\mathcal{O}_{K}$, $(L,q)$ it is a quadratic lattice. For
$a\in\mathcal{O}_{K}$ we set

$r(L,a)=\lvert\\{x\in L\mid q(x)=a\\}\rvert.$

An element $a\in\mathcal{O}_{K}$ is locally represented everywhere by $(L,q)$
if it is represented by the completion
$L_{v}=L\otimes_{\mathcal{O}_{K}}\mathcal{O}_{K,v}$ for all places $v$ of $K$.
The following result is a special case of Theorem 5.1 in [50].

######  Proposition .

Let $(L,q)$ be a quadratic $\mathcal{O}_{K}$-lattice of rank four and suppose
that $q$ is totally positive definite. Then, for every $\eta>0$ and
$s\in\mathbb{N}_{0}$, there exists a constant $C_{\eta,s}>0$, such that for
all $a\in\mathcal{O}_{K}$ that are locally represented everywhere by $L$, with
$\lvert\operatorname{N}_{K/\mathbb{Q}}(a)\rvert$ sufficiently large and
$\mathfrak{p}^{s}\nmid a\mathcal{O}_{K}$ if $L_{\mathfrak{p}}$ is anisotropic,
the asymptotic formula

$r(L,a)\,=\,r(\operatorname{gen}L,a)+O(\lvert\operatorname{N}_{K/\mathbb{Q}}(a)\rvert^{\frac{11}{18}+\varepsilon})$

holds with

$r(\operatorname{gen}L,a)\,\geq\,C_{\eta,s}\cdot\operatorname{N}_{K/\mathbb{Q}}\big{(}a\,({}_{\mathcal{O}_{K}}\langle
q(L)\rangle)^{-1}\big{)}^{1-\eta}.$

In particular, $r(L,a)$ is of order of magnitude
$\lvert\operatorname{N}_{K/\mathbb{Q}}(a)\rvert^{1-\eta}$. If $T$ is any
classical $\mathcal{O}_{K}$-order and $I$ any left $T$-ideal (in particular if
$I=T$), then the restriction of the reduced norm to $I$ makes
$(I,\operatorname{nr}\mid I)$ into a quadratic $\mathcal{O}_{K}$-lattice of
rank four and this is the situation that we will apply this result to.

_Ideal theory in $R$:_ Let $\alpha$ be the set of all classical maximal
$\mathcal{O}_{K}$-orders in $A$ (i.e., the equivalence class of the maximal
order $R$). Conjugation extends to ideals: For $I\in\mathcal{F}_{v}(\alpha)$
define $\overline{I}=\\{\,\overline{x}\mid x\in I\,\\}$. Then $\overline{I}$
is a fractional $(\mathcal{O}_{r}(I),\mathcal{O}_{l}(I))$-ideal,
$I\cdot\overline{I}=\mathcal{O}_{l}(I)\operatorname{nr}(I)$, and
$\overline{I}\cdot I=\mathcal{O}_{r}(I)\operatorname{nr}(I)$, and hence
$I^{-1}=\overline{I}\cdot(\mathcal{O}_{l}(I)\operatorname{nr}(I))^{-1}=(\operatorname{nr}(I)\mathcal{O}_{r}(I))^{-1}\cdot\overline{I}$.

$\mathcal{F}_{v}(\alpha)$ takes a particularly simple form: If
$\mathfrak{p}\mid\mathfrak{D}$ then there exists a maximal two-sided $R$-ideal
$\mathfrak{P}$ with $\mathfrak{P}^{2}=\mathfrak{p}$,
$\operatorname{nr}(\mathfrak{P})=\mathfrak{p}$ and if $I$ is a left or right
$R$-ideal with $\operatorname{nr}(I)=\mathfrak{p}^{k}$, then
$I=\mathfrak{P}^{k}$.

If $\mathfrak{p}\nmid\mathfrak{D}$, then $\mathfrak{P}=\mathfrak{p}R$ is the
maximal two-sided $R$-ideal lying above $\mathfrak{p}$, and
$R_{\mathfrak{p}}/\mathfrak{P}_{\mathfrak{p}}\cong
M_{2}(\mathcal{O}_{K,\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}})\cong
M_{2}(\mathbb{F}_{\operatorname{N}_{K/\mathbb{Q}}(\mathfrak{p})})$. In
particular there are $\operatorname{N}_{K/\mathbb{Q}}(\mathfrak{p})+1$ maximal
left $R$-ideals (respectively maximal right $R$-ideals) with reduced norm
$\mathfrak{p}$. If $M,N$ are two distinct maximal left $R$-ideals with
$\operatorname{nr}(M)=\operatorname{nr}(N)=\mathfrak{p}$, then $M\cap
N=\mathfrak{P}$ (since the composition length of
$M_{2}(\mathbb{F}_{\operatorname{N}_{K/\mathbb{Q}}(\mathfrak{p})})$ is two).
This implies that if $M\cdot M^{\prime}=N\cdot N^{\prime}$ with maximal
integral $M^{\prime},N^{\prime}\in\mathcal{F}_{v}(\alpha)$, then $M\cdot
M^{\prime}=N\cdot N^{\prime}=\mathfrak{P}$, and thus necessarily
$M^{\prime}=\overline{M}$, $N^{\prime}=\overline{N}$.

We therefore explicitly know all relations between maximal integral elements
of $\mathcal{F}_{v}(\alpha)$: From Section 4 we know that all relations are
generated from those between pairs of products of two elements and it also
characterizes the only relation between maximal integral elements of coprime
reduced norm. With the discussion above we now also know the relations between
two maximal integral elements of the same reduced norm: Either there are none,
or the product is $\mathfrak{P}$.

A left $R$-ideal $I$ is _primitive_ if it is not contained in an ideal of the
form $R\mathfrak{a}$ with $\mathfrak{a}\,\triangleleft\,\mathcal{O}_{K}$. If
$\operatorname{nr}(I)=\mathfrak{p}^{k}$ with
$\mathfrak{p}\in\max(\mathcal{O}_{K})$ and $I$ is primitive, then it has a
unique rigid factorization in $\mathcal{I}_{v}(\alpha)$.

###  7.3. Proofs of Section 7.1 and Section 7.1

We start with some lemmas.

######  Lemma .

Let $T$ be a classical $\mathcal{O}_{K}$-order in $A$. For all but finitely
many associativity classes of totally positive prime elements
$q\in\mathcal{O}_{K}$ we have: If $x\in T$ with $\operatorname{nr}(x)=q$ and
$x^{2}=\varepsilon q$ for some $\varepsilon\in T^{\times}$, then
$\varepsilon=-1$.

###### Proof.

$x$ satisfies the polynomial equation
$x^{2}-\operatorname{tr}(x)x+\operatorname{nr}(x)=0$. Substituting
$x^{2}=\varepsilon q$ and $\operatorname{nr}(x)=q$ yields

$\operatorname{tr}(x)x=(1+\varepsilon)q.$ (2)

It will thus suffice to show that for all but finitely many
$\mathcal{O}_{K}q$, we have $\operatorname{tr}(x)=0$.

Assume that $q\in\mathcal{O}_{K}$ is a totally positive prime element, $x\in
T$ with $x^{2}=\varepsilon q$ and $\operatorname{tr}(x)\neq 0$. Then
$K(x)=K(\varepsilon)$ by (2). Let $L=K(\varepsilon)$. Since $\varepsilon\in
L\cap T^{\times}\subset\mathcal{O}_{L}^{\times}$,

$\mathcal{O}_{L}q=\mathcal{O}_{L}\varepsilon q=(\mathcal{O}_{L}x)^{2},$

and therefore $q$ ramifies in $\mathcal{O}_{L}$, implying
$\mathcal{O}_{K}q\mid d_{L/K}$. Hence, for fixed $L$ there are only finitely
many possibilities for $\mathcal{O}_{K}q$, and moreover there are only
finitely many possibilities for $L=K(\varepsilon)$ because
$[T^{\times}:\mathcal{O}_{K}^{\times}]$ is finite since $A$ is totally
definite. 666It should be pointed out that an even stronger statement is true.
For any fixed totally real field $K$, there are only finitely many totally
imaginary quadratic extensions that have larger unit group (i.e., weak unit
defect), while all other totally imaginary quadratic extensions $L/K$ have
$\mathcal{O}_{L}^{\times}=\mathcal{O}_{K}^{\times}$ (i.e., strong unit
defect). This follows from
$[\mathcal{O}_{L}^{\times}:\mu(L)\mathcal{O}_{K}^{\times}]\in\\{1,2\\}$ ([54,
Theorem 4.12]): For $\varepsilon\in\mathcal{O}_{L}^{\times}$ we have
$\varepsilon^{2}=\eta\zeta$ with $\zeta\in\mu(L)$ and
$\eta\in\mathcal{O}_{K}^{\times}$. But
$\operatorname{ord}(\zeta)\mid[L:\mathbb{Q}]=2[K:\mathbb{Q}]$ and if
$\gamma\in(\mathcal{O}_{K}^{\times})^{2}$, then $\eta\gamma\zeta$ yields the
same extension. Since
$[\mathcal{O}_{K}^{\times}:(\mathcal{O}_{K}^{\times})^{2}]<\infty$ there are
therefore only finitely many such extensions. This argument is due to Remak in
[48, §3].  Thus there are, up to associativity, only finitely many such $q$. ∎

######  Lemma .

Let $T$ be a classical $\mathcal{O}_{K}$-order in $A$. For every
$M\in\mathbb{N}$ there exists a $C\in\mathbb{N}$ such that for all totally
positive prime elements $q\in\mathcal{O}_{K}$ with
$q\in\operatorname{nr}_{A_{\mathfrak{p}}/K_{\mathfrak{p}}}(T_{\mathfrak{p}})$
for all $\mathfrak{p}\in\max(\mathcal{O}_{K})$ and
$\operatorname{N}_{K/\mathbb{Q}}(q)\geq C$

$\lvert\\{a\in T\mid\operatorname{nr}(a)=q\text{ and }a^{2}\neq-q\\}\rvert\geq
M.$

###### Proof.

Let $q$ be a totally positive prime element of $\mathcal{O}_{K}$. We derive an
upper bound with order of magnitude
$\sqrt{\operatorname{N}_{K/\mathbb{Q}}(q)}\log(\operatorname{N}_{K/\mathbb{Q}}(q))^{2[K:\mathbb{Q}]-1}$
on the number of elements $a\in T$ with $a^{2}=-q$ (based on counting optimal
embeddings). Comparing this to the lower bound of order of magnitude
$\operatorname{N}_{K/\mathbb{Q}}(q)^{1-\eta}$ for the number of elements $a\in
T$ with $\operatorname{nr}(a)=q$ obtained from Section 7.2 will give the
result.

If $a\in T$ with $\operatorname{nr}(a)=q$ and $a^{2}=-q$, then
$\mathcal{O}_{K}[a]\subset T$ is isomorphic to the order
$\mathcal{O}_{K}[\sqrt{-q}]$ in the relative quadratic extension
$K(\sqrt{-q})$. We determine an upper bound the number of embeddings of
$\mathcal{O}_{K}[\sqrt{-q}]$ into $T$. For this we may without loss of
generality assume that $T$ is a classical Eichler order of squarefree level,
for otherwise we may replace it by a classical Eichler order of squarefree
level in which it is contained (e.g., a classical maximal order), and bound
the number of embeddings there.

Let $L=K(\sqrt{-q})$. Since $f=X^{2}+q$ is the minimal polynomial of
$\sqrt{-q}$ over $K$, we get for the different
$\delta_{L/K}(\sqrt{-q})=f^{\prime}(\sqrt{-q})=2\sqrt{-q}$. Therefore, we find
for the conductor of $\mathcal{O}_{K}[\sqrt{-q}]$ in $\mathcal{O}_{L}$,

$\mathfrak{f}_{\mathcal{O}_{K}[\sqrt{-q}]}=\delta_{L/K}(\sqrt{-q})D_{L/K}^{-1}\;\mid\;2\mathcal{O}_{L}.$

(cf. [43, Proposition 4.12 and Theorem 4.8]). Since
$2\mathcal{O}_{L}\subset\mathcal{O}_{K}[\sqrt{-q}]\subset\mathcal{O}_{L}$ and
$\lvert\mathcal{O}_{L}/2\mathcal{O}_{L}\rvert=2^{[L:\mathbb{Q}]}$, there are
at most $2^{2^{[L:\mathbb{Q}]}}$ orders in $\mathcal{O}_{L}$ that contain
$\mathcal{O}_{K}[\sqrt{-q}]$. For any such order $\mathcal{O}$ with
$\mathcal{O}_{K}[\sqrt{-q}]\subset\mathcal{O}\subset\mathcal{O}_{L}$, we have

$h(\mathcal{O})=h(\mathcal{O}_{L})\frac{\lvert(\mathcal{O}_{L}/\mathfrak{f}_{\mathcal{O}})^{\times}\rvert}{\lvert(\mathcal{O}/\mathfrak{f}_{\mathcal{O}})^{\times}\rvert}\,\leq\,h(\mathcal{O}_{L})2^{[L:\mathbb{Q}]}$

(cf. [44, §I.12.9 and §I.12.11]). The number of optimal embeddings of
$\mathcal{O}$ into $T$ is bounded by a constant times $h(\mathcal{O})$, and
hence the total number of embeddings of $\mathcal{O}_{K}[\sqrt{-q}]$ into $T$
is bounded by a constant times $h(\mathcal{O}_{L})$, where the constant does
not depend on $q$. Combining the upper bound

$h(\mathcal{O}_{L})\,\ll\,\sqrt{\lvert d_{L}\rvert}\,\log(\lvert
d_{L}\rvert)^{[L:\mathbb{Q}]-1}$

(cf. [43, Theorem 4.4]), with

$d_{L}=\operatorname{N}_{L/\mathbb{Q}}(D_{L/\mathbb{Q}})=\operatorname{N}_{L/\mathbb{Q}}(D_{K/\mathbb{Q}})\operatorname{N}_{L/\mathbb{Q}}(D_{L/K})\leq
d_{K}^{2}2^{[L:\mathbb{Q}]}\lvert\operatorname{N}_{L/\mathbb{Q}}(\sqrt{-q})\rvert=d_{K}^{2}2^{[L:\mathbb{Q}]}\operatorname{N}_{K/\mathbb{Q}}(q).$

(here $D_{L/K}\mid 2\sqrt{-q}\mathcal{O}_{L}$ was used), we obtain

$h(\mathcal{O}_{L})\,\ll\,\sqrt{\operatorname{N}_{K/\mathbb{Q}}(q)}\,\log{(\operatorname{N}_{K/\mathbb{Q}}(q))}^{[L:\mathbb{Q}]-1},$

and thus an upper bound of the same order for $\lvert\\{a\in
T\mid\operatorname{nr}(a)=q\text{ and }a^{2}=-q\\}\rvert$.

By Section 7.2, for every $\eta>0$ and sufficiently large (in norm) $q$ with
$q$ being locally represented everywhere by the norm form, $\lvert\\{a\in
T\mid\operatorname{nr}(a)=q\\}\rvert$ grows with order of magnitude
$\operatorname{N}_{K/\mathbb{Q}}(q)^{1-\eta}$, and the claim follows, by
choosing $\eta$ small enough, say $\eta<\frac{1}{4}$. ∎

######  Remark .

For any classical $\mathcal{O}_{K}$-order $T$ of $A$ there are infinitely many
pairwise non-associated totally positive primes $q\in\mathcal{O}_{K}$ that are
locally represented everywhere by $\operatorname{nr}_{A/K}$ on $T$. This can
easily be seen as follows: Let
$\mathfrak{D}\mathfrak{N}\,\triangleleft\,\mathcal{O}_{K}$ be the discriminant
of $T$. If $\mathfrak{p}\in\max(\mathcal{O}_{K})$ with
$\mathfrak{p}\nmid\mathfrak{D}\mathfrak{N}$, then $T_{\mathfrak{p}}\cong
M_{2}(\mathcal{O}_{K,\mathfrak{p}})$ and thus
$\operatorname{nr}_{A_{\mathfrak{p}}/K_{\mathfrak{p}}}(T_{\mathfrak{p}})=\mathcal{O}_{K,\mathfrak{p}}$.
If $\mathfrak{p}\mid\mathfrak{D}\mathfrak{N}$, since
$\operatorname{center}(T_{\mathfrak{p}})=\mathcal{O}_{K,\mathfrak{p}}$,
certainly still every square of $\mathcal{O}_{K,\mathfrak{p}}$ is represented
by $\operatorname{nr}_{A_{\mathfrak{p}}/K_{\mathfrak{p}}}$ on
$T_{\mathfrak{p}}$ (in fact, if $\mathfrak{p}\mid\mathfrak{D}$ but
$\mathfrak{p}\nmid\mathfrak{N}$ then $T_{\mathfrak{p}}$ is isomorphic to the
unique classical maximal $\mathcal{O}_{K,\mathfrak{p}}$-order in the unique
quaternion division algebra over $K_{\mathfrak{p}}$, for which
$\operatorname{nr}(T_{\mathfrak{p}})=\mathcal{O}_{K,\mathfrak{p}}$ also
holds). By Hensel’s Lemma therefore every totally positive prime element
$q\in\mathcal{O}_{K}$ with $q\equiv 1\mod 4\mathfrak{D}\mathfrak{N}$ is
locally represented everywhere by $\operatorname{nr}_{A/K}$ on $T$. But there
are infinitely many pairwise non-associated such primes, because every class
of the ray class group
$\operatorname{\mathcal{C}}_{4\mathfrak{D}\mathfrak{N}}^{+}(\mathcal{O}_{K})$
contains infinitely many pairwise distinct maximal ideals, and primes $q$ of
the required form correspond exactly to the trivial class in
$\operatorname{\mathcal{C}}_{4\mathfrak{D}\mathfrak{N}}^{+}(\mathcal{O}_{K})$.

######  Lemma .

Let $q$ be a totally positive prime element of $\mathcal{O}_{K}$. Let $I$ be a
non-principal right $R$-ideal with $\operatorname{nr}(I)=q^{m}\mathcal{O}_{K}$
for some $m\in\mathbb{N}$, and $J$ be a left $S=\mathcal{O}_{l}(I)$-ideal with
$\operatorname{nr}(J)=q^{n}\mathcal{O}_{K}$ for some $n\in\mathbb{N}$ such
that: $I\cong J$ (as left $S$-ideals) and $I$ (respectively $J$) is not
contained in any principal left $S$-ideal except $S$ itself, and not contained
in any principal right $\mathcal{O}_{r}(I)$-ideal (respectively right
$\mathcal{O}_{r}(J)$-ideal) except $\mathcal{O}_{r}(I)$ (respectively
$\mathcal{O}_{r}(J)$) itself.

Assume further that $a\in S$ with $\operatorname{nr}(a)=q$ and $a^{2}S\neq
qS$.

  1. 1.

For all $l\in\mathbb{N}$, $(a^{l}\overline{J}a^{-l})a^{l}I$ is a principal
right $R$-ideal and an atom of $\mathcal{H}_{R^{\bullet}}$. In particular,
$a^{l}q^{m}\in\mathcal{A}(R^{\bullet})$ for all $l\in\mathbb{N}$.

  2. 2.

$\overline{J}I\in\mathcal{A}(\mathcal{H}_{R^{\bullet}})$ if it is primitive.
In particular if $m=n=1$ and $I\neq J$, then
$\overline{J}I\in\mathcal{A}(\mathcal{H}_{R^{\bullet}})$.

###### Proof.

Since $I$ is not contained in any principal right $R$-ideal, it is in
particular not contained in $qR$, hence primitive. Similarly, $J$ is
primitive. Let
$M_{1}*\ldots*M_{m}\in\mathsf{Z}^{*}_{\mathcal{I}_{v}(\alpha)}(I)$ and
$N_{1}*\ldots*N_{n}\in\mathsf{Z}^{*}_{\mathcal{I}_{v}(\alpha)}(J)$, with
$M_{1},\ldots,M_{m},N_{1},\ldots,N_{n}\in\mathcal{M}_{v}(\alpha)$, be the
unique rigid factorizations of $I$ and $J$.

  1. 1.

Since $I\cong J$ as left $S$-ideals,
$(a^{l}\overline{J}a^{-l})a^{l}I=a^{l}\overline{J}I$ is principal. A rigid
factorization of it is given by

$(a^{l}\overline{N_{n}}a^{-l})*\ldots*(a^{l}\overline{N_{1}}a^{-l})*(a^{l}Sa^{-l})a*(a^{l-1}Sa^{-(l-1)})a*\ldots*(aSa^{-1})a*M_{1}*\ldots*M_{m}$

with
$M_{i},\,a^{l}\overline{N_{j}}a^{-l},\,(a^{l-k}Sa^{-(l-k)})a\in\mathcal{M}_{v}(\alpha)$
for $i\in[1,m]$, $j\in[1,n]$ and $k\in[0,l-1]$. By the restrictions imposed on
$I$, $J$ and $a$, this is the only rigid factorization of
$a^{l}\overline{J}I$. Since any non-empty proper subproduct starting from the
left (or the right) is non-principal, it is an atom in
$\mathcal{H}_{R^{\bullet}}$. The “in particular” statement follows by setting
$J=I$, as then
$a^{l}\overline{J}I=a^{l}q^{m}R\in\mathcal{A}(\mathcal{H}_{R^{\bullet}})$ and
because of Section 5, therefore $a^{l}q^{m}\in\mathcal{A}(R^{\bullet})$.

  2. 2.

By primitivity,

$\overline{N_{n}}*\ldots*\overline{N_{1}}*M_{1}*\ldots*M_{m}\in\mathsf{Z}^{*}_{\mathcal{I}_{v}(\alpha)}(\overline{J}I)$

is the unique rigid factorization of $\overline{J}I$, and since as before no
non-empty proper subproduct from the left (or the right) is principal, it is
an atom in $\mathcal{H}_{R^{\bullet}}$. For the “in particular” statement,
note that if $m=n=1$ (i.e., $I$ and $J$ are both maximal left $S$-ideals),
then $\overline{J}I=qR$ if and only if $I=J$, and otherwise $\overline{J}I$ is
necessarily primitive.

∎

######  Lemma .

Let $I$ be a left $R$-ideal, $S=\mathcal{O}_{r}(I)$, and $a\in R\cap S$. Then

$\prod_{i=1}^{l}(a^{l-i+1}Ra^{-(l-i+1)})a\cdot
I=a^{l}I=(a^{l}Ia^{-l})a^{l}=(a^{l}Ia^{-l})\cdot\prod_{i=1}^{l}(a^{l-i+1}Sa^{-(l-i+1)})a$

with the left-most and the right-most expressions being proper products of

$(a^{l-i+1}Ra^{-(l-i+1)})a,\;I,\;(a^{l-i+1}Sa^{-(l-i+1)})a,\;a^{l}Ia^{-l}\,\in\,\mathcal{I}_{v}(\alpha).$

(The products have to be read in ascending order with “$i=1$” to the very
left.)

###### Proof.

The formulas are clear, and so is that the products are proper ones. The key
point is that these one-sided ideals are indeed integral. But this is so
because $a\in S$, hence $a\in a^{k}Sa^{-k}$ for all $k\in\mathbb{N}_{0}$,
implying that $(a^{k}Sa^{-k})a\in\mathcal{I}_{v}(\alpha)$, and similarly $a\in
R$, thus $(a^{k}Ra^{-k})a\in\mathcal{I}_{v}(\alpha)$. ∎

######  Lemma .

Let $M$ be a maximal left $R$-ideal, and $N$ a maximal left
$\mathcal{O}_{r}(M)$-ideal. If $M\cdot N=N^{\prime}\cdot M^{\prime}$, then
$\overline{M}\cdot N^{\prime}=N\cdot\overline{M^{\prime}}$.

###### Proof.

Since
$\mathcal{O}_{r}(\overline{M})=\mathcal{O}_{l}(M)=\mathcal{O}_{l}(N^{\prime})$
and
$\mathcal{O}_{r}(N)=\mathcal{O}_{r}(M^{\prime})=\mathcal{O}_{l}(\overline{M^{\prime}})$
the product is proper. We have

$\overline{M}\cdot N^{\prime}\cdot M^{\prime}=\overline{M}\cdot M\cdot
N=\operatorname{nr}(M)\mathcal{O}_{l}(N)\cdot
N=N\cdot\mathcal{O}_{r}(N)\operatorname{nr}(M)=N\cdot\overline{M^{\prime}}\cdot
M^{\prime},$

and thus $\overline{M}\cdot N^{\prime}=N\cdot\overline{M^{\prime}}$. ∎

###### Proof of Section 7.1.

Let $p\in\mathcal{O}_{K}$ be a totally positive prime element with
$p\mathcal{O}_{K}\nmid d_{K}\mathfrak{D}\mathfrak{N}$ and with
$\operatorname{nr}(p)$ satisfying the bound of Section 7.2 for the classical
maximal order $R$ (with $C=1$). Then there exists a maximal right $R$-ideal
$U$ with $\operatorname{nr}(U)=p\mathcal{O}_{K}$ that is non-free (i.e., non-
principal) but is stably free (i.e., $[\operatorname{nr}(U)]=\mathbf{0}$ in
$\operatorname{\mathcal{C}}^{+}(\mathcal{O}_{K})$). Let $U=U_{0},\ldots,U_{r}$
be the maximal left $\mathcal{O}_{l}(U)$-ideals of reduced norm
$p\mathcal{O}_{K}$ (of which there are
$r+1=\operatorname{N}_{K/\mathbb{Q}}(p)+1$). By Section 7.3, there exists a
totally positive prime element $q\in\mathcal{O}_{K}$, $q\mathcal{O}_{K}\nmid
pd_{K}\mathfrak{D}\mathfrak{N}$, and an element

$a\,\in\,\mathcal{O}_{l}(U)\,\cap\,\bigcap_{j=0}^{r}\mathcal{O}_{r}(U_{j})\quad\text{with}\quad\operatorname{nr}(a)=q\text{
and }a^{2}\neq-q,$

and in fact, by Section 7.3, we may make this choice such that
$a^{2}\neq\varepsilon q$ for any $\varepsilon\in R^{\times}$. In addition, we
may take $\operatorname{N}_{K/\mathbb{Q}}(q)$ to be sufficiently large to
satisfy the bound of Section 7.2 for $\mathcal{O}_{l}(U)$ (with $C=2$). Then
there exist distinct left $\mathcal{O}_{l}(U)$-ideals $I$ and $J$ such that
$I\cong J\cong U$ and
$\operatorname{nr}(I)=\operatorname{nr}(J)=q\mathcal{O}_{K}$.

Set $S=\mathcal{O}_{r}(I)$, and observe that $S\cong R$, because $U\cong I$.
By Section 7.3,
$(a^{l}\overline{J}a^{-l})a^{l}I\in\mathcal{A}(\mathcal{H}_{S^{\bullet}})$ for
all $l\in\mathbb{N}_{0}$, say $(a^{l}\overline{J}a^{-l})a^{l}I=y_{l}S$ with
$y_{l}\in\mathcal{A}(S^{\bullet})$. We consider the principal right $S$-ideal
$X_{l}=(a^{l}\overline{J}a^{-l})a^{l}Ip\subset S$, say $X_{l}=x_{l}S$ with
$x_{l}\in S^{\bullet}$. We will first determine all possible rigid
factorizations of $X_{l}$ in $\mathcal{I}_{v}(\alpha)$. As in Section 7.3, the
right $S$-ideal $(a^{l}\overline{J}a^{-l})a^{l}I$ has reduced norm
$q^{l+2}\mathcal{O}_{K}$, is primitive, and thus possesses a unique rigid
factorization,

$(a^{l}\overline{J}a^{-l})*(a^{l}\mathcal{O}_{l}(I)a^{-l})a*(a^{l-1}\mathcal{O}_{l}(I)a^{-(l-1)})a*\ldots*(a\mathcal{O}_{l}(I)a^{-1})a*I\;\in\;\mathsf{Z}^{*}_{\mathcal{I}_{v}(\alpha)}((a^{l}\overline{J}a^{-l})a^{l}I),$

with the $l+2$ factors
$a^{l}\overline{J}a^{-l},\,(a^{l-k}\mathcal{O}_{l}(I)a^{-(l-k)})a$ for
$k\in[0,l-1]$ and $I$ all in $\mathcal{M}_{v}(\alpha)$.

For an element with a unique rigid factorization we make the convention of
identifying the element and its factorization when this is notationally
convenient. For principal ideals we omit the order and only write the
generator if it is clear from the neighboring elements in the factorization
what the order must be. For example, we can write the previous rigid
factorization as $a^{l}\overline{J}a^{-l}*a^{l}*I$.

$X_{l}$ has $(r+1)\binom{l+4}{2}$ rigid factorizations: They arise from the
different rigid factorizations
$U_{i}*\overline{U_{i}}\in\mathsf{Z}^{*}_{\mathcal{I}_{v}(\alpha)}(\mathcal{O}_{l}(U)p)$
for $i\in[0,r]$ and the possible transpositions of $U_{i}$ and
$\overline{U_{i}}$. We denote the rigid factorization of $X_{l}$ that arises
from $a^{l}Ja^{-l}*a^{l}*U_{i}*\overline{U_{i}}*I$ by transposing the one-
sided ideals of norm $p$ to the positions $m\in[-1,l+1]$ and $n\in[m,l+1]$ in
the factorization by $F_{i,m,n}$: Here, the left-most position in the rigid
factorization is denoted by $-1$, the right-most by $l+1$. So, by “the rigid
factorization obtained by transposing $U_{i}$ to the position $-1$ and
$\overline{U_{i}}$ to $l+1$” we mean the unique rigid factorization of $X_{l}$
that has a factor of norm $p\mathcal{O}_{K}$ as the first factor and as the
last factor, and that can be transformed into
$a^{l}\overline{J}a^{-l}*a^{l}*U_{i}*\overline{U_{i}}*I$ by transposition of
maximal integral elements with coprime norm.

For $i\in[0,r]$ let $V_{i}\in\mathcal{M}_{v}(\alpha)$ and
$M_{i}\in\mathcal{M}_{v}(\alpha)$ be defined by
$\overline{U_{i}}I=M_{i}\overline{V_{i}}$ under transposition, and let
$W_{i}\in\mathcal{M}_{v}(\alpha)$ and $N_{i}\in\mathcal{M}_{v}(\alpha)$ be
defined by $W_{i}\overline{N_{i}}=\overline{J}U_{i}$ under transposition.
($\\{V_{i}\mid i\in[0,r]\\}$ is then the set of all $r+1$ left
$S=\mathcal{O}_{r}(I)$-ideals of reduced norm $p$. Similarly $\\{W_{i}\mid
i\in[0,r]\\}$ is then the set of all $r+1$ left $\mathcal{O}_{r}(J)$-ideals of
reduced norm $p$, and since $\mathcal{O}_{r}(I)\cong\mathcal{O}_{r}(J)$ the
sets are actually conjugate under conjugation by an element of $A^{\times}$.)
By Section 7.3 we then also have $U_{i}M_{i}=IV_{i}$ and
$\overline{W_{i}}\,\overline{J}=\overline{N_{i}}\,\overline{U_{i}}$ under
transposition. Using Section 7.3 to see that $a$ transposes “nicely” with
$U_{i}$ and $\overline{U_{i}}$, we can explicitly describe all $F_{i,m,n}$ as
follows:

  1. (Case 1)

If $m=n=-1$:

$F_{i,m,n}=a^{l}W_{i}a^{-l}*a^{l}\overline{W_{i}}a^{-l}*a^{l}\overline{J}a^{-l}*a^{l}*I.$

  2. (Case 2)

If $m=-1$ and $0\leq n\leq l$:

$F_{i,m,n}=a^{l}W_{i}a^{-l}*a^{l}\overline{N}_{i}a^{-l}*a^{n}*a^{l-n}\overline{U_{i}}a^{-(l-n)}*a^{l-n}*I.$

  3. (Case 3)

If $0\leq m\leq n\leq l$:

$F_{i,m,n}=a^{l}\overline{J}a^{-l}*a^{m}*a^{l-m}U_{i}a^{-(l-m)}*a^{n-m}*a^{l-n}\overline{U_{i}}a^{-(l-n)}*a^{l-n}*I.$

  4. (Case 4)

If $m=-1$ and $n=l+1$:

$F_{i,m,n}=a^{l}W_{i}a^{-l}*a^{l}\overline{N_{i}}a^{-l}*a^{l}*M_{i}*\overline{V_{i}}.$

  5. (Case 5)

If $0\leq m\leq l$ and $n=l+1$:

$F_{i,m,n}=a^{l}\overline{J}a^{-l}*a^{m}*a^{l-m}U_{i}a^{-(l-m)}*a^{l-m}*M_{i}*\overline{V_{i}}.$

  6. (Case 6)

If $m=n=l+1$:

$F_{i,m,n}=a^{l}\overline{J}a^{-l}*a^{l}*I*V_{i}*\overline{V_{i}}.$

For each of these rigid factorizations of the ideal $X_{l}$ in
$\mathcal{I}_{v}(\alpha)$ we can form minimal subproducts of principal one-
sided ideals (starting from the left or the right) to obtain a representation
of $X_{l}$ as a product in $\mathcal{H}_{S^{\bullet}}$ (and hence a
representation of $x_{l}$ as a product of elements of $S^{\bullet}$). But only
when each of these minimal principal subproducts is an atom of
$\mathcal{H}_{S^{\bullet}}$ this gives rise to an actual rigid factorization
of $x_{l}$ into atoms. We discuss the individual cases one-by-one:

  1. Case 1.

If $m=n=-1$: If $W_{i}$ is non-principal, then this does not give rise to a
rigid factorization into atoms, as the first principal factor is
$a^{l}(W_{i}\overline{W_{i}})a^{-l}=a^{l}(p\mathcal{O}_{r}(J))a^{-l}$, and
this is not an atom (since there is at least one element in $\\{W_{i}\mid
i\in[0,r]\\}$ that is principal by Section 7.2). If on the other hand $W_{i}$
is principal, then this gives rise to a rigid factorization of $X_{l}$ in
$\mathcal{H}_{S^{\bullet}}$ of length $3$, with atomic factors
$a^{l}W_{i}a^{-l}$, $a^{l}\overline{W_{i}}a^{-l}$ and
$(a^{l}\overline{J}a^{-l})a^{l}I$, which in turn gives rise to a rigid
factorization of length $3$ of $x_{l}\in S$.

  2. Case 2.

If $m=-1$ and $0\leq n\leq l$:

    1. Case 2a

If $U_{i}\cong I$: Then the last principal factor is necessarily
$(a^{l-n}\overline{U_{i}}a^{-(l-n)})a^{l-n}I$. If $n<l$, then transposing
$\overline{U_{i}}$ to the right shows that this is not an atom in
$\mathcal{H}_{S^{\bullet}}$. If $n=l$ then
$\overline{U_{i}}I=M_{i}\overline{V_{i}}$ is an atom if and only if $V_{i}$ is
non-principal. Since also $U_{i}\cong J$, the factor
$W_{i}\overline{N_{i}}=\overline{J}U_{i}$ is principal, and, because
$\overline{J}$ and $\overline{U_{i}}$ are non-principal, this is either an
atom (if $W_{i}$ is non-principal), or a product of two atoms (if $W_{i}$ is
principal). So if $V_{i}$ is non-principal we get a rigid factorization of
length either $l+2$ or $l+3$, and if $V_{i}$ is principal we get no rigid
factorization into atoms.

    2. Case 2b

If $U_{i}\not\cong I$: Then either there are no non-trivial principal factors
(if $W_{i}$ is non-principal), or the first factor is $W_{i}$ and the
remaining product does not factor into non-trivial principal factors. But then
this second factor is not an atom, because after transposition of
$\overline{U_{i}}$ to the very left of the second factor (i.e, position $0$),
we have a principal factor $\overline{W_{i}}$. So in any case, this does not
give rise to a rigid factorization into atoms.

  3. Case 3.

If $0\leq m\leq n\leq l$: If $I\not\cong U_{i}$, then there are no non-trivial
principal factors, and hence no rigid factorization into atoms is obtained.

If $I\cong U_{i}$, then the first principal factor is
$(a^{l}\overline{J}a^{-l})a^{m}(a^{l-m}U_{i}a^{-(l-m)})$, and the last one is
$(a^{l-n}\overline{U_{i}}a^{-(l-n)})a^{l-n}I$. If $m>0$ (or $n<l$), then by
transposing $U_{i}$ to the left in the first factor (or $\overline{U_{i}}$ to
the right in the second factor) once, we see that this does not give rise to a
rigid factorization into atoms. Consider now $m=0$ and $n=l$. If $V_{i}$ is
principal, then $\overline{J}U_{i}=V_{i}\overline{K_{i}}$ implies that the
first factor $a^{l}\overline{J}U_{i}a^{-l}$ is no atom, and hence again we get
no rigid factorization into atoms. Analogously we get no rigid factorization
into atoms if $W_{i}$ is principal. If on the other hand $V_{i}$ and $W_{i}$
are both non-principal then $\overline{J}U_{i}$ is an atom, and so is
$\overline{U_{i}}I$. Thus we obtain a rigid factorization of $X_{l}$ (and
hence of $x_{l}$) of length $l+2$. (It is then in fact the same one as the one
obtained from Case 2 in the same situation.)

  4. Case 4.

If $m=-1$ and $n=l+1$:

    1. Case 4a

If $U_{i}\cong I$: Then $W_{i}\overline{N_{i}}$ and $M_{i}\overline{V_{i}}$
are both principal. If $W_{i}$ is non-principal, then
$W_{i}\overline{N_{i}}=\overline{J}U_{i}$ is an atom since $J$ is non-
principal. If $W_{i}$ is principal, then $W_{i}\overline{N_{i}}$ is a product
of two atoms. Similarly, $M_{i}\overline{V_{i}}$ is either an atom or a
product of two atoms. So in this case we get a rigid factorization of length
$l+2$, $l+3$ or $l+4$. (In the case that $V_{i}$ is non-principal, it is the
same one as in Case 2. In the case that $V_{i}$ and $W_{i}$ are both non-
principal it is the same as in Case 3 in the same situation.)

    2. Case 4b

If $U_{i}\not\cong I$: Then $W_{i}\overline{N_{i}}$ and
$M_{i}\overline{V_{i}}$ are both non-principal. If $W_{i}$ is principal, but
$\overline{V_{i}}$ is not, then the second principal factor is necessarily
$(a^{l}\overline{N_{i}}a^{-l})a^{l}M_{i}\overline{V_{i}}$, and this cannot be
split as a non-trivial product of principal factors. But transposing
$\overline{V_{i}}$ to the very left in this factor gives a principal factor
$\overline{W_{i}}$, hence
$(a^{l}\overline{N_{i}}a^{-l})a^{l}M_{i}\overline{V_{i}}$ is not an atom.
Arguing analogously, if $\overline{V_{i}}$ is principal but $W_{i}$ is not, no
rigid factorization into atoms is obtained.

Finally, if $W_{i}$ and $V_{i}$ are both principal, we get a rigid
factorization into $3$ atoms.

  5. Case 5.

If $0\leq m\leq l$ and $n=l+1$: This is analogous to Case 2.

  6. Case 6.

If $m=n=l+1$: This is analogous to Case 1.

Since there is at least one $i\in[0,r]$ for which $W_{i}$ is principal, we get
at least one rigid factorization of $x_{l}$ of length $3$ from Case 1. For
$i=0$, $U_{i}\cong I$, so Case 4 gives at least one factorization with length
in $[l+2,l+4]$. Note that which of the lengths in $[l+2,l+4]$ occur in Case 2,
Case 3, and Case 4 depends only on the principality of certain one-sided
ideals, and not on $l$. Thus we have shown that there exists a set
$\emptyset\neq E\subset\\{\,2,3,4\,\\}$ such that, for any choice of
$l\in\mathbb{N}_{0}$,

$\mathsf{L}_{S^{\bullet}}(x_{l})=\\{\,3\,\\}\;\cup\;(l+E),$

and $x_{l}$ has the claimed form $x_{l}=y_{l}p$ with
$y_{l}\in\mathcal{A}(S^{\bullet})$ and $p$ a totally positive prime element of
$\mathcal{O}_{K}$. Since $S\cong R$, the same is true for $R$. ∎

######  Remark .

  1. 1.

In the proof, the classical $\mathcal{O}_{K}$-order

$T=\mathcal{O}_{l}(U)\,\cap\,\bigcap_{j=0}^{r}\mathcal{O}_{r}(U_{j})$

is maximal at every prime $\mathfrak{r}\in\max(\mathcal{O}_{K})$ with
$\mathfrak{r}\neq p\mathcal{O}_{K}$ (thus $T_{\mathfrak{r}}\cong
M_{2}(\mathcal{O}_{K,\mathfrak{r}})$ if $\mathfrak{r}\nmid p\mathfrak{D}$ and
$T_{\mathfrak{r}}$ is isomorphic to the unique classical maximal
$\mathcal{O}_{K,\mathfrak{r}}$-order in the unique quaternion division algebra
over $K_{\mathfrak{r}}$ if $\mathfrak{r}\mid\mathfrak{D}$). At
$\mathfrak{p}=p\mathcal{O}_{K}$, it is not hard too see by local calculations
that

$T_{\mathfrak{p}}\cong\left\\{\begin{pmatrix}a&b\\\
p^{2}c&a+pd\end{pmatrix}\;\bigg{\lvert}\;a,b,c,d\in\mathcal{O}_{K,\mathfrak{p}}\right\\}.$

But $T_{\mathfrak{p}}$ is not a classical Eichler order, and so neither is
$T$.

  2. 2.

If $\mathfrak{r}\in\max(\mathcal{O}_{K})$ we can find infinitely many pairwise
non-associated totally positive prime elements $q\in\mathcal{O}_{K}$ such that
$\mathfrak{r}$ splits in $K(\sqrt{-q})$, and infinitely many pairwise non-
associated totally positive prime elements $q\in\mathcal{O}_{K}$ such that
$\mathfrak{r}$ is inert in $K(\sqrt{-q})$: We may restrict ourselves to $q$
with $\operatorname{N}_{K/\mathbb{Q}}(q)$ odd, and
$q\mathcal{O}_{K}\neq\mathfrak{r}$. Let
$\mathfrak{r}^{\prime}=\mathfrak{r}^{1+\mathsf{v}_{\mathfrak{r}}(4)}$. If
$-q\equiv 1\mod\mathfrak{r}^{\prime}$, then $-q$ is a square in
$\mathcal{O}_{K}/\mathfrak{r}^{\prime}$ and hence $\mathfrak{r}$ splits in
$K(\sqrt{-q})$. If $-q\equiv a\mod\mathfrak{r}^{\prime}$, with $a$ a non-
square in $\mathcal{O}_{K}/\mathfrak{r}^{\prime}$, then $\mathfrak{r}$ is
inert in $K(\sqrt{-q})$. It therefore suffices to show that in every class of
$(\mathcal{O}_{K}/\mathfrak{r}^{\prime})^{\times}$ there are infinitely many
pairwise non-associated totally positive prime elements of $\mathcal{O}_{K}$.
Since we have the exact sequence

$\mathbf{1}\to\mathcal{O}_{K}^{\times,+}/\\{x\in\mathcal{O}_{K}^{\times,+}\mid
x\equiv_{\mathfrak{r}^{\prime}}1\\}\to(\mathcal{O}_{K}/\mathfrak{r}^{\prime})^{\times}\to\operatorname{\mathcal{C}}_{\mathfrak{r}^{\prime}}^{+}(\mathcal{O}_{K})\to\operatorname{\mathcal{C}}^{+}(\mathcal{O}_{K})\to\mathbf{1},$

(cf. [43, Lemma 3.2], [44, Exercises VI.1.12, VI.1.13]), it suffices that
every class in the kernel of
$\operatorname{\mathcal{C}}_{\mathfrak{r}^{\prime}}^{+}(\mathcal{O}_{K})\to\operatorname{\mathcal{C}}^{+}(\mathcal{O}_{K})$
contains infinitely many pairwise non-associated prime elements. But this is
so, because in fact every class of
$\operatorname{\mathcal{C}}_{\mathfrak{r}^{\prime}}^{+}(\mathcal{O}_{K})$
contains infinitely many pairwise distinct maximal ideals (cf. [43, Corollary
7 to Proposition 7.9]).

  3. 3.

Using the previous observation to find a suitable element $a$ in the proof, we
can replace Section 7.3 by a simpler one if $\mathfrak{D}\neq\mathcal{O}_{K}$:
Choosing the totally positive prime element $q\in\mathcal{O}_{K}$ such that a
prime divisor $\mathfrak{r}\mid\mathfrak{D}$ splits in $K(\sqrt{-q})$, the
field $K(\sqrt{-q})$ does not embed into $A$ at all (see e.g. [53, Theoreme
III.3.8] or [41, Theorem 7.3.3]).

If $\mathfrak{D}=\mathcal{O}_{K}$, we may make use of the fact that the
particular classical order $T$ in the proof is contained in a classical
Eichler order of squarefree level $\mathfrak{p}$. Taking $q$ such that
$\mathfrak{p}$ is inert in $K(\sqrt{-q})$, the formulas for counting optimal
embeddings ([53, Corollaire III.5.12]) show that no order of $K(\sqrt{-q})$
embeds into $T$.

For this approach we only need the qualitative statement of Section 7.2, but
not the order of magnitude.

###### Proof of Section 7.1.

It suffices to prove the claim for $n=1$. Let $a$ in $R^{\bullet}$, and let

$\\{I_{1}^{(1)}*\ldots*I_{k}^{(1)},\;\ldots,\;I_{1}^{(l)}*\ldots*I_{k}^{(l)}\\}=\mathsf{Z}^{*}_{\mathcal{I}_{v}(\alpha)}(Ra)\subset\mathcal{F}(\mathcal{M}_{v}(\alpha))$

be the set of all rigid factorizations of $Ra$ in $\mathcal{I}_{v}(\alpha)$.
Using Section 7.2, we can choose a totally positive prime element
$q\in\mathcal{O}_{K}$ with $q\nmid\operatorname{nr}(a)$ and such that there
exists an $x\in A^{\times}$ with $\operatorname{nr}(x)=q$ and

$x\;\in\;T=\bigcap_{i=1}^{l}\bigcap_{j=1}^{k}\mathcal{O}_{l}(I_{j}^{(i)})\cap\mathcal{O}_{r}(I_{j}^{(i)}).$

We claim $\mathsf{L}(xa)=1+\mathsf{L}(a)$. The rigid factorizations of $Rxa$
in $\mathcal{I}_{v}(\alpha)$ are given by all possible transpositions of $x$
to any position in

$I_{1}^{(i)}*\ldots*I_{m}^{(i)}*x,$

for all $i\in[1,l]$. But, since $x\in T$, it follows from Section 7.3 that any
such rigid factorization is of the form

$I_{1}^{(i)}*\ldots*I_{m}^{(i)}*x*x^{-1}I_{m+1}^{(i)}x*\ldots*x^{-1}I_{k}^{(i)}x.$

for $m\in[0,k]$. We see that for the principal subproducts this does not
change anything except insert one additional factor (corresponding to $x$) at
some position. Thus, for each $i\in[1,l]$,
$I_{1}^{(i)}*\ldots*I_{m}^{(i)}*x*x^{-1}I_{m+1}^{(i)}x*\ldots*x^{-1}I_{k}^{(i)}x$
gives rise to a rigid factorization of $ax$ in $R^{\bullet}$ of length $l+1$
if and only if $I_{1}^{(i)}*\ldots*I_{k}^{(i)}$ gives rise to a rigid
factorization of $a$ of length $l$. ∎

##  8\. Acknowledgments

I thank Alfred Geroldinger for suggesting the topic, and him and Franz Halter-
Koch for reading preliminary versions of this manuscript and providing many
valuable comments.

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# Emmy’s letter to Santa Claus (and a reply):
Vaidya geometries and scalar fields with null gradients

Valerio Faraoni, Andrea Giusti, and Bardia H. Fahim

Department of Physics & Astronomy, Bishop’s University
2600 College Street, Sherbrooke Québec, Canada J1M 1Z7

###### Abstract

Emmy, a child prodigy, writes a letter to Santa Claus asking for a Vaidya
solution sourced by a massless scalar field with null gradient for Christmas.
Santa writes back explaining why such a toy cannot be built and offers an
exact plane wave sourced by a scalar field with lightlike gradient as a
replacement.

##  1 Emmy’s letter to Santa Claus

Dear Santa,
Can I please have a Vaidya solution of the Einstein equations $G_{ab}=8\pi
T_{ab}$ sourced by a scalar field $\phi$ with null gradient this Christmas? I
really would like to play with it. My friends in string theory and holography
class play with Vaidya toys [1, 2]. Being just a first grader, I play with
simple scalar fields instead, but I like null vectors, null dust sprinkled all
over spacetime, and flashy surfaces moving at the speed of light.

As you certainly remember, an outgoing Vaidya solution [3, 4] (see also [5,
6]) is given by the line element111We follow the notation of Ref. [7] and use
units in which the speed of light and Newton’s constant are unity.

$ds^{2}=-\left(1-\frac{2m(u)}{r}\right)du^{2}-2dudr+r^{2}d\Omega_{(2)}^{2}\,,$
(1)

where $u$ is a retarded null coordinate, $m(u)$ is a regular mass function,
and $d\Omega_{(2)}^{2}\equiv d\vartheta^{2}+\sin^{2}\vartheta\,d\varphi^{2}$
is the line element on the unit 2-sphere. The only non-vanishing component of
the Ricci tensor is

$R_{uu}=-\,\frac{2m^{\prime}(u)}{r^{2}}\,,$ (2)

where a prime denotes differentiation with respect to $u$. The stress-energy
tensor of the null dust sourcing the Vaidya geometry is

$T_{ab}=-\,\frac{m^{\prime}(u)}{4\pi r^{2}}\,l_{a}l_{b}\,,$ (3)

where the vector $\ell^{a}$ is null and must be geodesic, as follows from the
conservation equation for the null dust $\nabla^{b}T_{ab}=0$ (see [8] for a
review on the null dust). For the ougoing Vaidya metric it is
$\ell_{a}=-\partial_{a}u$ [6, 5].

A null dust appears frequently in classical and quantum gravity, in Vaidya
spacetimes [3, 4], $pp$-waves [8, 9, 10], Robinson-Trautman geometries [11],
twisting solutions of the Einstein-Maxwell equations [11, 12, 13], in studies
of classical and quantum gravitational collapse, horizon formation, mass
inflation [14, 15, 16, 17, 18, 19, 20, 21, 22, 23], black hole evaporation
[24, 25, 26], and canonical Hamiltonians [8, 27, 1].

I really would like this null dust to be realized by an outgoing scalar field
$\phi$ with null gradient, $\nabla_{c}\phi\nabla^{c}\phi=0$, say
$\phi=\phi(u)$. I would also be happy with an ingoing Vaidya solution of the
form

$ds^{2}=-\left(1-\frac{2m(v)}{r}\right)dv^{2}+2dvdr+r^{2}d\Omega_{(2)}^{2}\,,$
(4)

where $v$ is the advanced null coordinate, with $\phi(v)$ going with it.

Can you please bring any of these two toys this Christmas? Many (advanced)
thanks,

Emmy

##  2 Santa’s reply

Dear Emmy,

Sorry for this retarded reply. Unfortunately, the toy that you are asking for
cannot be built in my workshop or anywhere else in this universe. Here is why.
First, we need to talk about the null dust sprinkled all over Vaidya’s
spacetime and whether a scalar field can be equivalent to a null dust.

It is well known [28, 29, 30] that a scalar field $\phi$ with timelike
gradient, $\nabla^{c}\phi\nabla_{c}\phi<0$ is equivalent to an irrotational
perfect fluid with four-velocity

$u_{c}=\frac{\nabla_{c}\phi}{\sqrt{-\nabla^{a}\phi\nabla_{a}\phi}}\,.$ (1)

Scalar fields with spacelike gradients are less interesting and seldom
considered from this point of view [31]. But what if
$\nabla^{c}\phi\nabla_{c}\phi=0$ instead? Well, the stress-energy tensor of a
massless, minimally coupled, scalar field $\phi$ is simply
$T_{ab}=\nabla_{a}\phi\nabla_{b}\phi-\frac{1}{2}\,g_{ab}\nabla^{c}\phi\nabla_{c}\phi$,
which matches that of a null dust if $\ell_{a}\equiv\nabla_{a}\phi$ is null.
Because $\ell^{a}$ is a gradient, only an irrotational dust can be reproduced
this way. Now, the divergence of $\ell^{a}$ is

$\nabla^{c}\ell_{c}=\nabla^{c}\nabla_{c}\phi\equiv\Box\phi=0$ (2)

by virtue of the Klein-Gordon equation of motion for $\phi$. The covariant
conservation equation $\nabla^{b}T_{ab}=0$ satisfied by any fluid, including
the null dust, gives

$\ell^{b}\nabla_{b}\ell_{c}=-\left(\nabla^{a}\ell_{a}\right)\ell_{c}$ (3)

which is the non-affinely parametrized geodesic equation. However, since
$\Box\phi=0$, the right hand side vanishes and $\ell^{a}$ must be null and
tangent to affinely-parametrized null geodesics.

There is a subtlety here: since a null vector is defined up to a
multiplicative (positive) factor, one could choose a different parametrization
of the null dust in which its stress-energy tensor is instead written as
$T_{ab}=\rho\,\ell_{a}\ell_{b}$ [8, 32]. For a general null dust, with this
choice the covariant conservation equation $\nabla^{b}T_{ab}=0$ would then
give non-affinely parametrized geodesics [8]. However, because the scalar
field null dust is irrotational, also this parametrization ends up giving
affine parametrization [8, 32].

Now consider the congruence of null geodesics tangent to the null vector
$\ell^{a}$ (outgoing null geodesics for the geometry (1), or ingoing null
geodesics for the geometry (4)). Let us focus on the outgoing Vaidya spacetime
(1) first: by comparing the relation $\ell_{a}=-\nabla_{a}u$ valid in Vaidya’s
geometry with $\ell_{a}=\nabla_{a}\phi$ specifying that the null dust is
generated by a scalar field, we obtain (apart from an irrelevant additive
integration constant) $\phi=-u$, so the scalar field is outgoing at the speed
of light. However, this result is in contradiction with the Klein-Gordon
equation $\Box\phi=0$ because

$\displaystyle\Box\phi$ $\displaystyle=$
$\displaystyle\nabla^{c}\ell_{c}=-\nabla^{c}\nabla_{c}u=-\frac{1}{\sqrt{-g}}\,\partial_{\mu}\left(\sqrt{-g}\,g^{\mu\nu}\partial_{\nu}u\right)$
(4) $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{-g}}\,\partial_{\mu}\left(\sqrt{-g}\,g^{\mu
u}\right)=-\frac{1}{r^{2}\sin\vartheta}\,\partial_{r}\left(r^{2}\sin\vartheta\,g^{ur}\right)=\frac{2}{r}\,,$

where we used the fact that

$\left(g^{\mu\nu}\right)=\left(\begin{array}[]{cccc}0&-1&0&0\\\ &&&\\\
-1&1-\frac{2m}{r}&0&0\\\ &&&\\\ 0&0&\frac{1}{r^{2}}&0\\\ &&&\\\
0&0&0&\frac{1}{r^{2}\sin\vartheta}\\\ \end{array}\right)$ (5)

and $\sqrt{-g}=r^{2}\sin\vartheta$. Clearly, $\Box\phi=-2/r\neq 0$ and the
Klein-Gordon equation cannot be satisfied. Although a scalar field with
lightlike gradient is a null dust realization, a scalar field-sourced outgoing
Vaidya geometry is impossible. More generally, if one considers radial
outgoing and ingoing null geodesics with tangents $\ell^{a}$ and $n^{a}$ in
the outgoing Vaidya geometry (1), their expansions are easily found to be
(e.g., [6])

$\theta_{(\ell)}=\nabla_{c}\ell^{c}=\frac{2}{r}\,,\;\;\;\;\;\;\;\;\;\;\;\theta_{(n)}=\frac{2m(u)-r}{r^{2}}\,.$
(6)

Clearly, both expansions are non-vanishing for $r>2m$. Put in other words, the
congruence of outgoing radial null geodesics that is deemed to have
$\ell^{c}=\nabla^{c}\phi$ as its four-tangent field is twist-free (because
$\ell^{c}$ is a gradient), shear-free (because of spherical symmetry), and
cannot also be expansion-free.

The same argument, adapted to the replacement $u\rightarrow v$ demonstrates
the impossibility of ingoing Vaidya solutions sourced by a scalar field (that
would have to be $\phi=-v$). So, it is impossible to build your toy! But don’t
be disheartened: since you like null dust and surfaces travelling at light
speed I can offer you, as a replacement, an exact plane wave sourced by a
scalar field with null four-gradient. This plane wave, described in [33], is
given in null coordinates by the Szekeres form [34]

$ds^{2}=-2\,\mbox{e}^{-M(u)}dudv+\mbox{e}^{-U(u)}\left(\mbox{e}^{V(u)}dx^{2}+\mbox{e}^{-V(u)}dy^{2}\right)\,,$
(7)

where [33]

$\displaystyle\mbox{e}^{-U}$ $\displaystyle=$ $\displaystyle
f(u)+\frac{1}{2}\,,$ (8) $\displaystyle\mbox{e}^{V}$ $\displaystyle=$
$\displaystyle\left(\frac{1+\sqrt{\frac{1}{2}-f}}{1-\sqrt{\frac{1}{2}-f}}\right)^{\lambda_{1}/2}\,,$
(9) $\displaystyle\mbox{e}^{-M}$ $\displaystyle=$ $\displaystyle
kf^{\prime}(u)\,\frac{\left(\frac{1}{2}+f\right)^{\frac{\alpha^{2}-1}{2}}}{\left(\frac{1}{2}-f\right)^{\alpha^{2}}}\,,$
(10) $\displaystyle\phi$ $\displaystyle=$
$\displaystyle\phi_{0}+\frac{\lambda_{2}}{2}\ln\left(\frac{1+\sqrt{\frac{1}{2}-f}}{1-\sqrt{\frac{1}{2}-f}}\right)\,,$
(11)

and where $f(u)$ is an arbitrary (but regular) function of $u$,
$\lambda_{1,2}$ and $k$ are constants, and [33]

$\alpha^{2}=\frac{\lambda_{1}^{2}}{4}+\lambda_{2}^{2}\,.$ (12)

In this case we have

$\nabla_{\mu}\phi=-\frac{\lambda_{2}f^{\prime}(u)}{\sqrt{\frac{1}{2}-f}\left(1+2\,f\right)}\,\delta_{\mu
u}$ (13)

and

$\left(g^{\mu\nu}\right)=\left(\begin{array}[]{cccc}0&-\mbox{e}^{M}&0&0\\\
&&&\\\ -\mbox{e}^{M}&0&0&0\\\ &&&\\\ 0&0&\mbox{e}^{U-V}&0\\\ &&&\\\
0&0&0&\mbox{e}^{U+V}\\\ \end{array}\right)\,,$ (14)

using which one obtains

$\nabla^{c}\phi\nabla_{c}\phi=g^{uu}\phi^{\prime}(u)=0\,.$ (15)

In this case $\ell^{c}\equiv\nabla^{c}\phi$ can be a null vector without
contradicting the vanishing of the divergence $\nabla^{c}\ell_{c}=\Box\phi=0$
because we have planar, instead of spherical, symmetry and null geodesics with
tangents $\ell^{c}$ can form a non-expanding congruence.

Yours,
Santa

##  3 Epilogue

Santa has demonstrated that Vaidya geometries sourced by a massless scalar
field acting as a null dust cannot be solutions of the Einstein equations. The
physical reason is that these spherically symmetric geometries must
necessarily have non-vanishing expansion, which contradicts the Klein-Gordon
equation. In fact, an (irrotational) null dust created from a lighlike scalar
field gradient $\ell_{a}=\nabla_{a}\phi$ must necessarily have zero divergence
$\nabla^{c}\ell_{c}=\Box\phi$, which is incompatible with the expanding (or
contracting) nature of the Vaidya geometries.

Emmy, who likes to play with symmetries, is happy anyway about the new plane
wave toy that she will receive this Christmas (which can be built because,
contrary to Vaidya, it can simultaneously have zero expansion and satify the
Klein-Gordon equation). Along the way, the child has also learned something
interesting from her correspondence with Santa.

## Acknowledgments

This work is supported, in part, by the Natural Sciences & Engineering
Research Council of Canada (Grant No. 2016-03803 to V.F.) and by Bishop’s
University. The work of A.G. has been carried out in the framework of the
activities of the Italian National Group for Mathematical Physics [Gruppo
Nazionale per la Fisica Matematica (GNFM), Istituto Nazionale di Alta
Matematica (INdAM)].

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