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Abstract Vergleichsstellensätze
for preordered semifields and semirings II

Tobias Fritz Department of Mathematics, University of Innsbruck [email protected]
Abstract.

The present paper continues our foundational work on real algebra with preordered commutative semifields and semirings. We prove two abstract Vergleichsstellensätze for preordered commutative semirings of polynomial growth. These generalize the results of Part I by no longer assuming 101\geq 0. Such a generalization comes with substantial technical complications: our Vergleichsstellensätze now also need to take into account infinitesimal information encoded in the form of monotone derivations in addition to the monotone homomorphisms to the nonnegative reals and tropical reals. The auxiliary technical results we develop along the way include surprising implications between inequalities in preordered semifields and a type classification for multiplicatively Archimedean fully preordered semifields.

Among other applications, two companion papers use these results in order to derive limit new results in probability and information theory; one on asymptotics of random walks on topological abelian groups, and the other on the asymptotics of matrix majorization.

2020 Mathematics Subject Classification:
Primary: 06F25; Secondary: 16W80, 16Y60, 12K10, 14P10
Acknowledgements. We thank Erkka Theodor Haapasalo, Xiaosheng Mu, Tim Netzer and Péter Vrana for useful feedback, discussion and suggestions. We also thank Luciano Pomatto, Markus Schweighofer and Omer Tamuz for further discussions.

1.  Introduction

This paper is part of an emerging research program on real algebra with commutative preordered semirings and semifields. Here, semirings and semifields are like rings and fields, but without the assumption of additive inverses. In Part I [3], detailed motivation for this project was given. It comes in the form of two main points:

  • \triangleright

    Our Vergleichsstellensätze have entirely new applications which are not covered by the classical Positivstellensätze. For example, take any class of mathematical structures—such as representations of a Lie group—for which notions of direct sum and tensor product exist, and the tensor products distribute over direct sums. Then the isomorphism classes form a semiring. In many cases, this semiring carries a canonical preorder with respect to one structure being included (up to isomorphism) in another. One then obtains a preordered semiring, and results such as our Vergleichsstellensätze are useful tools in understanding its structure.

    The results of the present paper have found their first applications in probability and information theory. This comprises new formulas for the asymptotics of random walks on topological abelian groups [4] as well on the asymptotics of matrix majorization [2].

  • \triangleright

    From the perspective of real algebra itself, semifields have a number of advantages over fields that make them of intrinsic interest. One such advantage is that if +(X)\mathbb{R}_{+}(X) is the semifield of rational functions with nonnegative coefficients, then the evaluation maps are well-defined homomorphisms +(X)+\mathbb{R}_{+}(X)\to\mathbb{R}_{+}, in stark contrast to the lack of evaluation homomorphisms on fields of rational functions. Another advantage is that semifields can combine the convenience of multiplicative inverses with the presence of nilpotency. For example, there is a semifield F(+)[X]/(X2)F\coloneqq\mathbb{R}_{(+)}[X]/(X^{2}) whose nonzero elements are the linear polynomials r+sXr+sX with r>0r>0. Since its enveloping ring is the ring of dual numbers, F[X]/(X2)F\otimes\mathbb{Z}\cong\mathbb{R}[X]/(X^{2}), it is clear that FF cannot be embedded into a field.

As in real algebra generally, to understand the structure of a particular preordered semiring, it is useful to probe this structure through homomorphisms to test objects such as the real numbers. These homomorphisms can be thought of in geometrical terms as points of a spectrum. By analogy with Positivstellensätze, Part I has introduced the term Vergleichsstellensatz for a type of result which relates the given algebraic preorder to a spectral preorder. It provides sufficient conditions for the existence of an algebraic certificate witnessing a spectral preorder relation.

More concretely, in Part I we first proved a separation theorem for preordered semifields, which states that every semifield preorder is the intersection of its total semifield preorder extensions. From this, we derived a Vergleichsstellensatz for a certain class of preordered semirings SS in which 1>01>0. In this result, the role of the spectrum is played by the monotone homomorphisms ϕ:S+\phi:S\to\mathbb{R}_{+} and ϕ:S𝕋+\phi:S\to\mathbb{TR}_{+}, where 𝕋+([0,),max,)\mathbb{TR}_{+}\coloneqq([0,\infty),\max,\cdot) is the semifield of tropical reals. Roughly speaking, for nonzero x,ySx,y\in S our Vergleichsstellensatz considers two kinds of algebraic certificates:

  • \triangleright

    A catalytic certificate, stating that there is nonzero aSa\in S with

    axay.ax\leq ay. (1.1)
  • \triangleright

    An asymptotic certificate, which states that

    xnynn1.x^{n}\leq y^{n}\qquad\forall n\gg 1. (1.2)

These are certificates for the spectral preorder relation in the sense that if either of them holds, then ϕ(x)ϕ(y)\phi(x)\leq\phi(y) for all ϕ\phi is easily implied. Our Vergleichsstellensatz now provides an almost converse: if we have strict inequality ϕ(x)<ϕ(y)\phi(x)<\phi(y) for all such ϕ\phi, then both algebraic certificates hold. The fully formal statement with precise assumptions will be recalled as Theorem 2.4.

The goal of the present paper is to dig deeper and develop abstract Vergleichsstellensätze that apply even if 101\not\geq 0. This may seem like an artificial problem on first look, because why would one want to do this? However, there actually are many applications, for example to probability and information theory, where preordered semirings with 101\not\geq 0 appear. To see why, it is enough to note that probability measures are by definition normalized to 11, which indicates that one only wants measures of the same normalization to be comparable at all. In particular, the zero measure will not be comparable to any normalized measure, resulting in 101\not\geq 0 and 101\not\leq 0. Two such applications of the results of this paper have been worked out in companion papers [4, 2], and we now give a brief sketch of what the relevant preordered semiring is111We do so in a watered-down version where only finitely supported measures are considered; the preordered semiring actually considered in [4] has arbitrary measures (of compact support) on a topological abelian group as elements. and why one has 101\not\geq 0. The elements of the polynomial semiring +[X1,,Xd]\mathbb{R}_{+}[X_{1},\ldots,X_{d}] can be identified with finitely supported measures on d\mathbb{N}^{d}, where addition of polynomials corresponds to addition of measures and multiplication of polynomials corresponds to convolution of measures. The semiring preorder generated by

X11,,Xd1X_{1}\geq 1,\quad\ldots,\quad X_{d}\geq 1

then matches exactly the so-called first-order stochastic dominance at the level of measures. In algebraic terms, for p,q+[X1,,Xd]p,q\in\mathbb{R}_{+}[X_{1},\ldots,X_{d}] we have pqp\leq q if and only if one can increase the exponents in some of the terms of pp so as to obtain qq. Then in order for p,q+[X1,,Xd]p,q\in\mathbb{R}_{+}[X_{1},\ldots,X_{d}] to be comparable at all, it is necessary for the sum of the coefficients to be the same, i.e. we must have p(1,,1)=q(1,,1)p(1,\ldots,1)=q(1,\ldots,1). In particular, we indeed have 101\not\geq 0 and 101\not\leq 0 in this preordered semiring.

Nevertheless, it is of interest to know when a catalytic preordering certificate (1.1) and/or an asymptotic preordering certificate (1.2) exist. Since +[X1,,Xd]\mathbb{R}_{+}[X_{1},\ldots,X_{d}] is isomorphic to the semiring of finitely supported measures on d\mathbb{N}^{d} with convolution as multiplication, these are equivalently questions about random walks on d\mathbb{N}^{d}. In particular, the asymptotic preordering detects when one random walk will dominate another one at late times, in the sense of its distribution being further “upwards” componentwise in d\mathbb{N}^{d}. Our Vergleichsstellensätze are exactly the right tool to detect when this dominance occurs by relating it to the spectral preorder, which can be calculated very concretely. Of course, among the relevant spectral preorder relations are the point evaluations like

p(r1,,rd)>q(r1,,rd)r1,,rd>1.p(r_{1},\ldots,r_{d})>q(r_{1},\ldots,r_{d})\qquad\forall r_{1},\ldots,r_{d}>1.

But something special happens at the point (1,,1)(1,\ldots,1): due to p(1,,1)=q(1,,1)p(1,\ldots,1)=q(1,\ldots,1), which is the normalization of probability, it turns out that infinitesimal information around the evaluation homomorphism ff(1,,1)f\mapsto f(1,\ldots,1) must be taken into account. This comes in the form of the inequalities

pXi(1,,1)<qXi(1,,1)i=1,,d.\frac{\partial p}{\partial X_{i}}(1,\ldots,1)<\frac{\partial q}{\partial X_{i}}(1,\ldots,1)\qquad\forall i=1,\ldots,d. (1.3)

In terms of our formalism, these arise because the map

Xi|(1,,1):+[X1,,Xd]\frac{\partial}{\partial X_{i}}|_{(1,\ldots,1)}\>:\>\mathbb{R}_{+}[X_{1},\ldots,X_{d}]\longrightarrow\mathbb{R}

is a monotone derivation with respect to the evaluation homomorphism at (1,,1)(1,\ldots,1). For more detail, we refer to Examples 7.3 and 8.7 and the companion paper [4], where this is done more generally for random walks with compactly supported steps on topological abelian groups in general.

As this example may already indicate, the technical challenges that arise in dealing with preordered semirings with 101\not\geq 0 are substantially greater than in the earlier case with 101\geq 0 that was considered in Part I. A cleaner example illustrating some of the difficulty is the semifield F(+)[X]/(X2)F\coloneqq\mathbb{R}_{(+)}[X]/(X^{2}) mentioned above, when equipped with the semiring preorder defined as

r1+s1Xr2+s2X:r1=r2s1s2.r_{1}+s_{1}X\leq r_{2}+s_{2}X\quad:\Longleftrightarrow\quad r_{1}=r_{2}\>\>\land\>\>s_{1}\leq s_{2}.

Note that this is exactly the semiring preorder generated by X1X\geq 1. Now the only homomorphism F+F\to\mathbb{R}_{+} is the projection r+sXrr+sX\mapsto r, and the only homomorphism F𝕋+F\to\mathbb{TR}_{+} to the tropical reals 𝕋+\mathbb{TR}_{+} is the degenerate one mapping every nonzero element to 11. Therefore if we were to use the same definition of the spectral preordering as in Part I, where this involved only monotone homomorphisms with values in +\mathbb{R}_{+} and 𝕋+\mathbb{TR}_{+}, then this spectral preordering would degenerate completely and could not display any interesting relation to the algebraic preordering on FF.

As indicated already by (1.3), our solution to this problem is to enlarge the spectrum by infinitesimal information in the form of monotone derivations. In fact, we will consider the preordered semifield (+)[X]/(X2)\mathbb{R}_{(+)}[X]/(X^{2}) semifield itself as another test object, so that the structure of other preordered semirings can also be probed through monotone homomorphisms with values in (+)[X]/(X2)\mathbb{R}_{(+)}[X]/(X^{2}). The two components of such a homomorphism form a pair (ϕ,D)(\phi,D), where ϕ:S+\phi:S\to\mathbb{R}_{+} is a homomorphism that is degenerate in the sense that

xyϕ(x)=ϕ(y),x\leq y\quad\Longrightarrow\quad\phi(x)=\phi(y),

and a monotone ϕ\phi-derivation D:SD:S\to\mathbb{R}, which is an additive monotone map satisfying the Leibniz rule with respect to ϕ\phi,

D(xy)=ϕ(x)D(y)+D(x)ϕ(y).D(xy)=\phi(x)D(y)+D(x)\phi(y).

Furthermore, we will also have to consider the opposite semifields +op\mathbb{R}_{+}^{\mathrm{op}} and 𝕋+op\mathbb{TR}_{+}^{\mathrm{op}} as test objects. This is not so surprising in light of the fact that reversing the preorder on a preordered semiring SS results in another preordered semiring SopS^{\mathrm{op}}.

In full technical detail, and using notions that will be introduced in the main text, our Vergleichsstellensatz for catalytic certificates is then the following.

Theorem 7.1.

Let SS be a zerosumfree preordered semidomain with a power universal pair u,u+Su_{-},u_{+}\in S and such that:

  • \triangleright

    S/S/\!\sim has quasi-complements and quasi-inverses.

  • \triangleright

    𝖥𝗋𝖺𝖼(S/)\operatorname{\mathsf{Frac}}(S/\!\sim)\otimes\mathbb{Z} is a finite product of fields.

Let nonzero x,ySx,y\in S with xyx\sim y satisfy the following:

  • \triangleright

    For every nondegenerate monotone homomorphism ϕ:S𝕂\phi:S\to\mathbb{K} with trivial kernel and 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\},

    ϕ(x)<ϕ(y).\phi(x)<\phi(y). (1.4)
  • \triangleright

    For every monotone additive map D:SD:S\to\mathbb{R}, which is a ϕ\phi-derivation for some degenerate homomorphism ϕ:S+\phi:S\to\mathbb{R}_{+} with trivial kernel and satisfies D(u+)=D(u)+1D(u_{+})=D(u_{-})+1,

    D(x)<D(y).D(x)<D(y). (1.5)

Then there is nonzero aSa\in S such that axayax\leq ay.

Moreover, if SS is also a semialgebra, then it is enough to consider +\mathbb{R}_{+}-linear derivations DD in the assumptions.

Conversely, if axayax\leq ay holds for some nonzero aa, then the same spectral conditions (1.4, 1.5) are trivially implied with non-strict inequalities \leq in place of <<. The first two itemized assumptions on SS can be thought of as saying that although we do not require 101\geq 0 to be the case, this should nevertheless not fail too badly, in the sense that the preorder relation \leq must still be suitably large. In the final part of the theorem statement, the term semialgebra refers to the case in which SS comes equipped with a scalar multiplication by +\mathbb{R}_{+}.

We do not currently have a Vergleichsstellensatz for asymptotic certificates that would apply at the same level of generality, but we do have one that applies under somewhat stronger assumptions on SS (which still fall short of assuming 101\geq 0). This takes the following form, which in its conclusions precisely matches our main result of Part I.

Theorem 8.6.

Let SS be a preordered semiring with a power universal element uSu\in S. Suppose that for some dd\in\mathbb{N}, there is a surjective homomorphism :S>0d{0}\|\cdot\|:S\to\mathbb{R}_{>0}^{d}\cup\{0\} with trivial kernel and such that

aba=bab.a\leq b\quad\Longrightarrow\quad\|a\|=\|b\|\quad\Longrightarrow\quad a\sim b.

Let x,ySx,y\in S be nonzero with x=y\|x\|=\|y\|. Then the following are equivalent:

  1. (a)
    • \triangleright

      For every nondegenerate monotone homomorphism ϕ:S𝕂\phi:S\to\mathbb{K} with trivial kernel and 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\},

      ϕ(x)ϕ(y).\phi(x)\leq\phi(y).
    • \triangleright

      For every i=1,,di=1,\ldots,d and monotone i\|\cdot\|_{i}-derivation D:SD:S\to\mathbb{R} with D(u)=1D(u)=1,

      D(x)D(y).D(x)\leq D(y).
  2. (b)

    For every ε>0\varepsilon>0, we have

    xnuεnynn1.x^{n}\leq u^{\lfloor\varepsilon n\rfloor}y^{n}\qquad\forall n\gg 1.

Moreover, suppose that the inequalities in (a) are all strict. Then also the following hold:

  1. (c)

    There is kk\in\mathbb{N} such that

    ukxnukynn1.u^{k}x^{n}\leq u^{k}y^{n}\qquad\forall n\gg 1.
  2. (d)

    If yy is power universal as well, then

    xnynn1.x^{n}\leq y^{n}\qquad\forall n\gg 1.
  3. (e)

    There is nonzero aSa\in S such that

    axay.ax\leq ay.

    Moreover, there is kk\in\mathbb{N} such that aukj=0nxjynja\coloneqq u^{k}\sum_{j=0}^{n}x^{j}y^{n-j} for any n1n\gg 1 does the job.

Finally, if SS is also a semialgebra, then all statements also hold with only +\mathbb{R}_{+}-linear derivations DD in (a).

We again refer to [4] for the application to random walks, which gives much stronger results than what has been achieved with purely probabilistic methods so far, and to [2] for another application to information theory.

Overview

We now give some indication of the content of each section.

  • \triangleright

    Section 2 summarizes the main definitions and results of Part I, so that the present paper can be read independently of [3].222The proof of Theorem 8.6, which refers back to the proof of our Vergleichsstellensatz from Part I, is an exception to this.

  • \triangleright

    Section 3 introduces a few additional relevant definitions and makes some basic observations that will be used in the remainder of the paper.

The next few sections are devoted to developing some structure theory of preordered semifields. This builds the technical groundwork for our main results, but we also expect it to be relevant for future work in the area.

  • \triangleright

    Section 4 proves a number of important and surprisingly strong inequalities in preordered semirings and semifields by elementary means. Although we will not dwell on this relation further, readers with a good background in probability theory will be able to interpret many of those inequalities in terms of second-order stochastic dominance.

  • \triangleright

    Section 5 introduces multiplicatively Archimedean fully preordered semifields and analyzes their structure, resulting in a classification into five types. Remark 5.2 provides a sense in which these preordered semifields are the building blocks of all totally preordered semifields. And since the latter are a stepping stone for analyzing the structure of any preordered semifield by our first Vergleichsstellensatz from Part I (Theorem 2.1), the results of Section 5 are an important part of the structure theory of preordered semifields in general.

  • \triangleright

    Section 6 introduces a certain derived preorder relation on any preordered semifield, the so-called ambient preorder. The main use of this construction is that even if 11 and 0 are not preordered relative to each other in a given preordered semifield, they often will be with regards to the ambient preorder. With some effort, this then allows us to leverage the results of Part I, making them apply in particular to multiplicatively Archimedean fully preordered semifields as studied in Section 5. This results in two embeddings theorems, namely Proposition 6.7 and Proposition 6.8, which constitute the main results of this section.

The final two sections contain our two main results, the proofs of which crucially rely on the auxiliary results of Sections 4, 5 and 6.

  • \triangleright

    Section 7 combines the results obtained so far and derives a catalytic Vergleichsstellensatz, namely the Theorem 7.1 quoted above. While the main applications appear elsewhere, Example 7.3 showcases the application to polynomials mentioned above.

  • \triangleright

    Section 8 then proves an asymptotic Vergleichsstellensatz, namely the Theorem 8.6 quoted above. The proof is based on Theorem 7.1 and the compactness of the test spectrum (Proposition 8.5), as introduced in Definition 8.3. Example 8.7 briefly illustrates the statement in the same polynomial semiring example as before.

2.  Background from Part I

Here, we recall the basic definitions around semirings together with the main new definitions and results developed in Part I [3] as far as they are relevant to this paper. In the following, all unreferenced definitions and results are standard. Those of Part I are referenced with their theorem number prefixed by “I”.

A commutative semiring is a set SS together with two commutative monoid structures (S,+,0)(S,+,0) and (S,,1)(S,\cdot,1) such that the multiplication \cdot distributes over the addition ++ and 10=01\cdot 0=0. Since we will not consider the noncommutative case at all, we simply use the term semiring as short for “commutative semiring”. The set of multiplicatively invertible elements in a semiring SS is denoted S×S^{\times}. There is a unique semiring homomorphism S\mathbb{N}\to S, given by nn1n\mapsto n1, and we often abuse notation by writing nn instead of n1n1. A semiring SS is zerosumfree if x+y=0x+y=0 in SS implies x=y=0x=y=0. It has quasi-complements if for every xSx\in S there are ySy\in S and nn\in\mathbb{N} such that x+y=nx+y=n (I.2.19).

A semifield FF is a semiring such that F×=F{0}F^{\times}=F\setminus\{0\}. A semifield is strict if 11 has no additive inverse, or equivalently if F×F^{\times} is closed under addition, or yet equivalently if FF is zerosumfree. Every semifield that is not a field is strict. A semidomain is a semiring without zero divisors and such that 101\neq 0. A semidomain SS has a semifield of fractions 𝖥𝗋𝖺𝖼(S)\operatorname{\mathsf{Frac}}(S) together with a homomorphism S𝖥𝗋𝖺𝖼(S)S\to\operatorname{\mathsf{Frac}}(S) that is the initial semiring homomorphism from SS into a semifield. 𝖥𝗋𝖺𝖼(S)\operatorname{\mathsf{Frac}}(S) is a strict semifield if and only if SS is zerosumfree.

A preorder is a binary relation that is reflexive and transitive. We typically denote a preorder by \leq, where the symbols \geq, << and >> have their standard induced meaning. We write \preceq when another preorder symbol is needed. We also use the following two derived relations:

  • \triangleright

    \approx is the smallest equivalence relation contained in \leq. In other words, xyx\approx y is shorthand for (xy)(yx)(x\leq y)\>\land\>(y\leq x).

  • \triangleright

    \sim is the equivalence relation generated by \leq.

A preorder is total if xyx\leq y or yxy\leq x for all xx and yy. A map ff between preordered sets is monotone if xyx\leq y implies f(x)f(y)f(x)\leq f(y). A monotone map ff is an order embedding if the converse holds as well. If XX is a preordered set, then XopX^{\mathrm{op}} denotes the preordered set with the opposite preorder in which xyx\leq y holds if and only if yxy\leq x in XX.

A preordered semiring is a semiring together with a preorder relation such that for every aSa\in S, both addition by aa and multiplication by aa are monotone maps SSS\to S (I.3.7). Note that 101\geq 0 is not required. If SS is a preordered semiring, then so is SopS^{\mathrm{op}}. Preordered semifields and preordered semidomains (I.3.16/21) are preordered semirings that are semifields, respectively semidomains, with the additional condition that x0x\approx 0 implies x=0x=0; for a preordered semifield, this means equivalently that 101\not\leq 0 or 101\not\geq 0.

The following is a central result in the theory of preordered semifields.

2.1 Theorem (I.6.6).

Let FF be a preordered semifield. Then the preorder on FF is the intersection of all its total semifield preorder extensions.

In other words, if xyx\not\leq y in FF, then the preorder on FF can be extended to a total semifield preorder \preceq such that still xyx\not\preceq y, which by totality in particular implies xyx\succ y.

A preordered semifield FF is multiplicatively Archimedean if for all nonzero x,y>1x,y>1 in FF there is kk\in\mathbb{N} with yxky\leq x^{k} (Definition I.4.1). The paradigmatic examples of multiplicatively Archimedean semifields are +\mathbb{R}_{+} with its usual algebraic structure as well as the tropical reals 𝕋+\mathbb{TR}_{+}, defined as

𝕋+(+,max,)({},max,+),\mathbb{TR}_{+}\,\coloneqq\,(\mathbb{R}_{+},\max,\cdot)\,\cong\,(\mathbb{R}\cup\{-\infty\},\max,+),

where the first equation is the multiplicative picture of 𝕋+\mathbb{TR}_{+} and the second the additive picture. The isomorphism between them is given by the natural logarithm333While the logarithm with respect to any base will work, we have found it convenient to fix a particular choice.. As far as it matters, we always specify explicitly which picture of 𝕋+\mathbb{TR}_{+} we use, with the multiplicative picture often being preferred. The tropical reals also contain the Boolean semifield 𝔹={0,1}\mathbb{B}=\{0,1\}, in which 1+1=11+1=1. It is the terminal object in the category of strict semifields and semiring homomorphisms. All the strict semifields mentioned in this paragraph are preordered semifields with respect to the standard preorder relation.

2.2 Theorem (I.4.2).

Let FF be a multiplicatively Archimedean preordered semifield. Then FF order embeds into one of the following:

+,+op,𝕋+,𝕋+op.\mathbb{R}_{+},\quad\mathbb{R}_{+}^{\mathrm{op}},\quad\mathbb{TR}_{+},\quad\mathbb{TR}_{+}^{\mathrm{op}}.

A surprising construction on preordered semifields is the categorical product (I.3.19). If F1F_{1} and F2F_{2} are preordered semifields, then their categorical product has underlying set

(F1××F2×){(0,0)}(F_{1}^{\times}\times F_{2}^{\times})\cup\{(0,0)\}

and carries the componentwise algebraic operations and the componentwise preorder, and it is straightforward to see that it is a preordered semifield again (having the universal property of a categorical product).

Next, we recall the relevant definitions around polynomial growth (I.3.27). If SS is a preordered semiring, then a pair of nonzero elements u,u+Su_{-},u_{+}\in S is a power universal pair if uu+u_{-}\leq u_{+} and for every nonzero x,ySx,y\in S with xyx\leq y, there is kk\in\mathbb{N} such that

yukxu+k.yu_{-}^{k}\leq xu_{+}^{k}. (2.1)

It follows that the same property holds already if merely xyx\sim y (I.3.28). We say that SS is of polynomial growth if it has a power universal pair. A power universal element is uSu\in S such that (1,u)(1,u) is a power universal pair. A preordered semifield of polynomial growth has a power universal element given by uu+u1u\coloneqq u_{+}u_{-}^{-1}. Therefore when working with preordered semifields of polynomial growth, we will always work with a power universal element. On the other hand, there are preordered semirings of polynomial growth that do not have a power universal element but merely a power universal pair, such as the polynomial semiring [X1,,Xd]\mathbb{N}[X_{1},\ldots,X_{d}] equipped with the coefficientwise preorder (I.3.37).

Theorem 2.2 implies the following:

2.3 Corollary (I.4.3).

Let FF be a totally preordered semifield of polynomial growth. Then there is a monotone homomorphism ϕ:F𝕂\phi:F\to\mathbb{K} with 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\} such that ϕ(u)>1\phi(u)>1 for every power universal element u>1u>1.

Based on all of these results, we had also developed an abstract Vergleichsstellensatz for preordered semirings SS with 1>01>0 and having a power universal element uu. This uses the test spectrum (I.7.1), which under these assumptions444In Section 7, we will introduce a different definition of test spectrum applying to a different class of preordered semirings. A suitable general definition of spectrum of a preordered semiring remains to be found. is the disjoint union of the monotone homomorphisms to +\mathbb{R}_{+} and 𝕋+\mathbb{TR}_{+}, where the latter are suitably normalized,

𝖳𝖲𝗉𝖾𝗋(S)\displaystyle\mathsf{TSper}(S)\coloneqq { monotone hom ϕ:S+}\displaystyle\{\textrm{ monotone hom }\phi:S\to\mathbb{R}_{+}\}
{ monotone hom ϕ:S𝕋+ with ϕ(u)=e},\displaystyle\sqcup\{\textrm{ monotone hom }\phi:S\to\mathbb{TR}_{+}\textrm{ with }\phi(u)=e\},

and equipped with a certain topology that turns this set into a compact Hausdorff space (I.7.9). The resulting Vergleichsstellensatz reads as follows.

2.4 Theorem (I.7.15).

Let SS be a preordered semiring with 1>01>0 and a power universal element uu. Then for nonzero x,ySx,y\in S, the following are equivalent:

  1. (a)

    ϕ(x)ϕ(y)\phi(x)\leq\phi(y) for all ϕ𝖳𝖲𝗉𝖾𝗋(S)\phi\in\mathsf{TSper}(S).

  2. (b)

    For every ε>0\varepsilon>0 we have

    xnuεnynn1.x^{n}\leq u^{\lfloor\varepsilon n\rfloor}y^{n}\qquad\forall n\gg 1.

Moreover, if ϕ(x)<ϕ(y)\phi(x)<\phi(y) for all ϕ𝖳𝖲𝗉𝖾𝗋(S)\phi\in\mathsf{TSper}(S), then also the following hold:

  1. (c)

    There is kk\in\mathbb{N} such that

    ukxnukynn1.u^{k}x^{n}\leq u^{k}y^{n}\qquad\forall n\gg 1.
  2. (d)

    If yy is a power universal element itself, then also

    xnynn1.x^{n}\leq y^{n}\qquad\forall n\gg 1.
  3. (e)

    There is nonzero aSa\in S such that

    axay.ax\leq ay.

    More concretely, a=ukj=0nxjynja=u^{k}\sum_{j=0}^{n}x^{j}y^{n-j} works for some kk and all n1n\gg 1.

This theorem specializes to a version of Strassen’s Vergleichsstellensatz [7, Corollary 2.6] for u=2u=2. In Part I, we also showed how the classical Positivstellensatz of Krivine–Kadison–Dubois can be derived from the latter, and therefore also from our Vergleichsstellensatz, in an elementary way (I.8.4).

3.  Further basic definitions

Later in the paper, and in particular in the statements of our main results, we will use a few additional concepts that we introduce now.

3.1 Definition.

A semidomain SS has quasi-inverses if for every nonzero xSx\in S there are n>0n\in\mathbb{N}_{>0} and ySy\in S with xy=nxy=n.

As for some basic examples, every semifield trivially has quasi-inverses. \mathbb{N} also has quasi-inverses. So does every semidomain of the form >0n{(0,,0)}\mathbb{N}_{>0}^{n}\cup\{(0,\ldots,0)\}, where the semiring structure is given by the componentwise operations.

The relevance of quasi-inverses is explained by the following observation.

3.2 Lemma.

Let SS be a semidomain with quasi-complements and quasi-inverses. Then the semifield of fractions 𝖥𝗋𝖺𝖼(S)\operatorname{\mathsf{Frac}}(S) has quasi-complements too.

Proof.

We first construct quasi-complements for fractions of the form 1b𝖥𝗋𝖺𝖼(S)\frac{1}{b}\in\operatorname{\mathsf{Frac}}(S) for nonzero bSb\in S. We choose a quasi-inverse cSc\in S with bc=n>0bc=n\in\mathbb{N}_{>0}. Then 1b=cn\frac{1}{b}=\frac{c}{n} in 𝖥𝗋𝖺𝖼(S)\operatorname{\mathsf{Frac}}(S). Choosing a quasi-complement dd for cc makes cn+dn\frac{c}{n}+\frac{d}{n} into a rational number. By further adding a suitable additional rational number, we therefore obtain a natural number, as was to be shown.

For arbitrary ab𝖥𝗋𝖺𝖼(S)\frac{a}{b}\in\operatorname{\mathsf{Frac}}(S), we choose a quasi-complement cSc\in S for aa, resulting in a+c=ma+c=m\in\mathbb{N}. Then

ab+cb=mb,\frac{a}{b}+\frac{c}{b}=\frac{m}{b},

and the claim follows by the previous case upon multiplying any quasi-complement of 1b\frac{1}{b} by mm. ∎

3.3 Definition.

Let SS and TT be preordered semirings. A monotone homomorphism ϕ:ST\phi:S\to T is degenerate if it factors through S/S/\!\sim, or equivalently if for all x,ySx,y\in S,

xyϕ(x)=ϕ(y).x\leq y\quad\Longrightarrow\quad\phi(x)=\phi(y).

Otherwise ϕ\phi is nondegenerate.

The next definition refers to unordered structures only, but will later be used in the context of preordered semirings SS with monotone DD and degenerate ϕ\phi.

3.4 Definition.

Let SS be a semiring and ϕ:S+\phi:S\to\mathbb{R}_{+} a homomorphism. Then a ϕ\phi-derivation is a map D:SD:S\to\mathbb{R} such that the Leibniz rule

D(xy)=ϕ(x)D(y)+D(x)ϕ(y)D(xy)=\phi(x)D(y)+D(x)\phi(y)

holds for all x,ySx,y\in S.

Note that the set of ϕ\phi-derivations is a vector space over \mathbb{R}. And if SS is a preordered semiring, then the set of monotone ϕ\phi-derivations is a convex cone inside this vector space. In either case, the ϕ\phi-derivations can be thought of geometrically as tangent vectors to the spectral point ϕ\phi.

Our results will take a slightly stronger form for semirings which have a scalar multiplication by +\mathbb{R}_{+}.

3.5 Definition.

A semialgebra555Since we will not consider scalar multiplication by anything other than +\mathbb{R}_{+}, we prefer omitting explicit mention of the semifield of scalars. SS is a semiring together with a scalar multiplication

+×S\displaystyle\mathbb{R}_{+}\times S S\displaystyle\longrightarrow S
(r,x)\displaystyle(r,x) rx\displaystyle\longmapsto rx

that is a commutative monoid homomorphism in each argument and satisfies the following additional laws:

  • \triangleright

    1x=x1x=x,

  • \triangleright

    (rx)(sy)=(rs)(xy)(rx)(sy)=(rs)(xy).

In other words, a semialgebra is a semiring which at the same time is a semimodule over +\mathbb{R}_{+} [5, Chapter 14] in such a way that the multiplication of SS is bilinear.666The definition of semimodule includes the additional law r(sx)=(rs)xr(sx)=(rs)x, but this is implied by the ones we have assumed. As with algebras over commutative rings, a semialgebra is equivalently a semiring SS equipped with a semiring homomorphism +S\mathbb{R}_{+}\to S.

3.6 Definition.

For a semiring SS, we write SS\otimes\mathbb{Z} for the ring generated by SS, i.e. the initial object in the category of commutative rings equipped with a semiring homomorphism SSS\to S\otimes\mathbb{Z}.

As is well-known, one obtains SS\otimes\mathbb{Z} by applying the Grothendieck construction to SS, which means taking the elements of SS\otimes\mathbb{Z} to be formal differences of elements of SS. If SS is a semialgebra, then SS\otimes\mathbb{Z} is an \mathbb{R}-algebra in the obvious way.

Changing topic, we will also need a piece of terminology for preorders that are not necessarily total, but merely total on connected components.

3.7 Definition.
  1. (a)

    A preorder relation \leq is full if

    xyxyxyx\sim y\quad\Longrightarrow\quad x\leq y\enspace\lor\enspace x\geq y

    for all xx and yy.

Here is an equivalent characterization.

3.8 Lemma.

A preorder is full if and only if the following holds for all a,x,ya,x,y:

ax,yx,yaxyxy.a\leq x,y\enspace\lor\enspace x,y\leq a\quad\Longrightarrow\quad x\leq y\enspace\lor\enspace x\geq y.
Proof.

This condition is clearly necessary. For sufficiency, let us assume that the condition holds. We temporarily write xyx\simeq y as shorthand for xyxyx\leq y\>\lor\>x\geq y. In order to prove that this is indeed the equivalence relation generated by \leq, which is \sim, we need to show that it is transitive. So let xyzx\simeq y\simeq z. If xyzx\leq y\leq z, then we can conclude xzx\leq z and hence xzx\simeq z by transitivity of \leq, and likewise if xyzx\geq y\geq z. If xyzx\leq y\geq z instead, then applying the assumption with aya\coloneqq y results in the desired xzx\simeq z, and similarly if xyzx\geq y\leq z. ∎

Clearly every total preorder is full, but not conversely. For example, the trivial preorder on any set is full.

4.  Inequalities in preordered semirings and semifields

In this section, we prove some elementary but nontrivial results on implications between inequalities in preordered semirings and semifields (with the focus on the latter). These will form an important building block for the deeper results that we develop in the subsequent sections.

Chaining inequalities in preordered semirings

We start with some observations on chaining inequalities. In this subsection, everything takes place in a preordered semiring SS, without any additional hypotheses.

4.1 Lemma.

If a+xa+ya+x\leq a+y, then also a+nxa+nya+nx\leq a+ny for every nn\in\mathbb{N}.

Proof.

The claim is trivial for n=0n=0. For the induction step, we use

a+(n+1)x\displaystyle a+(n+1)x =(a+x)+nx(a+y)+nx\displaystyle=(a+x)+nx\leq(a+y)+nx
=y+(a+nx)y+(a+ny)=a+(n+1)y.\displaystyle=y+(a+nx)\leq y+(a+ny)=a+(n+1)y.\qed

We will routinely use this trick in the rest of the paper and simply call it chaining. A stronger statement along the same lines is as follows.

4.2 Lemma.

Let p=iriXi[X]p=\sum_{i}r_{i}X^{i}\in\mathbb{N}[X] be any polynomial with coefficients ri>0r_{i}>0 for all i=0,,deg(p)i=0,\ldots,\deg(p). If x+1y+1x+1\leq y+1 in SS, then also p(x)p(y)p(x)\leq p(y),

Proof.

We first prove that

xp(y)+1yp(y)+1xp(y)+1\leq yp(y)+1

for any such polynomial pp. Using well-founded induction, it is enough to show that if this holds for pp, then it also holds for p+1p+1 and for Xp+1Xp+1. Indeed for the former,

x(p(y)+1)+1xp(y)+y+1y(p(y)+1)+1,x(p(y)+1)+1\leq xp(y)+y+1\leq y(p(y)+1)+1,

where we first use the overall assumption and then the induction assumption, whereas for the latter similarly

x(yp(y)+1)+1\displaystyle x(yp(y)+1)+1 =xyp(y)+x+1xyp(y)+y+1\displaystyle=xyp(y)+x+1\leq xyp(y)+y+1
=y(xp(y)+1)+1y(yp(y)+1)+1,\displaystyle=y(xp(y)+1)+1\leq y(yp(y)+1)+1,

as was to be shown.

Getting to the claim itself, we use the same type of induction on pp. Now the first case is trivial, while the second case has induction assumption p(x)p(y)p(x)\leq p(y) and proves that

1+xp(x)1+xp(y)1+yp(y),1+xp(x)\leq 1+xp(y)\leq 1+yp(y),

where the first step is by induction assumption and the second by the auxiliary statement above. ∎

Some inequalities in preordered semifields

In this subsection and the following ones, everything takes place in a preordered semifield FF. The next few results will be a working horse for us in Section 5.

4.3 Lemma.

Let xF×x\in F^{\times}. If x+x12x+x^{-1}\geq 2, then also the following hold for all m,n1m,n\geq 1:

  1. (i)

    mxn+nxmm+nmx^{n}+nx^{-m}\geq m+n.

  2. (ii)

    2n1(xn+xn)(x+x1)n2^{n-1}(x^{n}+x^{-n})\geq(x+x^{-1})^{n}.

  3. (iii)

    xm+n+1xm+xnx^{m+n}+1\geq x^{m}+x^{n}.

If x+x2>2x+x^{-2}>2, then these inequalities also hold strictly.

Proof.

We focus on the non-strict inequality case, since the strict one follows the same way upon noting that at least one inequality in each chain of inequalities will be strict. We first prove a few auxiliary statements that are special cases of the above claims. We routinely use the assumption in the form xj+1+xj12xjx^{j+1}+x^{j-1}\geq 2x^{j}.

  1. (a)

    x2+x2x+x1x^{2}+x^{-2}\geq x+x^{-1}.

    Indeed, repeatedly applying the assumption gives

    2(x+x1)(x2+x2)\displaystyle 2(x+x^{-1})(x^{2}+x^{-2}) =2x3+2x+2x1+2x3\displaystyle=2x^{3}+2x+2x^{-1}+2x^{-3}
    2x3+x+2+x1+2x3\displaystyle\geq 2x^{3}+x+2+x^{-1}+2x^{-3}
    x3+2x2+2+2x2+x3\displaystyle\geq x^{3}+2x^{2}+2+2x^{-2}+x^{-3}
    x3+x2+2x+2x1+x2+x3\displaystyle\geq x^{3}+x^{2}+2x+2x^{-1}+x^{-2}+x^{-3}
    x3+x2+x+2+x1+x2+x3\displaystyle\geq x^{3}+x^{2}+x+2+x^{-1}+x^{-2}+x^{-3}
    3x2+2+3x2\displaystyle\geq 3x^{2}+2+3x^{-2}
    2x2+2x+2x1+2x2\displaystyle\geq 2x^{2}+2x+2x^{-1}+2x^{-2}
    2(x2+2+x2)\displaystyle\geq 2(x^{2}+2+x^{-2})
    =2(x+x1)(x+x1),\displaystyle=2(x+x^{-1})(x+x^{-1}),

    so that the claim follows upon cancelling 2(x+x1)2(x+x^{-1}).

  2. (b)

    The map nxn+xnn\mapsto x^{n}+x^{-n} is monotone in nn\in\mathbb{N}.

    Indeed using induction on nn, the inequality

    xn+1+x(n+1)xn+xnx^{n+1}+x^{-(n+1)}\geq x^{n}+x^{-n}

    holds by assumption in the base case n=0n=0. For the induction step from nn to n+1n+1, we compute

    (x+x1)(xn+2+x(n+2))\displaystyle(x+x^{-1})(x^{n+2}+x^{-(n+2)}) =xn+3+xn+1+x(n+1)+x(n+3)\displaystyle=x^{n+3}+x^{n+1}+x^{-(n+1)}+x^{-(n+3)}
    xn+3+xn1+x(n1)+x(n+3)\displaystyle\geq x^{n+3}+x^{n-1}+x^{-(n-1)}+x^{-(n+3)}
    xn+2+xn+xn+x(n+2)\displaystyle\geq x^{n+2}+x^{n}+x^{-n}+x^{-(n+2)}
    =(x+x1)(xn+1+x(n+1)),\displaystyle=(x+x^{-1})(x^{n+1}+x^{-(n+1)}),

    and again cancel the term x+x1x+x^{-1} from both sides. Here, the first inequality holds by induction assumption and the second by the previous item.

  3. (c)

    In particular, we therefore have xn+xn2x^{n}+x^{-n}\geq 2 for all nn\in\mathbb{N}.

  4. (d)

    xn+1+n(n+1)xx^{n+1}+n\geq(n+1)x for all nn\in\mathbb{N}.

    This claim is trivial for n=0n=0 and holds by assumption for n=1n=1. For all other nn we use induction, distinguishing the case of even exponent,

    x2m+2+(2m+1)\displaystyle x^{2m+2}+(2m+1) =xm+1(xm+1+x(m+1))+2m\displaystyle=x^{m+1}(x^{m+1}+x^{-(m+1)})+2m
    2xm+1+2m\displaystyle\geq 2x^{m+1}+2m
    (2m+2)x,\displaystyle\geq(2m+2)x,

    where the first inequality is by the previous item and the second by the induction assumption for n=mn=m. The case of odd exponent is slightly more difficult: with m1m\geq 1,

    (x+1)(x2m+1+2m)\displaystyle(x+1)(x^{2m+1}+2m) =x2m+2+x2m+1+2mx+2m\displaystyle=x^{2m+2}+x^{2m+1}+2mx+2m
    2xm+1+x2m+1+2mx+(2m1)\displaystyle\geq 2x^{m+1}+x^{2m+1}+2mx+(2m-1)
    2xm+1+2mx2+x+(2m1)\displaystyle\geq 2x^{m+1}+2mx^{2}+x+(2m-1)
    xm+1+2mx2+(m+2)x+(m1)\displaystyle\geq x^{m+1}+2mx^{2}+(m+2)x+(m-1)
    3mx2+3x+(m1)\displaystyle\geq 3mx^{2}+3x+(m-1)
    (2m+1)x2+(2m+1)x\displaystyle\geq(2m+1)x^{2}+(2m+1)x
    =(x+1)(2m+1)x,\displaystyle=(x+1)(2m+1)x,

    where the first inequality uses xm+1+x(m+1)2x^{m+1}+x^{-(m+1)}\geq 2 as an instance of (c), the subsequent three use the induction assumption, and the final one is just the assumed x+x12x+x^{-1}\geq 2.

We now prove the actual three claims.

  1. (i)

    This is trivial for n=0n=0 or m=0m=0. For the induction step in nn assuming fixed m1m\geq 1, we use

    (m+n)(mxn+1+(n+1)xm)\displaystyle(m+n)\mathopen{}\mathclose{{}\left(mx^{n+1}+(n+1)x^{-m}}\right) m(m+n+1)xn+n(m+n+1)xm\displaystyle\geq m\mathopen{}\mathclose{{}\left(m+n+1}\right)x^{n}+n\mathopen{}\mathclose{{}\left(m+n+1}\right)x^{-m}
    (m+n)(m+n+1),\displaystyle\geq(m+n)(m+n+1),

    where the first inequality holds because of (m+n)xn+1+xm(m+n+1)xn(m+n)x^{n+1}+x^{-m}\geq(m+n+1)x^{n} as a consequence of the previous item, and the second by the induction assumption.

  2. (ii)

    For any n1n\geq 1 and j=0,,nj=0,\ldots,n, the inequality

    (nj)xn+jxnnxn2j(n-j)x^{n}+jx^{-n}\geq nx^{n-2j}

    holds by the previous item. Together with standard identities for binomial coefficients, it gives

    2n1(xn+xn)\displaystyle 2^{n-1}\mathopen{}\mathclose{{}\left(x^{n}+x^{-n}}\right) =j=0n1(n1nj1)xn+j=1n(n1j1)xn\displaystyle=\sum_{j=0}^{n-1}\binom{n-1}{n-j-1}x^{n}+\sum_{j=1}^{n}\binom{n-1}{j-1}x^{-n}
    =n1j=0n(nj)((nj)xn+jxn)\displaystyle=n^{-1}\sum_{j=0}^{n}\binom{n}{j}\mathopen{}\mathclose{{}\left((n-j)x^{n}+jx^{-n}}\right)
    j=0n(nj)xn2j=(x+x1)n,\displaystyle\geq\sum_{j=0}^{n}\binom{n}{j}x^{n-2j}=(x+x^{-1})^{n},

    as was to be shown.

  3. (iii)

    We derive this from (i),

    (m+n)(xm+n+1)\displaystyle(m+n)(x^{m+n}+1) =(mxn+nxm)xm+(nxm+mxn)xn\displaystyle=(mx^{n}+nx^{-m})x^{m}+(nx^{m}+mx^{-n})x^{n}
    (m+n)(xm+xn),\displaystyle\geq(m+n)(x^{m}+x^{n}),

    which already gives the claim upon dividing by m+nm+n. ∎

Inequalities of a similar flavour are now quite easy to derive.

4.4 Lemma.

Let xF×x\in F^{\times}. If x+x12x+x^{-1}\geq 2, then also

(n+22)xn(n+12)xn+1+j=0nxj\binom{n+2}{2}x^{n}\leq\binom{n+1}{2}x^{n+1}+\sum_{j=0}^{n}x^{j}

for all nn\in\mathbb{N}.

Proof.

By Lemma 4.3(i), we have

(nj+1)xn(nj)xn+1+xj,(n-j+1)x^{n}\leq(n-j)x^{n+1}+x^{j},

for all j=0,,nj=0,\ldots,n, and this gives the claim upon summation over jj. ∎

One of the auxiliary inequalities used in the proof of Lemma 4.3 is worth noting separately.

4.5 Corollary.

Let xF×x\in F^{\times}.

  1. (i)

    If x+x12x+x^{-1}\geq 2, then xn+xn2x^{n}+x^{-n}\geq 2 for all nn\in\mathbb{N}.

  2. (ii)

    If x+x1>2x+x^{-1}>2, then xn+xn>2x^{n}+x^{-n}>2 for all n>0n\in\mathbb{N}_{>0}.

Alternatively, this can also be regarded as a weakening of Lemma 4.3(ii). The following two lemmas are again important technical results, the significance of which will become clearer in Section 5.

4.6 Lemma.

Let x1x\geq 1 in FF be such that

xn+1+1xn+1x^{n+1}+1\leq x^{n}+1

for some nn\in\mathbb{N}. Then also (x+1)m2mxn(x+1)^{m}\leq 2^{m}x^{n} for all mm\in\mathbb{N}.

Proof.

We start with an auxiliary statement similar to Lemma 4.2: whenever p=iriXi[X]p=\sum_{i}r_{i}X^{i}\in\mathbb{N}[X] is any polynomial with coefficients r0,,rdeg(p)0r_{0},\ldots,r_{\deg(p)}\neq 0, then

p(x)irixmin(i,n).p(x)\leq\sum_{i}r_{i}x^{\min(i,n)}. (4.1)

To prove this, we use well-founded induction similar to the one in the proof of Lemma 4.2. The statement is trivial whenever deg(p)n\deg(p)\leq n. When deg(p)>n\deg(p)>n, we can write

p=Xdeg(p)+Xdeg(p)(n+1)+p^p=X^{\deg(p)}+X^{\deg(p)-(n+1)}+\hat{p}

for some p^[X]\hat{p}\in\mathbb{N}[X], where by the induction assumption the claim can be assumed to hold for the “smaller” polynomial Xdeg(p)1+Xdeg(p)(n+1)+p^X^{\deg(p)-1}+X^{\deg(p)-(n+1)}+\hat{p}, which differs from pp only in the exponent of the leading term. But then

p(x)=(xn+1+1)xdeg(p)(n+1)+p^(x)\displaystyle p(x)=(x^{n+1}+1)x^{\deg(p)-(n+1)}+\hat{p}(x) (xn+1)xdeg(p)(n+1)+p^(x)\displaystyle\leq(x^{n}+1)x^{\deg(p)-(n+1)}+\hat{p}(x)
=xdeg(p)1+xdeg(p)(n+1)+p^(x).\displaystyle=x^{\deg(p)-1}+x^{\deg(p)-(n+1)}+\hat{p}(x).

Applying the induction assumption now proves the claim (4.1) upon using deg(p)>n\deg(p)>n.

Upon bounding the right-hand side of (4.1) further using x1x\geq 1, we obtain the somewhat weaker bound

p(x)(iri)xn=p(1)xn,p(x)\leq\mathopen{}\mathclose{{}\left(\sum_{i}r_{i}}\right)x^{n}=p(1)x^{n},

which is more convenient since now the right-hand side is a mere monomial. The claim follows upon applying this statement to the polynomial p(X+1)mp\coloneqq(X+1)^{m}. ∎

4.7 Lemma.

Let x1x\geq 1 in FF be such that x2+23xx^{2}+2\geq 3x. Then also

xn+1+12xnx^{n+1}+1\geq 2x^{n}

for all nn\in\mathbb{N}.

Proof.

We use induction on nn. The base case n=0n=0 is trivial by x1x\geq 1. The induction step from nn to n+1n+1 is

(2x+1)(xn+2+1)\displaystyle(2x+1)(x^{n+2}+1) =2xn+3+xn+2+2x+1\displaystyle=2x^{n+3}+x^{n+2}+2x+1
2xn+3+2xn+1+x+1\displaystyle\geq 2x^{n+3}+2x^{n+1}+x+1
xn+3+3xn+2+x+1\displaystyle\geq x^{n+3}+3x^{n+2}+x+1
xn+3+2xn+2+2xn+1+1\displaystyle\geq x^{n+3}+2x^{n+2}+2x^{n+1}+1
xn+3+2xn+2+xn+1+2xn\displaystyle\geq x^{n+3}+2x^{n+2}+x^{n+1}+2x^{n}
xn+3+xn+2+4xn+1\displaystyle\geq x^{n+3}+x^{n+2}+4x^{n+1}
4xn+2+2xn+1\displaystyle\geq 4x^{n+2}+2x^{n+1}
=(2x+1)2xn+1,\displaystyle=(2x+1)2x^{n+1},

where each inequality step uses either the induction assumption or the assumed inequality x2+23xx^{2}+2\geq 3x. ∎

Applying Lemma 4.3 to both FF and FopF^{\mathrm{op}} gives a result which is worth stating separately, since it has some relevance to the arctic case in Section 5.

4.8 Lemma.

Let xF×x\in F^{\times}. If x+x12x+x^{-1}\approx 2, then also the following hold for all m,nm,n\in\mathbb{N}:

  1. (i)

    xn+xn2x^{n}+x^{-n}\approx 2.

  2. (ii)

    mxn+nxmm+nmx^{n}+nx^{-m}\approx m+n.

  3. (iii)

    xm+xnxm+n+1x^{m}+x^{n}\approx x^{m+n}+1.

Note that the first equation is a special case of the second for m=nm=n, by invertibility of positive integers, but it nevertheless seems worth stating separately.

Proof.

By Lemma 4.3. ∎

A supermodularity inequality in preordered semifields

So far, most of our inequality results have been concerned with polynomial expressions involving only a single element of FF. We now move beyond that case.

4.9 Lemma.

For xF×x\in F^{\times} with x+x12x+x^{-1}\geq 2 and any aFa\in F, the function

F,na+xn\mathbb{Z}\longrightarrow F,\qquad n\longmapsto a+x^{n}

is multiplicatively supermodular: for all m,nm,n\in\mathbb{N} and \ell\in\mathbb{Z},

(a+x+m+n)(a+x)(a+x+m)(a+x+n).(a+x^{\ell+m+n})(a+x^{\ell})\geq(a+x^{\ell+m})(a+x^{\ell+n}).

Moreover, this holds with \approx under the stronger assumption x+x12x+x^{-1}\approx 2.

Proof.

Since aa is arbitrary, it is sufficient to consider the case =0\ell=0 for simplicity. Then we have

(a+xm+n)(a+1)\displaystyle(a+x^{m+n})(a+1) =a2+a(xm+n+1)+xm+n\displaystyle=a^{2}+a(x^{m+n}+1)+x^{m+n}
a2+a(xm+xn)+xm+n\displaystyle\geq a^{2}+a(x^{m}+x^{n})+x^{m+n}
=(a+xn)(a+xm),\displaystyle=(a+x^{n})(a+x^{m}),

where the inequality step is by Lemma 4.3(iii). The claim about \approx holds likewise by Lemma 4.8. ∎

Towards cancellation criteria in preordered semifields

The following results will later be strengthened, under additional hypotheses, to useful cancellation criteria.

4.10 Lemma.

Suppose that x,yF×x,y\in F^{\times} satisfy x+x12x+x^{-1}\geq 2 and y1y\geq 1. Then

x+1y+1xnyn+1n.x+1\leq y+1\quad\Longrightarrow\quad x^{n}\leq y^{n+1}\enspace\forall n\in\mathbb{N}.
Proof.

By Lemma 4.2, the assumption implies that for given nn\in\mathbb{N},

(n+12)xn+1+j=0nxj(n+12)yn+1+j=0nyj.\binom{n+1}{2}x^{n+1}+\sum_{j=0}^{n}x^{j}\leq\binom{n+1}{2}y^{n+1}+\sum_{j=0}^{n}y^{j}.

Together with Lemma 4.4, we get

(n+22)xn(n+12)xn+1+j=0nxj(n+12)yn+1+j=0nyj(n+22)yn+1,\binom{n+2}{2}x^{n}\leq\binom{n+1}{2}x^{n+1}+\sum_{j=0}^{n}x^{j}\leq\binom{n+1}{2}y^{n+1}+\sum_{j=0}^{n}y^{j}\leq\binom{n+2}{2}y^{n+1},

where the final step is simply by using y1y\geq 1. ∎

4.11 Lemma.

Suppose that a,x,yF×a,x,y\in F^{\times} satisfy x+x12x+x^{-1}\geq 2 and y1y\geq 1. Then

a+xa+ya1+xa1+yxnyn+1n.\begin{matrix}a+x\leq a+y\\[2.0pt] a^{-1}+x\leq a^{-1}+y\end{matrix}\quad\Longrightarrow\quad x^{n}\leq y^{n+1}\enspace\forall n\in\mathbb{N}.
Proof.

Adding the first inequality of the main assumption to aa times the second gives

(a+1)(x+1)=a+x+1+axa+y+1+ay=(a+1)(y+1),(a+1)(x+1)=a+x+1+ax\leq a+y+1+ay=(a+1)(y+1),

so that the claim follows from the previous lemma upon cancelling a+1a+1. ∎

5.  Type classification of multiplicatively Archimedean
fully preordered semifields

Theorem 2.2 is an embedding theorem for multiplicatively Archimedean totally preordered semifields. We now aim at generalizing this statement to a substantially more difficult case, namely from total semifield preorders to those that are merely full in the sense of Definition 3.7, i.e. total on connected components. In particular, this does not require 0 and 11 to be ordered relative to one another. We now explain why an embedding theorem as simple as Theorem 2.2 cannot be expected to hold.

5.1 Example.

Let F(+)[X]/(X2)F\coloneqq\mathbb{R}_{(+)}[X]/(X^{2}) be the semifield of all linear functions r+sXr+sX with r>0r>0 or r=s=0r=s=0 modulo X2X^{2}. This is a semifield because (r+sX)1=r2(rsX)(r+sX)^{-1}=r^{-2}(r-sX) for r>0r>0, and it becomes a preordered semifield if we put

r1+s1Xr2+s2X:r1=r2s1s2.r_{1}+s_{1}X\leq r_{2}+s_{2}X\qquad:\Longleftrightarrow\qquad r_{1}=r_{2}\enspace\land\enspace s_{1}\leq s_{2}.

It is clear that this preorder is full. However, this fully preordered semifield has the counterintuitive feature that a1a\sim 1 in FF implies a+a1=2a+a^{-1}=2. Note that these are exactly the elements of the form a=1+sXa=1+sX for any ss\in\mathbb{R}.

Moreover, since x+x1=2x+x^{-1}=2 in +\mathbb{R}_{+} or 𝕋+\mathbb{TR}_{+} only happens for x=1x=1, it follows that every monotone homomorphism ϕ:F+\phi:F\to\mathbb{R}_{+} or ϕ:F𝕋+\phi:F\to\mathbb{TR}_{+} satisfies ϕ(a)=1\phi(a)=1 for a1a\sim 1. Therefore there is no order embedding of FF into +\mathbb{R}_{+} or 𝕋+\mathbb{TR}_{+}.

5.2 Remark.

Let FF be a totally preordered semifield, and let aF×a\in F^{\times} with a1a\geq 1 be given. Then the layer preorder a\leq_{a} is the relation defined by

xay:(k:xyakyxak)(n:xnyna)x\leq_{a}y\quad:\Longleftrightarrow\quad\mathopen{}\mathclose{{}\left(\exists k\in\mathbb{N}:x\leq ya^{k}\>\land\>y\leq xa^{k}}\right)\>\land\>\mathopen{}\mathclose{{}\left(\forall n\in\mathbb{N}:x^{n}\leq y^{n}a}\right)

Then the layer preorder a\leq_{a} makes FF into a multiplicatively Archimedean fully preordered semifield.

To show this, we first prove transitivity of a\leq_{a}, where the nontrivial part is to show that xnynax^{n}\leq y^{n}a and ynznay^{n}\leq z^{n}a for all nn imply xnznax^{n}\leq z^{n}a. This is because xnynznx^{n}\sim y^{n}\sim z^{n} and a1a\sim 1 give xnznax^{n}\sim z^{n}a, and xn>znax^{n}>z^{n}a would imply x2n>z2na2x^{2n}>z^{2n}a^{2} and therefore be in contradiction with

x2ny2naz2na2.x^{2n}\leq y^{2n}a\leq z^{2n}a^{2}.

The a\leq_{a}-monotonicity of multiplication is obvious, while the a\leq_{a}-monotonicity of addition follows by a1a\geq 1 and the binomial expansion. Fullness follows by totality of \leq and xayx\sim_{a}y being equivalent to the first condition involving kk only. It remains to establish multiplicative Archimedeanicity. So let x,y>a1x,y>_{a}1. Then there are k,nk,n\in\mathbb{N} with yaky\leq a^{k} and xn>ax^{n}>a, and hence yxkny\leq x^{kn}, which produces the claim.

Perhaps surprisingly, there is a way to associate real numbers to elements of multiplicatively Archimedean fully preordered semifields in such a way that the assigned numbers measure the “size” of the elements.

5.3 Lemma.

Fix nonzero u>1u>1 in a multiplicatively Archimedean fully preordered semifield FF. Then for every nonzero x1x\sim 1 in FF, there is a unique rr\in\mathbb{R} such that the following hold for all pq\frac{p}{q}\in\mathbb{Q} with q>0q\in\mathbb{N}_{>0}:

  1. (a)

    If pq<r\frac{p}{q}<r, then xq>upx^{q}>u^{p}.

  2. (b)

    If pq>r\frac{p}{q}>r, then xq<upx^{q}<u^{p}.

Proof.

It is clear that there can be at most one such rr, since otherwise xqx^{q} and upu^{p} would be strictly ordered in both directions for any pq\frac{p}{q} that lies between them. Also the set of fractions pq\frac{p}{q} with xq>upx^{q}>u^{p} is easily seen to be downwards closed, and similarly the set of fractions pq\frac{p}{q} with xq<upx^{q}<u^{p} is upwards closed. Moreover, if some fraction pq\frac{p}{q} is strictly smaller than all fractions in the first set, then it clearly must belong to the second set, and vice versa.

To see that these sets make up a Dedekind cut, it thus remains to show that they are both nonempty. But indeed xq<upx^{q}<u^{p} holds for sufficiently large pq\frac{p}{q} since uu is a power universal element, and similarly xq>upx^{q}>u^{p} holds for small enough (sufficiently negative) pq\frac{p}{q}. ∎

Before diving further into the classification theory, here is a useful criterion for deriving ordering relations.

5.4 Lemma.

Let FF be a multiplicatively Archimedean fully preordered semifield, and suppose that x,y,zF×x,y,z\in F^{\times} satisfy xyx\sim y and z1z\sim 1. Suppose that for every nn\in\mathbb{N} we have xnynzx^{n}\leq y^{n}z. Then xyx\leq y.

Proof.

Assume x>yx>y for contradiction. Then there is there is kk\in\mathbb{N} such that z<(xy1)kz<(xy^{-1})^{k} by multiplicative Archimedeanicity, and hence ykz<xkykzy^{k}z<x^{k}\leq y^{k}z, a contradiction. ∎

The five basic types and the type classification

In Theorem 2.2, we had distinguished the real and the tropical case as well as their two opposites. In our present more general context, it will be useful to distinguish five cases, where the additional case (iii) below corresponds to the situation of Example 5.1.

5.5 Definition.

A preordered semifield FF is

  1. (i)

    max-tropical if

    x+x12x,x+x^{-1}\approx 2x,
  2. (ii)

    max-temperate if

    2<x+x1<2x,2<x+x^{-1}<2x,
  3. (iii)

    arctic if

    x+x12,x+x^{-1}\approx 2,
  4. (iv)

    min-temperate if

    2x1<x+x1<2,2x^{-1}<x+x^{-1}<2,
  5. (v)

    min-tropical if

    x+x12x1,x+x^{-1}\approx 2x^{-1},

holds for all xF×x\in F^{\times} with x>1x>1.

We say that FF is tropical if it is min-tropical or max-tropical, and similarly temperate if it is min-temperate or max-temperate.

For any FF and any nonzero x>1x>1, the element x+x1x+x^{-1} must lie somewhere in the order interval [2x1,2x][2x^{-1},2x]. These five types thus make a distinction depending on where x+x1x+x^{-1} lands in that interval, using the three elements

2x1<2<2x2x^{-1}<2<2x

for comparison. Since the answer is required to be the same for all x>1x>1, it follows that if FF is any preordered semifield for which the preorder on F×F^{\times} is nontrivial, then it can be of at most one of these five types.

5.6 Example.

+\mathbb{R}_{+} is max-temperate and 𝕋+\mathbb{TR}_{+} is max-tropical. Similarly, +op\mathbb{R}_{+}^{\mathrm{op}} is min-temperate and 𝕋+op\mathbb{TR}_{+}^{\mathrm{op}} is min-tropical.777More generally, reversing the preorder from FF to FopF^{\mathrm{op}} also “reverses” the type. Example 5.1 is arctic, and is isomorphic to its own opposite via r+sXrsXr+sX\mapsto r-sX. Hence all five types do occur.

A general preordered semifield does not need to be of either type. For example, taking the categorical product of two preordered semifields of different types produces a preordered semifield which does not have a type.

5.7 Remark.

Our choice of terminology tropical, temperate and arctic is based on the historical contingency of the established term tropical. Arguably an intrinsically more reasonable choice would be to use tropical and arctic in the exactly opposite manner, for two reasons. First, there are two tropical cases but only one arctic case, and the latter is sandwiched in between the two temperate cases. This is exactly opposite to how the tropics are sandwiched in between the arctic regions in geographical reality. Second, in terms of an analogy with thermodynamics, the tropical cases correspond to zero temperature [6], which is indeed rather cold; in contrast to this, we conjecture that the arctic case can be associated with infinite temperature.

5.8 Remark.

While we will not do this in the present work, it may also be of interest to refine the above definition so as to assign a type to any strictly ordered pair of elements x<yx<y in any preordered semiring SS, by similarly considering where the element x2+y2x^{2}+y^{2} falls relative to the elements

2x2<2xy<2y2.2x^{2}<2xy<2y^{2}.

Here, x<yx<y implies that x2+y2x^{2}+y^{2} must be somewhere in the order interval [2x2,2y2][2x^{2},2y^{2}].

Here is why the five types are relevant in our context.

5.9 Proposition.

Suppose that FF is a multiplicatively Archimedean fully preordered semifield with nontrivial preorder on F×F^{\times}. Then FF is of exactly of one of the five types of Definition 5.5.

Proof.

By the nontriviality assumption, there must be some x>1x>1 in F×F^{\times}, implying that the five conditions are mutually exclusive. We therefore only need to show that if some fixed nonzero x>1x>1 satisfies one of them, then any other nonzero y>1y>1 also satisfies exactly the same condition. To do so, we use multiplicative Archimedeanicity to choose kk\in\mathbb{N} with y<xky<x^{k}.

First, we tackle the max-tropical case by showing that x+x12xx+x^{-1}\approx 2x implies y+y12yy+y^{-1}\approx 2y. Since y+y12yy+y^{-1}\leq 2y is trivial by y>1y>1, it is enough to prove y+y12yy+y^{-1}\geq 2y. We have

2(x2+x2)x2+1+2x2x2+2+x2(x+x1)24x2,2(x^{2}+x^{-2})\approx x^{2}+1+2x^{-2}\approx x^{2}+2+x^{-2}\approx(x+x^{-1})^{2}\approx 4x^{2},

and hence by iteration xk+xk2xkx^{k}+x^{-k}\approx 2x^{k} whenever kk is a power of two, which we can assume for the above kk without loss of generality. Thus upon replacing xx by xkx^{k}, we can also assume y<xy<x. But then

y+y1=y(1+y2)y(1+x2)2y,y+y^{-1}=y(1+y^{-2})\geq y(1+x^{-2})\approx 2y,

as was to be shown.

Second, for the max-temperate case, it is thus enough to show that the inequality x+x1>2x+x^{-1}>2 implies y+y1>2y+y^{-1}>2. We have xk+xk>2x^{k}+x^{-k}>2 by Corollary 4.5, so that replacing xx by xkx^{k} lets us again assume y<xy<x without loss of generality. By Lemma 4.3(ii) and multiplicative Archimedeanicity, we can find nn\in\mathbb{N} such that

2n1(xn+xn)(x+x1)n>2nx,2^{n-1}(x^{n}+x^{-n})\geq(x+x^{-1})^{n}>2^{n}x,

resulting in xn+xn>2xx^{n}+x^{-n}>2x. By x>1x>1, we can weaken this to xn+x(n+1)>2x^{n}+x^{-(n+1)}>2. Now let mm\in\mathbb{N} be the smallest integer with xnymx^{n}\leq y^{m}. Then also ym<xn+1y^{m}<x^{n+1}, since xn+1ymx^{n+1}\leq y^{m} would contradict the minimality of mm by y<xy<x. This gives

ym+ymxn+x(n+1)>2.y^{m}+y^{-m}\geq x^{n}+x^{-(n+1)}>2.

Since y+y12y+y^{-1}\leq 2 would imply ym+ym2y^{m}+y^{-m}\leq 2 again by Corollary 4.5, this proves that indeed y+y1>2y+y^{-1}>2 due to fullness.

Third, x+x12x+x^{-1}\approx 2 implies y+y12y+y^{-1}\approx 2. For since FF is fully preordered, y>1y>1 implies that y+y12y+y^{-1}\leq 2 or y+y12y+y^{-1}\geq 2 or both. It is indeed both, since by the previous paragraph a strict inequality would also imply a strict inequality between y+y1y+y^{-1} and 22.

The other cases follow by symmetry upon replacing FF by FopF^{\mathrm{op}}. ∎

A cancellation criterion

Throughout this subsection and the following ones, FF is still a multiplicatively Archimedean fully preordered semifield.

Thanks to multiplicative Archimedeanicity, we can improve on the inequalities derived in Section 4. In particular, we can turn Lemma 4.11 into an actual cancellation criterion.

5.10 Proposition.

Let a,x,yF×a,x,y\in F^{\times} with x1x\sim 1 be such that x+x12x+x^{-1}\geq 2 and y1y\geq 1. Then

a+xa+ya1+xa1+yxy.\begin{matrix}a+x\leq a+y\\[4.0pt] a^{-1}+x\leq a^{-1}+y\end{matrix}\quad\Longrightarrow\quad x\leq y.
Proof.

Combine Lemma 4.11 with Lemma 5.4. ∎

Over the course of the next few short subsections, we will sharpen the type classification by deriving further inequalities for FF under type hypotheses.

The max-tropical case

The following justifies the term “max-tropical” further by clarifying in what sense addition on max-tropical FF is analogous to addition in the tropical semifield 𝕋+\mathbb{TR}_{+}.

5.11 Lemma.

Let FF be max-tropical and x,yF×x,y\in F^{\times}. If xyx\sim y, then x+y2max(x,y)x+y\approx 2\max(x,y).

Note that this obviously holds in 𝕋+\mathbb{TR}_{+}, where we have x+y=max(x,y)x+y=\max(x,y) and 2=12=1.

Proof.

We can assume x>1x>1 and y=1y=1 without loss of generality, in which case we need to show x+12xx+1\approx 2x. We already know x2+12x2x^{2}+1\approx 2x^{2} by max-tropicality, and hence

(x+1)2=x2+1+2x2x2+2x=2x(x+1),(x+1)^{2}=x^{2}+1+2x\approx 2x^{2}+2x=2x(x+1),

which implies the claim by invertibility of x+1x+1. ∎

Of course, if FF is min-tropical, then we similarly get x+y2min(x,y)x+y\approx 2\min(x,y) for xyx\sim y.

The arctic case

Something analogous works in the arctic case. It is an instructive exercise to verify the following explicitly for Example 5.1.

5.12 Lemma.

Let FF be arctic and x,yF×x,y\in F^{\times}. If x,y1x,y\sim 1, then also

x+yxy+1.x+y\approx xy+1.
Proof.

We first show that the equation holds with x2x^{2} and y2y^{2} in place of xx and yy, in which case

x2+y2=xy(xy1+(xy1)1)2xyx2y2+1x^{2}+y^{2}=xy(xy^{-1}+(xy^{-1})^{-1})\approx 2xy\approx x^{2}y^{2}+1

proves the claim. This gives the general case via

(x+y)2=x2+y2+2xyx2y2+2xy+1=(xy+1)2,(x+y)^{2}=x^{2}+y^{2}+2xy\approx x^{2}y^{2}+2xy+1=(xy+1)^{2},

since the squares can be cancelled: a strict inequality in either direction would likewise hold for their squares. ∎

Given a polynomial or Laurent polynomial pp, we write pp^{\prime} for its derivative and obtain the following formulas for evaluating pp.

5.13 Lemma.

Suppose that FF is arctic and let xF×x\in F^{\times} with x1x\sim 1 and p,q[X,X1]p,q\in\mathbb{N}[X,X^{-1}] nonzero. Then:

  1. (i)

    p(x)xp(1)+(p(1)1)p(x)\approx x^{p^{\prime}(1)}+(p(1)-1).

  2. (ii)

    If x1x\geq 1, then p(x)q(x)p(x)\leq q(x) if and only if p(1)=q(1)p(1)=q(1) and p(1)q(1)p^{\prime}(1)\leq q^{\prime}(1).

  3. (iii)

    If x1x\not\approx 1, then p(x)q(x)p(x)\approx q(x) if and only if p(1)=q(1)p(1)=q(1) and p(1)=q(1)p^{\prime}(1)=q^{\prime}(1).

It is instructive to consider how these formulas manifest themselves in the case of Example 5.1.

Proof.

For (i), we use induction on the sum of coefficients p(1)p(1), where the base case p(1)=1p(1)=1 is trivial since then pp is necessarily a single monomial. For the induction step, we write p=xdeg(p)+p^p=x^{\deg(p)}+\hat{p} for some p^[X,X1]\hat{p}\in\mathbb{N}[X,X^{-1}], and obtain by the induction assumption

p(x)xp^(1)+(p^(1)1)+xdeg(p)xp^(1)+deg(p)+p^(1)xp(1)+(p(1)1),p(x)\approx x^{\hat{p}^{\prime}(1)}+(\hat{p}(1)-1)+x^{\deg(p)}\approx x^{\hat{p}^{\prime}(1)+\deg(p)}+\hat{p}(1)\approx x^{p^{\prime}(1)}+(p(1)-1),

where the second step is by Lemma 5.12.

For (ii), the “if” part is immediate from (i). For the “only if” part, note first that p(x)q(x)p(x)\sim q(x) is equivalent to p(1)=q(1)p(1)=q(1) since p(x)p(1)p(x)\sim p(1) and q(x)q(1)q(x)\sim q(1). Taking x>1x>1 without loss of generality, assuming p(x)q(x)p(x)\leq q(x) amounts to

xp(1)+(p(1)1)xq(1)+(p(1)1)x^{p^{\prime}(1)}+(p(1)-1)\leq x^{q^{\prime}(1)}+(p(1)-1)

by (i). This implies xp(1)+1xq(1)+1x^{p^{\prime}(1)}+1\leq x^{q^{\prime}(1)}+1 by chaining. But then xp(1)xq(1)x^{p^{\prime}(1)}\leq x^{q^{\prime}(1)} because of Proposition 5.10, which implies p(1)q(1)p^{\prime}(1)\leq q^{\prime}(1) thanks to x>1x>1.

Finally, (iii) follows directly from (ii). ∎

Away from the tropical case

If FF is max-tropical, then we have xn+12x^{-n}+1\approx 2 for every nonzero x>1x>1 and nn\in\mathbb{N}. If FF is min-tropical, then we similarly have xn+12x^{n}+1\approx 2. The following result can be thought of as providing converse statements.

5.14 Lemma.

Let xF×x\in F^{\times} with x>1x>1.

  1. (a)

    If FF is not min-tropical, then

    xn+1<xn+1+1x^{n}+1<x^{n+1}+1

    for all nn\in\mathbb{N}.

  2. (b)

    If FF is not max-tropical, then

    x(n+1)+1<xn+1x^{-(n+1)}+1<x^{-n}+1

    for all nn\in\mathbb{N}.

Thus if FF is not tropical, then the map nxn+1n\mapsto x^{n}+1 is strictly increasing across all nn\in\mathbb{Z}.

Proof.

These two cases become equivalent upon reversing the order and replacing xx by x1x^{-1}. We therefore only treat the first case.

If we had xn+1+1xn+1x^{n+1}+1\leq x^{n}+1 for some nn, then we would get (x+1)m2mxn(x+1)^{m}\leq 2^{m}x^{n} for all mm\in\mathbb{N} by Lemma 4.6 and therefore x+12x+1\leq 2 by Lemma 5.4. But then also

x2+3x2+2x+1=(x+1)24,x^{2}+3\leq x^{2}+2x+1=(x+1)^{2}\leq 4,

so that chaining gives 3(x2+1)63(x^{2}+1)\leq 6, or equivalently x+x12x1x+x^{-1}\leq 2x^{-1}, which contradicts the assumption that FF is not min-tropical. ∎

5.15 Lemma.

Suppose that FF is not max-tropical and let xF×x\in F^{\times}. If x>1x>1, then for every n>0n\in\mathbb{N}_{>0} there is kk\in\mathbb{N} such that

xn+k(k+1)x.x^{n}+k\leq(k+1)x.
Proof.

We show this first for n=2n=2. If x+x12x+x^{-1}\leq 2, then this holds with k=1k=1, so assume x+x1>2x+x^{-1}>2, meaning that FF is max-temperate.

Using x+1<2xx+1<2x from Lemma 5.14, Lemma 5.4 shows that there is m2m\geq 2 such that

(x+1)m+12m+1xm.(x+1)^{m+1}\leq 2^{m+1}x^{m}.

Expanding the left-hand side and using x1x\geq 1 gives the weaker bound

xm+1+(2m+11)2m+1xm.x^{m+1}+(2^{m+1}-1)\leq 2^{m+1}x^{m}.

Thus there are m,km,k\in\mathbb{N} such that

xm+1+k(k+1)xm.x^{m+1}+k\leq(k+1)x^{m}. (5.1)

We now claim that if this holds for some m2m\geq 2, then it also holds with m1m-1 in place of mm. Indeed the following estimates show that it is enough to increase kk by 11,

(x+1)(xm+(k+1))\displaystyle(x+1)(x^{m}+(k+1)) =xm+1+xm+(k+1)x+k+1\displaystyle=x^{m+1}+x^{m}+(k+1)x+k+1
(k+2)xm+(k+1)x+1\displaystyle\leq(k+2)x^{m}+(k+1)x+1
(k+2)xm+(k+2)xm1\displaystyle\leq(k+2)x^{m}+(k+2)x^{m-1}
=(x+1)(k+2)xm1,\displaystyle=(x+1)(k+2)x^{m-1},

where the first inequality step uses the assumption, and the second one uses merely x1x\geq 1 and m2m\geq 2. Upon iterating this argument, we therefore conclude that (5.1) holds even with m=1m=1, meaning that there is kk such that

x2+k(k+1)x,x^{2}+k\leq(k+1)x,

as was to be shown for n=2n=2.

We now show that if the claim holds for n2n\geq 2, then it also holds for 2n2n,

x2n+(k+1)3\displaystyle x^{2n}+(k+1)^{3} x2n+2kxn+k2+(k3+2k2+k+1)\displaystyle\leq x^{2n}+2kx^{n}+k^{2}+(k^{3}+2k^{2}+k+1)
=(xn+k)2+(k3+2k2+k+1)\displaystyle=(x^{n}+k)^{2}+(k^{3}+2k^{2}+k+1)
(k+1)2x2+(k3+2k2+k+1)\displaystyle\leq(k+1)^{2}x^{2}+(k^{3}+2k^{2}+k+1)
=(k+1)2(x2+k)+1\displaystyle=(k+1)^{2}\mathopen{}\mathclose{{}\left(x^{2}+k}\right)+1
((k+1)3+1)x,\displaystyle\leq\mathopen{}\mathclose{{}\left((k+1)^{3}+1}\right)x,

where the first inequality step uses only x1x\geq 1 and the other two use the assumption. In particular the claim holds whenever nn is a power of two. This is enough for the general case by the monotonicity in nn proven just before. ∎

We also derive a further statement which makes explicit use of positive linear combinations with rational coefficients. Recall that these exist in any strict semifield.

5.16 Lemma.

Suppose that FF is not tropical and xF×x\in F^{\times}. If x>1x>1, then for every rational r(0,1)r\in(0,1), we have

1<rx+(1r)<x.1<rx+(1-r)<x.
Proof.

By reversing the order and replacing xx by x1x^{-1} and rr by 1r1-r, the second inequality reduces to the first. We therefore only prove the first.

Since the expression rx+(1r)rx+(1-r) is obviously non-strictly monotone in rr, it is enough to prove the claim for r=2nr=2^{-n} with n>0n\in\mathbb{N}_{>0}, in which case it amounts to

1<2nx+(12n).1<2^{-n}x+(1-2^{-n}).

We indeed have 1+1<1+x1+1<1+x by Lemma 5.14, which is the n=1n=1 case. For the induction step from nn to n+1n+1, we apply this same inequality 1+1<1+y1+1<1+y with y2nx+(12n)y\coloneqq 2^{-n}x+(1-2^{-n}), which satisfies y>1y>1 by the induction assumption. ∎

Away from the arctic case

While the previous lemmas were concerned with FF not being max-tropical or min-tropical, we now consider a similar statement for FF not arctic.

5.17 Lemma.

Suppose that FF is not arctic and xF×x\in F^{\times}. If x>1x>1 and x+x1>2x+x^{-1}>2, then for every \ell\in\mathbb{N} there is nn\in\mathbb{N} such that for all mm\in\mathbb{N}.

xn+1+xm>xn+.x^{n+1}+\ell x^{-m}>x^{n}+\ell.

While this quite clear in the max-tropical case, the main difficulty lies in proving it in the max-temperate case (but restricting to this case explicitly would not simplify the proof).

Proof.

We prove a number of auxiliary statements first before getting to the claim itself.

  1. (a)

    There is nn\in\mathbb{N} such that xn+xn2xx^{n}+x^{-n}\geq 2x.

    Indeed by Lemma 5.4, there is nn\in\mathbb{N} such that (x+x1)n2nx(x+x^{-1})^{n}\geq 2^{n}x. Hence by Lemma 4.3,

    2n1(xn+xn)(x+x1)n2nx,2^{n-1}(x^{n}+x^{-n})\geq(x+x^{-1})^{n}\geq 2^{n}x,

    as was to be shown.

  2. (b)

    For every ε<1\varepsilon<1 in \mathbb{R} there are m,n>0m,n\in\mathbb{N}_{>0} such that m>εnm>\varepsilon n and

    xn+xn2xm.x^{n}+x^{-n}\geq 2x^{m}. (5.2)

    Indeed if this inequality holds for given nn and mm, then it also holds for all multiples, since for every >0\ell\in\mathbb{N}_{>0},

    xn+xn21(xn+xn)2xm,x^{\ell n}+x^{-\ell n}\geq 2^{1-\ell}(x^{n}+x^{-n})^{\ell}\geq 2x^{\ell m},

    where the first step is by Lemma 4.3 and the second by assumption. Now let ε\varepsilon be the supremum of all fractions mn\frac{m}{n} for which the inequality (5.2) holds; our goal is to show that ε=1\varepsilon=1, whereas what we know by (a) is ε>0\varepsilon>0. Indeed we claim that ε3ε22ε\varepsilon\geq\frac{3-\varepsilon^{2}}{2}\varepsilon, which then implies ε=1\varepsilon=1 because of 0<ε10<\varepsilon\leq 1. In order to prove this claim, suppose that a fraction mn\frac{m}{n} satisfies the inequality. Then also

    2(x2n3+x2n3)\displaystyle 2(x^{2n^{3}}+x^{-2n^{3}}) (x2n3+x2mn2)+(x2mn2+x2n3)\displaystyle\geq(x^{2n^{3}}+x^{2mn^{2}})+(x^{2mn^{2}}+x^{-2n^{3}})
    =xn2(n+m)(xn2(nm)+xn2(nm))+xn2(nm)(xn2(n+m)+xn2(n+m))\displaystyle=x^{n^{2}(n+m)}(x^{n^{2}(n-m)}+x^{-n^{2}(n-m)})+x^{-n^{2}(n-m)}(x^{n^{2}(n+m)}+x^{-n^{2}(n+m)})
    2(xn2(n+m)xmn(nm)+xn2(nm)xmn(n+m))\displaystyle\geq 2\mathopen{}\mathclose{{}\left(x^{n^{2}(n+m)}x^{mn(n-m)}+x^{-n^{2}(n-m)}x^{mn(n+m)}}\right)
    =2(xn(n2+2mnm2)+xn(n22mnm2))\displaystyle=2\mathopen{}\mathclose{{}\left(x^{n(n^{2}+2mn-m^{2})}+x^{-n(n^{2}-2mn-m^{2})}}\right)
    =2x2mn2(xn(n2m2)+xn(n2m2))\displaystyle=2x^{2mn^{2}}(x^{n(n^{2}-m^{2})}+x^{-n(n^{2}-m^{2})})
    4x2mn2xm(n2m2)\displaystyle\geq 4x^{2mn^{2}}x^{m(n^{2}-m^{2})}
    =4x3mn2m3\displaystyle=4x^{3mn^{2}-m^{3}}

    where all inequality steps are per the above. Therefore

    ε3mn2m32n3=3(mn)22mn\varepsilon\geq\frac{3mn^{2}-m^{3}}{2n^{3}}=\frac{3-\mathopen{}\mathclose{{}\left(\frac{m}{n}}\right)^{2}}{2}\cdot\frac{m}{n}

    Thus as mnε\frac{m}{n}\nearrow\varepsilon, we get the claimed ε3ε22ε\varepsilon\geq\frac{3-\varepsilon^{2}}{2}\,\varepsilon.

  3. (c)

    There is nn\in\mathbb{N} such that

    xn+xm2x^{n}+x^{-m}\geq 2

    for all mm\in\mathbb{N}.

    Taking ε=12\varepsilon=\frac{1}{2} in (b), we have nn such that

    x2n+x2n2xn.x^{2n}+x^{-2n}\geq 2x^{n}.

    There is no loss in replacing xx by xnx^{n}, so that we can assume x4+12x3x^{4}+1\geq 2x^{3} without loss of generality. But then also

    x4+22x3+13x2,x^{4}+2\geq 2x^{3}+1\geq 3x^{2},

    where the second step is by Lemma 4.3. Therefore x2x^{2} satisfies the hypotheses of Lemma 4.7, and we get that

    x2(m+1)+12x2mx^{2(m+1)}+1\geq 2x^{2m}

    for all mm\in\mathbb{N}. Therefore also

    x2+x2m2x^{2}+x^{-2m}\geq 2

    for all mm\in\mathbb{N}, which is enough.

  4. (d)

    For every \ell\in\mathbb{N} there is nn\in\mathbb{N} such that

    xn+xm1+x^{n}+\ell x^{-m}\geq 1+\ell

    for all mm\in\mathbb{N}.

    Indeed for =1\ell=1, this is exactly (c). Moreover if the inequality holds for some \ell, then it also holds for all <\ell^{\prime}<\ell, since multiplying the inequality by \ell^{\prime} and adding \ell-\ell^{\prime} times the inequality xn1x^{n}\geq 1 results in

    xn+xm+,\ell x^{n}+\ell\ell^{\prime}x^{-m}\geq\ell+\ell\ell^{\prime},

    which is equivalent to the desired inequality with \ell^{\prime} in place of \ell. Therefore it is enough to show that if the statement holds for given \ell\in\mathbb{N}, then it also holds for 2+12\ell+1.

    This step from \ell to 2+12\ell+1 works as follows,

    x2n+(2+1)xm\displaystyle x^{2n}+(2\ell+1)x^{-m} =xn(xn+xmn)+(+1)xm\displaystyle=x^{n}(x^{n}+\ell x^{-m-n})+(\ell+1)x^{-m}
    (+1)xn+(+1)xm\displaystyle\geq(\ell+1)x^{n}+(\ell+1)x^{-m}
    =(+1)(xn+xm)\displaystyle=(\ell+1)(x^{n}+x^{-m})
    2(+1),\displaystyle\geq 2(\ell+1),

    where we have assumed that the given nn is large enough to work both for the given \ell and for =1\ell=1.

  5. (e)

    The actual claim is then the k=1k=1 case of the following: for every k>0k\in\mathbb{N}_{>0} and \ell\in\mathbb{N} there is nn\in\mathbb{N} such that

    xn+k+xmxn+x^{n+k}+\ell x^{-m}\geq x^{n}+\ell

    for all mm\in\mathbb{N}.

    Indeed (d) shows that this holds for some kk with n=0n=0. Since it automatically holds for all larger kk, it is enough to show that if the statement holds for a given even kk, then it also holds with k2\frac{k}{2} in place of kk. Assuming kk to be even without loss of generality and replacing xx by xkx^{k}, it is enough888Note that xk+xk>2x^{k}+x^{-k}>2 by Lemma 4.3. to show that the k=2k=2 case implies the k=1k=1 case, at the cost of replacing \ell by 22\ell and nn by n+3n+3,

    (x+x1)(xn+4+xm)\displaystyle(x+x^{-1})(x^{n+4}+\ell x^{-m}) =xn+5+xn+3+xm+1+xm1\displaystyle=x^{n+5}+x^{n+3}+\ell x^{-m+1}+\ell x^{-m-1}
    xn+5+x(xn+2+2xm2)\displaystyle\geq x^{n+5}+x(x^{n+2}+2\ell x^{-m-2})
    xn+5+x(xn+2)\displaystyle\geq x^{n+5}+x(x^{n}+2\ell)
    xn+4+xn+2+2x\displaystyle\geq x^{n+4}+x^{n+2}+2\ell x
    (x+x1)(xn+3+),\displaystyle\geq(x+x^{-1})(x^{n+3}+\ell),

    where the first and fourth inequality step use merely x1x\geq 1, the second is by assumption, and the third by x2+x2x+x1x^{2}+x^{-2}\geq x+x^{-1} from Lemma 4.3.∎

It may also be of interest to know under which conditions a semifield FF can support any nontrivial full semifield preorder at all that is multiplicatively Archimedean and arctic. The following result provides one relevant criterion.

5.18 Proposition.

Let FF be a strict semifield with quasi-complements such that the ring FF\otimes\mathbb{Z} is absolutely flat999Recall that a ring RR is absolutely flat if every ideal in RR is idempotent. For example, every product of fields is absolutely flat.. Then every multiplicatively Archimedean full semifield preorder on FF is temperate or tropical.

Proof.

Let \leq be a multiplicatively Archimedean full semifield preorder on FF. Then

I{xyxy in F}I\coloneqq\{x-y\mid x\sim y\textrm{ in }F\}

is an ideal in FF\otimes\mathbb{Z}, which is idempotent by assumption. Therefore for a>1a>1, there are elements xi,yi,ciFx_{i},y_{i},c_{i}\in F for i=1,,i=1,\ldots,\ell such that xi,yi1x_{i},y_{i}\sim 1 and

a1=i=1ci(xi1)(yi1)a-1=\sum_{i=1}^{\ell}c_{i}(x_{i}-1)(y_{i}-1)

holds in FF\otimes\mathbb{Z}. But this means that there is dFd\in F such that

a+i=1ci(xi+yi)+d=1+i=1ci(xiyi+1)+d.a+\sum_{i=1}^{\ell}c_{i}(x_{i}+y_{i})+d=1+\sum_{i=1}^{\ell}c_{i}(x_{i}y_{i}+1)+d.

Now if \leq was arctic, then we would have xi+yixiyi+1x_{i}+y_{i}\sim x_{i}y_{i}+1 by Lemma 5.12, and therefore a+z1+za+z\approx 1+z with zi=1ci(xiyi+1)+dz\coloneqq\sum_{i=1}^{\ell}c_{i}(x_{i}y_{i}+1)+d. Upon adding a quasi-complement of zz on both sides, we obtain further a+n1+na+n\approx 1+n for some nn\in\mathbb{N}. By chaining, we can reduce to the case n=1n=1 without loss of generality. But then applying the cancellation criterion of Proposition 5.10 to 1+a1+11+a\leq 1+1 shows a1a\leq 1, contradicting the initial assumption a>1a>1. ∎

For example, the categorical product of +\mathbb{R}_{+} with itself (any number of times) is a semifield that satisfies the assumptions, and therefore does not support any multiplicatively Archimedean full semifield preorder of arctic type.

6.  The ambient preorder

Perhaps surprisingly, every multiplicatively Archimedean fully preordered semifield can be equipped with a canonical total semifield preorder which extends the given preorder, and often does so in such a way that this induces an order embedding into some 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\}. This derived preorder is defined as follows, for preordered semifields in general.

6.1 Definition.

Let FF be a preordered semifield. Given fixed elements a,bFa,b\in F, the ambient preorder a,b\leq\leq_{a,b} is the relation on FF defined by

xa,by:ay+bxax+by.x\leq\leq_{a,b}y\quad:\Longleftrightarrow\quad ay+bx\leq ax+by.
6.2 Lemma.

If aba\not\approx b, then the ambient preorder a,b\leq\leq_{a,b} also makes FF into a preordered semifield, and 1a,b01\geq\geq_{a,b}0 if and only if aba\leq b.

Proof.

The condition aba\not\approx b clearly guarantees 1  01\,\,\cancel{\approx\approx}\,\,0. All other required properties are also straightforward to verify, apart from the transitivity of a,b\leq\leq_{a,b}. The latter is where the assumption that FF is a semifield (rather than a mere semiring) comes in. Indeed assuming xa,bya,bzx\leq\leq_{a,b}y\leq\leq_{a,b}z, we have

ay+bxax+by,az+byay+bz.ay+bx\leq ax+by,\qquad az+by\leq ay+bz.

We then obtain

(x+y)(az+bx)\displaystyle(x+y)(az+bx) =bx2+(az+by)x+ayz\displaystyle=bx^{2}+(az+by)x+ayz
bx2+(ay+bz)x+ayz\displaystyle\leq bx^{2}+(ay+bz)x+ayz
=(x+z)(ay+bx)\displaystyle=(x+z)(ay+bx)
(x+z)(ax+by)\displaystyle\leq(x+z)(ax+by)
=ax2+(az+by)x+byz\displaystyle=ax^{2}+(az+by)x+byz
ax2+(ay+bz)x+byz\displaystyle\leq ax^{2}+(ay+bz)x+byz
=(x+y)(ax+bz).\displaystyle=(x+y)(ax+bz).

Thus if x+y0x+y\neq 0, then the desired xa,bzx\leq\leq_{a,b}z follows. The complementary case is x=y=0x=y=0, in which case the claim holds trivially by x=yx=y. ∎

6.3 Example.

The same definition of ambient preorder does not extend to general preordered semirings, since the transitivity may fail. For an explicit example, consider the semiring S/(56)S\coloneqq\mathbb{N}/(5\simeq 6), which is \mathbb{N} with all numbers 5\geq 5 identified with 55. Equip SS with either the trivial preorder or the total preorder inherited from \mathbb{N}. In either case, we have 21,251,212\leq\leq_{1,2}5\leq\leq_{1,2}1 but 21,212\cancel{\leq\leq}_{1,2}1.

6.4 Remark.

An interesting feature of the ambient preorder is its behaviour under reversing \leq: we have xa,byx\leq\leq_{a,b}y in FF if and only if xb,ayx\leq\leq_{b,a}y in FopF^{\mathrm{op}}.

While the ambient preorder makes sense on any preordered semifield, we now return to the assumption that FF is a multiplicatively Archimedean fully preordered semifield, where the ambient preorder will facilitate the proof of our separation results. Given such an FF, we fix an arbitrary uF×u\in F^{\times} with u>1u>1. As the notation indicates, uu is power universal (by the definition of multiplicative Archimedeanicity). We suspect that the ambient preorder 1,u\leq\leq_{1,u} is independent of the particular choice of uu, but we have not been able to prove this so far, and we will not need it in the following. We nevertheless suppress the dependence on uu from our notation of the ambient preorder by writing \leq\leq as shorthand for 1,u\leq\leq_{1,u}. In other words, we put

xy:xu+yx+yu,x\leq\leq y\quad:\Longleftrightarrow\quad xu+y\leq x+yu,

and this is what we will use in the rest of this section. By Lemma 5.17, we can find nn\in\mathbb{N} such that un+1+umun+1u^{n+1}+u^{-m}\geq u^{n}+1 for all mm\in\mathbb{N}. Hence upon replacing uu by un+1u^{n+1} if necessary, we can achieve in particular that

u+um2u+u^{-m}\geq 2 (6.1)

for all mm\in\mathbb{N}. We assume from now on that such uu has been fixed.

6.5 Lemma.

\leq\leq is a total preorder.

Proof.

Since u1u\sim 1, we have xu+yx+yuxu+y\sim x+yu. Hence this follows from the assumption that the preorder on FF is full. ∎

The following auxiliary results will play a key technical role in the proofs of our main theorems.

6.6 Lemma.

Let FF be a multiplicatively Archimedean fully preordered semifield of arctic, max-temperate or max-tropical type and with u>1u>1. Suppose that the quotient semifield F/F/\!\sim has quasi-complements. Then the following holds for a suitable choice of uu:

  1. (i)

    \leq\leq extends \leq.

  2. (ii)

    If \leq is max-temperate or max-tropical, then for all xyx\sim y in FF, we have

    xyxy,x\leq y\quad\Longleftrightarrow\quad x\leq\leq y,

    and every x>1x>1 is power universal with respect to \leq\leq.

Furthermore, with v2uv\coloneqq 2u we have:

  1. (iii)

    vv is a power universal element for \leq\leq.

  2. (iv)

    v+v1>>2v+v^{-1}>>2.

  3. (v)

    If \leq is arctic, then also v+v1<<2vv+v^{-1}<<2v.

Since reversing the original preorder \leq keeps the ambient preorder \leq\leq invariant, these statements hold similarly in the min-temperate and min-tropical cases, where reversing the preorder also entails that uu needs to be replaced by u1u^{-1}.

Proof.

First, u1u\geq 1 shows that 101\geq\geq 0.

  1. (i), (ii)

    We need to show that for all x,yFx,y\in F,

    xyxy,x\leq y\quad\Longrightarrow\quad x\leq\leq y,

    as well as the converse in the max-tropical and max-temperate cases, assuming that xyx\sim y.

    All of this is trivial if x=y=0x=y=0. If exactly one is nonzero, then for xyx\sim y to hold we would need to have 101\sim 0, making FF totally preordered, in which case FF or FopF^{\mathrm{op}} embeds into +\mathbb{R}_{+} or 𝕋+\mathbb{TR}_{+} by Theorem 2.2, where the claims follow by a straightforward computation. We thus assume that x,yF×x,y\in F^{\times}, and put y=1y=1 without loss of generality.

    In the arctic case, we thus are assuming x1x\leq 1 and need to show that

    xu+1x+u.xu+1\leq x+u.

    This holds even with \approx by Lemma 5.12. We thus turn to the max-temperate and max-tropical cases. If x<1x<1, then we can find kk\in\mathbb{N} such that xku<1x^{k}u<1. Then

    xku+12xk+u,x^{k}u+1\leq 2\leq x^{k}+u,

    where the second inequality holds by power universality of uu and the assumed (6.1). The first inequality is strict by Lemma 5.14 in the max-temperate case, while the second inequality is strict in the max-tropical case by Lemma 5.11. Thus xku+1<xk+ux^{k}u+1<x^{k}+u in both cases, and we conclude xk<<1x^{k}<<1. This implies x<<1x<<1 by totality of \leq\leq.

    The case x>1x>1 is analogous, resulting in x>>1x>>1. And finally if x1x\approx 1, then of course we also have

    xu+1u+1=1+ux+u,xu+1\approx u+1=1+u\approx x+u,

    resulting in x1x\approx\approx 1.

    The final claim on power universality of x>1x>1 holds because some power of xx dominates uu since FF is multiplicatively Archimedean, and we will prove uu to be power universal in the upcoming proof of (iii).

  2. (iii)

    The definition of the ambient preorder shows that u1u\geq\geq 1 is equivalent to u+u12u+u^{-1}\geq 2, which we have assumed. Since 212\geq\geq 1, this implies vu1v\geq\geq u\geq\geq 1.

    For power universality, suppose first that \leq is not arctic, and therefore is max-temperate or max-tropical. We then show that even uu itself is a power universal element for \leq\leq, which means that for all xF×x\in F^{\times} there is nn\in\mathbb{N} with xunx\leq\leq u^{n}. This latter inequality amounts to

    xu+unx+un+1.xu+u^{n}\leq x+u^{n+1}.

    We consider three subcases.

    • \triangleright

      Suppose x1x\sim 1.

      We then choose kk\in\mathbb{N} with ukxuku^{-k}\leq x\leq u^{k} and apply Lemma 5.17, which gives the middle inequality in

      xu+unuk+1+unuk+un+1x+un+1xu+u^{n}\leq u^{k+1}+u^{n}\leq u^{-k}+u^{n+1}\leq x+u^{n+1}

      for sufficiently large nn.

    • \triangleright

      Suppose x=x=\ell\in\mathbb{N}.

      Then we need to find nn\in\mathbb{N} such that

      u+un+un+1.\ell u+u^{n}\leq\ell+u^{n+1}.

      Multiplying both sides by u1u^{-1} shows that this is again covered by Lemma 5.17.

    • \triangleright

      Now for general xx, let yFy\in F be a quasi-complement for xx in F/F/\!\sim, so that x+y>0x+y\sim\ell\in\mathbb{N}_{>0}. But what we have already shown is therefore that both \ell and 1(x+y)\ell^{-1}(x+y) are upper bounded with respect to \leq\leq by some power of uu, say unu^{n}. Hence

      xx+y=1(x+y)u2n,x\leq\leq x+y=\ell\cdot\ell^{-1}(x+y)\leq\leq u^{2n},

      as was to be shown.

    Second, suppose that \leq is arctic. We then need to show that for every xF×x\in F^{\times}, there is nn\in\mathbb{N} such that

    xu+2nunx+2nun+1,xu+2^{n}u^{n}\leq x+2^{n}u^{n+1},

    and we do so using the same case distinctions as above.

    • \triangleright

      Suppose x1x\sim 1.

      We take n=1n=1 and apply Lemma 5.12 in order to obtain

      xu+2uxu2+2uxu2+1+u2x+2u2.xu+2u\leq xu^{2}+2u\approx xu^{2}+1+u^{2}\approx x+2u^{2}.
    • \triangleright

      Suppose x=x=\ell\in\mathbb{N}.

      Now choose nn such that 2n\ell\leq 2^{n} in \mathbb{N}. Then again using Lemma 5.12,

      u+2nun\displaystyle\ell u+2^{n}u^{n} =(u+un)+(2n)un\displaystyle=\ell(u+u^{n})+(2^{n}-\ell)u^{n}
      (1+un+1)+(2n)un\displaystyle\approx\ell(1+u^{n+1})+(2^{n}-\ell)u^{n}
      +2nun+1,\displaystyle\leq\ell+2^{n}u^{n+1},

      as was to be shown.

    • \triangleright

      The case of general xx reduces to the two previous ones just as above.

  3. (iv)

    What we need to prove is that 4u+u1>>44u+u^{-1}>>4, which unfolds to

    4u+(4u+u1)<4+(4u+u1)u.4u+(4u+u^{-1})<4+(4u+u^{-1})u.

    This holds non-strictly because of

    8u2+14u3+4u+14u3+5u8u^{2}+1\leq 4u^{3}+4u+1\leq 4u^{3}+5u

    and dividing by uu. In the arctic case and the max-temperate one, the second inequality is strict by Lemma 5.16, which implies the claim. Strict inequality holds also in the max-tropical case, since then

    8u2+19u2<9u3=4u3+5u.8u^{2}+1\approx 9u^{2}<9u^{3}=4u^{3}+5u.
  4. (v)

    We need to prove 4u+u1<<8u4u+u^{-1}<<8u, or equivalently

    4u3+8u2+u<8u3+4u2+1,4u^{3}+8u^{2}+u<8u^{3}+4u^{2}+1,

    which is indeed true by Lemma 5.12 and 43+82+1<83+424\cdot 3+8\cdot 2+1<8\cdot 3+4\cdot 2. ∎

6.7 Proposition.

Let FF be a multiplicatively Archimedean fully preordered semifield with u>1u>1 such that:

  • \triangleright

    FF is of max-temperate or max-tropical type.

  • \triangleright

    F/F/\!\sim has quasi-complements.

Then there is a homomorphism ϕ:F𝕂\phi:F\to\mathbb{K} with 𝕂{+,𝕋+}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{TR}_{+}\} such that for all xyx\sim y in FF, we have

xyϕ(x)ϕ(y).x\leq y\quad\Longleftrightarrow\quad\phi(x)\leq\phi(y). (6.2)
Proof.

By Lemma 6.6, the ambient preorder \leq\leq turns FF into a totally preordered semifield with power universal element v>>1v>>1. Therefore Corollary 2.3 produces a \leq\leq-monotone homomorphism ϕ:F𝕂\phi:F\to\mathbb{K} for 𝕂{+,𝕋+}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{TR}_{+}\}, where the other two cases are excluded by v+v1>>2v+v^{-1}>>2.

The desired equivalence is again trivial when x=0x=0 or y=0y=0, so we assume x,yF×x,y\in F^{\times} and put y=1y=1 without loss of generality. Then if x>1x>1, we also obtain that x>>1x>>1 is power universal by Lemma 6.6, resulting in ϕ(x)>1\phi(x)>1 by Corollary 2.3. Similarly x<1x<1 implies ϕ(x)<1\phi(x)<1. Finally, x1x\approx 1 yields x1x\approx\approx 1, and therefore ϕ(x)=1\phi(x)=1. ∎

We next aim at an analogous statement for the arctic case. This is formulated in terms of (+)[X]/(X2)\mathbb{R}_{(+)}[X]/(X^{2}), the preordered semifield of arctic type introduced in Example 5.1. It plays a similarly paradigmatic role as +\mathbb{R}_{+} does in the max-temperate case and 𝕋+\mathbb{TR}_{+} in the max-tropical case.

6.8 Proposition.

Let FF be a multiplicatively Archimedean fully preordered semifield with a power universal element u>1u>1 such that:

  • \triangleright

    FF is of arctic type.

  • \triangleright

    F/F/\!\sim has quasi-complements.

  • \triangleright

    (F/)(F/\!\sim)\otimes\mathbb{Z} is a finite product of fields.

Then there exists a homomorphism ψ:F(+)[X]/(X2)\psi:F\to\mathbb{R}_{(+)}[X]/(X^{2}) such that for all xyx\sim y in FF, we have

xyψ(x)ψ(y).x\leq y\quad\Longleftrightarrow\quad\psi(x)\leq\psi(y). (6.3)

Moreover, if FF is also a semialgebra, then ψ\psi can be chosen so as to preserve scalar multiplication by +\mathbb{R}_{+}.

As the proof will show, the assumption on (F/)(F/\!\sim)\otimes\mathbb{Z} can in fact be weakened to formal smoothness over \mathbb{Q} (and probably further). We nevertheless phrase the statement in terms of the stronger assumption stating that this ring should be a finite product of fields, since this is how we will use the statement later and is more elementary.

Proof.

The monotone homomorphisms ψ:F(+)[X]/(X2)\psi:F\to\mathbb{R}_{(+)}[X]/(X^{2}) are precisely the maps of the form

ψ(a)=ϕ(a)+D(a)X\psi(a)=\phi(a)+D(a)X

for a degenerate homomorphism ϕ:F+\phi:F\to\mathbb{R}_{+} and a monotone additive map D:FD:F\to\mathbb{R} satisfying the Leibniz rule with respect to ϕ\phi, which is

D(ab)=ϕ(a)D(b)+D(a)ϕ(b)D(ab)=\phi(a)D(b)+D(a)\phi(b)

for all a,bFa,b\in F. If FF is a semialgebra, then ϕ\phi automatically preserves scalar multiplication since the identity map is the only homomorphism ++\mathbb{R}_{+}\to\mathbb{R}_{+}, and therefore ψ\psi preserves scalar multiplication if and only if DD is +\mathbb{R}_{+}-linear. In either case, we will construct such ψ\psi by constructing its components ϕ\phi and DD in the following.

By quasi-complements and Lemma 6.6, the ambient preorder \leq\leq has a power universal element v>>1v>>1 with 2<<v+v1<<2v2<<v+v^{-1}<<2v, so that Corollary 2.3 provides us with a \leq\leq-monotone homomorphism ϕ:F+\phi:F\to\mathbb{R}_{+}. Consider next the multiplicative group

G{xF×x1}.G\coloneqq\{x\in F^{\times}\mid x\sim 1\}.

For xGx\in G, let D(x)D(x) be the number associated to it by Lemma 5.3. Then D:GD:G\to\mathbb{R} is an order embedding by construction. We prove a few auxiliary statements.

  1. (a)

    For x,y1x,y\sim 1,

    D(xy)=D(x)+D(y),D(xy)=D(x)+D(y), (6.4)

    which in particular implies D(1)=0D(1)=0.

    This follows easily from the definition of DD.

  2. (b)

    Moreover, we also have

    D(rx+(1r)y)=rD(x)+(1r)D(y)D\mathopen{}\mathclose{{}\left(rx+(1-r)y}\right)=rD(x)+(1-r)D(y) (6.5)

    for all rational r[0,1]r\in[0,1], and for all r[0,1]r\in[0,1] if FF is a semialgebra.

    To see this for rational rr, we assume x>1x>1 and y1y\approx 1 without loss of generality. The claim then follows by an application of Lemma 5.13. In the semialgebra case, the claim for general rr follows by monotonicity of rrx+(1r)yr\mapsto rx+(1-r)y and rational approximation.

  3. (c)

    The map

    F×F,aau+1a+1F^{\times}\longrightarrow F,\qquad a\longmapsto\frac{au+1}{a+1}

    is \leq\leq-to-\leq-monotone.

    Indeed aba\leq\leq b means exactly that au+ba+buau+b\leq a+bu, and therefore the claim follows by

    (au+1)(b+1)\displaystyle(au+1)(b+1) =abu+au+b+1\displaystyle=abu+au+b+1
    abu+bu+a+1\displaystyle\leq abu+bu+a+1
    =(bu+1)(a+1)\displaystyle=(bu+1)(a+1)

    and dividing.

  4. (d)

    For all a,bF×a,b\in F^{\times} and x,y1x,y\sim 1, we have

    D(ax+bya+b)=ϕ(a)D(x)+ϕ(b)D(y)ϕ(a)+ϕ(b).D\mathopen{}\mathclose{{}\left(\frac{ax+by}{a+b}}\right)=\frac{\phi(a)D(x)+\phi(b)D(y)}{\phi(a)+\phi(b)}. (6.6)

    For the proof, we put b=y=1b=y=1 without loss of generality. We then show the desired equation

    D(ax+1a+1)=ϕ(a)ϕ(a)+1D(x)D\mathopen{}\mathclose{{}\left(\frac{ax+1}{a+1}}\right)=\frac{\phi(a)}{\phi(a)+1}D(x)

    first for x=ux=u. For a>0a\in\mathbb{Q}_{>0} it follows by (6.5). For general aa, we can use rational approximation in the ambient preorder \leq\leq, which implies the claim by (c). Therefore the desired equation holds with a=ua=u for all rr.

    We now argue that for any n>0n\in\mathbb{N}_{>0}, the desired equation (6.6) holds for x1x\sim 1 if and only if it holds for xnx^{n} in place of xx. Using xxn+(n1)nx\approx\frac{x^{n}+(n-1)}{n}, we obtain

    D(ax+1a+1)\displaystyle D\mathopen{}\mathclose{{}\left(\frac{ax+1}{a+1}}\right) =D(axn+(n1)n+1a+1)\displaystyle=D\mathopen{}\mathclose{{}\left(\frac{a\cdot\frac{x^{n}+(n-1)}{n}+1}{a+1}}\right)
    =D(1naxn+1a+1+n1na+1a+1)\displaystyle=D\mathopen{}\mathclose{{}\left(\frac{1}{n}\cdot\frac{ax^{n}+1}{a+1}+\frac{n-1}{n}\cdot\frac{a+1}{a+1}}\right)
    =1nD(axn+1a+1)\displaystyle=\frac{1}{n}D\mathopen{}\mathclose{{}\left(\frac{ax^{n}+1}{a+1}}\right)

    by (6.5) and D(1)=0D(1)=0. This implies the claim since the right-hand side of (6.6) receives the same factor of 1n\frac{1}{n} from D(x)=1nD(xn)D(x)=\frac{1}{n}D(x^{n}).

    Finally, another approximation argument shows that the equation therefore holds for all x1x\sim 1, based on the facts established in the previous paragraphs together with monotonicity in xx.

We now turn to a number of considerations involving rings. By assumption we have a ring isomorphism (F/)i=1nKi(F/\!\sim)\otimes\mathbb{Z}\cong\prod_{i=1}^{n}K_{i}, where the KiK_{i} are fields. Then every KiK_{i} is of characteristic zero, since the image of FF in KiK_{i} is a subsemifield that is a quotient semifield of FF, and every quotient of a strict semifield is again a strict semifield (or the zero ring, which is covered by our assumptions in case that the number of factors is n=0n=0).

The set

J{baab in F}J\coloneqq\{b-a\mid a\sim b\textrm{ in }F\}

is an ideal in R(F/)R\coloneqq(F/\!\approx)\otimes\mathbb{Z}, namely precisely the kernel of the canonical projection homomorphism (F/)(F/)(F/\!\approx)\otimes\mathbb{Z}\longrightarrow(F/\!\sim)\otimes\mathbb{Z}. We have J2=0J^{2}=0, since aba\sim b and cdc\sim d imply (ba)(dc)=0(b-a)(d-c)=0 in RR via Lemma 5.12 and

bd+ac=ac(ba1dc1+1)ac(ba1+dc1)=bc+ad,bd+ac=ac(ba^{-1}dc^{-1}+1)\approx ac(ba^{-1}+dc^{-1})=bc+ad,

assuming a,c0a,c\neq 0 without loss of generality. Hence RR is a square-zero extension of the quotient ring R/Ji=1nKiR/J\cong\prod_{i=1}^{n}K_{i}. Each KiK_{i} is formally smooth over \mathbb{Q} [8, Corollary 9.3.7], and therefore also their product is formally smooth over \mathbb{Q}. (A finite product of formally smooth algebras is formally smooth by lifting of idempotents.) In particular, the square-zero extension RR/JR\to R/J is split by a ring homomorphism β:RR/J\beta:R\to R/J. In the semialgebra case, RR is an \mathbb{R}-algebra in a canonical way, and in this case we have formal smoothness over \mathbb{R} for the same reason, so that we can choose β\beta to be \mathbb{R}-linear.

The universal property of (F/)(F/\approx)\otimes\mathbb{Z} implies that ϕ:F+\phi:F\to\mathbb{R}_{+} uniquely extends to a ring homomorphism ϕ^:R\hat{\phi}:R\to\mathbb{R}. In the following, we will construct a \mathbb{Q}-linear map

D^:R\hat{D}:R\longrightarrow\mathbb{R}

which similarly extends the DD defined above, in two stages.

  1. (e)

    On the nilpotent ideal JJ, taking

    D^(ba)ϕ(a)D(ba)\hat{D}(b-a)\coloneqq\phi(a)D\mathopen{}\mathclose{{}\left(\frac{b}{a}}\right)

    for aba\sim b in F×F^{\times} produces a well-defined map D^:J\hat{D}:J\to\mathbb{R}.

    Indeed, for well-definedness it is enough to show that adding some cF×c\in F^{\times} to both terms leaves the right-hand side invariant. This will be the special case obtained by taking d=cd=c in the additivity proof of the next item.

  2. (f)

    The map D^:J\hat{D}:J\to\mathbb{R} is additive. If FF is a semialgebra, then it is \mathbb{R}-linear.

    Indeed taking a,b,c,dF×a,b,c,d\in F^{\times} with aba\sim b and cdc\sim d, we obtain

    D^((ba)+(dc))\displaystyle\hat{D}\mathopen{}\mathclose{{}\left((b-a)+(d-c)}\right) =ϕ(a+c)D(b+da+c)\displaystyle=\phi(a+c)D\mathopen{}\mathclose{{}\left(\frac{b+d}{a+c}}\right)
    =(ϕ(a)+ϕ(c))D(aa+cba+ca+cdc)\displaystyle=\mathopen{}\mathclose{{}\left(\phi(a)+\phi(c)}\right)\,D\mathopen{}\mathclose{{}\left(\frac{a}{a+c}\cdot\frac{b}{a}+\frac{c}{a+c}\cdot\frac{d}{c}}\right)
    =ϕ(a)D(ba)+ϕ(c)D(dc)\displaystyle=\phi(a)D\mathopen{}\mathclose{{}\left(\frac{b}{a}}\right)+\phi(c)D\mathopen{}\mathclose{{}\left(\frac{d}{c}}\right)
    =D^(ba)+D^(dc),\displaystyle=\hat{D}(b-a)+\hat{D}(d-c),

    where the third step uses (6.6). In the semialgebra case, it is enough to verify preservation of scalar multiplication by positive scalars, which holds since ϕ\phi is necessarily a semialgebra homomorphism (because the identity map is the only semiring homomorphism ++\mathbb{R}_{+}\to\mathbb{R}_{+}).

  3. (g)

    The map D^:J\hat{D}:J\to\mathbb{R} satisfies D^(ra)=ϕ^(r)D^(a)\hat{D}(ra)=\hat{\phi}(r)\hat{D}(a) for all aJa\in J and rRr\in R.

    Indeed, writing a=cba=c-b for bcb\sim c in F×F^{\times} and plugging in the definition of D^\hat{D} shows that this holds for all rF×r\in F^{\times}. But this is nough by linearity in rr since R=F×F×R=F^{\times}-F^{\times}.

  4. (h)

    Extending by

    D^(a)D^(aβ(a))\hat{D}(a)\coloneqq\hat{D}(a-\beta(a))

    where β:R/JR\beta:R/J\to R is the splitting obtained above, defines an additive map D^:R\hat{D}:R\to\mathbb{R} satisfying the Leibniz rule with respect to ϕ^\hat{\phi}. If FF is a semialgebra, then it is \mathbb{R}-linear.

    Since aβ(a)Ja-\beta(a)\in J for all aRa\in R, this element indeed lies in the domain of D^\hat{D}. And since β(a)=0\beta(a)=0 for aJa\in J, it recovers the D^:J\hat{D}:J\to\mathbb{R} already defined above, and in particular there is no ambiguity in notation. The additivity follows by the additivity of β\beta and (f). For the Leibniz rule, we take a,bRa,b\in R and compute

    D^(ab)\displaystyle\hat{D}(ab) =D^(abβ(ab))\displaystyle=\hat{D}(ab-\beta(ab))
    =D^(a(bβ(b))+β(b)(aβ(a)))\displaystyle=\hat{D}(a(b-\beta(b))+\beta(b)(a-\beta(a)))
    =ϕ^(a)D^(bβ(b))+ϕ^(β(b))D^(aβ(a))\displaystyle=\hat{\phi}(a)\hat{D}(b-\beta(b))+\hat{\phi}(\beta(b))\hat{D}(a-\beta(a))
    =ϕ^(a)D^(b)+ϕ^(b)D^(a),\displaystyle=\hat{\phi}(a)\hat{D}(b)+\hat{\phi}(b)\hat{D}(a),

    where the second step uses multiplicativity of β\beta, the third additivity of DD as well as (g), and the fourth simply the general definition of D^\hat{D} as well as the fact that ϕ^\hat{\phi} factors across R/JR/J.

    The claimed \mathbb{R}-linearity in the semialgebra case holds by the \mathbb{R}-linearity in (f) and since β\beta is \mathbb{R}-linear.

Overall, we can therefore define the desired D:FD:F\to\mathbb{R} as the composite map

F(F/)D^.F\longrightarrow(F/\approx)\otimes\mathbb{Z}\stackrel{{\scriptstyle\hat{D}}}{{\longrightarrow}}\mathbb{R}.

The properties of D^\hat{D} proven above imply that this map is indeed a ϕ\phi-derivation. Moreover, the definition of D^\hat{D} shows that it restricts to our original DD as defined on the multiplicative group GG. This implies the desired equivalence (6.3) upon taking y=1y=1 without loss of generality. ∎

6.9 Remark.

It is interesting to ask how unique the derivation DD constructed in the proof is. Clearly DD can be replaced by any positive multiple ot itself, but is there more freedom? We answer this question now.

In terms of the data from the proof, the factored homomorphism ϕ^:R/J\hat{\phi}:R/J\to\mathbb{R} makes \mathbb{R} into an R/JR/J-module. If Δ:R/J\Delta:R/J\longrightarrow\mathbb{R} is now any derivation (over \mathbb{Z}), then we can take any derivation DD as in the proof and modify it via

DD+Δ,D^{\prime}\coloneqq D+\Delta,

obtaining another derivation DD^{\prime} that works just as well. In particular, DD^{\prime} is still monotone, and in fact xyx\sim y implies

D(y)D(x)=D(y)D(x).D^{\prime}(y)-D^{\prime}(x)=D(y)-D(x).

Conversely, if DD and DD^{\prime} are two ϕ\phi-derivations that take the same values on the multiplicative group G={aF×a1}G=\{a\in F^{\times}\mid a\sim 1\}, then the arguments given in the proof show that D=D+ΔD^{\prime}=D+\Delta as above.

Therefore we can say that DD, when normalized to D(u)=1D(u)=1, is unique up to the R/JR/J-module of derivations Der(R/J,)\mathrm{Der}_{\mathbb{Z}}(R/J,\mathbb{R}). Using R/Ji=1nKiR/J\cong\prod_{i=1}^{n}K_{i} lets us identify this module with the real vector space

i=1nDer(Ki,).\bigoplus_{i=1}^{n}\mathrm{Der}_{\mathbb{Q}}(K_{i},\mathbb{R}). (6.7)

Thus there can be a large ambiguity in the construction of DD, for example already if F/+F/\!\sim{}\cong\mathbb{R}_{+}.

If FF is a semialgebra, then this ambiguity is attenuated by the additional +\mathbb{R}_{+}-linearity condition. By the same arguments, the derivation is then unique up to elements of i=1nDer(Ki,)\bigoplus_{i=1}^{n}\mathrm{Der}_{\mathbb{R}}(K_{i},\mathbb{R}). For example if FF is a semialgebra with F/+F/\!\sim{}\cong\mathbb{R}_{+}, then DD is unique.

7.  A stronger catalytic Vergleichsstellensatz

Theorem 2.4, as developed in Part I, concludes both an “asymptotic” ordering of the form

ukxnukynn1u^{k}x^{n}\leq u^{k}y^{n}\qquad\forall n\gg 1

and a “catalytic” ordering of the form

axayax\leq ay

for some nonzero aSa\in S from the assumption that ϕ(x)<ϕ(y)\phi(x)<\phi(y) for all ϕ𝖳𝖲𝗉𝖾𝗋(S)\phi\in\mathsf{TSper}(S). Although this is a useful and quite broadly applicable result, the relevant assumption 1>01>0 makes this result not strong enough for the applications that have been mentioned in the introduction. The goal of this section is to prove a deeper Vergleichsstellensatz that applies more generally, obtained by putting together the auxiliary results developed in the previous sections. Here it is.

7.1 Theorem.

Let SS be a zerosumfree preordered semidomain with a power universal pair u,u+Su_{-},u_{+}\in S and such that:

  • \triangleright

    S/S/\!\sim has quasi-complements and quasi-inverses.

  • \triangleright

    𝖥𝗋𝖺𝖼(S/)\operatorname{\mathsf{Frac}}(S/\!\sim)\otimes\mathbb{Z} is a finite product of fields.

Let nonzero x,ySx,y\in S with xyx\sim y satisfy the following:

  • \triangleright

    For every nondegenerate monotone homomorphism ϕ:S𝕂\phi:S\to\mathbb{K} with trivial kernel and 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\},

    ϕ(x)<ϕ(y).\phi(x)<\phi(y).
  • \triangleright

    For every monotone additive map D:SD:S\to\mathbb{R}, which is a ϕ\phi-derivation for some degenerate homomorphism ϕ:S+\phi:S\to\mathbb{R}_{+} with trivial kernel and satisfies D(u+)=D(u)+1D(u_{+})=D(u_{-})+1,

    D(x)<D(y).D(x)<D(y).

Then there is nonzero aSa\in S such that axayax\leq ay.

Moreover, if SS is also a semialgebra, then it is enough to consider +\mathbb{R}_{+}-linear derivations DD in the assumptions.

Of course, if nonzero aa with axayax\leq ay exists, then this conversely implies the non-strict inequalities ϕ(x)ϕ(y)\phi(x)\leq\phi(y) and D(x)D(y)D(x)\leq D(y) for all ϕ\phi and DD as in the statement.

Proof.

As sketched in the proof of Proposition 6.8, the monotone ϕ\phi-derivations DD for degenerate ϕ:S+\phi:S\to\mathbb{R}_{+} are in canonical bijection with the monotone homomorphisms S(+)[X]/(X2)S\to\mathbb{R}_{(+)}[X]/(X^{2}) whose composition with the projection (+)[X]/(X2)+\mathbb{R}_{(+)}[X]/(X^{2})\to\mathbb{R}_{+} coincides with ϕ\phi.

We start the proof with the case where SS is a preordered semifield FF of polynomial growth. In this case, the two additional hypotheses amount to FF having quasi-complements and that F/F/\!\sim\mathop{\otimes}\mathbb{Z} is a finite product of fields. We fix a power universal element u1u\geq 1 in FF and also assume y=1y=1 without loss of generality. Considering this case will take the bulk of the proof; we will generalize from there in the final two paragraphs.

If x1x\not\leq 1, then by Theorem 2.1 we can find a total semifield preorder \preceq which extends \leq and satisfies x1x\succ 1. We fix such \preceq from now on. Now consider the layer preorder u\preceq_{u}, as defined in Remark 5.2. As we have seen, it makes FF into a multiplicatively Archimedean fully preordered semifield. Moreover, u\preceq_{u} still extends \leq, for the following reason. If a1a\geq 1, then the first condition in the definition of u\preceq_{u} in Remark 5.2 holds by the power universality of uu for \leq, and the second one holds trivially by a1a\succeq 1 and u1u\succeq 1. Hence au1a\geq_{u}1, as was to be shown. We also note uu1u\succ_{u}1 as well as the important inequality

xu1x\succeq_{u}1 (7.1)

for future use, which follows from the definition of u\preceq_{u} using ukxuku^{-k}\leq x\leq u^{k} for some kk and x1x\succeq 1. We now distinguish two cases.

  1. (1)

    u\preceq_{u} is tropical or temperate.

By reversing all preorders and replacing xx by x1x^{-1} if necessary, we can assume without loss of generality that u\preceq_{u} is max-tropical or max-temperate. Applying Proposition 6.7 to (F,u)(F,\preceq_{u}) produces a u\preceq_{u}-monotone homomorphism ϕ:F𝕂\phi:F\to\mathbb{K} for 𝕂{+,𝕋+}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{TR}_{+}\} with ϕ(u)>1\phi(u)>1. The inequality (7.1) produces ϕ(x)1\phi(x)\geq 1. On the other hand, since u\preceq_{u} extends \leq, it is clear that ϕ\phi is also \leq-monotone, and ϕ(u)>1\phi(u)>1 implies that ϕ\phi is nondegenerate. Therefore the assumption applies and gives ϕ(x)<1\phi(x)<1, a contradiction.

  1. (2)

    u\preceq_{u} is arctic.

In this case, we apply Proposition 6.8 to (F,u)(F,\preceq_{u}). Writing \simeq for the equivalence relation generated by u\preceq_{u}, we need to verify that the semifield F/F/\!\simeq has quasi-complements and that F/F/\!\simeq\mathop{\otimes}\mathbb{Z} is a finite product of fields. But these are both true since F/F/\!\simeq is a quotient of F/F/\!\sim (because u\preceq_{u} extends \leq) and these statements descend to quotients. We therefore obtain a degenerate homomorphism ϕ:F+\phi:F\to\mathbb{R}_{+} and a u\preceq_{u}-monotone ϕ\phi-derivation D:FD:F\to\mathbb{R} with D(u)>0D(u)>0. These properties in particular imply D(x)0D(x)\geq 0 by (7.1). By rescaling, we can assume D(u)=1D(u)=1 without loss of generality. But again since u\preceq_{u} extends \leq, we know that DD is in particular \leq-monotone, so that the assumption applies. This results in D(x)<0D(x)<0, a contradiction.

This proves the desired statement in the semifield case. It remains to reduce the general case to this one. We thus verify that the preordered semifield F𝖥𝗋𝖺𝖼(S)F\coloneqq\operatorname{\mathsf{Frac}}(S) satisfies the relevant assumptions. Clearly since 𝖥𝗋𝖺𝖼(S)/=𝖥𝗋𝖺𝖼(S/)\operatorname{\mathsf{Frac}}(S)/\!\sim\>=\operatorname{\mathsf{Frac}}(S/\!\sim), the existence of quasi-complements in F/F/\!\sim follows from the existence of quasi-complements and quasi-inverses in S/S/\!\sim by Lemma 3.2. Similarly, F/F/\!\sim\mathop{\otimes}\mathbb{Z} is also a finite product of fields.

Applying the statement to FF then produces the desired result: the relevant inequalities ϕ(x1)<ϕ(y1)\phi\mathopen{}\mathclose{{}\left(\frac{x}{1}}\right)<\phi\mathopen{}\mathclose{{}\left(\frac{y}{1}}\right) and D(x1)<D(y1)D\mathopen{}\mathclose{{}\left(\frac{x}{1}}\right)<D\mathopen{}\mathclose{{}\left(\frac{y}{1}}\right) hold because every such map restricts along the homomorphism S𝖥𝗋𝖺𝖼(S)S\to\operatorname{\mathsf{Frac}}(S) to a map of the corresponding type on SS, where the assumed inequalities apply. Since a homomorphism ϕ:𝖥𝗋𝖺𝖼(S)𝕂\phi:\operatorname{\mathsf{Frac}}(S)\to\mathbb{K} necessarily has trivial kernel, therefore so does its restriction to SS. ∎

7.2 Remark.

In order to determine the ring 𝖥𝗋𝖺𝖼(S/)\operatorname{\mathsf{Frac}}(S/\!\sim)\otimes\mathbb{Z} in practice, it is worth noting that the three types of constructions involved in its definition commute: we can perform the quotient by \sim, the localization at nonzero elements of SS and the Grothendieck construction in either order. This is most easily seen by the universal property: 𝖥𝗋𝖺𝖼(S/)\operatorname{\mathsf{Frac}}(S/\!\sim)\otimes\mathbb{Z} is initial in the category of rings RR equipped with a semiring homomorphism SRS\to R which identifies all \sim-related elements and maps all nonzero elements to the group of units R×R^{\times}.

7.3 Example.

Consider [X¯]\mathbb{N}[\underline{X}], the free semiring in dd variables X¯=(X1,,Xd)\underline{X}=(X_{1},\ldots,X_{d}), equipped with the semiring preorder generated by

X11,,Xd1.X_{1}\geq 1,\quad\ldots,\quad X_{d}\geq 1.

This preordered semiring is of polynomial growth, since u=1u_{-}=1 and u+=iXiu_{+}=\prod_{i}X_{i} form a power universal pair. We also have [X¯]/\mathbb{N}[\underline{X}]/\!\sim{}\cong\mathbb{N}, so that the other assumptions on SS in Theorem 7.1 obviously hold as well. The preorder can be characterized as fgf\leq g if and only if there is a finitely supported family of polynomials (pα)αd(p_{\alpha})_{\alpha\in\mathbb{N}^{d}} such that101010For a multiindex α\alpha, we use the shorthand notation X¯αi=1dXiαi\underline{X}^{\alpha}\coloneqq\prod_{i=1}^{d}X_{i}^{\alpha_{i}}.

f=αdpα,g=αdpαX¯α.f=\sum_{\alpha\in\mathbb{N}^{d}}p_{\alpha},\qquad g=\sum_{\alpha\in\mathbb{N}^{d}}p_{\alpha}\underline{X}^{\alpha}.

This works because this relation is a semiring preorder, and can be seen to be the smallest semiring preorder with Xi1X_{i}\geq 1 for all ii. Indeed the only part of this statement that is not straightforward is the transitivity. Given fghf\leq g\leq h, the desired fhf\leq h follows upon choosing a common refinement of the two given decompositions of gg by virtue of the Riesz decomposition property111111If (pi)iI(p_{i})_{i\in I} and (qj)jJ(q_{j})_{j\in J} are finitely supported families of polynomials with ipi=jqj\sum_{i}p_{i}=\sum_{j}q_{j}, then there is a doubly indexed family (rij)iI,jJ(r_{ij})_{i\in I,j\in J} with pi=jrijp_{i}=\sum_{j}r_{ij} and qj=jrijq_{j}=\sum_{j}r_{ij}. One possible proof of this is by induction on the size of the support..

In order to apply Theorem 7.1, we then first classify the monotone homomorphisms [X¯]𝕂\mathbb{N}[\underline{X}]\to\mathbb{K} for 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\}. Using the fact that [X¯]\mathbb{N}[\underline{X}] is the free semiring, it follows that the homomorphisms to +\mathbb{R}_{+} with trivial kernel are precisely the evaluation maps ff(r)f\mapsto f(r) for r>0dr\in\mathbb{R}_{>0}^{d}. Similarly, the homomorphisms [X¯]𝕋+\mathbb{N}[\underline{X}]\to\mathbb{TR}_{+} are given by optimization over the Newton polytope in a fixed direction βd\beta\in\mathbb{R}^{d},121212See I.2.9 for more detail.

[X¯]𝕋+,fmaxs𝖭𝖾𝗐𝗍𝗈𝗇(f)β,s.\mathbb{N}[\underline{X}]\longrightarrow\mathbb{TR}_{+},\qquad f\longmapsto\max_{s\,\in\,\operatorname{\mathsf{Newton}}(f)}\langle\beta,s\rangle.

By checking when the generating preorder relations Xi1X_{i}\geq 1 are preserved, we therefore obtain the following classification of the nondegenerate monotone homomorphisms ϕ:[X¯]𝕂\phi:\mathbb{N}[\underline{X}]\to\mathbb{K} with trivial kernel:

  • \triangleright

    For 𝕂=+\mathbb{K}=\mathbb{R}_{+}, the evaluation maps ff(r)f\mapsto f(r) with r>0dr\in\mathbb{R}_{>0}^{d} having components ri1r_{i}\geq 1, not all 11.

  • \triangleright

    For 𝕂=+op\mathbb{K}=\mathbb{R}_{+}^{\mathrm{op}}, the evaluation maps ff(r)f\mapsto f(r) with r>0dr\in\mathbb{R}_{>0}^{d} having components ri1r_{i}\leq 1, not all 11.

  • \triangleright

    For 𝕂=𝕋+\mathbb{K}=\mathbb{TR}_{+}, the optimization maps fmaxs𝖭𝖾𝗐𝗍𝗈𝗇(f)β,sf\mapsto\max_{s\,\in\,\operatorname{\mathsf{Newton}}(f)}\langle\beta,s\rangle for βd\beta\in\mathbb{R}^{d} having components βi0\beta_{i}\geq 0 not all 0.

  • \triangleright

    For 𝕂=𝕋+op\mathbb{K}=\mathbb{TR}_{+}^{\mathrm{op}}, the optimization maps fmaxs𝖭𝖾𝗐𝗍𝗈𝗇(f)β,sf\mapsto\max_{s\,\in\,\operatorname{\mathsf{Newton}}(f)}\langle\beta,s\rangle for βd\beta\in\mathbb{R}^{d} having components βi0\beta_{i}\leq 0 not all 0.

Concerning the derivations, there is exactly one degenerate homomorphism ϕ:[X¯]+\phi:\mathbb{N}[\underline{X}]\to\mathbb{R}_{+}, namely the evaluation map ff(1¯)f\mapsto f(\underline{1}). The monotone ϕ\phi-derivations [X¯]\mathbb{N}[\underline{X}]\to\mathbb{R} are parametrized by γ+d\gamma\in\mathbb{R}_{+}^{d} and given by

  • \triangleright

    fγ,ff\mapsto\langle\gamma,\nabla f\rangle, where f+d[X¯]\nabla f\in\mathbb{R}_{+}^{d}[\underline{X}] is the gradient of ff.

The normalization condition D(u+)=D(u)+1D(u_{+})=D(u_{-})+1 amounts to the constraint iγi=1\sum_{i}\gamma_{i}=1. Since every such γ\gamma is a convex combination of the standard basis vectors, it is sufficient to impose the assumed inequalities on these.

Hence Theorem 7.1 instantiates to the following result. Suppose that f,g[X¯]f,g\in\mathbb{N}[\underline{X}] satisfy the following conditions:

  • \triangleright

    f(1¯)=g(1¯)f(\underline{1})=g(\underline{1}).

  • \triangleright

    f(r)<g(r)f(r)<g(r) for all r[1,)d{1¯}r\in[1,\infty)^{d}\setminus\{\underline{1}\}.

  • \triangleright

    f(r)>g(r)f(r)>g(r) for all r(0,1]d{1¯}r\in(0,1]^{d}\setminus\{\underline{1}\}.

  • \triangleright

    For every β+d{0}\beta\in\mathbb{R}_{+}^{d}\setminus\{0\},

    maxs𝖭𝖾𝗐𝗍𝗈𝗇(f)β,s\displaystyle\max_{s\,\in\,\operatorname{\mathsf{Newton}}(f)}\langle\beta,s\rangle <maxs𝖭𝖾𝗐𝗍𝗈𝗇(g)β,s,\displaystyle<\max_{s\,\in\,\operatorname{\mathsf{Newton}}(g)}\langle\beta,s\rangle,
    mins𝖭𝖾𝗐𝗍𝗈𝗇(f)β,s\displaystyle\min_{s\,\in\,\operatorname{\mathsf{Newton}}(f)}\langle\beta,s\rangle <mins𝖭𝖾𝗐𝗍𝗈𝗇(g)β,s.\displaystyle<\min_{s\,\in\,\operatorname{\mathsf{Newton}}(g)}\langle\beta,s\rangle.
  • \triangleright

    For all i=1,,di=1,\ldots,d,

    fXi(1¯)<gXi(1¯)\frac{\partial f}{\partial X_{i}}(\underline{1})<\frac{\partial g}{\partial X_{i}}(\underline{1})

Then there is a nonzero polynomial h[X¯]h\in\mathbb{N}[\underline{X}] and a family (pα)αd(p_{\alpha})_{\alpha\in\mathbb{N}^{d}} of polynomials pα[X¯]p_{\alpha}\in\mathbb{N}[\underline{X}] such that

hf=αdpα,hg=αdpαX¯α.hf=\sum_{\alpha\in\mathbb{N}^{d}}p_{\alpha},\qquad hg=\sum_{\alpha\in\mathbb{N}^{d}}p_{\alpha}\underline{X}^{\alpha}.

Conversely, this property is clearly sufficient to imply the above conditions with non-strict inequality.

Using the fact that the identity map is the only semiring homomorphism ++\mathbb{R}_{+}\to\mathbb{R}_{+}, it is straightforward to see that the same result holds with +\mathbb{R}_{+} coefficients instead of \mathbb{N} coefficients.

8.  A stronger asymptotic Vergleichsstellensatz

Throughout this section, we also assume that SS is a zerosumfree semidomain, and of polynomial growth with respect to a power universal element uSu\in S.

Theorem 2.4 also makes statements about “asymptotic” ordering relations of the form ukxnukynu^{k}x^{n}\leq u^{k}y^{n} for all n1n\gg 1. Its proof conducted in Part I was based on a suitable definition of test spectrum and a proof that the test spectrum is a compact Hausdorff space. We do not yet have a sufficiently general definition of test spectrum to achieve a similar feat at the level of generality of Theorem 7.1, but we do under the stronger assumption that

S/>0d{0}S/\!\sim{}\!\cong\mathbb{R}_{>0}^{d}\cup\{0\} (8.1)

for some dd\in\mathbb{N}.131313Note that this amounts to S/+S/\!\sim{}\!\cong\mathbb{R}_{+} in the case d=1d=1, and we interpret it as S/𝔹S/\!\sim{}\!\cong\mathbb{B} for d=0d=0. If this is the case, then S/S/\!\sim obviously has quasi-complements and quasi-inverses and is such that 𝖥𝗋𝖺𝖼(S/)d\operatorname{\mathsf{Frac}}(S/\!\sim)\otimes\mathbb{Z}\cong\mathbb{R}^{d} is a finite product of fields, so that the algebraic assumptions on S/S/\!\sim in Theorem 7.1 are clearly satisfied.

The assumption (8.1) is most conveniently formulated a little differently: we assume that SS is a preordered semiring equipped with a fixed surjective141414The surjectivity requirement can arguably be relaxed and replaced by a suitable arithmetical conditions in the image of \|\cdot\| in >0d{0}\mathbb{R}_{>0}^{d}\cup\{0\}, but we have not worked this out in detail since the surjectivity holds in our applications [2, 4]. homomorphism

:S>0d{0}\|\cdot\|:S\longrightarrow\mathbb{R}_{>0}^{d}\cup\{0\}

with trivial kernel and such that

xyx=yxy.x\leq y\quad\Longrightarrow\quad\|x\|=\|y\|\quad\Longrightarrow\quad x\sim y.

Under these assumptions, we obtain the desired (8.1), where the isomorphism is implemented by \|\cdot\| itself. The components of \|\cdot\| are degenerate homomorphisms

1,,d:S+,\|\cdot\|_{1},\>\ldots,\>\|\cdot\|_{d}:S\longrightarrow\mathbb{R}_{+},

and there are no other ones.151515To see this for d1d\geq 1, note that every homomorphism >0d{0}+\mathbb{R}_{>0}^{d}\cup\{0\}\longrightarrow\mathbb{R}_{+} extends uniquely to a ring homomorphism d\mathbb{R}^{d}\to\mathbb{R}, and the only such homomorphisms are the projections. For d=0d=0, the statement is that there is no homomorphisms 𝔹+\mathbb{B}\to\mathbb{R}_{+} at all, which is clear.,161616This would not necessarily be the case without the surjectivity requirement on \|\cdot\|. For example if d=1d=1, and if the image of \|\cdot\| is a semiring isomorphic to [X]\mathbb{N}[X], then this fails in a particularly bad way due to the abundance of homomorphisms [X]+\mathbb{N}[X]\to\mathbb{R}_{+}.

8.1 Remark.

The notation \|\cdot\| is modelled after norms in functional analysis, since these sometimes have homomorphism properties when restricted to a semiring of suitably “positive” elements. For example if GG is an abelian group and +[G]\mathbb{R}_{+}[G] the associated group semialgebra, then the 1\ell^{1}-norm is a semiring homomorphism +[G]+\mathbb{R}_{+}[G]\to\mathbb{R}_{+}, making it an instance of the above with d=1d=1.

  The test spectrum

We now move towards the relevant notion of spectrum, using the multiplicative picture of 𝕋+\mathbb{TR}_{+}. We need a little more preparation to deal with the derivations, and in particular with the ambiguity discussed in Remark 6.9.

8.2 Definition.

For i=1,,di=1,\ldots,d, two i\|\cdot\|_{i}-derivations D,D:SD,D^{\prime}:S\to\mathbb{R} are interchangeable if their difference factors through S/S/\!\sim.

In other words, interchangeability of derivations DD and DD^{\prime} means that there is a \mathbb{Q}-linear derivation Δ:\Delta:\mathbb{R}\to\mathbb{R} such that171717To see this, note that every i\|\cdot\|_{i}-derivation uniquely extends to an additive map d\mathbb{R}^{d}\to\mathbb{R} that is a derivation with respect to the ii-projection map d\mathbb{R}^{d}\to\mathbb{R}. But every such derivation itself factors through the ii-th projection map, and hence our DDD^{\prime}-D actually factors through i\|\cdot\|_{i}.

D(x)=D(x)+Δ(xi)xS.D^{\prime}(x)=D(x)+\Delta(\|x\|_{i})\qquad\forall x\in S.

The real vector space of such Δ\Delta’s is exactly Der(,)\mathrm{Der}_{\mathbb{Q}}(\mathbb{R},\mathbb{R}), which is infinite-dimensional181818This is by the standard fact that the module of Kähler differentials for a transcendental field extension in characteristic zero is free with basis given by a transcendence basis of the extension [1, Theorem 16.14].. Since two interchangeable derivations satisfy

D(y)D(x)=D(y)D(x)D^{\prime}(y)-D^{\prime}(x)=D(y)-D(x)

whenever xyx\sim y, it is sufficient to consider only one of these derivations in our Vergleichsstellensätze in the spectral preordering. In the definition of the test spectrum below, this is why we only consider interchangeability classes of derivations. If SS is a semialgebra, then every interchangeability class has a canonical representative given by an +\mathbb{R}_{+}-linear derivation, which is a very convenient feature worth keeping in mind.

8.3 Definition.

The test spectrum of SS is the disjoint union

𝖳𝖲𝗉𝖾𝗋(S)\displaystyle\mathsf{TSper}(S)\>\coloneqq\> {monotone homs S+ or S+op with trivial kernel}{1,,d}\displaystyle\>\{\text{{monotone homs }}S\to\mathbb{R}_{+}\text{{ or }}S\to\mathbb{R}_{+}^{\mathrm{op}}\text{{ with trivial kernel}}\}\setminus\{\|\cdot\|_{1},\ldots,\|\cdot\|_{d}\}
{monotone homs S𝕋+ with ϕ(u)=e and trivial kernel}\displaystyle\sqcup\{\text{{monotone homs }}S\to\mathbb{TR}_{+}\text{{ with }}\phi(u)=e\text{{ and trivial kernel}}\}
{monotone homs S𝕋+op with ϕ(u)=e1 and trivial kernel}\displaystyle\sqcup\{\text{{monotone homs }}S\to\mathbb{TR}_{+}^{\mathrm{op}}\text{{ with }}\phi(u)=e^{-1}\text{{ and trivial kernel}}\}
i=1d{monotone i-derivations S with D(u)=1 modulo interchangeability}.\displaystyle\sqcup\bigsqcup_{i=1}^{d}\{\text{{monotone }}\|\cdot\|_{i}\text{{-derivations }}S\to\mathbb{R}\text{{ with }}D(u)=1\text{{ modulo interchangeability}}\}.

There are thus five types of points of 𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S), and these five types match the five types of Definition 5.5.

As in the simpler case treated in Part I, our goal is to turn 𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S) into a compact Hausdorff space. The relevant topology will again be the weak topology with respect to a certain class of maps. In the present case, these are the logarithmic comparison maps, defined for nonzero x,ySx,y\in S with xyx\sim y as

lcx,y(ϕ)logϕ(x)ϕ(y)logϕ(u),lcx,y(D)D(y)D(x)xi,\mathrm{lc}_{x,y}(\phi)\coloneqq\frac{\log\frac{\phi(x)}{\phi(y)}}{\log\phi(u)},\quad\qquad\mathrm{lc}_{x,y}(D)\coloneqq\frac{D(y)-D(x)}{\|x\|_{i}}, (8.2)

where the first equation applies in all four non-derivation cases and the second equation in the i\|\cdot\|_{i}-derivation case for i=1,,di=1,\ldots,d, respectively, where one may want to keep in mind that xi=yi\|x\|_{i}=\|y\|_{i}.

A few further clarifying comments on this definition are in order:

  • \triangleright

    The definition in the derivation case clearly respects interchangeability, making lcx,y\mathrm{lc}_{x,y} well-defined on 𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S).

  • \triangleright

    The denominator in the definition of lcx,y(ϕ)\mathrm{lc}_{x,y}(\phi) does not vanish for any ϕ\phi, since ϕ(u)=1\phi(u)=1 would imply by power universality that ϕ\phi is degenerate, and therefore equal to one of the i\|\cdot\|_{i}, which we have assumed not to be the case. In fact, the denominator is positive by ϕ(u)>1\phi(u)>1 for ϕ:S+\phi:S\to\mathbb{R}_{+} and ϕ:S𝕋+\phi:S\to\mathbb{TR}_{+}, and it is likewise negative for ϕ:S+op\phi:S\to\mathbb{R}_{+}^{\mathrm{op}} and ϕ:S𝕋+op\phi:S\to\mathbb{TR}_{+}^{\mathrm{op}}.

  • \triangleright

    It follows that lcx,y0\mathrm{lc}_{x,y}\geq 0 if xyx\leq y.

  • \triangleright

    The denominator also results in the convenient normalization lcu,1=1\mathrm{lc}_{u,1}=1.

  • \triangleright

    The denominator x\|x\| in the definition of lcx,y(D)\mathrm{lc}_{x,y}(D) is what makes the equation

    lcax,ay=lcx,y\mathrm{lc}_{ax,ay}=\mathrm{lc}_{x,y} (8.3)

    hold on all of 𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S) for all nonzero aSa\in S.

  • \triangleright

    We also have the following cocycle equation: for nonzero xyzx\sim y\sim z,

    lcx,z=lcx,y+lcy,z.\mathrm{lc}_{x,z}=\mathrm{lc}_{x,y}+\mathrm{lc}_{y,z}.
8.4 Definition.

𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S) carries the weakest topology which makes the logarithmic comparison maps

lcx,y:𝖳𝖲𝗉𝖾𝗋(S)\mathrm{lc}_{x,y}\>:\>\mathsf{TSper}(S)\longrightarrow\mathbb{R}

continuous for all nonzero x,ySx,y\in S with xyx\sim y.

The following compactness statement is now the analogue of I.7.9, where also the proof is conceptually similar.

8.5 Proposition.

With these definitions, 𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S) is a compact Hausdorff space.

Proof.

We first note that 𝖳𝖲𝗉𝖾𝗋(S)=𝖳𝖲𝗉𝖾𝗋(𝖥𝗋𝖺𝖼(S))\mathsf{TSper}(S)=\mathsf{TSper}(\operatorname{\mathsf{Frac}}(S)) by (8.3) and the fact that interchangeability of derivations on SS is equivalent to that on 𝖥𝗋𝖺𝖼(S)\operatorname{\mathsf{Frac}}(S), since S/𝖥𝗋𝖺𝖼(S)/S/\!\sim{}\!\cong\operatorname{\mathsf{Frac}}(S)/\!\sim by (8.1). We therefore assume without loss of generality that SS is a semifield FF. Again by (8.3), on 𝖳𝖲𝗉𝖾𝗋(F)\mathsf{TSper}(F) we have

lcx,y=lc1,yx.\mathrm{lc}_{x,y}=\mathrm{lc}_{1,\frac{y}{x}}.

The topology on 𝖳𝖲𝗉𝖾𝗋(F)\mathsf{TSper}(F) is therefore equivalently generated by the logarithmic evaluation maps

levx(ϕ)logϕ(x)logϕ(u),levx(D)D(x),\mathrm{lev}_{x}(\phi)\coloneqq\frac{\log\phi(x)}{\log\phi(u)},\qquad\quad\mathrm{lev}_{x}(D)\coloneqq D(x),

and these are parametrized by x1x\sim 1 in FF. These maps satisfy a degenerate form of the Leibniz rule,

levxy=levx+levy,\mathrm{lev}_{xy}=\mathrm{lev}_{x}+\mathrm{lev}_{y},

as well as monotonicity in xx and

levu=1,lev1=0.\mathrm{lev}_{u}=1,\qquad\mathrm{lev}_{1}=0.

We then start the proof of the claim by showing Hausdorffness first. Indeed the logarithmic evaluation map

levu+12:𝖳𝖲𝗉𝖾𝗋(F)\mathrm{lev}_{\frac{u+1}{2}}\>:\>\mathsf{TSper}(F)\longrightarrow\mathbb{R}

nicely distinguishes the types as follows:

  • \triangleright

    For ϕ:F𝕋+\phi:F\to\mathbb{TR}_{+}, we have levu+12(ϕ)=1\mathrm{lev}_{\frac{u+1}{2}}(\phi)=1.

  • \triangleright

    For ϕ:F+\phi:F\to\mathbb{R}_{+}, we have 12<levu+12(ϕ)<1\frac{1}{2}<\mathrm{lev}_{\frac{u+1}{2}}(\phi)<1, where the former is by r+12>r1/2\frac{r+1}{2}>r^{1/2} for all r>1r>1.

  • \triangleright

    For D:FD:F\to\mathbb{R} a i\|\cdot\|_{i}-derivation, we have levu+12(D)=12\mathrm{lev}_{\frac{u+1}{2}}(D)=\frac{1}{2}.

  • \triangleright

    For ϕ:F+op\phi:F\to\mathbb{R}_{+}^{\mathrm{op}}, we have 0<levu+12(ϕ)<120<\mathrm{lev}_{\frac{u+1}{2}}(\phi)<\frac{1}{2}, where the latter is by r+12<r1/2\frac{r+1}{2}<r^{1/2} for all r(0,1)r\in(0,1).

  • \triangleright

    For ϕ:F𝕋+op\phi:F\to\mathbb{TR}_{+}^{\mathrm{op}}, we have levu+12(ϕ)=0\mathrm{lev}_{\frac{u+1}{2}}(\phi)=0.

In particular, this shows that any two points of distinct types can be separated, namely by the continuous function levu+12C(𝖳𝖲𝗉𝖾𝗋(F))\mathrm{lev}_{\frac{u+1}{2}}\in C(\mathsf{TSper}(F)). To separate two points of the same type, we consider each one of the five cases separately, and show in each case that if two points cannot be separated, then they are equal:

  • \triangleright

    For ϕ,ψ:F𝕋+\phi,\psi:F\to\mathbb{TR}_{+}, suppose that levx(ϕ)=levx(ψ)\mathrm{lev}_{x}(\phi)=\mathrm{lev}_{x}(\psi) for all x1x\sim 1.

    Then ϕ(u)=ψ(u)=e\phi(u)=\psi(u)=e implies ϕ(x)=ψ(x)\phi(x)=\psi(x) for all x1x\sim 1 by definition of the logarithmic evaluation maps. For arbitrary aF×a\in F^{\times}, let kk\in\mathbb{N} be such that ekϕ(a),ψ(a)eke^{-k}\leq\phi(a),\psi(a)\leq e^{k}. Then

    ekϕ(a)=ϕ(a+uka+uk)=ψ(a+uka+uk)=ekψ(a),e^{-k}\phi(a)=\phi\mathopen{}\mathclose{{}\left(\frac{a+u^{-k}}{a+u^{k}}}\right)=\psi\mathopen{}\mathclose{{}\left(\frac{a+u^{-k}}{a+u^{k}}}\right)=e^{-k}\psi(a),

    implying that ϕ=ψ\phi=\psi.

  • \triangleright

    For ϕ,ψ:F+\phi,\psi:F\to\mathbb{R}_{+}, suppose that levx(ϕ)=levx(ψ)\mathrm{lev}_{x}(\phi)=\mathrm{lev}_{x}(\psi) for all x1x\sim 1.

    Then using the above levu+12\mathrm{lev}_{\frac{u+1}{2}} shows ϕ(u)=ψ(u)\phi(u)=\psi(u) after a short calculation. Therefore again ϕ(x)=ψ(x)\phi(x)=\psi(x) for all x1x\sim 1. Then for arbitrary aF×a\in F^{\times}, the desired ϕ(a)=ψ(a)\phi(a)=\psi(a) follows by

    ϕ(a)+ϕ(u)ϕ(a)+1=ϕ(a+ua+1)=ψ(a+ua+1)=ψ(a)+ψ(u)ψ(a)+1\frac{\phi(a)+\phi(u)}{\phi(a)+1}=\phi\mathopen{}\mathclose{{}\left(\frac{a+u}{a+1}}\right)=\psi\mathopen{}\mathclose{{}\left(\frac{a+u}{a+1}}\right)=\frac{\psi(a)+\psi(u)}{\psi(a)+1}

    and some calculation, using ϕ(u)=ψ(u)1\phi(u)=\psi(u)\neq 1 by nondegeneracy.

  • \triangleright

    For D,D:SD,D^{\prime}:S\to\mathbb{R}, where DD is an i\|\cdot\|_{i}-derivation and DD^{\prime} is a j\|\cdot\|_{j}-derivation for some i,j{1,,d}i,j\in\{1,\ldots,d\}, suppose that levx(D)=levx(D)\mathrm{lev}_{x}(D^{\prime})=\mathrm{lev}_{x}(D) for all x1x\sim 1.

    We first show that this requires j=ij=i. Indeed every derivation “remembers” the degenerate homomorphism i\|\cdot\|_{i} at which it is defined: using additivity and the Leibniz rule together with ui=1\|u\|_{i}=1, we obtain that for every aFa\in F,

    D(ua+1a+u)\displaystyle D\mathopen{}\mathclose{{}\left(\frac{ua+1}{a+u}}\right) =D(ua)(ai+1)D(a+u)(ai+1)(ai+1)2\displaystyle=\frac{D(ua)\mathopen{}\mathclose{{}\left(\|a\|_{i}+1}\right)-D(a+u)\mathopen{}\mathclose{{}\left(\|a\|_{i}+1}\right)}{\mathopen{}\mathclose{{}\left(\|a\|_{i}+1}\right)^{2}}
    =D(a)+aiD(u)D(a)D(u)ai+1\displaystyle=\frac{D(a)+\|a\|_{i}D(u)-D(a)-D(u)}{\|a\|_{i}+1}
    =ai1ai+1D(u).\displaystyle=\frac{\|a\|_{i}-1}{\|a\|_{i}+1}D(u).

    Since this can be solved uniquely for ai\|a\|_{i}, and the same applies to DD^{\prime} with respect to aj\|a\|_{j}, the assumptions on DD and DD^{\prime} imply that ai=aj\|a\|_{i}=\|a\|_{j} for all aFa\in F, and hence j=ij=i.

    But then DDD^{\prime}-D is a i\|\cdot\|_{i}-derivation that factors across F/F/\!\sim{}. Therefore DD^{\prime} and DD are interchangeable and represent the same point of 𝖳𝖲𝗉𝖾𝗋(F)\mathsf{TSper}(F).

  • \triangleright

    The remaining two cases involving +op\mathbb{R}_{+}^{\mathrm{op}} and 𝕋+op\mathbb{TR}_{+}^{\mathrm{op}} work similarly as the first two.

This completes the proof of Hausdorffness.

For compactness, we characterize 𝖳𝖲𝗉𝖾𝗋(F)\mathsf{TSper}(F) as a closed subspace of the product space x1[kx,kx]\prod_{x\sim 1}[-k_{x},k_{x}], which is compact by Tychonoff’s theorem. Here, kxk_{x}\in\mathbb{N} is such that the power universality inequalities

xukx,xukx1x\leq u^{k_{x}},\qquad xu^{k_{x}}\geq 1

hold. Given an element νx1[kx,kx]\nu\in\prod_{x\sim 1}[-k_{x},k_{x}], we claim that it corresponds under the logarithmic evaluation maps to a point of 𝖳𝖲𝗉𝖾𝗋(F)\mathsf{TSper}(F) if and only if the following conditions hold for all x,y1x,y\sim 1 and all nonzero aa:

  1. (a)

    νxy=νx+νy\nu_{xy}=\nu_{x}+\nu_{y}.

  2. (b)

    νu=1\nu_{u}=1.

  3. (c)

    If x1x\geq 1, then νx0\nu_{x}\geq 0.

  4. (d)

    We have

    (0νx+a1+aνx)(νxνx+a1+a0).\mathopen{}\mathclose{{}\left(0\leq\nu_{\frac{x+a}{1+a}}\leq\nu_{x}}\right)\quad\lor\quad\mathopen{}\mathclose{{}\left(\nu_{x}\leq\nu_{\frac{x+a}{1+a}}\leq 0}\right).

The proof is complete once this claim is established, since all of these conditions are clearly closed.

It is straightforward to verify that these conditions hold for a spectral point, so we focus on the converse. For given ν\nu, the task is to extend it to a spectral point ϕ\phi or DD defined on all of FF. To this end, consider the new preorder relation defined for x,yF×x,y\in F^{\times} by

xy:(xyνyx10).x\preceq y\quad:\Longleftrightarrow\quad(x\sim y\>\>\land\>\>\nu_{yx^{-1}}\geq 0).

Declaring additionally 000\preceq 0, the above properties (a)(d) then imply that \preceq makes FF into a preordered semifield, where the monotonicity of addition in particular relies on (d). Note that \preceq extends \leq and the quotient of FF by the equivalence relation generated by \preceq is still >0d{0}\mathbb{R}_{>0}^{d}\cup\{0\}. Moreover, (F,)(F,\preceq) is clearly a multiplicatively Archimedean fully preordered semifield with u1u\succ 1.

We once more distinguish types:

  • \triangleright

    If νu+12>12\nu_{\frac{u+1}{2}}>\frac{1}{2}, then we obtain u2+2u+14uu^{2}+2u+1\succ 4u from

    νu2+2u+14=2νu+12>1=νu,\nu_{\frac{u^{2}+2u+1}{4}}=2\nu_{\frac{u+1}{2}}>1=\nu_{u},

    which implies that \preceq is max-tropical or max-temperate. Therefore Proposition 6.7 applies and produces a homomorphism ϕ:F𝕂\phi:F\to\mathbb{K} with 𝕂{+,𝕋+}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{TR}_{+}\} such that for all xyx\sim y,

    xyϕ(x)ϕ(y).x\preceq y\quad\Longleftrightarrow\quad\phi(x)\leq\phi(y).

    In particular, ϕ\phi is still \leq-monotone and nondegenerate. For x1x\sim 1, the desired equation νx=logϕ(x)logϕ(u)\nu_{x}=\frac{\log\phi(x)}{\log\phi(u)} then holds as a consequence of the definition of \preceq and Lemma 5.3 using rational approximation for the real number νx\nu_{x}.

  • \triangleright

    If νu+12=12\nu_{\frac{u+1}{2}}=\frac{1}{2}, then ν\nu will correspond to a derivation. Indeed u2+2u+1u^{2}+2u+1 and 4u4u are now \preceq-equivalent, and if u+u12u+u^{-1}\succ 2 or u+u12u+u^{-1}\prec 2 was the case, then the cancellation criterion of Proposition 5.10 would imply u2+2u+14uu^{2}+2u+1\succ 4u or u2+2u+14uu^{2}+2u+1\prec 4u. Hence \preceq is arctic and Proposition 6.8 applies, and the resulting homomorphism ψ:F(+)[X]/(X2)\psi:F\to\mathbb{R}_{(+)}[X]/(X^{2}) must have components given by i\|\cdot\|_{i} for some i{1,,d}i\in\{1,\ldots,d\}, since these are the only degenerate homomorphisms F+F\to\mathbb{R}_{+}, together with some i\|\cdot\|_{i}-derivation D:FD:F\to\mathbb{R}. The claim νx=D(x)\nu_{x}=D(x) for nonzero xx now follows as in the previous item, while also using Lemma 5.13 again.

  • \triangleright

    If νu+12<12\nu_{\frac{u+1}{2}}<\frac{1}{2}, we can proceed as in the first case. The only difference is that Proposition 6.7 needs to be applied to op\preceq^{\mathrm{op}} since \preceq is now min-temperate or min-tropical. ∎

Here is now our second main result.

8.6 Theorem.

Let SS be a preordered semiring with a power universal element uSu\in S. Suppose that for some dd\in\mathbb{N}, there is a surjective homomorphism :S>0d{0}\|\cdot\|:S\to\mathbb{R}_{>0}^{d}\cup\{0\} with trivial kernel and such that

aba=bab.a\leq b\quad\Longrightarrow\quad\|a\|=\|b\|\quad\Longrightarrow\quad a\sim b.

Let x,ySx,y\in S be nonzero with x=y\|x\|=\|y\|. Then the following are equivalent:

  1. (a)
    • \triangleright

      For every nondegenerate monotone homomorphism ϕ:S𝕂\phi:S\to\mathbb{K} with trivial kernel and 𝕂{+,+op,𝕋+,𝕋+op}\mathbb{K}\in\{\mathbb{R}_{+},\mathbb{R}_{+}^{\mathrm{op}},\mathbb{TR}_{+},\mathbb{TR}_{+}^{\mathrm{op}}\},

      ϕ(x)ϕ(y).\phi(x)\leq\phi(y).
    • \triangleright

      For every i=1,,di=1,\ldots,d and monotone i\|\cdot\|_{i}-derivation D:SD:S\to\mathbb{R} with D(u)=1D(u)=1,

      D(x)D(y).D(x)\leq D(y).
  2. (b)

    For every ε>0\varepsilon>0, we have

    xnuεnynn1.x^{n}\leq u^{\lfloor\varepsilon n\rfloor}y^{n}\qquad\forall n\gg 1.

Moreover, suppose that the inequalities in (a) are all strict. Then also the following hold:

  1. (c)

    There is kk\in\mathbb{N} such that

    ukxnukynn1.u^{k}x^{n}\leq u^{k}y^{n}\qquad\forall n\gg 1.
  2. (d)

    If yy is power universal as well, then

    xnynn1.x^{n}\leq y^{n}\qquad\forall n\gg 1.
  3. (e)

    There is nonzero aSa\in S such that

    axay.ax\leq ay.

    Moreover, there is kk\in\mathbb{N} such that aukj=0nxjynja\coloneqq u^{k}\sum_{j=0}^{n}x^{j}y^{n-j} for any n1n\gg 1 does the job.

Finally, if SS is also a semialgebra, then all statements also hold with only +\mathbb{R}_{+}-linear derivations DD in (a).

Putting d=0d=0 recovers Theorem 2.4 as a much simpler special case.

Proof.

The implication from (b) to (a) is again straightforward. For the other implications, note first that the assumed S/>0{0}S/\!\sim{}\!\cong\mathbb{R}_{>0}\cup\{0\} implies that SS is a zerosumfree semidomain. Hence by Theorem 7.1, we know that (a) with strict inequalities implies that there is nonzero bSb\in S with bxbybx\leq by by Theorem 7.1. Choosing any cSc\in S with c=b1\|c\|=\|b\|^{-1} by the surjectivity of \|\cdot\| and taking abca\coloneqq bc results in a=1\|a\|=1, and therefore we have

axay,a1.ax\leq ay,\qquad a\sim 1.

The proof can now be completed by verbatim the same arguments as the proof of Theorem 2.4 conducted in Part I, where the compactness of 𝖳𝖲𝗉𝖾𝗋(S)\mathsf{TSper}(S) now enters through the logarithmic comparison function lcx,y\mathrm{lc}_{x,y} being strictly bounded away from zero since it already is strictly positive everywhere. ∎

8.7 Example.

Consider the polynomial semiring +[X¯]\mathbb{R}_{+}[\underline{X}] with the semiring preorder generated by

X11,,Xd1,X_{1}\geq 1,\quad\ldots,\quad X_{d}\geq 1,

as briefly discussed at the end of Example 7.3. Since +[X¯]/+\mathbb{R}_{+}[\underline{X}]/\!\sim\mathop{\cong}\mathbb{R}_{+}, Theorem 8.6 applies with d=1d=1 and proves the following. If f,g+[X¯]f,g\in\mathbb{R}_{+}[\underline{X}] satisfy the same conditions as those listed in Example 7.3, then we can conclude that for every n1n\gg 1 there is a family of polynomials (pα)αd(p_{\alpha})_{\alpha\in\mathbb{N}^{d}} such that

fn=αdpα,gn=αdpαX¯α.f^{n}=\sum_{\alpha\in\mathbb{N}^{d}}p_{\alpha},\qquad g^{n}=\sum_{\alpha\in\mathbb{N}^{d}}p_{\alpha}\underline{X}^{\alpha}.

This is by Theorem 8.6(c) together with the fact that factors of u=iXiu=\prod_{i}X_{i} can be cancelled from inequalities in this preordered semiring.

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