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# A Point-Free Look at Ostrowski’s Theorem and Absolute Values

Ming Ng and Steven Vickers

###### Abstract.

This paper investigates the absolute values on $\mathbb{Z}$ valued in the
upper reals (i.e. reals for which only a right Dedekind section is given).
These necessarily include multiplicative seminorms corresponding to the finite
prime fields $\mathbb{F}_{p}$. As an Ostrowski-type Theorem, the space of such
absolute values is homeomorphic to a space of prime ideals (with co-Zariski
topology) suitably paired with upper reals in the range $[-\infty,1]$, and
from this is recovered the standard Ostrowski’s Theorem for absolute values on
$\mathbb{Q}$.

Our approach is fully constructive, using, in the topos-theoretic sense,
geometric reasoning with point-free spaces, and that calls for a careful
distinction between Dedekinds vs. upper reals. This forces attention on
topological subtleties that are obscured in the classical treatment. In
particular, the admission of multiplicative seminorms points to connections
with Berkovich and adic spectra. The results are also intended to contribute
to characterising a (point-free) space of places of $\mathbb{Q}$.

_Thanks:_ Research partially supported by EPSRC Grant EP/V028812/1.

Classically, an absolute value on $\mathbb{Q}$ is defined as a multiplicative
norm $|\cdot|\colon\mathbb{Q}\rightarrow[0,\infty)$, and we have a complete
classification of all such absolute values via Ostrowski’s Theorem: up to
equivalence, they are all either Euclidean, or $p$-adic, or trivial. On a
basic level, this paper can be read as just a piece of constructive
mathematics. Our work involves sharpening Ostrowski’s theorem in various
sensible ways – e.g. by rephrasing Ostrowski’s Theorem as a representation
result (instead of just a classification result), or by adhering to an
inherently topological sense of geometric reasoning (instead of working
classically), extending previous work from [NV22] .

However, rather more interestingly, our investigations also bring to light a
subtle connection between topology and algebra previously elided by classical
assumptions, raising challenging implications for the foundations of
arithmetic geometry.

Let us elaborate. First, geometric mathematics presents us with various
_classically_ equivalent but _constructively_ inequivalent notions of the
reals. As such, there exists several options for reworking the notion of an
absolute value geometrically – e.g. should the map be valued in Dedekinds or
one-sided reals (see Section 1.2)? Further, since
$|\cdot|\colon\mathbb{Q}\rightarrow[0,\infty)$ is determined by its values on
the integers, one may also define an absolute value as a map from either
$\mathbb{Z}$ or $\mathbb{Q}$. This selection of topological and algebraic
options have a curious interaction, which we summarise in the following
observation.

######  Observation 0.1.

  1. (i)

An absolute value on $\mathbb{Q}$ (or indeed any field) must be valued in
Dedekinds, and not the one-sideds.

  2. (ii)

An absolute value on $\mathbb{Z}$ can be valued in the one-sided reals.

The justification for this will be seen in Proposition 2.10, after the
relevant definitions have been introduced. For now, the statement of
Observation 0.1 sets the scope for our present task: work towards an explicit
description of the space of absolute values, first by proving a modified
version of Ostrowski’s theorem for upper-valued absolute values on
$\mathbb{Z}$

$|\cdot|\colon\mathbb{Z}\rightarrow\overleftarrow{[0,\infty)},$

before recovering the standard Ostrowski’s Theorem for absolute values on
$\mathbb{Q}$.

Before proceeding, a few obligatory remarks regarding motivation. The decision
to use upper reals rather than Dedekinds may strike the lay reader as
constructivist hair-splitting, but in fact it ties together two a priori
unrelated mathematical threads:

  1. (a)

Vickers [Vic05]: To provide a geometric account of the completions of a
(generalised) metric space, it suffices for the metric to be valued in non-
negative upper reals (as opposed to the Dedekinds).

  2. (b)

Berkovich [Ber90]: For a suitable111By suitable, let us mean: non-Archimedean,
non-trivially valued, algebraically closed and complete. field $K$, every
point $x$ of the Berkovich affine line $\mathbb{A}_{\mathrm{Berk}}^{1}$
corresponds to a nested descending sequence of closed discs in $K$:

$D_{1}\supseteq D_{2}\supseteq\dots$

This connection is made precise in [Ng23] (and in a planned forthcoming
paper), which gives an example of how the structural gap between trivially vs.
non-trivially valued fields in Berkovich geometry can (surprisingly) be
eliminated via point-free techniques.222The reader familiar with Berkovich
geometry should notice the suggestive parallel between “a point of
$\mathbb{A}_{\mathrm{Berk}}^{1}$ = a nested sequence of discs” and “a point of
the upper reals = a rounded ideal” (see e.g. [NV22, Remark 1.30]).

There is a more fundamental question: why even rework these algebraic ideas
geometrically in the first place? Our answer lies in our wish to exploit the
tight connections between geometric logic and toposes. A key structure theorem
in topos theory tells us that the models of any geometric theory are
classified by some topos, and that any topos classifies the models of some
geometric theory.333 Technically: “topos” here means a Grothendieck topos,
i.e. a bounded $\mathcal{S}$-topos where the fixed base $\mathcal{S}$ can be
any elementary topos with natural numbers object.  In other words, if absolute
values can be described as models of some geometric theory, then we have at
our disposal a deep collection of topos-theoretic tools that allows us to
extract topological information from our logical setup. This insight has been
developed in [Ng23], which utilised descent techniques to reveal some striking
differences between the Archimedean vs. non-Archimedean places of
$\mathbb{Q}$. Again, the upper reals play a key role here, and their presence
raises challenging questions regarding certain longstanding assumptions in
number theory.

###### Contents

  1. 1 Preliminaries
  2. 2 Geometric Theories of Absolute Values
  3. 3 Preliminary lemmas for absolute values on $\mathbb{Z}$
  4. 4 Ostrowski’s Theorem for $\mathbb{Z}$
  5. 5 Absolute Values on $\mathbb{Q}$
  6. 6 Further Remarks and Discussion

##  1\. Preliminaries

Methodologically, the present paper is an exercise in point-free analysis,
developed using techniques of geometric logic.

Already in [NV22] we have described those techniques and used them to
construct real exponentiation and logarithms, so for a detailed introduction
we shall refer the reader to that paper and to other references mentioned
there. For the present we shall content ourselves with noting the more
prominent oddities that might otherwise distract the reader.

###  1.1. Point-free topological spaces

Geometric mathematics is a particular regime of constructive mathematics tied
to the (logical) perspective that a topos can be regarded as a “point-free
space” in the following sense:

######  Definition 1.1.

A _(point-free) space_ is described by a geometric theory $\mathbb{T}$, whose
models are the points of the space. We write $[\mathbb{T}]$ for the space of
models of $\mathbb{T}$.

A map $f\colon X\rightarrow Y$ is defined by a geometric construction of
points $f(x)\in Y$ out of points $x\in X$.

One way to read Definition 1.1 is that it pulls the usual notion of a _space_
away from its underlying set theory. Points of a space are now defined as
_models_ of a theory as opposed to _elements_ of a set (equipped with some
chosen topology); continuous maps are now defined as _geometric
transformations_ of such models as opposed to _functions_ on sets that respect
the chosen topology.

Note that the points, the models, may be sought _in any Grothendieck topos._
Hence $[\mathbb{T}]$ is not in a literal sense the collection of all models.
Rather, it is a “(pseudo-)functor of models”, by analogy with the “functor of
points” in algebraic geometry, used to replace a set of points in situations
where there are insufficient global points. (Compare this with the definition
of theory in [Joh02a, B4.2].) Technically, it is represented as the
classifying topos $\mathcal{S}[\mathbb{T}]$, the category of sheaves, but we
make the notational distinction so as to differentiate between maps (geometric
morphisms) $[\mathbb{T}]\to[\mathbb{T}^{\prime}]$ and functors
$\mathcal{S}[\mathbb{T}]\to\mathcal{S}[\mathbb{T}^{\prime}]$.

######  Definition 1.2 (Geometric Theories).

There are various ways to define what a geometric theory is; compare, for
example, [Joh02b, D1.1.6] and [Joh02a, B4.2.7]. We set out here some key
features. For further details of our usage, see [NV22] or [Vic22].

At its most basic, a geometric theory is a many sorted, first-order theory.
However, there are some special features and restrictions in how its
extralogical axioms may be expressed.

First, there is a two-level distinction between _formulae_ and _sequents_.

  * •

A _geometric formula_ is a logical formula built up from the symbols in the
signature and a context of finitely many free variables, using truth $\top$,
equality $=$, finite conjunctions $\land$, arbitrary (possibly infinite)
disjunctions $\bigvee$, and $\exists$.

  * •

A _geometric sequent_ is an expression of the form $\forall
xyz\ldots(\phi\rightarrow\psi)$, where $\phi$ and $\psi$ are geometric
formulae in the same context $\\{x,y,z,\ldots\\}$.

The theory’s axioms are in the form of sequents. Thus negation, implication,
and universal quantification are expressed in the theory at a single level –
they cannot be nested.

Adding extra axioms adds extra constraints on the models, and this provides
the point-free notion of _subspace._ Of course, we cannot simply talk about a
sub _set_ of the points.

Second, disjunctions (but not conjunctions) are allowed to be infinite. This
provides the link with topology, with the propositional formulae (those
without free variables) corresponding to the opens. The signature of a theory
supplies subbasic opens, and for the rest the finite $\land$ and the arbitrary
$\bigvee$ correspond to intersections and unions of opens.

Finally, the _geometric constructions_ , by which maps are defined, are
allowed in a theory as sort constructors. Colimits, finite limits (note the
analogy with disjunctions and finite conjunctions) and free algebra
constructions are all geometric, and geometric constructions include the
natural numbers $\mathbb{N}$, the integers $\mathbb{Z}$ and the rationals
$\mathbb{Q}$, along with their usual arithmetic structure (e.g. addition,
multiplication, strict order etc.).

The basic deal is that if you reason geometrically, then all your
constructions are continuous. You also find the very notion of topological
space is greatly generalised, to that of topos, and proper classes such as
those of sets or rings can be described as generalised spaces. For the
purposes of the present paper, however, we shall restrict ourselves to the
category $\mathrm{Loc}$ of _localic_ spaces, i.e. spaces of models of a theory
equivalent a propositional one. (For more background, see e.g. [Vic07a] or
[Ng23, Ch. 2]).

On the other hand, the price of that deal is a greatly reduced ability to
construct sets. For instance, powersets, the real line $\mathbb{R}$, and sets
of functions, are all non-geometric constructions, i.e. they are not
geometrically definable _as sets._ These constructions still enter into our
geometric reasoning, but they must be described as spaces (via geometric
theories), and have intrinsic non-discrete topologies that cannot be stripped
away. A phrase “the space of …” will presuppose the existence of some
geometric theory whose models are the “…”s.

######  Discussion 1.3 (Decidability and Topology).

Let $\phi$ be a propositional formula (no free variables) definable in
$\mathbb{T}$, and $M$ be a model of $\mathbb{T}$. Classically, the Law of
Excluded Middle $\phi\vee\lnot\phi$ automatically gives that $\phi$ evaluates
in $M$ as either $\top$ or $\bot$. However, in geometric logic, we are unable
to even express this principle as stated (much less affirm its validity) since
our syntax lacks negation. We must therefore look for alternative
formulations.

In our setting, the validity of non-constructive principles (such as LEM) gets
reframed as a question of topology. To illustrate, suppose $U$ is a
propositional formula, written “$U$” because, in point-free topology, we
regard “the collection of models in $[\mathbb{T}]$ satisfying $U$” as an open
subspace of $[\mathbb{T}]$. Hence, the question of whether the logical axiom

$\top\to U\vee\lnot U$

holds in $\mathbb{T}$ should properly be understood as asking if there is an
open $V$ that is a Boolean complement of $V$ in $[\mathbb{T}]$ ($U\wedge
V\rightarrow\bot$, $\top\rightarrow U\vee V$). We call $U$ _decidable_ in this
case. As we shall later see, in situations where a desired property is not
decidable, this often creates subtleties when trying to prove analogues of
classical results in our setting. On the other hand, if the relevant opens are
decidable, then the definitions can be made by case splits, $U$ versus $V$.
(Categorically, $[\mathbb{T}]$ is then a coproduct of $U$ and $V$.) For
example, equality on the rationals is decidable.

More generally, we must be more careful, but it is often helpful to look at
sub _spaces_ of 1. There is a lattice of subspaces (for an account from the
geometric point of view see [Vic07b]), and there every open $U$ of a space $X$
(with sequent $\top\rightarrow U$) does have a _closed complement_
$U^{\mathsf{c}}$, with sequent $U\rightarrow\bot$. This is a Boolean
complement in the lattice of subspaces. In other words, decidability of $U$
means it is _clopen_ , both open and closed (because it is $V^{\mathsf{c}}$).

Although we cannot define maps from $X$ by cases based on $U$ and
$U^{\mathsf{c}}$, the following lemma shows we can use cases to prove
properties corresponding to subspaces. That’s because if an open $U$ and its
closed complement $U^{\mathsf{c}}$ are both contained in some subspace, then
so must be their join in the subspace lattice, and that is the whole space
$X$. We emphasise that this does not allow us to say that every point in $X$
is either in $U$ or in $U^{\mathsf{c}}$ – that may fail to be (constructively)
true.

######  Lemma 1.4 (Case-Splitting Lemma).

Consider the following cospan in $\mathrm{Loc}$:

${Y}$${X}$${Z}$$\scriptstyle{i}$$\scriptstyle{f}$ (1)

where $i$ is a subspace embedding. Further, suppose that:

  * •

$X=U\vee U^{\mathsf{c}}$, where $U$ is an open subspace of $X$ and
$U^{\mathsf{c}}$ is its closed complement, with subspace embeddings
$i_{1}\colon U\hookrightarrow X$ and $i_{2}\colon
U^{\mathsf{c}}\hookrightarrow X$.

  * •

There are maps $f_{1}\colon U\rightarrow Y$ and $f_{2}\colon
U^{\mathsf{c}}\rightarrow Y$, with $f\circ i_{1}=i\circ f_{1}$ and $f\circ
i_{2}=i\circ f_{2}$.

Then the pullback $P$ of the cospan in Equation (1) is isomorphic to $X$.

###### Proof.

We have pullbacks of spaces, so the pullback $P$ of Diagram (1) exists:

${P}$${Y}$${X}$${Z}$$\scriptstyle{\widehat{f}}$$\scriptstyle{p}$
$\scriptstyle{i}$$\scriptstyle{f}$ (2)

Further (see e.g. [Joh82, §II.2.3]), embeddings of (localic) spaces are
precisely the regular monics in $\mathrm{Loc}$. Since regular monics are
preserved by pullback, this implies the map $p\colon P\rightarrow X$ of
Diagram (2) is a regular monic as well. In English, this means: the pullback
$P$ is a subspace of $X$.

Exploiting the universal pullback property, we obtain the following diagrams:

${U}$${P}$${Y}$${X}$${Z}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}f_{1}}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}i_{1}}$$\scriptstyle{\theta_{1}}$$\scriptstyle{\widehat{f}}$$\scriptstyle{p}$
$\scriptstyle{i}$$\scriptstyle{f}$${U^{\mathsf{c}}}$${P}$${Y}$${X}$${Z}$$\scriptstyle{\definecolor[named]{.}{rgb}{0,0,1}\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}f_{2}}$$\scriptstyle{\definecolor[named]{.}{rgb}{0,0,1}\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}i_{2}}$$\scriptstyle{\theta_{2}}$$\scriptstyle{\widehat{f}}$$\scriptstyle{p}$
$\scriptstyle{i}$$\scriptstyle{f}$ (3)

Since $i_{1}=p\circ\theta_{1}$ and $i_{2}=p\circ\theta_{2}$ are regular
monics, this implies $\theta_{1}$ and $\theta_{2}$ are regular monics as well,
i.e. $U$ and $U^{\mathsf{c}}$ are subspaces of $P$. Hence in the lattice of
subspaces of $X$, we have $X=U\vee U^{\mathsf{c}}\leq P$, so $p$ is an
isomorphism. ∎

###  1.2. One-sided reals

As is standard in point-free topology, we shall take the real numbers to be
Dedekind sections of the rationals. The two halves $(L,U)$ of a section are
taken to be _rounded_ , for instance if $q\in L$ then there is some
$q^{\prime}>q$ with $q^{\prime}\in L$. This has the effect that the definition
of real numbers as Dedekind sections also defines the Euclidean topology: each
rational $q$ defines the opens $(q,\infty)$ as those sections for which $q\in
L$, and $(-\infty,q)$ for $q\in U$. Of course, this is typical of point-free
spaces: the points are the models of some geometric theory, and then the
geometric formulae are the opens.

Having two halves $L$ and $U$ corresponds to approximating reals from below
and above, and classically, each can be derived from the other. Geometrically,
however, we also have “one-sided” approximation, lower ($L$) or upper ($U$),
and they are not (constructively) equivalent to Dedekind sections.
Qualitatively they correspond to the topologies of lower or upper
semicontinuity.

First, a series of remarks on order and topology in the upper reals.

######  Convention 1.5.

In any space we have a _specialisation order_ , $x\sqsubseteq y$ if every open
containing $x$ also contains $y$. For upper reals this comes down to $\forall
q\in\mathbb{Q}\,(x<q\rightarrow y<q)$. This says that, numerically, $x\geq y$,
so we shall write $\geq$ and $\leq$ to denote the specialisation order and its
opposite. For lower reals it is the other way round: specialisation order is
$\leq$.

In recognition of this, we use notation with arrows to show the upward
direction of the specialisation order, for example,
$\overleftarrow{[-\infty,\infty]}$ for the upper reals from $-\infty$ to
$\infty$ (inclusive). Thus the pairs $(x,y)$ such that $x\leq y$ form a
subspace of
$\overleftarrow{[-\infty,\infty]}\times\overleftarrow{[-\infty,\infty]}$.

######  Remark 1.6 (The Existence of Infima/Suprema).

  1. (i)

An important fact about point-free spaces is that they have all sups of
families of points directed with respect to the specialisation order.444 In
point-set topology this is an aspect of what is known as _sobriety._ For upper
reals, these are directed _infima._ In fact, since the rationals are totally
ordered, this implies the existence of arbitrary infima – at least if $\infty$
is included, as the nullary $\inf$.

  2. (ii)

Classically, it is taken as characteristic of reals that they have all bounded
infima and suprema. Constructively this is not the case. By item (i) we have
$\inf$s of upper reals and $\sup$s of lower reals. For Dedekind reals, we have
binary $\min$ and $\max$, but for an infinitary $\inf$ or $\sup$ of Dedekinds
we get, in general, only an upper or lower real respectively.

######  Remark 1.7.

It is clear from Remark 1.6 that we cannot capture strict inequality $x<y$ for
upper reals. Fixing $x$, there cannot be a space $\overleftarrow{(x,\infty]}$
because it would have to contain all the rationals bigger than $x$, and hence
also their infimum, which is $x$ itself. However, we do have a special case
for when $y$ is rational, as, for each $q$, $x<q$ defines an open
$\overleftarrow{[-\infty,q)}$ of $\overleftarrow{[-\infty,\infty]}$: it
corresponds to the geometric sequent $\top\rightarrow x<q$. Its closed
complement, which we shall write $\overleftarrow{[q,\infty)}$, corresponds to
the sequent (not a geometric formula) $x<q\rightarrow\bot$. It is easy to show
that it is equivalent to $q\leq x$, i.e. $\forall
r\in\mathbb{Q}(x<r\rightarrow q<r)$.

The one-sided reals have rich arithmetic structure, developed in [NV22] as far
as exponentiation and logarithms. However, there is one big proviso. All our
maps are continuous, and continuous maps preserve specialisation order. Hence
all the arithmetic operations, to be definable on one-sided reals, have to be
monotone (or antitone if they map from upper to lower or vice versa). For
example, upper reals can be added, but not subtracted (which would be antitone
in the second argument). Non-negative upper reals can be multiplied, but this
cannot be extended to negatives; and even non-negative upper reals cannot be
divided.

For exponentiation and logarithms these conditions can get quite intricate.
The main result we shall take from [NV22] is the following. Note the
hypothesis $1\leq x$. For $0<x<1$, the exponential $x^{\lambda}$ would be
antitone in $\lambda$.

######  Theorem 1.8 (Upper Real Exponents).

Fix a Dedekind $x\in[1,\infty)$. There exists the following exponentiation map
on the upper reals

$\displaystyle x^{\text{(---)}}\colon\overleftarrow{[-\infty,\infty)}$
$\displaystyle\longrightarrow\overleftarrow{[0,\infty)}$

satisfying the Basic Equations:

$\displaystyle
x^{\lambda+\lambda^{\prime}}=x^{\lambda}x^{\lambda^{\prime}}\text{, }\quad$
$\displaystyle x^{0}=1$ $\displaystyle
x^{\lambda\cdot\lambda^{\prime}}=(x^{\lambda})^{\lambda^{\prime}}\text{,
}\quad$ $\displaystyle x^{1}=x$
$\displaystyle(xy)^{\lambda}=x^{\lambda}y^{\lambda}\text{, }\quad$
$\displaystyle 1^{\lambda}=1.$

Further, the map $x^{\text{(---)}}$ is (weakly) monotonic, i.e.
$\lambda\leq\lambda^{\prime}$ implies $x^{\lambda}\leq x^{\lambda^{\prime}}$.

###### Proof.

The fact that $x^{\text{(---)}}$ is well-defined and satisfies the listed
basic equations is [NV22, Prop. 3.4]. The fact that $x^{\text{(---)}}$ is
monotonic come for free since $x^{\text{(---)}}$, being a continuous map, must
respect the specialisation order of $\overleftarrow{[-\infty,\infty)}$. ∎

The exponential map on upper reals also has an inverse $\log_{x}$.

Finally, exponentiation and logarithms on the usual Dedekind reals behave as
expected in the geometric setting. This was worked out explicitly in Sections
3 and 4 in [NV22], and will become relevant in Section 5.

###  1.3. Spectra

Classically, Ostrowski’s Theorem holds that all non-Archimedean absolute
values are equivalent to the $p$-adic absolute values $|\cdot|_{p}$, for some
prime $p$. As such, we need a way of talking about the primes in our setting.
Classically, one considers the set of prime ideals, perhaps equipped with an
additional topology (e.g. Zariski, coZariski, constructible). In the point-
free setting, however, the topology and points must be defined simultaneously
and cannot be thus separated. The following three examples develop this
remark, following the notation and topos-theoretic methods of Cole [Col16] and
Johnstone [Joh77b, §6.5]. In all three examples, $R$ denotes a commutative
ring with 1, and $\mathrm{spec}(R)$ denotes the classical set of prime ideals
of $R$.

######  Example 1.9 (The Zariski Spectrum).

The _Zariski Spectrum_ $\mathrm{LSpec}(R)$ for $R$ is the space whose points
are the _prime filters_ of $R$. More explicitly, they are the subsets $S$ of
$R$ satisfying the axioms:

  * •

$1\in S$, and $0\notin S$;

  * •

$(\forall a,a^{\prime}\in R).\,\,aa^{\prime}\in S\leftrightarrow a\in S\land
a^{\prime}\in S$;

  * •

$(\forall a,a^{\prime}\in R).\,\,a+a^{\prime}\in S\rightarrow a\in S\vee
a^{\prime}\in S$.

Notice: in $\mathrm{Set}$, these axioms say that $S$ is the complement of a
prime ideal of $R$.555 For the avoidance of doubt, we do not accept $R$ as a
prime ideal of itself.  Hence, working classically, we get a point-set
topological space equivalent to $\mathrm{spec}(R)$ equipped with the _Zariski
topology_ , generated by the basic Zariski open sets
$D(a)=\\{\mathfrak{p}\in\mathrm{spec}(R)\,|\,a\notin\mathfrak{p}\\}$.

######  Example 1.10 (The coZariski Spectrum).

The _coZariski Spectrum_ $\mathrm{ISpec}(R)$ for $R$ is the space whose points
are the _prime ideals_ of $R$. The axioms can be obtained using
contrapositives of those for $\mathrm{LSpec}(R)$. Regarded as a point-set
space, $\mathrm{ISpec}(R)$ is $\mathrm{spec}(R)$ equipped with the _coZariski
topology_ , which is generated by the sub-basic open set
$V(a)=\\{\mathfrak{p}\in\mathrm{spec}(R)\,|\,a\in\mathfrak{p}\\}$.

######  Example 1.11 (The Constructible Spectrum).

The _Constructible Spectrum_ $\mathrm{FSpec}(R)$ for $R$ is the space whose
points are the _complemented prime ideals_ of $R$. More explicitly, the points
of $\mathrm{FSpec}(R)$ are pairs $(P,S)$ where $P$ is a prime ideal, $S$ is a
prime filter and $P$ and $S$ are complements of each other (as subobjects of
$R$). Regarded as a point-set space, $\mathrm{FSpec}(R)$ is $\mathrm{spec}(R)$
equipped with the _constructible topology_ , which is the join of the Zariski
and coZariski topologies.

######  Discussion 1.12.

Examples 1.9 \- 1.11 combine to give the following picture. Viewed as _point-
free spaces_ , the points of the three spectral spaces are (constructively)
different. However, when viewed classically as _point-set spaces_ , then the
topology and points begin to separate. On the one hand, their underlying set
of points become classically equivalent: since every subset in $\mathrm{Set}$
is complementable, there is no longer a meaningful difference between, e.g. a
complementable prime ideal vs. a prime ideal in $\mathrm{Set}$. On the other
hand, their respective topologies (Zariski, coZariski, constructible) are
still different, although the algebraic reasons for this difference are now
obscured.

When proving Ostrowski’s Theorem, it is most natural to regard the $p$-adic
absolute values $|\cdot|_{p}$ as corresponding to the prime ideals of
$\mathbb{Z}$. In light of our above discussion, this indicates that the
correct space for us to use is the coZariski spectrum
$\mathrm{ISpec}(\mathbb{Z})$. Note that this is different from the standard
choice of the Zariski spectrum in classical algebraic geometry, where we are
interested in localising by inverting all the elements of a prime _filter_.

######  Definition 1.13.

We write $\mathrm{spec}^{+}(\mathbb{Z})$ for the set of non-zero prime
elements of $\mathbb{Z}$. It is geometrically isomorphic to $\mathbb{N}$, and
we have an obvious map from $\mathrm{spec}^{+}(\mathbb{Z})$ to
$\mathrm{FSpec}(\mathbb{Z})$ and hence also to $\mathrm{ISpec}(\mathbb{Z})$
and to $\mathrm{LSpec}(\mathbb{Z})$.

_Classically,_ the three spectra for $\mathbb{Z}$ can be understood as three
topologies on $\mathrm{spec}^{+}(\mathbb{Z})\cup\\{0\\}$.

  * •

For $\mathrm{ISpec}(\mathbb{Z})$, the opens are arbitrary subsets of
$\mathrm{spec}^{+}(\mathbb{Z})$, together with the whole space. $0$ is bottom
in the specialisation order.

  * •

For $\mathrm{LSpec}(\mathbb{Z})$, the opens are the empty set, and the sets
containing $0$ and all but finitely many of the non-zero primes. $0$ is top in
the specialisation order.

  * •

$\mathrm{FSpec}(\mathbb{Z})$ is the one-point compactification of
$\mathrm{spec}^{+}(\mathbb{Z})$, with topology generated by the other two. It
is $T_{1}$ (specialisation order is discrete).

For $\mathrm{ISpec}$, we can extract part of that picture geometrically.

######  Lemma 1.14.

Let $\mathrm{ISpec}^{+}(\mathbb{Z})$ be the (open) subspace of
$\mathrm{ISpec}^{+}(\mathbb{Z})$ comprising those prime ideals that contain a
non-zero element. Then
$\mathrm{ISpec}^{+}(\mathbb{Z})\cong\mathrm{spec}^{+}(\mathbb{Z})$.

###### Proof.

Let $\mathfrak{p}\in\mathrm{ISpec}(\mathbb{Z})$ be a prime ideal containing a
non-zero element $a$. We show that $\mathfrak{p}=(p)$ for some unique prime
number $p\in\mathrm{spec}^{+}(\mathbb{Z})$. Since also $-a\in\mathfrak{p}$, we
can assume without loss of generality that $a$ is positive. By unique prime
factorisation, we can represent $a$ as

$a=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{n}^{\alpha_{n}}\text{.}$

Now by primeness of $\mathfrak{p}$, we know there is some
$p\in\mathrm{spec}^{+}(\mathbb{Z})$ dividing $a$ with
$(p)\subseteq\mathfrak{p}$.

To show that $\mathfrak{p}\subseteq(p)$, suppose $b\in\mathfrak{p}$. Recall
that Bézout’s Identity can be (constructively) obtained from inverting the
Euclidean Algorithm and performing the relevant substitutions. One can thus
verify that there exist $m,n\in\mathbb{Z}$ such that

$\gcd(b,p)=mb+np\in\mathfrak{p}\text{.}$ (4)

Because $p$ is prime, that gcd must be either 1 or $p$, but 1 is not in
$\mathfrak{p}$. Hence $\gcd(b,p)=p$ and $b\in(p)$. ∎

######  Remark 1.15.

Classically, one typically proves that $\mathfrak{p}=(p)$ for some prime
$p\in\mathbb{Z}$ by obtaining it as an easy corollary of the more general
result that all ideals of $\mathbb{Z}$ are principal. However, proving the
latter typically invokes the assumption that we can pick the least element of
any non-trivial ideal $I\subset\mathbb{Z}$ (see, e.g. [vdW91a, §3.7]), which
is a non-geometric assumption since membership of $I$ is not decidable.

######  Remark 1.16.

Since a prime ideal of $\mathbb{Z}$ is determined by its positive elements, we
might also consider a notion of prime ideal of $\mathbb{N}_{+}$. In fact, an
analogue of Lemma 1.14 also exists for prime ideals of the positive integers
$\mathbb{N}_{+}$, with now:

  * •

Instead of non-trivial prime ideals of $\mathbb{Z}$ we consider inhabited
prime ideals of $\mathbb{N}_{+}$;

  * •

We shall need to explicitly require that these prime ideals
$\mathfrak{p}\in\mathrm{ISpec}(\mathbb{N}_{+})$ are also closed under formal
subtraction, that is:

$\forall m,n\in\mathbb{N}_{+}\text{ . }m\in\mathfrak{p}\wedge
m+n\in\mathfrak{p}\rightarrow n\in\mathfrak{p}\text{.}$

Then, the same argument works, so long as we avoid using the negative
coefficients. The Euclidean algorithm, for finding the $\gcd$ of two natural
numbers $a_{0}$ and $a_{1}$, relies on repeated integer division to produce a
sequence of natural numbers with $a_{i-1}=ma_{i}+a_{i+1}$ and $0\leq
a_{i+1}<a_{i}$. This eventually hits $0$ at some $a_{n+1}$, with $a_{n}$
dividing $a_{n-1}$, and then $a_{n}=\gcd(a_{0},a_{1})$. The formal subtraction
rule ensures that if $\mathfrak{p}$ contains $a_{0}$ and $a_{1}$ then it
contains all the $a_{i}$s up to $a_{n}$. If $a_{1}$ is the prime $p$, then
$a_{n}\in\mathfrak{p}$ must be either 1 – which is impossible – or $p$, so $p$
divides $a_{0}$.

##  2\. Geometric Theories of Absolute Values

Our aim is to examine the notion of absolute value when using upper reals
instead of Dedekinds, and our basic general notion will be as follows.

######  Definition 2.1 (Absolute Values, valued in upper reals).

Let $R$ be a commutative ring (discrete, not topological). An _absolute value_
on $R$ is a map $|\cdot|\colon R\to\overleftarrow{[0,\infty)}$ satisfying the
following conditions:

$\begin{array}[]{ll}|0|=0\\\ |1|=1\\\ |xy|=|x||y|&\text{(Multiplicativity)}\\\
|x+y|\leq|x|+|y|&\text{(Triangle Inequality)}\end{array}$

We write $R\text{-}\overleftarrow{\mathsf{av}}$ for the geometric theory of
absolute values on $R$, $[R\text{-}\overleftarrow{\mathsf{av}}]$ for the
corresponding point-free space. We also write $\overleftarrow{\mathsf{av}}$
for $\mathbb{Z}\text{-}\overleftarrow{\mathsf{av}}$.

We have expressed this in a form that brings out the mathematics of absolute
values rather than the geometric logic. Nonetheless, there is a geometric
theory implicit in the definition. The signature has two sorts, for $R$ and
for the positive rationals $\mathbb{Q}_{+}$, and a predicate $\phi$ on
$R\times\mathbb{Q}_{+}$ with $\phi(x,q)$ expressing $|x|<q$. Then the
conditions listed, together with the requirement for $|x|$ to be an upper
real, can all be expressed geometrically.

From the conditions, we can also derive the following.

  * •

$|-1|=1$.

[Why? We have $|-1|^{2}=|(-1)^{2}|=1$, and the only solution in the non-
negative upper reals is 1.]

  * •

$|-x|=|-1||x|=|x|$.

######  Definition 2.2.

A _Dedekind_ absolute value $|\cdot|$ on $R$ is one that factors via
$[0,\infty)$. We write $R\text{-}\mathsf{av}$ for the theory of Dedekind
absolute values on $R$.

A Dedekind absolute value is _positive definite_ if it satisfies the axiom

$\forall x\in R\,(\top\rightarrow x=0\vee|x|>0)\text{.}$

We write $R\text{-}\mathsf{av}^{+}$ for the theory of positive definite
Dedekind absolute values on $R$.

######  Discussion 2.3 (Embeddings vs. Monics).

Geometrically, we need to distinguish between properties defined by geometric
sequents and those that also rely on uniquely defined structure. The former
define subspaces, and corresponding embeddings, while the latter define maps
of a more general kind: monics, because of uniqueness of the extra structure,
but not embeddings, because the extra structure changes the topology by
introducing more opens.

Thus $[R\text{-}\mathsf{av}^{+}]$ is a subspace of $[R\text{-}\mathsf{av}]$.
By contrast, to say an upper real is Dedekind relies on _structure,_ the lower
part. The map
$[R\text{-}\mathsf{av}]\to[R\text{-}\overleftarrow{\mathsf{av}}]$ that forgets
the lower parts is monic, because two Dedekind reals are equal iff their right
sections are equal, but not an embedding, because $[R\text{-}\mathsf{av}]$ has
a stronger topology, with extra opens provided by those lower parts.

######  Observation 2.4.

The choice of axioms in $R\text{-}\overleftarrow{\mathsf{av}}$ reflects the
topological constraints the upper reals puts on the algebra. In particular,
notice:

  1. (i)

Without the Dedekind property, $|x|>0$ is not available as a geometric
formula, and our description of positive definiteness is not geometric. We
might try instead to formulate it as

$\forall x\in R((\forall q\in\mathbb{Q}_{+}\,|x|<q)\rightarrow x=0)\text{,}$

but the $\forall q$ there, nested inside a sequent, is not a geometric
formula. Hence our Definition 2.1 does not embody positive definiteness.

Put otherwise, the topology of the upper reals forces us to work with the
multiplicative _seminorms_ for $R$, as opposed to the usual norms.

  2. (ii)

In the classical setting, the property $|1|=1$ can be derived from the others
using positive definiteness. For multiplicativity gives us $|1|\cdot|1|=|1|$,
which implies $|1|\cdot(|1|-1)=0$. Positive definiteness tells us that
$|1|\neq 0$, and thus $|1|=1$.

The standard examples of absolute values can all be reworked to obey
Definition 2.1, which we sketch below. The syntactic details have been
suppressed for readability, but one easily checks that the definitions only
use arithmetic operations that are well-defined on the upper reals and satisfy
the required axioms.

######  Example 2.5.

We give some standard examples on $\mathbb{Z}$. Since equality there is
decidable, it suffices to define $|\cdot|$ for $n\neq 0$ since we already
require that $|0|=0$ by definition.

  1. (i)

The _trivial absolute value_ on $\mathbb{Z}$, denoted $|\cdot|_{0}$, is
defined as

$|n|_{0}=1,\qquad\qquad\text{for all}\,\,n\neq 0.$

  2. (ii)

The _Euclidean absolute value_ on $\mathbb{Z}$, denoted $|\cdot|_{\infty}$, is
defined as the usual norm

$|n|_{\infty}=n,\qquad\qquad\text{for all}\,\,n\in\mathbb{N}_{+},$

which is extended to the negative integers by the fact $|n|=|-n|$ for all
$n\in\mathbb{Z}$.

  3. (iii)

Fix some prime $p\in\mathbb{N}_{+}$. By unique prime factorisation, any non-
zero integer $n\in\mathbb{Z}_{\neq 0}$ can be represented as $n=p^{r}z$, where
$r\in\mathbb{N}$, $z\in\mathbb{Z}$ and $\gcd(p,z)=1$. As such, define the
$p$-adic ordinal

$\mathrm{ord}_{p}(n):=\max\\{r\in\mathbb{N}\,\big{|}\,p^{r}\,\text{divides}\,\,n\\}.$

The canonical _$p$ -adic absolute value_ on $\mathbb{Z}$ is then defined as

$|n|_{p}=p^{-\mathrm{ord}_{p}(n)}\,\,\,\,\qquad\text{for all}\,\,n\neq 0.$

  4. (iv)

Our final example here is admitted by our use of seminorms. Let
$p\in\mathbb{N}_{+}$ be prime. The _$p$ -characteristic absolute value_ on
$\mathbb{Z}$ is then defined as

$|n|_{\mathbb{F}_{p}}=\begin{cases}0&\text{if $p$ divides $n$,}\\\ 1&\text{if
$p$ does not divide $n$.}\end{cases}$

Further examples can be obtained by exponentiating – in fact, the classical
Ostrowski’s Theorem says these are the only examples. For our setting the
details are explained more precisely in Theorem 4.1. For the moment, let us
say roughly that $|\cdot|_{\infty}^{\lambda}$ is an absolute value if
$0\leq\lambda\leq 1$, with $|\cdot|_{\infty}^{0}=|\cdot|_{0}$. Also,
$\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda}$ gives an absolute value for
$-\infty\leq\lambda\leq 0$, with $|\cdot|_{p}$ for $\lambda=-1$ and
$|\cdot|_{\mathbb{F}_{p}}$ for $\lambda=-\infty$.

######  Remark 2.6.

Why not go one step further and work with the upper-valued absolute values on
$\mathbb{N}$? After all, as Proposition 2.7 will show, upper-valued absolute
values on $\mathbb{Z}$ themselves are determined by their values on the
positive integers $\mathbb{N}_{+}$. A fuller answer will be presented in
Discussion 4.4. For now, let us say that proving Ostrowski’s Theorem without
additive inverses results in certain technical difficulties which incline us
to stick with $\mathbb{Z}$.

######  Proposition 2.7.

Absolute values over $\mathbb{Z}$ can equivalently be defined as maps
$|\cdot|\colon\mathbb{N}\to\overleftarrow{[0,\infty)}$ satisfying $|1|=1$,
multiplicativity, and the Triangle Inequality (quantified over $\mathbb{N}$,
of course), together with the _Subtractive Triangle Inequality_

$|m|\leq|m+n|+|n|\quad\text{ for all $n\in\mathbb{N}$}\text{.}$

###### Proof.

If an absolute value on $\mathbb{Z}$ is given, then it must satisfy the
subtractive inequality when restricted to $\mathbb{N}$. (In fact, the usual
and subtractive inequalities are _equivalent_ when applied to the whole of
$\mathbb{Z}$.)

Now suppose we are given $|\cdot|$ on $\mathbb{N}$ satisfying the subtractive
law. Clearly $|\cdot|$ must be extended to $\mathbb{Z}$ by $|0|=0$ and, for
$n>0$, $|-n|=|n|$. The notation will be less confusing if we temporarily write
$\mathop{\mathrm{mag}}n$ for the _magnitude_ of $n$, i.e. $|n|_{\infty}$ from
Examples 2.5. Then $|n|=|\mathop{\mathrm{mag}}n|$.

The only non-obvious question is whether it satisfies the Triangle Inequality
on the whole of $\mathbb{Z}$. Suppose, then, we are given $m,n\in\mathbb{Z}$.

We claim that, of the three magnitudes of $m,n,m+n$, one is the sum of the
other two. Note that if it is true for $m$ and $n$, then it is also true for
$-m$ and $-n$: because the three magnitudes are unchanged.

Let us consider how many of $m,n,m+n$ are negative.

If none, then the magnitudes also are $m,n,m+n$, so the claim holds. Hence it
also holds if all three are negative.

Now suppose just one is negative. It cannot be $m+n$, so, without loss of
generality, we can take it to be $m$. Then the three magnitudes are $-m,n,m+n$
and the claim holds. Hence it also holds if two are negative.

Now let us return to the Triangle Inequality for $\mathbb{Z}$, which reduces
to

$|\mathop{\mathrm{mag}}(m+n)|\leq|\mathop{\mathrm{mag}}m|+|\mathop{\mathrm{mag}}n|\text{.}$

If $\mathop{\mathrm{mag}}(m+n)$ is the sum of the other two, then this is an
instance of the Triangle Inequality for $\mathbb{N}$, while if
$\mathop{\mathrm{mag}}m$ or $\mathop{\mathrm{mag}}n$ is the sum of the other
two then it is the Subtractive Inequality. ∎

Next, let us recall the definition of a geometric field:

######  Definition 2.8.

A discrete ring $R$ is called a geometric field if it satisfies the following:

  1. (i)

$0\neq 1$

  2. (ii)

For any $x\in R$, either $x=0$ or $\exists x^{-1}\in R$ such that $x\cdot
x^{-1}=1$.

######  Remark 2.9.

Classically, Definition 2.8 is a plain statement of what it means to be a
field. Geometrically, the requirement for $R$ to be discrete is important, as
the definition does not work for a topological field such as $\mathbb{R}$. In
fact, it is easy to show that a geometric field must have decidable equality
[Joh77a, Lemma 2.1].

######  Proposition 2.10.

Let $R$ be a geometric field. Any absolute value $|\cdot|$ on $R$ is Dedekind
and positive definite.

In fact,
$[R\text{-}\overleftarrow{\mathsf{av}}]\cong[R\text{-}\mathsf{av}^{+}]$.

###### Proof.

Suppose $x\in R$. If $q$ is rational, then we can define $q<|x|$ when $q<0$,
or there are $y\in R$ with $xy=1$, _and_ $r^{\prime}\in\mathbb{Q}$ with
$|y|<r^{\prime}$ (so $0<r^{\prime}$) and $q<r^{\prime-1}$. Clearly this
defines a lower real; we must show that, together with the given relation
$|x|<r$, it defines a Dedekind.

For disjointness, suppose $q<|x|$ (with $y$ and $r$ as above), and $|x|<q$.
Then $1=|x||y|<qr^{\prime}<1$, a contradiction.

For locatedness, suppose $a<b$ are rationals. If $a<0$ then $a<|x|$ by
definition. If $a=0$, then replace $a$ by $b/2$. Hence we can assume $0<a$.
Now if $x=0$ then $|x|=0<a<b$, while if $x$ has an inverse $y$, then
$|x||y|=1<b/a$, so we can find $r,r^{\prime}$ with $|x|<r$, $|y|<r^{\prime}$,
and $rr^{\prime}=b/a$. If $r\leq b$ then $|x|<b$ and we are done. If $b<r$
then $rr^{\prime}<r/a$, so $a<1/r^{\prime}$ and $a<|x|$.

Positive definiteness follows from the fact that if $xy=1$ then
$|x|=|y|^{-1}>0$.

We have now shown that, from an absolute value, we can define the unique
structure needed to make it Dedekind and positive definite. Roughly speaking
the extra opens $q<|x|$ are provided using $|x^{-1}|<q^{-1}$. This gives the
map $[R\text{-}\overleftarrow{\mathsf{av}}]\to[R\text{-}\mathsf{av}^{+}]$, and
it is clearly inverse to the forgetful map in the opposite direction. ∎

###  2.1. Archimedean vs. Non-Archimedean Absolute Values on $\mathbb{Z}$

######  Definition 2.11.

Let $|\cdot|$ be an absolute value on $\mathbb{Z}$. We shall write
$\mathbb{Z}_{\neq 0}$ for the set of non-negative integers.

  1. (i)

$|\cdot|$ is _non-Archimedean (NA)_ if $|n|<1$ for some $n\in\mathbb{Z}_{\neq
0}$.

Immediately we see that the non-Archimedeans form an open subspace of
$[\overleftarrow{\mathsf{av}}]$ – it’s the join of the open subspaces $|n|<1$.
We write $[\overleftarrow{\mathsf{av}}_{NA}]$ for it.

  2. (ii)

$|\cdot|$ is _Archimedean (A)_ if $1\leq|n|$ for every $n\in\mathbb{Z}_{\neq
0}$.

This is a meet of closed subspaces for $1\leq|n|$, and in fact is the closed
complement of $[\overleftarrow{\mathsf{av}}_{NA}]$. We write
$[\overleftarrow{\mathsf{av}}_{A}]$ for it.

  3. (iii)

$|\cdot|$ is _ultrametric_ if $|m+n|\leq\max\\{|m|,|n|\\}$ for every
$m,n\in\mathbb{Z}_{\neq 0}$.

The ultrametrics do form a subspace of $[\overleftarrow{\mathsf{av}}]$, which
we shall write $[\overleftarrow{\mathsf{av}}_{U}]$, but it is neither open nor
closed. It is the meet of an infinite family of subspaces described by the
sequent

$\forall
m,n\in\mathbb{Z},q\in\mathbb{Q}(|m|<q\wedge|n|<q\rightarrow|m+n|<q)\text{.}$

_We have departed from the usual convention here._ Some guiding observations:

  * •

Normally $|\cdot|$ would be considered Archimedean only if it has some
$|n|>1$, but we cannot express that for $|n|$ as upper real. For us, the
trivial absolute value has been reassigned as Archimedean, though this does
not make a huge difference to the classical account, insofar as that
frequently excludes the trivial absolute value as a special case.

  * •

Notice: the trivial absolute value is still ultrametric, so ultrametric and
non-Archimedean are no longer synonymous. (In Lemma 3.3, however, we shall
prove that every non-Archimedean is ultrametric, as expected.)

  * •

The fact that $[\overleftarrow{\mathsf{av}}_{NA}]$ and
$[\overleftarrow{\mathsf{av}}_{A}]$ are now open and closed complements makes
the mathematics significantly smoother for us. In particular, it enables us to
use (via Lemma 1.4) case-splitting techniques that, more generally, are
invalid geometrically.

##  3\. Preliminary lemmas for absolute values on $\mathbb{Z}$

In this Section we gather together some preliminary lemmas that are explicit
or implicit in the treatment of [vdW91b].

The first is adjusted for upper reals, lacking the ability to divide.

######  Lemma 3.1.

Let $\alpha,\beta$ be positive Dedekinds, and $\gamma,\gamma^{\prime}$ be non-
negative upper reals such that

$\gamma^{v}\leq(\alpha v+\beta)\cdot(\gamma^{\prime})^{v}$

for all $v\in\mathbb{N}_{+}$. Then $\gamma\leq\gamma^{\prime}$.

###### Proof.

First, a basic but key observation: if
$\gamma,\gamma^{\prime}\in\overleftarrow{[0,\infty]}$ such that
$\gamma\leq(1+\delta)\gamma^{\prime}$ for all positive rationals $\delta$,
then this implies $\gamma\leq\gamma^{\prime}$. It thus suffices to prove that
the Lemma’s hypothesis implies $\gamma\leq(1+\delta)\gamma^{\prime}$, for all
positive rationals $\delta$.

Fix such a rational $\delta>0$. Binomial expansion yields the inequality
$(1+\delta)^{v}\geq 1+v\delta+\frac{v(v-1)}{2}\delta^{2}$ for any integer
$v\geq 2$. It is clear that for sufficiently large $v$, we get

$v\delta>\beta\qquad\text{and}\qquad\frac{1}{2}(v-1)\delta^{2}>\alpha,$

and so

$\gamma^{v}\leq(\alpha
v+\beta)\cdot(\gamma^{\prime})^{v}<(1+\delta)^{v}\cdot(\gamma^{\prime})^{v}=\left((1+\delta)\gamma^{\prime}\right)^{v}.$
(5)

By [NV22, Lemma 2.12], $\text{(---)}^{v}$ reflects non-strict order on non-
negative upper reals: in other words, from Equation (5) we can deduce
$\gamma\leq(1+\delta)\gamma^{\prime}$ as required. ∎

The second Lemma encapsulates a formula in the main proof in [vdW91b]. We have
clarified its role by varying the base of the logarithms.

######  Lemma 3.2 (Fundamental Lemma for Ostrowski).

Let $a,b>1$ be any pair of integers greater than 1, and let $|\cdot|$ be any
absolute value on $\mathbb{Z}$.

Then:

  1. (i)

$\log_{a}|a|\leq\max\\{0,\log_{b}|b|\\}\leq 1$

  2. (ii)

$\max\\{0,\log_{b}|b|\\}=\max\\{0,\log_{a}|a|\\}$ for any $a,b>1$.

In particular, we can associate a constant upper real
$M_{|\cdot|}:=\max\\{0,\log_{b}|b|\\}\leq 1$ to any absolute value $|\cdot|$
since by (ii) we know $M_{|\cdot|}$ is independent of our choice of $b>1$.

###### Proof.

(i): First, note that for any $n\in\mathbb{N}_{+}$, the triangle inequality
yields

$|n|=|\underbrace{1+1+\dots+1}_{n}|\leq\underbrace{|1|+|1|+\dots+|1|}_{n}=n.$
(6)

Hence, if $b>1$, we have $\log_{b}|b|\leq 1$.

Now, given any pair of integers $a,b$ such that $a,b>1$, and given any
$v\in\mathbb{N}_{+}$, we may expand $a^{v}$ in powers of $b$ as follows:

$a^{v}=c_{0}+c_{1}b+...+c_{r}b^{r}$ (7)

where $0\leq c_{i}<b$ for $0\leq i\leq r$, and $c_{r}\neq 0$. It is obvious
that

$b^{r}\leq a^{v},$

which (taking $\log_{b}\text{(---)}$ on both sides) yields

$r\leq\log_{b}a^{v}=v\log_{b}a.$ (8)

Hence, Equation (7) and the triangle inequality give

$\displaystyle|a^{v}|$
$\displaystyle\leq|c_{0}|+|c_{1}||b|+...+|c_{n}||b|^{r}$ $\displaystyle\leq
b(1+|b|+\dots+|b|^{r})\leq b(r+1)B^{r}$

where $B=\max\\{1,|b|\\}$. By Equation (8), we get

$|a|^{v}=|a^{v}|<b(v\log_{b}a+1)\cdot(B^{\log_{b}a})^{v}\text{.}$

Applying Lemma 3.1, this yields

$|a|\leq B^{\log_{b}a},$

which in turn yields

$\displaystyle\log_{a}|a|$
$\displaystyle\leq\max\\{0,\log_{a}|b|^{\log_{b}a}\\}$
$\displaystyle=\max\\{0,\log_{b}a^{\log_{a}|b|}\\}=\max\\{0,\log_{b}|b|\\}.$

To prove (ii), note that (i) yields for any pair of integers $a,b>1$

$\max\\{0,\log_{a}|a|\\}\leq\max\\{0,\max\\{0,\log_{b}|b|\\}\\}=\max\\{0,\log_{b}|b|\\},$

and so by symmetry

$\max\\{0,\log_{a}|a|\\}=\max\\{0,\log_{b}|b|\\}.$

∎

Our third Lemma is that non-Archimedean absolute values have the ultrametric
property. It is important to keep in mind our Definition 2.11. The two terms
are no longer synonymous, because the trivial valuation is ultrametric without
being non-Archimedean.

######  Lemma 3.3.

If an absolute value $|\cdot|$ on $\mathbb{Z}$ is non-Archimedean then it is
ultrametric.

###### Proof.

For some $n$ we have $|n|<1$, so $\max\\{0,\log_{n}|n|\\}=0$. Then by Lemma
3.2, $|m|\leq 1$ for all $m$. We can now apply Lemma 3.1 to show
$|a+b|\leq\max\\{|a|,|b|\\}$. For any $v\in\mathbb{N}_{+}$, we have
$|{v\choose i}|\leq 1$ and so binomial expansion yields

$|a+b|^{v}\leq|a|^{v}+|a|^{v-1}|b|+\cdots+|b|^{v}\leq(v+1)\cdot\left(\max\\{|a|,|b|\\}\right)^{v}\text{.}$

∎

##  4\. Ostrowski’s Theorem for $\mathbb{Z}$

Ostrowski’s Theorem is typically phrased as a classification result — it
answers the question: ‘What are all the non-trivial places of $\mathbb{Q}$?’.
Here, we sharpen this to a representation result for absolute values on
$\mathbb{Z}$:

######  Theorem 4.1 (Ostrowski’s Theorem for $\mathbb{Z}$).

As our setup, denote:

  * •

$[\overleftarrow{\mathsf{av}}]:=$ The space of absolute values on
$\mathbb{Z}$, valued in upper reals.

  * •

$\mathrm{ISpec}(\mathbb{Z}):=$ The space of prime ideals of $\mathbb{Z}$.

  * •

$\mathbb{Z}_{\neq 0}:=$ The set of non-zero integers.

  * •

$\overleftarrow{[-\infty,1]}:=$ The space of upper reals bounded above by $1$.

Define

$\mathfrak{P}_{\overleftarrow{\Lambda}}:=\\{(\mathfrak{p},\lambda)\in\mathrm{ISpec}(\mathbb{Z})\times\overleftarrow{[-\infty,1]}\,\,\big{|}\,\,\lambda<0\leftrightarrow\exists
a\in\mathbb{Z}_{\neq 0}.(a\in\mathfrak{p})\\}.$

Then the following spaces are equivalent:

$[\overleftarrow{\mathsf{av}}]\cong\mathfrak{P}_{\overleftarrow{\Lambda}}.$

Informally, Theorem 4.1 says: any absolute value $|\cdot|$ of $\mathbb{Z}$ can
be canonically associated to a pair

$(\mathfrak{p},\lambda)\in\mathrm{ISpec}(\mathbb{Z})\times\overleftarrow{[-\infty,1]}$

satisfying certain compatibility conditions.

Before we begin the proof, some preparatory remarks. One, the standard proofs
of Ostrowski’s Theorem (spelled out in, e.g. [vdW91b]) can be adapted to our
setting, but they still only give us one direction of the isomorphism.
Additional work is therefore needed to construct the second direction, and to
show that the two directions are inverse to each other. Two, the decision to
work with upper reals (as opposed to Dedekinds) makes the algebra rather
delicate, for reasons already alluded to in Observation 0.1. Some care is
needed in order to maintain geometricity throughout the proof. For the expert
reader, let us remark that this (interestingly) results in a picture of
$[\overleftarrow{\mathsf{av}}]$ that is similar but slightly different from
the classical picture of the Berkovich spectrum
$\mathcal{M}(\mathbb{Z})$.666Recall: the Berkovich Spectrum on $\mathbb{Z}$ is
defined as the space of multiplicative seminorms on $\mathbb{Z}$, but is
equipped with the Gel’fand topology, unlike the non-Hausdorff topology that
emerges here due to our use of upper reals.

###  4.1. First Direction: Classification of Absolute Values

First we define a map

$\displaystyle\widehat{f}\colon[\overleftarrow{\mathsf{av}}]$
$\displaystyle\longrightarrow\mathfrak{P}_{\overleftarrow{\Lambda}}$
$\displaystyle|\cdot|$
$\displaystyle\longmapsto(\mathfrak{p}_{|\cdot|},\lambda_{|\cdot|})$

by

  * •

$\mathfrak{p}_{|\cdot|}:=\\{n\in\mathbb{Z}\big{|}|n|<1\\}$

  * •

$\lambda_{|\cdot|}:=\inf\\{\log_{m}|m|\big{|}\,\,m\in\mathrm{spec}^{+}(\mathbb{Z})\\}$.

Why is $\lambda_{|\cdot|}$ a geometric construction? Note that each $m$ is a
Dedekind real, so by [NV22] we have a map
$\log_{m}\colon\overleftarrow{[0,\infty)}\rightarrow\overleftarrow{[-\infty,\infty)}$.
Then $\lambda_{|\cdot|}$ is defined as an infimum of upper reals, and so (see
Remark 1.6) itself is also an upper real.

To show $\mathfrak{p}_{|\cdot|}$ is a prime ideal, the only point of
difficulty is to show closure under addition and multiplication.

  * •

_Closure under Addition._ Suppose $m,n\in\mathfrak{p}_{|\cdot|}$. If either is
zero, then immediately $m+n\in\mathfrak{p}_{|\cdot|}$. If both are non-zero,
then by definition $|\cdot|$ is non-Archimedean and hence (Lemma 3.3)
ultrametric, so $|m+n|\leq\max\\{|m|,|n|\\}<1$.

  * •

_Closure under Multiplication._ Suppose $m\in\mathfrak{p}_{|\cdot|}$, and
$n\in\mathbb{Z}$. If either is zero, then $m\cdot n\in\mathfrak{p}_{|\cdot|}$.
If $m\neq 0$, then $|\cdot|$ is non-Archimedean. Assume that $m>1$. If $n=1$,
it is obvious $m\cdot n\in\mathfrak{p}_{|\cdot|}$. If $n>1$, the Fundamental
Lemma 3.2 yields

$\log_{n}|n|\leq\max\\{0,\log_{m}|m|\\}=0,$

and so $|n|\leq 1$. As such, $|m\cdot n|=|m||n|<1$ and so $m\cdot
n\in\mathfrak{p}_{|\cdot|}$. Finally, if instead $m<-1$ and/or $n<-1$, the
same argument still holds since $|m|=|-m|$ and $|n|=|-n|$.

Now we know that $\mathfrak{p}_{|\cdot|}$ is prime, by Lemma 1.14 it has a
non-zero element iff there is a (unique) non-zero prime $p$ for which $|p|<1$,
i.e. $\log_{p}|p|<0$, and this holds iff $\lambda_{|\cdot|}<0$. Hence
$(\mathfrak{p}_{|\cdot|},\lambda_{|\cdot|})$ is in
$\mathfrak{P}_{\overleftarrow{\Lambda}}$ and we have completed the
construction of $\widehat{f}$.

In the Archimedean and non-Archimedean cases, we can recover $|\cdot|$ from
$\widehat{f}(|\cdot|)$.

######  Proposition 4.2.

Let $|\cdot|$ be an absolute value on $\mathbb{Z}$.

  1. (i)

$|\cdot|$ is non-Archimedean iff $\lambda_{|\cdot|}<0$. In that case,
$\mathfrak{p}_{|\cdot|}=(p)$ for some unique prime $p$,
$\lambda_{|\cdot|}=\log_{p}|p|$, and for all $n\in\mathbb{N}_{+}$ we have

$|n|=\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda_{|\cdot|}}\text{.}$

  2. (ii)

$|\cdot|$ is Archimedean iff $0\leq\lambda_{|\cdot|}$. In that case,
$\mathfrak{p}_{|\cdot|}=(0)$, and for all $n\in\mathbb{N}_{+}$ we have

$|n|=n^{\lambda_{|\cdot|}}\text{.}$

###### Proof.

In each case, the facts about $\mathfrak{p}_{|\cdot|}$ and the sign of
$\lambda_{|\cdot|}$ have already been noted.

(i): Suppose $|\cdot|$ is non-Archimedean. By multiplicativity,

$|n|=\prod_{q\text{ prime}}|q|^{\mathrm{ord}_{q}(n)}\text{.}$

If $q$ is a prime not equal to $p$, then $q\notin\mathfrak{p}_{|\cdot|}$ and
so $1\leq|q|$. On the other hand, $\log_{q}|q|\leq\max\\{0,\log_{p}|p|\\}=0$
by Lemma 3.2, so $|q|\leq 1$. We deduce that $|q|=1$, and it follows that
$\lambda_{|\cdot|}=\log_{p}|p|$ and

$|n|=|p|^{\mathrm{ord}_{p}(n)}=\left(p^{\log_{p}|p|}\right)^{\mathrm{ord}_{p}(n)}=p^{\lambda_{|\cdot|}\cdot\mathrm{ord}_{p}(n)}\text{.}$

(ii): Suppose $|\cdot|$ is Archimedean. Then for all $n\geq 2$ in
$\mathbb{Z}_{+}$ (the case $n=1$ is trivial) we have
$\log_{n}|n|=\max\\{0,\log_{n}|n|\\}$, which is the constant $M_{|\cdot|}$ of
Lemma 3.2. It follows that $\lambda_{|\cdot|}$ is this constant, so
$|n|=n^{\log_{n}|n|}=n^{\lambda_{|\cdot|}}$. ∎

######  Discussion 4.3 (Topological Constraints by Upper Reals).

Unlike the classical Ostrowski’s Theorem, Proposition 4.2 does not directly
prove that any non-Archimedean absolute value is equivalent to $|\cdot|_{p}$
for some prime $p\in\mathbb{N}_{+}$. How come? Suppose, given some non-
Archimedean $|\cdot|$, we want a real $\lambda_{|\cdot|}$ such that
$|\cdot|=|\cdot|^{\lambda_{|\cdot|}}_{p}$. Since $|p|_{p}=p^{-1}=\frac{1}{p}$,
the relevant exponent would be $\lambda_{|\cdot|}=\log_{\frac{1}{p}}|p|$,
which we know to be a positive lower real by [NV22, Remark 4.4].
Geometrically, it is more natural to use the signed upper real $\lambda$ to
give a uniform treatment.

######  Discussion 4.4 (Restricting to $\mathbb{N}$).

Let us return to Remark 2.6. The suggestion there is that we might be able to
deal with absolute values entirely through their restriction to positive
integers, so long as we use subtractive versions of the definition of ideals
(Remark 1.16) and the triangle inequality (Proposition 2.7).

By Proposition 2.7 it is indeed enough to define an absolute value by its
restriction to $\mathbb{N}$. It can immediately be extended to $\mathbb{Z}$,
after which we have our proof of Ostrowski. However, there still remains the
question whether that proof itself can be framed in terms of the absolute
values on $\mathbb{N}$.

There is a definite obstruction to making this work, and we do not know
whether it can be circumvented. In defining $\widehat{f}$, we need to show
that $\mathfrak{p}_{|\cdot|}$ is a prime ideal (in the sense of $\mathbb{N}$),
and for that we use Lemma 3.3 to show that if $\mathfrak{p}_{|\cdot|}$ is non-
trivial then $|\cdot|$ is ultrametric. But to work with $\mathbb{N}$, the
ultrametric property needs to have a subtractive version,

$|m|\leq\max\\{|m+n|,|n|\\}\text{.}$

Together with the usual version, they say that the two largest of $|m|$, $|n|$
and $|m+n|$ are equal, which is clear once Theorem 4.1 has been proved.
(Consider the two smallest of $\mathrm{ord}_{p}(m)$, $\mathrm{ord}_{p}(n)$,
and $\mathrm{ord}_{p}(m+n)$.) Before then, however, the proof of Lemma 3.3
does not readily adapt to the subtractive version.

###  4.2. Second Direction: $(\mathfrak{p},\lambda)$ determines an Absolute
Value

We now work to define a map

$\displaystyle\widehat{g}\colon\mathfrak{P}_{\overleftarrow{\Lambda}}$
$\displaystyle\longrightarrow[\overleftarrow{\mathsf{av}}]$
$\displaystyle(\mathfrak{p},\lambda)$
$\displaystyle\longmapsto|\cdot|_{\mathfrak{p},\lambda}$

inverse to $\widehat{f}$.

In fact, Proposition 4.2 already tells us how to do this in the complementary
Archimedean and non-Archimedean cases. However, the fact that we require this
construction to be geometric creates subtleties. In particular, it is not
decidable if $\mathfrak{p}$ is non-trivial or $(0)$, nor is it decidable if
$\lambda\leq\lambda^{\prime}$ for any
$\lambda,\lambda^{\prime}\in\overleftarrow{[-\infty,1]}$. Hence, given
$(\mathfrak{p},\lambda)\in(\mathfrak{P},\Lambda)$, our desired map
$\widehat{g}$ cannot be directly defined using the following case-splittings:

  * •

Case 1: $\lambda<0$, Case 2: $0\leq\lambda$; or

  * •

Case 1: $\exists a\in\mathbb{Z}_{\neq 0}$ such that $a\in\mathfrak{p}$, Case
2: $\mathfrak{p}=(0)$.

In other words, the natural faultlines along which one might split
$(\mathfrak{p},\lambda)\in\mathfrak{P}_{\overleftarrow{\Lambda}}$ into the
Archimedean vs. non-Archimedean case does not work in a naive way (i.e. by
using decidability to make the split as a coproduct before defining the map
separately on the two components).

As such we must first find a geometric construction that covers both cases in
a uniform way.

######  Construction 4.5.

Suppose $(\mathfrak{p},\lambda)\in\mathfrak{P}_{\overleftarrow{\Lambda}}$. We
define
$|\cdot|_{\mathfrak{p},\lambda}\colon\mathbb{Z}\rightarrow\overleftarrow{[0,\infty)}$
as

$|n|_{\mathfrak{p},\lambda}=\begin{cases}0&\text{if $n=0$}\\\
\min\left\\{1,\inf\\{\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda}\mid
p\text{ prime in
}\mathfrak{p}\\}\right\\}\boldsymbol{\cdot}\max\left\\{1,n^{\lambda}\right\\}&\text{if
$n>0$}\\\ |-n|_{\mathfrak{p},\lambda}&\text{if $n<0$}\end{cases}$

######  Discussion 4.6.

Although Construction 4.5 consists of many different components, each
component is geometric, and so the final construction is also geometric. To
see this, note:

  * •

The initial case split into $n=0$ vs. $n>0$ vs. $n<0$ is permitted since $<$
is decidable on $\mathbb{Z}$.

  * •

The $p$-adic ordinal $\mathrm{ord}_{p}(n)$ is also geometric. This essentially
follows from the Euclidean Algorithm, which gives a constructive account of
unique prime factorisation for any $n\in\mathbb{Z}_{\neq 0}$.

  * •

The construction

$\inf\\{(p^{\mathrm{ord}_{p}(n)})^{\lambda}\,\big{|}\,p\,\text{prime
in}\,\mathfrak{p}\\}$

is a geometric workaround the fact it is undecidable if $\mathfrak{p}=(0)$.
Its geometricity comes from the fact that upper reals possess arbitrary
$\inf$s (Remark 1.6). We shall see how this works in Proposition 4.7.

  * •

Finally, recall from Section 1.2 that $\min$, $\max$ and multiplication are
all well-defined operations on the non-negative upper reals.

We can now see how, in the two particular cases corresponding to the non-
Archimedean and Archimedean cases in Proposition 4.2, the definition splits in
the expected way.

######  Proposition 4.7.

Let $(\mathfrak{p},\lambda)$ be in $\mathfrak{P}_{\overleftarrow{\Lambda}}$.

  1. (i)

If $\lambda<0$ then there is a unique prime $p$ in $\mathfrak{p}$ and, for
$n\neq 0$,

$|n|_{\mathfrak{p},\lambda}=\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda}\text{.}$

  2. (ii)

If $0\leq\lambda$ then, for $n>0$,

$|n|_{\mathfrak{p},\lambda}=n^{\lambda}\text{.}$

###### Proof.

(i): By the $\mathfrak{P}_{\overleftarrow{\Lambda}}$ condition and Lemma 1.14,
there is a unique prime $p$ in $\mathfrak{p}$ and, for $n>0$, we have both
$\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda}$ and $n^{\lambda}$ less than
1. It follows that $|n|_{\mathfrak{p},\lambda}$ evaluates as stated. For $n<0$
the same formula works.

(ii): If $0\leq\lambda$, then $\mathfrak{p}$ must be $(0)$. Then the $\inf$ in
the expression is empty, which evaluates as $\infty$. We have $1\leq
n^{\lambda}$, so the entire expression evaluates as stated. ∎

With this observation we can use the Case-Splitting Lemma 1.4 to show that
$\widehat{g}$ maps to $[\overleftarrow{\mathsf{av}}]$.

######  Proposition 4.8.

If $(\mathfrak{p},\lambda)$ is in $\mathfrak{P}_{\overleftarrow{\Lambda}}$,
then $|\cdot|_{\mathfrak{p},\lambda}$ is an absolute value, and moreover
$\widehat{f}(|\cdot|_{\mathfrak{p},\lambda})=(\mathfrak{p},\lambda)$.

###### Proof.

We know that $|\cdot|_{\mathfrak{p},\lambda}$ is a map from $\mathbb{Z}$ to
$\overleftarrow{[-\infty,\infty)}$, and $[\overleftarrow{\mathsf{av}}]$ is a
subspace of the space of such maps. We are trying to show that the inverse
image of that subspace is the whole of
$\mathfrak{P}_{\overleftarrow{\Lambda}}$, and for that it suffices to show
that it contains both the open subspace where $\lambda<0$ and its closed
complement, where $0\leq\lambda$. This is what Lemma 1.4 tells us.

In the light of Proposition 4.7, the multiplicative property is clear for both
cases. It is the triangle inequality that we need to address.

Case 1, $\lambda<0$: In fact we expect, and shall prove, the ultrametric
inequality. Suppose $m,n\neq 0$ (it is trivial if either is zero), with
$m=z_{1}p^{r_{1}}$ and $n=z_{2}p^{r_{2}}$, where
$z_{1},z_{2}\in\mathbb{Z}_{\neq 0}$, $r_{1},r_{2}\in\mathbb{N}$ and
$\gcd(z_{1},p)=1=\gcd(z_{2},p)$. We might as well suppose $r_{1}\leq r_{2}$,
so

$m+n=(p^{r_{1}})\cdot(z_{1}+z_{2}p^{r_{2}-r_{1}})\text{,}$

and $\mathrm{ord}_{p}(m+n)\geq r_{1}$. It follows (using $\lambda<0$) that

$|m+n|_{\mathfrak{p},\lambda}\leq
p^{r_{1}\lambda}\leq\max(|m|_{\mathfrak{p},\lambda},|n|_{\mathfrak{p},\lambda})\text{.}$

Case 2, $0\leq\lambda$: By Proposition 4.7, we need to show that

$(m+n)^{\lambda}\leq m^{\lambda}+n^{\lambda}.$ (9)

Suppose we have positive rationals $0<q,t\leq 1$. Since $t^{\text{(---)}}$ is
antitonic with respect to rational exponents, this implies that $t\leq t^{q}$,
and so

$m^{q}t\leq m^{q}t^{q},\qquad\text{for any positive integer $m>0$.}$

This in turn implies

$\begin{split}(m+n)^{q}&=(m+n)^{q}\left(\left(\frac{m}{m+n}\right)+\left(\frac{n}{m+n}\right)\right)\\\
&\leq(m+n)^{q}\left(\frac{m}{m+n}\right)^{q}+(m+n)^{q}\left(\frac{n}{m+n}\right)^{q}=m^{q}+n^{q}\text{.}\end{split}$
(10)

The calculation here is intrinsically two-sided, relying on the assumption
that $q$ is rational: for an upper real $\lambda$ instead of $q$, we should be
trying to multiply, for example, an upper real $(m+n)^{\lambda}$ by a lower
real $\left(\frac{m}{m+n}\right)^{\lambda}$, and that is not well-defined (cf.
the remark on monotonicity in Section 1.2).

However, the _conclusion_ $(m+n)^{q}\leq m^{q}+n^{q}$ still makes sense for
upper reals, with $\leq$ now the (opposite of the) specialisation order. Hence
for an upper real $\lambda$ we have

$(m+n)^{\lambda}=\inf_{\lambda<q}(m+n)^{q}\leq\inf_{\lambda<q}(m^{q}+n^{q})=\inf_{\lambda<q}m^{q}+\inf_{\lambda<q}n^{q}=m^{\lambda}+n^{\lambda}\text{.}$

To summarise: $|\cdot|_{p,\lambda}$ is an absolute value in both Cases 1 and
2. By Lemma 1.4, this defines a map
$\hat{g}\colon\mathfrak{P}_{\overleftarrow{\Lambda}}\to[\overleftarrow{\mathsf{av}}]$.
For the rest of the result we look at the equaliser of
$\widehat{f}\circ\widehat{g}$ and the identity, a subspace of
$\mathfrak{P}_{\overleftarrow{\Lambda}}$. Again, to show it is the whole of
$\mathfrak{P}_{\overleftarrow{\Lambda}}$ it suffices to show the equation is
satisfied in both the two cases.

Case 1, $\lambda<0$: For $\mathfrak{p}$, we have

$\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda}<1\Leftrightarrow
p^{\mathrm{ord}_{p}(n)}>1\Leftrightarrow\mathrm{ord}_{p}(n)>0\Leftrightarrow
p\text{ divides }n\Leftrightarrow n\in\mathfrak{p}\text{.}$

For $\lambda$, consider that for a prime $q$ we have

$\log_{q}\left(\left(p^{\mathrm{ord}_{p}(q)}\right)^{\lambda}\right)=\begin{cases}\log_{p}\left(p^{\lambda}\right)=\lambda&\text{if
$q=p$,}\\\ \log_{q}1=0&\text{if $q\neq p$.}\end{cases}$

Hence the $\inf$ of these is $\lambda$.

Case 2, $0\leq\lambda$: For $\mathfrak{p}$, we have $n^{\lambda}<1$ iff $n=0$.

For $\lambda$,

$\log_{q}\left(q^{\lambda}\right)=\lambda\text{,}$

and the $\inf$ of these is $\lambda$. ∎

It remains to show that $\widehat{g}\circ\widehat{f}$ is the identity. Again
it suffices to verify this on the non-Archimedeans and the Archimedeans, and
that was done in Propositions 4.2 and 4.7.

This concludes the proof of Theorem 4.1.

Notice: the open-closed gluing of $[\overleftarrow{\mathsf{av}}_{NA}]$ to
$[\overleftarrow{\mathsf{av}}_{A}]$ in $[\overleftarrow{\mathsf{av}}]$
transfers to $\mathfrak{P}_{\overleftarrow{\Lambda}}$.

######  Corollary 4.9.

$\begin{split}[\overleftarrow{\mathsf{av}}_{NA}]&\cong\mathrm{spec}^{+}(\mathbb{Z})\times\overleftarrow{[-\infty,0)}\\\
[\overleftarrow{\mathsf{av}}_{A}]&\cong\overleftarrow{[0,1]}\end{split}$

###### Proof.

The main Theorem splits up using Propositions 4.2 and 4.7. ∎

######  Discussion 4.10 (Undecidability and Geometricity).

Given our previous remarks about decidability issues, our proof of Theorem 4.1
highlights an interesting subtlety. Namely, why is undecidability a barrier to
geometricity when defining constructions, but not when proving properties?
Examining the hypotheses of the Case-Splitting Lemma 1.4, a key tool in our
proof, reveals an interesting fine print. The Lemma only justifies analysing a
construction via case-splitting once the construction already exists (i.e. the
map $X\xrightarrow{f}Z$); it does not justify defining a new construction via
case-splitting, as one might have hoped.

##  5\. Absolute Values on $\mathbb{Q}$

We now turn to absolute values on $\mathbb{Q}$, with a view to recovering
Ostrowski’s theorem in its usual form. Straight away, Proposition 2.10 tells
us that
$[\mathbb{Q}\text{-}\overleftarrow{\mathsf{av}}]\cong[\mathbb{Q}\text{-}\mathsf{av}^{+}]$,
so, as far as $\mathbb{Q}$ is concerned, no extra generality is obtained by
using the upper reals.

However, we can still exploit Theorem 4.1 in a non-trivial way by observing
that an absolute value on $\mathbb{Q}$ can be derived from its restriction to
$\mathbb{Z}$ – which, again, is Dedekind and positive definite.

######  Proposition 5.1.

The restriction map
$[\mathbb{Q}\text{-}\mathsf{av}^{+}]\to[\mathbb{Z}\text{-}\mathsf{av}^{+}]$ is
an isomorphism.

###### Proof.

For the inverse, suppose $|\cdot|$ is a positive definite, Dedekind absolute
value on $\mathbb{Z}$. If $n\neq 0$ then $|n|$ is positive, and hence
invertible. Thus, to extend $|\cdot|$ to $\mathbb{Q}$ we must define
$|m/n|=|m|/|n|$. This is easily checked to be well defined and a positive
definite, Dedekind absolute value on $\mathbb{Q}$, restricting to the original
$|\cdot|$ on $\mathbb{Z}$. ∎

We then have the forgetful map to
$[\mathbb{Z}\text{-}\overleftarrow{\mathsf{av}}]$, and our aim is to exploit
its isomorphism with $\mathfrak{P}_{\overleftarrow{\Lambda}}$. Clearly, we
shall be interested in allowing the $\lambda$ components in
$\mathfrak{P}_{\overleftarrow{\Lambda}}$ to be Dedekind.

######  Definition 5.2 (cf. Theorem 4.1).

Define

$\mathfrak{P}_{\Lambda}:=\\{(\mathfrak{p},\lambda)\in\mathrm{ISpec}(\mathbb{Z})\times(-\infty,1]\mid\lambda<0\leftrightarrow\exists
a\in\mathbb{Z}_{\neq 0}.(a\in\mathfrak{p})\\}\text{.}$

Thus we have

${{[\mathbb{Q}\text{-}\overleftarrow{\mathsf{av}}]}}$${{[\mathbb{Q}\text{-}\mathsf{av}^{+}]}}$${{[\mathbb{Z}\text{-}\mathsf{av}^{+}]}}$${{[\mathbb{Z}\text{-}\overleftarrow{\mathsf{av}}]}}$${\mathfrak{P}_{\Lambda}}$${\mathfrak{P}_{\overleftarrow{\Lambda}}}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$$\scriptstyle{\cong}$
(11)

An obvious conjecture is that the isomorphism of Theorem 4.1 lifts to one
between $[\mathbb{Z}\text{-}\mathsf{av}^{+}]$ and $\mathfrak{P}_{\Lambda}$.
Encouragingly (Propositions 5.5 and 5.7), the isomorphisms of Corollary 4.9
_do_ lift, and in Theorem 5.8 we use this to obtain the standard Ostrowski
result for non-trivial Dedekind absolute values.

Unfortunately (Observation 5.10), the isomorphisms do not transfer across the
boundary at the trivial absolute value. The basic problem is that an infimum
of Dedekind reals has an upper part, but is not necessarily Dedekind.
Unexpectedly, we find that $[\mathbb{Z}\text{-}\mathsf{av}^{+}]$ and
$\mathfrak{P}_{\Lambda}$ are two distinct, non-isomorphic open-closed gluings
of $\mathrm{spec}^{+}(\mathbb{Z})\times(-\infty,0)$ to $[0,1]$. We do not
fully understand what is going on here, nor what it has to say about number
theory.

We are already writing $[\overleftarrow{\mathsf{av}}]$ for
$[\mathbb{Z}\text{-}\overleftarrow{\mathsf{av}}]$. Let us also write
$[\mathsf{av}]$ for the isomorphic spaces
$[\mathbb{Q}\text{-}\overleftarrow{\mathsf{av}}]$,
$[\mathbb{Q}\text{-}\mathsf{av}^{+}]$, and
$[\mathbb{Z}\text{-}\mathsf{av}^{+}]$. Recalling Definition 2.11,
$[\mathsf{av}]$ inherits three subspaces from $[\overleftarrow{\mathsf{av}}]$:
the open subspace of non-Archimedeans, its closed complement of Archimedeans,
and the space of ultrametrics. We shall write them as $[\mathsf{av}_{NA}]$,
$[\mathsf{av}_{A}]$, and $[\mathsf{av}_{U}]$. Keep in mind that our definition
of Archimedean is non-standard in that it includes the trivial absolute value.

######  Definition 5.3.

An absolute value $|\cdot|$ over $\mathbb{Q}$ is _non-trivial Archimedean_ if
there exists some $n\in\mathbb{Z}_{\neq 0}$ such that $|n|>1$. They form an
open subspace $[\mathsf{av}_{A^{+}}]$ of $[\mathsf{av}]$.

######  Proposition 5.4.

  1. (i)

$[\mathsf{av}_{NA}]$ and $[\mathsf{av}_{A^{+}}]$ are disjoint.

  2. (ii)

The closed complement of $[\mathsf{av}_{NA}]\vee[\mathsf{av}_{A^{+}}]$ is
$\\{|\cdot|_{0}\\}$.

(In the light of this, we shall call $|\cdot|$ in $[\mathsf{av}]$ _non-
trivial_ if it is in $[\mathsf{av}_{NA}]\vee[\mathsf{av}_{A^{+}}]$.)

  3. (iii)

The closed complement of $[\mathsf{av}_{A^{+}}]$ is $[\mathsf{av}_{U}]$.

###### Proof.

(i) follows from Lemma 3.2: if we have $|m|<1$, with $m\neq 0$, then we cannot
have any $|n|>1$.

(ii) For $n\neq 0$, if $|n|<1$ and $|n|>1$ are both impossible, then we must
have $|n|=1$.

(iii) By part (ii), the closed complement of $[\mathsf{av}_{A^{+}}]$ is the
subspace join of $[\mathsf{av}_{NA}]$ and $\\{|\cdot|_{0}\\}$. Using Lemma
3.3, both these are contained in $[\mathsf{av}_{U}]$.

Conversely, if $|\cdot|$ is ultrametric and non-trivial Archimedean, with
$|n|>1$, then from $|n|=|1+\cdots+1|\leq\max(1,\ldots,1)=1$ we get a
contradiction. Hence $[\mathsf{av}_{U}]$ is contained in the complement of
$[\mathsf{av}_{A^{+}}]$. ∎

The following two propositions extend our previous work in Section 4 to the
present setting. An important constructive issue is a consequence of Remark
1.6: even for a Dedekind absolute value, the upper real $\lambda_{|\cdot|}$ of
Section 4.1, calculated as an $\inf$, is not necessarily Dedekind. To prove it
is, we need to deal with the non-Archimedean and Archimedean cases separately.

######  Proposition 5.5.

We have an isomorphism

$[\mathsf{av}_{NA}]\cong\mathrm{spec}^{+}(\mathbb{Z})\times(-\infty,0)\text{.}$

###### Proof.

If $|\cdot|\in[\mathsf{av}_{NA}]$, consider
$\widehat{f}(|\cdot|)=(\mathfrak{p}_{|\cdot|},\lambda_{|\cdot|})$. Because
$|\cdot|$ is non-Archimedean, there is a unique
$p\in\mathrm{spec}^{+}(\mathbb{Z})$ such that $\mathfrak{p}_{|\cdot|}=(p)$,
and this gives a map $[\mathsf{av}_{NA}]\to\mathrm{spec}^{+}(\mathbb{Z})$. We
then have
$\lambda_{|\cdot|}=\inf_{q\in\mathrm{spec}^{+}(\mathbb{Z})}\log_{q}|q|=\log_{p}|p|$,
because $p$ is the unique prime for which $|p|<1$, and this
$\lambda_{|\cdot|}$ is Dedekind in $(-\infty,0)$. This gives a map

$F\colon[\mathsf{av}_{NA}]\rightarrow\mathrm{spec}^{+}(\mathbb{Z})\times(-\infty,0).$

In the opposite direction, suppose
$(\mathfrak{p},\lambda)\in\mathfrak{P}_{\Lambda}$ has $\mathfrak{p}=(p)$ (and
$\lambda<0$). Then, by Proposition 4.7, $\widehat{g}$ maps it to the absolute
value with $|n|_{(p),\lambda}=\left(p^{\mathrm{ord}_{p}(n)}\right)^{\lambda}$,
which is Dedekind and positive definite. Call the associated map of this
construction $G$. Assembling the data, this gives commutative diagrams

${{[\mathsf{av}_{NA}]}}$${{[\overleftarrow{\mathsf{av}}_{NA}]}}$${{[\overleftarrow{\mathsf{av}}]}}$${\mathrm{spec}^{+}(\mathbb{Z})\times(-\infty,0)}$${\mathfrak{P}_{\Lambda}}$${\mathfrak{P}_{\overleftarrow{\Lambda}}}$$\scriptstyle{F}$$\scriptstyle{\widehat{f}}$$\scriptstyle{G}$$\scriptstyle{\widehat{g}}$
(12)

Both rows are monic (using Discussion 2.3), so $F$ and $G$ are inverse to each
other because $\widehat{f}$ and $\widehat{g}$ are. ∎

######  Corollary 5.6.

For any $|\cdot|\in[\mathsf{av}_{NA}]$ with $|p|<1$,
$p\in\mathrm{spec}^{+}(\mathbb{Z})$, there exists $\alpha\in(0,\infty)$ such
that $|\cdot|=|\cdot|_{p}^{\alpha}$.

###### Proof.

Take $\alpha=-\lambda_{|\cdot|}$ and use Proposition 4.7. ∎

######  Proposition 5.7.

We have isomorphisms

$[\mathsf{av}_{A}]\cong[0,1]\text{,}$
$[\mathsf{av}_{A^{+}}]\cong(0,1]\text{.}$

In particular, for any $|\cdot|\in[\mathsf{av}_{A}]$, there exists
$\alpha\in[0,1]$ such that $|\cdot|=|\cdot|^{\alpha}_{\infty}$; if $|\cdot|$
is non-trivial then $\alpha>0$.

###### Proof.

The proof is similar to that of Proposition 5.5. Define the maps

$\displaystyle F\colon[\mathsf{av}_{A}]$ $\displaystyle\longrightarrow[0,1]$
$\displaystyle|\cdot|$ $\displaystyle\longmapsto\log_{2}|2|,$

and

$\displaystyle G\colon[0,1]$ $\displaystyle\longrightarrow[\mathsf{av}_{A}]$
$\displaystyle\lambda$
$\displaystyle\longmapsto|\cdot|^{\lambda}_{\infty}=\widehat{g}((0),\lambda).$

From the Fundamental Lemma 3.2, we see that $\log_{n}|n|=F(|\cdot|)$ for all
integers $n>1$, and so $\widehat{f}(|\cdot|)=((0),F(|\cdot|))$. Also,
$G(\lambda)$ is clearly Dedekind and positive definite.

We now have a diagram like that of (12), but with $[\mathsf{av}_{A}]$ and
$[0,1]$ on the left. The same argument shows that $F$ and $G$ are mutually
inverse. A similar proof shows that $[\mathsf{av}_{A^{+}}]\cong(0,1]$. ∎

######  Theorem 5.8 (Ostrowski’s Theorem for $\mathbb{Q}$).

Let $|\cdot|$ be a non-trivial absolute value on $\mathbb{Q}$. Then, one of
the following must hold:

  1. (i)

$|\cdot|=|\cdot|^{\alpha}_{\infty}$ for some $\alpha\in(0,1]$; or

  2. (ii)

$|\cdot|=|\cdot|_{p}^{\alpha}$ for some $\alpha\in(0,\infty)$ and some prime
$p\in\mathbb{N}_{+}$.

###### Proof.

Working with non-trivial absolute values, we are in
$[\mathsf{av}_{NA}]\vee[\mathsf{av}_{A^{+}}]$. As it is a join of two disjoint
opens, we can validly use case splitting in the geometric reasoning to combine
the $F$-maps of Propositions 5.5 and 5.7 to get a map from
$[\mathsf{av}_{NA}]\vee[\mathsf{av}_{A^{+}}]$ to $(0,\infty)\vee(0,1]$.
Categorically, $[\mathsf{av}_{NA}]\vee[\mathsf{av}_{A^{+}}]$ is the coproduct
of $[\mathsf{av}_{NA}]$ and $[\mathsf{av}_{A^{+}}]$. Similarly, we can combine
the $G$-maps to get an inverse.

For the second part, we then use Corollary 5.6. ∎

######  Discussion 5.9.

While negation inverts orientation on one-sided reals (and so, a negated upper
real becomes a lower real), a negated Dedekind real is still a Dedekind, so
Corollary 5.6 avoids the same issues mentioned in Discussion 4.3. Further,
unlike the one-sided reals, $-\infty,\infty$ are not Dedekinds, which is why
$\lambda=\log_{\frac{1}{p}}|p|\in(0,\infty)$, as opposed to
$\lambda\in(0,\infty]$. In other words, if we wish to view $\lambda$ as a
Dedekind, then we lose the ability to speak about the true seminorms, the
$p$-characteristic absolute values $|\cdot|_{\mathbb{F}_{p}}$, which
correspond to $|\cdot|^{\infty}_{p}$ for prime $p\in\mathbb{N}_{+}$. This,
combined with Observations 0.1 and 2.4, brings into focus the following
connection between the algebra (absolute values) and topology (the reals):

  * •

Multiplicative Seminorms $\leftrightsquigarrow$ Upper reals

  * •

(Positive Definite) Norms $\leftrightsquigarrow$ Dedekind reals.

Perhaps surprisingly, the assumption of non-triviality is essential in Theorem
5.8. As alluded to earlier, we cannot combine Propositions 5.5 and 5.7 to get
an isomorphism of $[\mathsf{av}]$ with $\mathfrak{P}_{\Lambda}$. In fact,
there is not even any map from $[\mathsf{av}]$ to $\mathfrak{P}_{\Lambda}$
that makes the obvious diagram commute with
$\widehat{f}\colon[\overleftarrow{\mathsf{av}}]\to\mathfrak{P}_{\overleftarrow{\Lambda}}$.
We prove this by restricting to $[\mathsf{av}_{U}]$ and considering the
projection from $\mathfrak{P}_{\Lambda}$ to $(-\infty,0]$.

######  Observation 5.10.

There is no map $F^{\prime}\colon[\mathsf{av}_{U}]\to(-\infty,0]$ compatible
with $\widehat{f}$ restricted to $[\overleftarrow{\mathsf{av}}_{U}]$:

${{[\mathsf{av}_{U}]}}$${{[\overleftarrow{\mathsf{av}}_{U}]}}$${{(-\infty,0]}}$${{\overleftarrow{[-\infty,0]}}}$$\scriptstyle{F^{\prime}}$$\scriptstyle{\widehat{f}}$
(13)

Consequently, there is no map from $[\mathsf{av}]$ (i.e.
$[\mathbb{Z}\text{-}\overleftarrow{\mathsf{av}}]$) to $\mathfrak{P}_{\Lambda}$
that makes diagram (11) commute.

###### Proof.

The conditions tell us that

$F^{\prime}(|\cdot|)=\inf_{p\text{ prime}}\log_{p}|p|\text{,}$

but also that this infimum is to be Dedekind (and not just upper). Let us
examine the inverse image $F^{\prime-1}(-1,0]$.

Using multiplicativity and the theory of logarithms, we see that a subbase of
opens for $[\mathsf{av}_{U}]$ is provided by conditions $q<\log_{p}|p|<r$,
where $q$ and $r$ are rationals and $p$ is prime. Since $|\cdot|$ is
ultrametric, we can assume $q<0$. Hence a base of opens is provided by finite
conjunctions

$\bigwedge_{i=0}^{n}q_{i}<\log_{p_{i}}|p_{i}|<r_{i}\text{.}$ (14)

If $F^{\prime}$ exists, then $F^{\prime-1}(-1,0]$ is an open containing
$|\cdot|_{0}$ (because $F^{\prime}(|\cdot|_{0})=0$), and so $|\cdot|_{0}$ is
in such a basic. We must then have $0<r_{i}$ for every $i$. However, for such
a basic it is impossible to be contained in $F^{\prime-1}(-1,0]$. For we can
define $|\cdot|$ so that $|p|<1$ for some prime $p$ distinct from all the
$p_{i}$s, and $\log_{p}|p|=-2$. Then that $|\cdot|$ is in our basic, but not
in $F^{\prime-1}(-1,0]$.

Now suppose we have a map from $[\mathsf{av}]$ to $\mathfrak{P}_{\Lambda}$ as
suggested. We then have the following commutative diagram.

${{[\mathsf{av}_{U}]}}$${{[\overleftarrow{\mathsf{av}}_{U}]}}$${{[\mathsf{av}]}}$${{[\overleftarrow{\mathsf{av}}]}}$${\mathfrak{P}_{\Lambda}}$${\mathfrak{P}_{\overleftarrow{\Lambda}}}$${{(-\infty,1]}}$${{\overleftarrow{[-\infty,1]}}}$

The vertical composites factor via $(-\infty,0]$ and
$\overleftarrow{[-\infty,0)}$, and we obtain a map $F^{\prime}$ as in Diagram
(13). ∎

######  Discussion 5.11.

To test our understanding, consider the following argument:

  * •

$[\mathsf{av}_{NA}]$ and $[\mathsf{av}_{A}]$ are open/closed complement
subspaces in $[\mathsf{av}]$.

  * •

Propositions 5.5 and 5.7 characterise $[\mathsf{av}_{NA}]$ and
$[\mathsf{av}_{A}]$.

  * •

Hence, by appealing to the Case-Splitting Lemma 1.4 or otherwise, conclude
that $[\mathsf{av}]\cong\mathfrak{P}_{\Lambda}$.

This proof strategy imitates what we did in Section 4, yet Observation 5.10
tells us that the argument is wrong. So where does it fail? Recall: in Section
4, we defined a map

$\displaystyle\hat{f}\colon[\overleftarrow{\mathsf{av}}]$
$\displaystyle\longrightarrow\mathfrak{P}_{\overleftarrow{\Lambda}}$ (15)
$\displaystyle|\cdot|$
$\displaystyle\longmapsto(\mathfrak{p}_{|\cdot|},\lambda_{|\cdot|})$

by verifying that
$\mathfrak{p}_{|\cdot|},\lambda_{|\cdot|}\in\mathfrak{P}_{\overleftarrow{\Lambda}}$
_without_ any initial assumptions on whether $|\cdot|$ is Archimedean or non-
Archimedean. However, our present setting requires $\lambda_{|\cdot|}$ to be a
Dedekind, not just an upper real. This was verified in Propositions 5.5 and
5.7 but notice this argument involves a case-splitting. This is why the
argument in Section 4 does not extend to yield a map
$[\mathsf{av}]\to\mathfrak{P}_{\Lambda}$, or indeed
$[\mathsf{av}_{U}]\to\mathfrak{P}_{\Lambda}$. In addition, the Dedekinds do
not form a subspace of the upper reals (cf. Discussion 2.3), so the Case-
Splitting Lemma cannot be applied to glue Propositions 5.5 and 5.7 together in
order to assemble the desired map.

A related question looks at the spectra. The reader may reasonably wonder:
does $\mathrm{ISpec}$ still play a role in the analysis of absolute values on
$\mathbb{Q}$? After all, $\mathrm{ISpec}$ is not mentioned in the statement of
Theorem 5.8 (unlike Theorem 4.1), and only features implicitly in its proof.
Might we not e.g. use the Zariski spectrum $\mathrm{LSpec}(\mathbb{Z})$
instead to denote the non-Archimedean places of $\mathbb{Q}$?

The answer comes from a closer examination of the maps from $[\mathsf{av}]$ to
the other spectra. We know already that $[\mathsf{av}]$ has a map to
$\mathrm{ISpec}$, via $[\overleftarrow{\mathsf{av}}]$ and
$\mathfrak{P}_{\overleftarrow{\Lambda}}$. Can we use the Dedekind property to
find a map to either $\mathrm{LSpec}$ or $\mathrm{FSpec}$? Again, we can find
a negative answer by restricting to $[\mathsf{av}_{U}]$. Of course, we already
know that there is a map from $[\mathsf{av}_{NA}]$ to
$\mathrm{spec}^{+}(\mathbb{Z})$, and hence to the other three spectra. The
problem comes when we try to bring in the trivial absolute value.

######  Observation 5.12 (coZariski vs. Zariski/Constructible Topology).

There is no map from $[\mathsf{av}_{U}]$ to $\mathrm{LSpec}(\mathbb{Z})$ that
maps $|\cdot|_{0}$ to $\mathbb{Z}-\\{0\\}$, and each $|\cdot|_{p}^{\alpha}$ to
the prime filter of integers not divisible by $p$.

It follows that there is no map to $\mathrm{FSpec}(\mathbb{Z})$ with the
corresponding properties.

###### Proof.

Suppose there did exist such a map

$f\colon[\mathsf{av}_{U}]\rightarrow\mathrm{LSpec}(\mathbb{Z}).$

Now consider the open in $\mathrm{LSpec}(\mathbb{Z})$ comprising all those
prime filters containing 2. It has the prime filters corresponding to all non-
zero primes except 2, as well as the top filter $\mathbb{Z}-\\{0\\}$. Let $U$
be its inverse image under $f$.

Again, $U$ is a join of basics corresponding to propositions as in condition
(14). Hence $|\cdot|_{0}$ is in one of those basics, which must have
$q_{i}<0<r_{i}$ for every $i$ – and the values $r_{i}$ are then irrelevant. We
can now find a 2-adic $|\cdot|$ such that

$\log_{2}|2|>\max_{i}q_{i}\text{,}$

and this $|\cdot|$ is in our basic open. That’s a contradiction, since its
image under $f$ is a (the) prime filter not containing 2. ∎

######  Discussion 5.13.

Informally, Observation 5.12 holds because such a map requires us to decide,
given some generic $|\cdot|\in[\mathsf{av}_{U}]$, whether $|\cdot|$ is trivial
or non-trivial, which we can only do after verifying that $|p|=1$ for all non-
trivial primes $p$. This, however, requires universal quantification and is
therefore non-geometric.

##  6\. Further Remarks and Discussion

Our discussion after Theorem 5.8 gives an overview of how Dedekind absolute
values interact with the “edge cases”. While perhaps unsurprising that
$[\mathsf{av}]$ does not include the $p$-characteristic absolute values
(Discussion 5.9), it was unexpected (at least to the authors) how the trivial
absolute value is enmeshed with topological subtleties. Extending Observation
5.10, we pose the obvious test problem:

######  Problem 6.1.

Characterise the (entire) space $[\mathsf{av}]$, including the trivial
absolute value.

######  Discussion 6.2.

Much has already been said regarding the difficulties reconciling Archimedean
vs. the non-Archimedean structures (see e.g. the introduction of [Maz93]),
which is often viewed as being bound up with the differences between the
analytic vs. the algebraic world. Observations 5.10 and 5.12 highlight a
different perspective: just reconciling the trivial vs. non-trivial non-
Archimedean absolute values itself has its own set of difficulties that
require some care (at least in the Dedekind setting).

Perhaps then before we start connecting the Archimedean and the non-
Archimedean, one should first figure out how the trivial absolute value
interacts with each component. An important clue comes from our Ostrowski’s
Theorem 4.1 for $\mathbb{Z}$: the trivial absolute value acts as a kind of
hinge point between the Archimedeans and non-Archimedeans so long as we work
with the (honest) upper reals. The challenge then is to adapt this picture to
the Dedekind absolute values (Discussion 2.3 may be relevant here). After
which, it will be very interesting to meditate on how our characterisation of
$[\mathsf{av}]$ may tell us something useful about connecting the analytic
with the algebraic.

######  Discussion 6.3.

As alluded to in the introduction, the issue of reconciling trivial vs. non-
trivial norms also persists in Berkovich geometry – see e.g. Example 1.4 in
[Ber90] or the case-split in §3.4 and 3.5 regarding GAGA-type results for
trivial vs. non-trivial valuations. A point-free perspective on some of these
issues was investigated in [Ng23, Ch. 5] – again, the upper reals play a
helpful role.

Might the results of the present paper be understood as suggesting that we
should abandon Dedekind absolute values and simply work with upper-valued
absolute values on $\mathbb{Z}$? We argue no. One important reason is that
part of what makes understanding the Dedekind absolute values so interesting
is their connection with Local-Global Principles from number theory, which is
less clear for the non-Dedekind case.

Let us elaborate. A polynomial $f$ with $\mathbb{Q}$-coefficients is said to
satisfy the _Local-Global Principle_ just in case that $f$ has rational
solutions iff $f$ has solutions over all non-trivial completions of
$\mathbb{Q}$ (up to equivalence). There are well-known examples of polynomials
that satisfy the Local-Global principle (e.g. quadratic forms) as well as
those that fail it (e.g. $3x^{3}+4y^{3}+5z^{3}=0$), and improving our
understanding of the obstructions to this principle is a deep and active area
of research.

Now, we say that two Dedekind absolute values $|\cdot|_{1},|\cdot|_{2}$ of
$\mathbb{Q}$ belong to the same _place_ iff there exists $\alpha\in(0,1]$ such
that $|\cdot|_{1}=|\cdot|^{\alpha}_{2}$ or $|\cdot|_{2}=|\cdot|^{\alpha}_{1}$.
In particular, one can verify that any two absolute values from the same place
define topologically equivalent completions of $\mathbb{Q}$. Notice then that
our Ostrowski’s Theorem 5.8 for $\mathbb{Q}$ classifies all non-trivial
completions of $\mathbb{Q}$ up to equivalence, as is relevant for the Local-
Global Principle. Notice also (e.g. by Discussion 4.3) that the language of
places does not adapt well to the setting of upper-valued absolute values on
$\mathbb{Z}$.

This sets up the following test problem, which will be the focus of a
subsequent paper by the authors.

######  Problem 6.4.

Characterise $[\mathrm{places}]$, the space of places of $\mathbb{Q}$.

######  Discussion 6.5.

Observation 5.12 gives us our first important clue. It is natural to expect
there to exist a quotient map

$\mathrm{quot}\colon[\mathsf{av}_{U}]\longrightarrow[\mathrm{places}_{U}]$

that sends an ultrametric absolute value to its corresponding place. However,
if $[\mathrm{places}_{U}]=\mathrm{LSpec}(\mathbb{Z})$ or
$\mathrm{FSpec}(\mathbb{Z})$, then Observation 5.12 tells us that no such map
exists. This gives another compelling reason for working with
$\mathrm{ISpec}(\mathbb{Z})$, as opposed to the other spectral spaces.

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$\Gamma(\vartheta):=\mathfrak{X}(X_{0})/\textnormal{im}(\rho)$. A section
$[v]\in\Gamma(\vartheta)$ is called _invariant_ if there exists a vector field
$\xi$ on $X_{1}$ which is both $s_{X}$-related and $t_{X}$-related to $v$. The
resulting space of invariant elements is denoted by
$\Gamma(\vartheta)^{\textnormal{inv}}$. Recall that for each section
$\alpha\in\Gamma(A_{X})$ we have an associated multiplicative vector field
$\delta(\alpha)=(\alpha^{r}-\alpha^{l},\rho(\alpha))$ on
$X_{1}\rightrightarrows X_{0}$. From Lemma 4.7 in [13] it follows that there
is a natural linear map from $\mathfrak{X}_{m}(X)/\textnormal{im}(\delta)$ to
$\Gamma(\vartheta)^{\textnormal{inv}}$ which associates to a multiplicative
vector field $\xi$ on $X_{1}$ the class modulo $\textnormal{im}(\rho)$ of the
vector field $v$ on $X_{0}$ associated with $\xi$. Furthermore, if
$X_{1}\rightrightarrows X_{0}$ is proper then the latter map induces an
isomorphism
$\mathfrak{X}_{m}(X)/\textnormal{im}(\delta)\cong\Gamma(\vartheta)^{\textnormal{inv}}$,
see Theorem 6.1 in [13].

Let $(X_{1}\rightrightarrows X_{0},\eta)$ be a proper Riemannian groupoid.
Motivated by the previous facts we define the space of _Killing invariant
sections_ $\Gamma_{\eta}(\vartheta)^{\textnormal{inv}}$ as the set of sections
$[v]\in\Gamma(\vartheta)$ for which there exists a projectable vector field of
Killing type $\xi$ on $X_{1}$ over $v$. That is, Conditions a. and b. in the
definition of projectable vector field of Killing type are satisfied with
$\xi$ and $v$. Therefore, as consequence of Lemma 4.7 and Theorem in [13] we
immediately get that:

######  Proposition 4.21.

There exists a natural isomorphism between
$\mathfrak{o}_{m}^{\textnormal{w}}(X,\eta)/\textnormal{im}(\delta)$ and
$\Gamma_{\eta}(\vartheta)^{\textnormal{inv}}$.

In particular, Theorem 4.20 provides us with a method to construct geometric
Killing vector fields over the quotient Riemannian stack
$([X_{0}/X_{1}],[\eta])$.

Conflict of interests. The authors declare that they have no conflict of
interests.

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