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Rieg theory

Semirings with exponentiation, connecting arithmetic, logic, and categorical counting.

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A rieg is a commutative semiring equipped with exponentiation satisfying the usual exponent laws, including x0=1x^0=1 and hence 00=10^0=1. This project studies the structure itself and the examples it connects.

The research programme is described in note 7. The current questions concern the subrieg generated by nonnegative rationals, game interpretations of the free rieg, functions represented by unary rieg terms, and finite-valued presheaf categories.

One concrete result narrows the topological problem: a compact Hausdorff topological rieg with cancellative addition is trivial. The question whether a nontrivial connected compact Hausdorff rieg exists therefore reduces to the noncancellative case, where it has a positive answer.

For finite-valued presheaves, essentially finite slices give a sufficient condition for a topos. A partial answer proposes an exact criterion stated in terms of finite presheaves. The right Kan extension approach is absorbed into that note, with the coreflection X↦XfinX\mapsto X_{\mathrm{fin}} onto locally finite presheaves as its comonad.

The next step is concrete: give a criterion for finite presheaf topoi in terms of the indexing category alone.

Work in progress. The Introduction is incomplete.

NOTES ON RIEG THEORY: SEMIRING WITH EXPONENTIALS IN LOGIC, PROFINITE ARITHMETIC, ENUMERATIVE COMBINATORICS, AND CATEGORY THEORY

RYUYA HORA

Abstract. In this paper, we introduce the notion of a rieg, that is, a semiring equipped with a binary operation called exponentiation, and present its basic theory along with numerous examples. We begin by reviewing the classification of quotient riegs over ℕ given by Burris and Lee. Then, through concrete examples, we propose how riegs can appear in (pro)finite arithmetic, category theory, enumerative combinatorics, and logic. Typical examples of a rieg include the rieg of natural numbers ℕ, the five cyclic models (ℕ1,1,ℕ1,2,ℕ1,6,ℕ1,42, and ℕ1,1806), other quotient riegs of ℕ ( = modulo arithmetic with exponentials), the canonical profinite rieg ℕ ⊔ℤ^, the topological rieg of 10-adic natural numbers ℕ∞,10∞ = ℕ ⊔ (ℤ2 ×ℤ5) (which does not work for the 5-adic natural numbers ℕ ⊔ℤ5), the cardinal arithmetic like 2ℵ0, the rieg of truth values {⊥,⊤ ⁡} with pq := (q → p), Heyting algebras including the set of open subsets 𝒪(X) of a topological space X, the rieg of finite directed graphs, the Burnside rieg of a finite group, the rieg of rooted trees, the rieg of isomorphism classes of finite groups, and other decategorified topoi.

2020 Mathematics Subject Classification. 16Y60, 03G10, 11D61, 20E18, 18D15, 18B25.

Key words and phrases. Semiring, rig, exponential, profinite set, Tarski highschool problem, objective number theory, cartesian closed category, topos, Burnside rig, Heyting algebra, equational theory.

1. Introduction

The aim of the present paper is to popularise an algebraic structure, which we will call riegs, by providing several theorems and examples. As a rig defined as a “ring without negatives” ([Sch90]) a rieg is a “rig with exponentials.” In other words, a rieg is a ring-like structure that has 0,1,+,×, and a binary operation ↑ (Definition 2.2.1). The prototypical example of a rieg is the rieg of natural numbers.

Of course, the idea of considering ring-like structures with exponentials is not new.

In the context of category theory, [Bir42]

[Sch00a] Objective number theory is the study of addition and multiplication (and eventually exponentiation) of objects in suitable categories.

Fiore [FDB06]

To the best of the author’s knowledge, the field in which rigs with (binary) exponentiation have been most actively studied is (finite) model theory. The Tarski High School problem concerns the theory of the positive integers with addition, multiplication, and exponentiation. Even after this problem was resolved, research on small models continued to progress. The story until 2005 is summarized in [BY05].

Acknowledgement. The author would like to thank his supervisor Ryu Hasegawa for helpful discussions and suggestions. He was supported by JSPS KAKENHI Grant Number JP24KJ0837 and FoPM, WINGS Program, the University of Tokyo. He is also grateful to Ivan Di Liberti for pointing out Remark 2.3.3.

Survey.

  • One can see the idea of an algebra of structures, with exponential in [Bir42].
  • Wilkie gives a solution to Tarski’s high school algebra problem, circulated as a preprint and published in 2000 [Wil00].
  • [BL92] (and its summary [BL93]) give what we call the Burris-Lee theorem
  • Schanuel publishes papers on objective number theory [Sch00a][Sch00b]
  • The survey paper [BY05] was published.
  • Fiore, Cosmo, and Balat give a type-theoretic view on this structure, with 0 [FDB06].

Notation 1.0.1. We adopt the following notations:

  • ℕ denotes the set of all non-negative integers. Sometimes, we regard it as a poset with the natural ordering.
  • ℕ∗ denotes the set of all positive integers. We may regard it as a poset with the divisibility order.

2. Riegs

2.1. Preliminaries: rigs. Since our theory of riegs is based on the theory of rigs ( = semirings), we start by recalling the notion of rigs. (The word “rig” comes from “ring without negatives” [Sch90].) We only need the definition and a few examples.

Definition 2.1.1 (Rig). A (possibly non-commutative) rig is a set R equipped with

  • two constants ( = nullary operations) 0,1 ∈ R and
  • two binary operations, +,×: R × R → R

such that the following equations hold for all x,y,z ∈ R:

(1)
0 + x = x,
(2)
(x + y) + z = x + (y + z),
(3)
x + y = y + x,
(4)
1 × x = x,
(5)
x × 1 = x,
(6)
(x × y) × z = x × (y × z),
(7)
x × 0 = 0,
(8)
x × (y + z) = (x × y) + (x × z),
(9)
0 × x = 0, and
(10)
(x + y) × z = (x × z) + (y × z).

If R satisfies an additional equation

(11)
x × y = y × x,

we call it a commutative rig. A rig homomorphism is a function that preserves the four operations 0,1,+,×. The category of rigs and rig homomorphisms is denoted by 𝐑𝐢𝐠𝐬.

For the sake of simplicity, we may write 𝑥𝑦 for x × y. Let us provide three examples, each of which plays a specific role in what follows.

Example 2.1.2 (Rings). Of course, every ring is a rig; the theory of rigs is obtained from that of rings by dropping the axiom that every element has an additive inverse. However, rings will play only a marginal role in this paper: we will see that the only rieg whose underlying rig is a ring is the degenerate one (Proposition 2.3.2).

Example 2.1.3 (Rig of natural numbers). The rig of natural numbers (with the usual addition and multiplication) ℕ is the initial object of the category 𝐑𝐢𝐠𝐬. Its quotients will be the main subject of Section 3.

Example 2.1.4 (Rig of endohomomorphisms on a commutative monoid). For a commutative monoid (A,e,∗), the set of monoid endohomomorphisms End(A,e,∗) admits a rig structure as follows:

  • The element 0 ∈End(A,e,∗) is the constant function to the neutral element A → A: a↦e.
  • The element 1 ∈End(A,e,∗) is the identity function idA: A → A.
  • The sum f + g of two elements f,g ∈End(A,e,∗) is given by (f + g)(a) := f(a) ∗ g(a). (Here, we use the commutativity of ∗.)
  • The product f × g of two elements f,g ∈End(A,e,∗) is given by the composite f ∘ g.

For the additive monoid (ℕ,0,+), the rig End(ℕ,0,+) is isomorphic to the rig of natural numbers (ℕ,0,1,+,×). A rig of endohomomorphisms is, however, typically non-commutative; for the additive group (ℤ2,0,+), the rig End(ℤ2,0,+) is isomorphic to the ring M2(ℤ) of 2 × 2 matrices over ℤ, which is not commutative. This is the example that we will use to define riegs (Definition 2.2.1).

2.2. Definition. This subsection aims to introduce the notion of riegs. Although it is possible to define riegs by familiar but numerous equations (Proposition 2.2.2), we adopt another definition, which turns out to be surprisingly simple. Note that End(R,1,×) in the definition below is the rig explained in Example 2.1.4 obtained from the multiplicative commutative monoid (R,1,×).

Definition 2.2.1 (Rieg). A rieg is a commutative rig (R,0,1,+,×) equipped with a rig homomorphism R →End(R,1,×).

At first glance, this definition may not seem to have anything to do with exponents. So let us unpack the definition into a sequence of equalities. The rig homomorphism R →End(R,1,×) sends an element x ∈ R to a (commutative) monoid homomorphism, which will be denoted by

−x: (R,1,×) → (R,1,×).

The condition that −x: R → R is a monoid homomorphism is equivalent to the familiar equations

  • 1x = 1, and
  • (𝑦𝑧)x = yxzx.

For the function R →End(R,1,×) to be a rig homomorphism, it must preserve 0, 1, +, and ×. Preservation of each operation corresponds to the equation listed in Table 1.

Preserved rig structure The required equation
0 x0 = 1
1 x1 = x
+ xy+z = xyxz
× x𝑦𝑧 = (xz)y

Table 1. Four of the rieg equations

Summarising the above observations, we obtain an equational definition of the notion of riegs. While it is longer than Definition 2.2.1, it consists only of familiar conditions.

Proposition 2.2.2 (Equational definition of riegs). The definition of riegs (Definition 2.2.1) can be rephrased by the five operations:

  • two constants ( = nullary operations), 0,1 ∈ R, and
  • three binary operations, +,×,↑: R × R → R,

and fourteen equations:

(1)
0 + x = x
(2)
(x + y) + z = x + (y + z)
(3)
x + y = y + x
(4)
1x = x
(5)
(𝑥𝑦)z = x(𝑦𝑧)
(6)
𝑥𝑦 = 𝑦𝑥
(7)
x0 = 0
(8)
x(y + z) = 𝑥𝑦 + 𝑥𝑧
(9)
1x = 1
(10)
(𝑦𝑧)x = yxzx.
(11)
x0 = 1
(12)
x1 = x
(13)
xy+z = xyxz
(14)
x𝑦𝑧 = (xz)y,

where 𝑥𝑦 denotes x × y, and xy denotes x ↑ y as usual.

If we forget the constant 0, these are exactly the eleven “high school identities” of Tarski [BY05]. The other three equations, 0 + x = x, x × 0 = 0, and x0 = 1, are the ones that involve 0.

Remark 2.2.3. As a special case of the equation x0 = 1, we have 00 = 1. This convention will be repeatedly interpreted in many ways in this paper.

From this equational (or algebraic) definition, we can define the notion of rieg homomorphisms in the usual way.

Definition 2.2.4. A rieg homomorphism is a function that preserves all (five) operations 0,1,+,×,↑. We write 𝐑𝐢𝐞𝐠𝐬 for the category of riegs.

As we will see numerous examples later, for now, we introduce only two trivial (but important!) examples of riegs: the initial object and the terminal object in the category of riegs.

Example 2.2.5 (The initial rieg: natural numbers). The set of all natural numbers ℕ = {0,1,2,…} equipped with the usual zero 0, one 1, addition +, multiplication ×, and exponentiation ↑, forms the rieg of natural numbers ℕ. Here are a few remarks about this example:

  • This rieg ℕ is an initial object in the category 𝐑𝐢𝐞𝐠𝐬. Indeed, for a rieg R, the unique rig homomorphism ϕ: ℕ → R automatically preserves ↑: for n,m ∈ ℕ, the equations xy+z = xyxz, x1 = x, and x0 = 1 give ϕ(n)ϕ(m) = ϕ(n)1+⋯+1 = ϕ(n)⋯ϕ(n) = ϕ(nm), where both products have m factors.
  • This is the unique rieg structure on the rig of natural numbers, since ℕ is the initial rig (Example 2.1.3) and a rieg structure is given by a rig homomorphism ℕ →End(ℕ,1,×).
  • We adopt the convention that 00 = 1 (see Remark 2.2.3).

Example 2.2.6 (The degenerate rieg). A singleton {∗} admits exactly one rieg structure, which we call the degenerate rieg 1. The degenerate rieg is a terminal object of the category 𝐑𝐢𝐞𝐠𝐬.

2.3. Digression: Why rigs, not rings?. The reader may wonder: why do we develop our theory using rigs, even though rings are so famous and successful? There are several possible answers, but perhaps the simplest is that subtraction and exponentiation are fundamentally incompatible—if we have subtraction, then we have to give a meaning to 0−1!

Claim 2.3.1. If a rieg R contains an additive inverse − 1 of 1, then it is degenerate, that is, 𝑅≅1 (cf. Example 2.2.6).

Proof. We have the equation 1 = 00 = 01+(−1) = 01 × 0−1 = 0 × 0−1 = 0. This implies that every element r ∈ R is equal to 0, since r = r × 1 = r × 0 = 0. □

Proposition 2.3.2 (No rieg is a ring). If a rieg R is a ring, then it is degenerate, that is, 𝑅≅1.

Proof. A ring contains an additive inverse of 1, so Claim 2.3.1 applies. □

Remark 2.3.3. A similar problem appears for division, and there we have a solution. A meadow is a commutative ring with a total inverse operation, and its axioms imply 0−1 = 0 [BHT09]. So we may hope to do the same thing here, that is, to choose a value of 0−1. This does not work. Whatever the value of 0−1 is, we have 0 × 0−1 = 0, and the proof of Claim 2.3.1 still gives 1 = 0.

For example, exponentiation can be defined on the rig ℕ , but it is impossible on the ring ℤ (Proposition 2.3.2). To the best of the author’s understanding, one reason why exponential structures—despite being present in many mathematical objects—have not received the attention they perhaps deserve is that the process of decategorification often involves introducing formal subtraction. For example, the ring of integers ℤ is obtained by introducing a formal subtraction to ℕ, which inherently admits the exponential structure of the cartesian closed category 𝐅𝐢𝐧𝐒𝐞𝐭 (Section 4). This subtraction does help clarify theories (sometimes by reducing the amount of information), but at the same time, it forces us to entirely forget the exponential structure.

3. Example 1: Riegs in finite and profinite arithmetic

3.1. Rieg theoretic phenomena. Before delving into the technical details, let us first take a look at a few phenomena that will be studied in the rest of this section.

3.1.1. Phenomenon 1: modular arithmetic with exponentials. Let us consider the following high school level problem from elementary number theory, which involves exponentiation.

Puzzle 3.1.1. Find all pairs of natural numbers (n,m) such that

2n + 5 = m2.

Answer. Consider the equation modulo 8. If n ≥ 3, the equation is simplified to

5 ≡ m2 𝑚𝑜𝑑8.

However, since the quadratic residues modulo 8 are 0, 1, and 4, it is impossible. Therefore, we only need to check n = 0, 1, and 2. We conclude that the only solution is (n,m) = (2,3).

In this example, when considering exponentiation modulo 8, we have to set aside the “first exceptions”, namely the three numbers n = 0,1,2.

3.1.2. Phenomenon 2: Profinite dynamical systems with exponentials. Let us consider the following puzzle.

Puzzle 3.1.2 (A puzzle provided by J.Koizumi). Find a four-digit natural number n such that the last four digits of 3n are equal to n.

There is a surprisingly simple way to obtain the solution — namely, iteration! First, take an arbitrary natural number n, say n = 2024. Then, repeatedly replace n by the last four digits of 3n1.

  • 32024 ≡ 6481
  • 36481 ≡ 6803
  • 36803 ≡ 2027
  • 32027 ≡ 4987
  • 34987 ≡ 9387
  • 39387 ≡ 5387
  • 35387 ≡ 5387

We obtain an answer n = 5387.

Answer. n = 5387 is the only answer.

Then, it is natural to ask whether we can go farther:

Question 3.1.3. Is there an infinite sequence of decimal digits

…a4a3a2a1

such that, for every integer k ≥ 1, the truncation nk := ∑ ⁡ i=1kai10i−1 satisfies the congruence

3nk ≡ nk(𝑚𝑜𝑑10k)?

The answer is Yes! (See Figure 1)

387 ≡ 87𝑚𝑜𝑑100 3387 ≡ 387𝑚𝑜𝑑1000 35387 ≡ 5387𝑚𝑜𝑑10000 395387 ≡ 95387𝑚𝑜𝑑100000 3195387 ≡ 195387𝑚𝑜𝑑1000000 34195387 ≡ 4195387𝑚𝑜𝑑10000000 364195387 ≡ 64195387𝑚𝑜𝑑100000000 3464195387 ≡ 464195387𝑚𝑜𝑑1000000000 ⋮

Figure 1. The unique solution of 3n = n𝑚𝑜𝑑10k for each k ≥ 2

Remark 3.1.4 (Attribution). The affirmative answer is not new. It is a special case of a theorem of Jiménez Urroz and Yebra [JY09, Theorem 3]: for every valid base b (an integer such that q∣b for every prime p∣b and every prime q∣p − 1) and every integer a not divisible by the squarefree part of b, there is a unique such digit sequence for the congruences ax ≡ x(𝑚𝑜𝑑bn) for all sufficiently large n. The base b = 10 is valid, and our digit sequence is recorded as [OEIa], known there as the final digits of Graham’s number; the truncations displayed in Figure 1 form [OEIb]. What rieg theory adds is not the statement but a conceptual proof: the sequence is the unique fixed point of the contraction 3x (Proposition 3.3.12).

This phenomenon is reminiscent of p-adic integers. Indeed, the stabilization of the digits of an iterated exponential (or tower) modulo a fixed number is classically known, and is even used as a computational tool [Hit20]. However, in this case, the exponential operation is involved not only as an iteration of a fixed function but as a binary operation on the number system itself.

In this section, we will construct profinite systems of numbers, which can be regarded as p-adic number systems with exponentials. Using that structure, we will be able to answer the above question easily.

The rough sketch of our proof is as follows:

Answer (Rough sketch).

In a topological rieg ℕ∞,10∞ (Definition 3.3.6), we can take the limit α of the sequence

3,33,333 ,3333 ,33333 ⋯ → α.

By the continuity of 3x: ℕ∞,10∞ → ℕ∞,10∞, we have 3α = α.

Furthermore, the fixed point α turns out to be unique, which explains why the iteration in the above puzzle reaches the same answer regardless of the starting value: we will later regard 3x as a contraction with respect to a natural metric and apply the Banach fixed-point theorem (Proposition 3.3.12).

3.1.3. Phenomenon 3: Well-defined modulo exponentials. Let us conclude this subsection by introducing the fact that first attracted the author to riegs. As far as the author knows, it was originally proven by Burris and Lee [BL92; BL93]. It is well-known that the addition and multiplications of natural numbers (or even of integers) are well-defined modulo n, for any positive integer n ∈ ℕ. It is just saying that ℤ∕𝑛ℤ is a well-defined ring. But how about exponentials?

Puzzle 3.1.5. Find all positive integer n > 0 such that, for any positive integers a,b,a′,b′, we have

(a ≡ a′𝑚𝑜𝑑n) ∧ (b ≡ b′modn)⟹(ab ≡ a′b′ modn).

For the obvious reason, n = 1 satisfies the condition. Since a (positive) power of an odd (resp. even) number is odd (resp. even), n = 2 also satisfies the condition. However, n = 3 does not, since we have a counter example 21 ≡ 2≢1 ≡ 21+3.

Answer (A special case of [BL92; BL93]). There are exactly 5 possibilities: n = 1,2,6,42,1806

3.2. Finite arithmetic: the Burris-Lee theorem. Ordinary modular arithmetic remembers only a residue. Exponentiation also needs a finite initial part in which small exponents remain distinct. The result is a finite arithmetic with a tail followed by a cycle.

3.2.1. The Burris–Lee classification.

Definition 3.2.1 (ℕa,b). For a natural number a ≥ 0 and a positive integer b > 0, let ℕa,b be the (a + b)-element quotient ℕ∕∼a,b, where

n ∼a,bm⟺(n = m) ∨ ((n,m ≥ a) ∧ (n ≡ m𝑚𝑜𝑑b)).

Thus the numbers below a form the initial exceptional part, and the numbers from a onward repeat with period b.

Figure 2 shows this tail–cycle picture for ℕ3,4.

Diagram 1 from the Rieg theory manuscript

Figure 2. ℕ3,4

By [BL92, Proposition 3.6], every finite quotient of the positive integers has this tail–cycle form. The same paper determines exactly when exponentiation descends to it.

Theorem 3.2.2 (Burris–Lee [BL92, Theorem 3.12]). For a ≥ 0 and b > 0, the canonical operations descend from ℕ to ℕa,b if and only if, for every prime p and every positive integer e,

pe∣b⟹e ≤ a,p∣b⟹p − 1∣b.

Burris and Lee work with the positive integers and the language {+,×,↑,1}.2

3.2.2. Examples and the five cyclic models.

Example 3.2.3 (The rieg of truth values). The quotient ℕ1,1 is the rieg of truth values 2 = {⊥,⊤} from Section 5.

Example 3.2.4 (Parity rieg). The quotient ℕ1,2 is

{0,positive odd,positive even},

which we call the parity rieg. Keeping 0 separate is essential: exponentiation does not descend to ℤ∕2ℤ, since 20 is odd but 22 is even.

Example 3.2.5 (Overflow riegs). For n ≥ 0, the quotient ℕn,1 is the overflow rieg with threshold n, denoted by OverFlown. Its elements are 0,1,…,n − 1,⋆, where every natural number at least n is sent to ⋆.

The theorem gives all finite quotient riegs at once. Figures 3 and 4 keep their distribution visible.

Diagram 2 from the Rieg theory manuscript

Figure 3. Finite quotient riegs of ℕ for small a and b

a = 0 : ℕ0,1 a = 1 : ℕ1,1, ℕ1,2, ℕ1,6, ℕ1,42, ℕ1,1806 a = 2 : ℕ2,1, ℕ2,2, ℕ2,4, ℕ2,6, ℕ2,12, ℕ2,18, ℕ2,20, ℕ2,36, ℕ2,42, ℕ2,60, ℕ2,84,… a = 3 : ℕ3,1, ℕ3,2, ℕ3,4, ℕ3,6, ℕ3,8, ℕ3,12, ℕ3,18, ℕ3,20, ℕ3,24, ℕ3,36, ℕ3,40,… a = 4 : ℕ4,1, ℕ4,2, ℕ4,4, ℕ4,6, ℕ4,8, ℕ4,12, ℕ4,16, ℕ4,18, ℕ4,20, ℕ4,24, ℕ4,36,… ⋮ “a = ∞” : ℕ

Figure 4. Finite quotient riegs of ℕ, arranged by a

When a = 1, the positive part has no tail: its successor operation is a single cycle. Following [AJ25], we call such positive algebras cyclic models. There are surprisingly few of them.

Corollary 3.2.6 ([BL92, Corollary 3.14], attributed there to D. Higgs). The only finite quotient riegs of ℕ with cyclic positive part are

ℕ1,1,ℕ1,2,ℕ1,6,ℕ1,42,ℕ1,1806.

Equivalently, if a rieg R has a least positive integer n such that

1 = 1 + 1 + ⋯ + 1n,

then n is one of 1,2,6,42,1806. This is precisely the puzzle in Section 3.1.3; no separate calculation is needed here.

Example 3.2.7. The number 8 = 23 satisfies both conditions in Theorem 3.2.2 with a = 3. Hence ℕ3,8 is a rieg. This is exactly the finite arithmetic used in the high-school calculation of Section 3.1.1.

Example 3.2.8. The number 10 fails the second Burris–Lee condition, since 5∣10 but 4 ∤ 10. For every n ≥ 2, however, 10n = 2n5n and 4∣10n, so ℕn,10n is a rieg. These riegs form the finite inverse system used below to construct the topological rieg of 10-adic natural numbers (Definition 3.3.6).

3.2.3. Characteristics of riegs. Recall that the characteristic of a ring R is the smallest positive integer n such that 0 = 1 + 1 + ⋯ + 1n times, if such an n exists; otherwise, the characteristic is defined to be 0. For quotient rigs and riegs, we need to record both the beginning of the cycle and its period. We order ℕ × ℕ∗ by

(a,b) ≤ (a′,b′)⟺a ≤ a′ and b∣b′.

Definition 3.2.9 (Characteristic). The characteristic of a rig or rieg R, denoted by charR, is the minimum pair (a,b) ∈ ℕ × ℕ∗ such that

1 + ⋯ + 1a times = 1 + ⋯ + 1a times + 1 + ⋯ + 1b times,

if it exists. Otherwise, we put charR = (∞,0), where (∞,0) is a formal maximum element.

Corollary 3.2.10 (Burris–Lee theorem in terms of rieg characteristics [BL92; BL93]). A pair (a,b) ∈ ℕ × ℕ∗ is the characteristic of a rieg if and only if

pe∣b⟹e ≤ a,p∣b⟹p − 1∣b

for every prime p and every positive integer e.

Remark 3.2.11 (Why this definition is natural). Let R be a rig or rieg. Factor the unique homomorphism from the initial object as

ℕ ↠Im(ℕ → R) ↣ R.

If this image is finite, then it is ℕa,b and charR = (a,b). If the map ℕ → R is injective, then charR = (∞,0). The usual definition for rings cannot be used directly: if a positive integer n satisfies n = 0 in a rieg, then n − 1 is an additive inverse of 1, and the rieg is degenerate by Claim 2.3.1.

This viewpoint makes sense for any equational theory: for an algebra R, the image of the canonical map from the initial algebra is a quotient of the initial algebra, and this quotient deserves to be called the characteristic of R in general. For rings, the initial algebra is ℤ, the image is ℤ∕nℤ, and we recover the usual characteristic n. For riegs, the image is ℕa,b, and charR = (a,b) is just a name for this quotient.

3.3. Profinite arithmetic: a total exponentiation. In this subsection, we take the profinite completion of the initial rieg ℕ and describe it explicitly. Its underlying set turns out to be ℕ ⊔ℤ^. The main point is not that the exponentiation is total—this holds in every rieg by definition—but where it is total: although no non-degenerate ring admits a rieg structure (Proposition 2.3.2), the completion contains the profinite ring ℤ^, and the exponentiation xy is still defined for all pairs.

3.3.1. Profinite natural numbers with total exponentiation. For 𝒞 = 𝐑𝐢𝐠𝐬 or 𝐑𝐢𝐞𝐠𝐬, let Quot𝒞(ℕ) denote the poset of quotient 𝒞-algebras of ℕ. We put Q ≤ Q′ if there is a quotient map Q′↠ Q compatible with the maps from ℕ. Let FinQuot𝒞(ℕ) denote its subposet of finite quotients. Thus we use four notations, which form a square of subposets of Quot𝐑𝐢𝐠𝐬(ℕ):

Diagram 3 from the Rieg theory manuscript

Going down-left in the square imposes the arithmetic condition of Theorem 3.2.2 (admitting a rieg structure); going down-right imposes finiteness. The smallest poset FinQuot𝐑𝐢𝐞𝐠𝐬(ℕ), placed at the bottom, satisfies both conditions. Every finite quotient rig is of the form ℕa,b. The finite quotient riegs are the ones that satisfy Theorem 3.2.2, and ℕ is the maximum element of each full quotient poset. For two finite quotients, the order is

ℕa,b ≤ ℕa′,b′⟺a ≤ a′ and b∣b′.

Recall that the profinite completion of an algebraic structure is the limit of its finite quotient algebras (Section A.2).

Definition 3.3.1. The rieg of profinite natural numbers ℕ^ is the profinite completion of the initial rieg ℕ.

Thus

ℕ^ := lim (FinQuot𝐑𝐢𝐞𝐠𝐬(ℕ)op →𝐑𝐢𝐞𝐠𝐬).

Lemma 3.3.2. The quotient rieg ℕn,n! exists for every n ∈ ℕ. Moreover, the family

{ℕn,n!∣n ∈ ℕ}

is cofinal in FinQuot𝐑𝐢𝐞𝐠𝐬(ℕ).

Proof. Suppose that p is prime and that pe∣n!, where e is maximal. Counting the factors divisible by pk gives

e = ∑ k=1∞⌊ n pk ⌋ ≤∑ k=1∞ n pk = n p − 1 ≤ n.

Moreover, if p∣n!, then p ≤ n, so p − 1 divides n!. Thus ℕn,n! is a rieg by Theorem 3.2.2.

Now let ℕa,b be a finite quotient rieg and take n = max(a,b). Then a ≤ n and b∣n!, so

ℕa,b ≤ ℕn,n!.

This proves cofinality. □

Proposition 3.3.3. The profinite rieg ℕ^ is the limit of the following diagram.

Diagram 4 from the Rieg theory manuscript

Proof. By Lemma 3.3.2, the order-preserving functor

ℕ→FinQuot𝐑𝐢𝐞𝐠𝐬(ℕ),n⟼ℕn,n!,

is final. Therefore, it gives the same limit. □

Theorem 3.3.4 (The underlying set of ℕ^). The underlying set of ℕ^ is

ℕ^ = ℕ ⊔ℤ^,

where ℤ^ denotes the usual set of profinite integers. The canonical projection πa,b: ℕ^ → ℕa,b restricts to the canonical maps

ℕ ↠ ℕa,b

and

ℤ^ ↠ ℤ∕bℤ ↣ ℕa,b.

Proof. By Proposition 3.3.3 and lemma A.2.1, an element of ℕ^ is a compatible sequence

(xn ∈ ℕn,n!)n≥0.

Suppose that one coordinate is an exceptional element k < n. Compatibility then gives

xm = { k (m > k), [k] (m ≤ k).

Hence the sequence comes from the natural number k. Otherwise, every coordinate belongs to the modular part ℤ∕n!ℤ ⊂ ℕn,n!, so the sequence is an element of

ℤ^ = limnℤ∕n!ℤ.

Conversely, every element of ℕ and every element of ℤ^ gives a compatible sequence of the indicated form. □

Diagram 5 from the Rieg theory manuscript

Figure 5. ℤ^ as limits of ℤ∕n!ℤ

Corollary 3.3.5 (The unique continuous extension of exponentiation). The exponentiation of ℕ extends to a continuous operation

(−)(−): ℕ^ ×ℕ^→ℕ^

defined for all pairs of elements, and such a continuous extension is unique.

Proof. The limit defining ℕ^ is taken in the category of topological riegs. Hence its operations, including exponentiation, are defined coordinatewise and are continuous by Lemma A.2.1.

For the uniqueness, it suffices to show that the image of ℕ is dense in ℕ^: then ℕ × ℕ has dense image in ℕ^ ×ℕ^, and two continuous maps into the Hausdorff space ℕ^ that agree on a dense subset must coincide. By Proposition 3.3.3, every non-empty basic open subset of ℕ^ is the preimage of a non-empty subset S ⊆ ℕn,n! under the projection ℕ^ → ℕn,n! for some n ∈ ℕ. Since the composite ℕ ↠ ℕn,n! is surjective, this preimage contains an element of ℕ. □

This is different from the usual p-adic exponentiation, which is not defined on all pairs in ℤp × ℤp. In fact, a non-degenerate ring cannot be a rieg by Proposition 2.3.2. The summand ℕ in Theorem 3.3.4 is what makes ℕ^ different from a ring.

By using a smaller diagram of finite quotient riegs, we obtain variants of ℕ^. The following one appeared in Phenomenon 2 (Section 3.1.2).

Definition 3.3.6 (The topological rieg of 10-adic natural numbers). The topological rieg of 10-adic natural numbers ℕ∞,10∞ is the limit of the following diagram in the category of topological riegs:

Diagram 6 from the Rieg theory manuscript

By Example 3.2.8, each ℕn,10n is a rieg for n ≥ 2.

By the same argument as in Theorem 3.3.4, the underlying set of ℕ∞,10∞ is given by

ℕ∞,10∞ = ℕ ⊔ ℤ10 = ℕ ⊔ (ℤ2 × ℤ5),

where ℤ10 := limn(ℤ∕10nℤ)≅ℤ2 × ℤ5 denotes the set of 10-adic integers.

Remark 3.3.7 (No 5-adic analogue). This construction does not work for the powers of a single odd prime, say 5: since 5∣5n but 4 = 5 − 1 ∤ 5n, the second condition in Theorem 3.2.2 fails. Hence the rig ℕa,5n admits no rieg structure for any a. In this sense, there is no topological rieg of “5-adic natural numbers” ℕ ⊔ ℤ5 obtained in this way.

3.3.2. Contractive exponential dynamics.

Definition 3.3.8 (Carmichael function [Car10]). For a positive integer n, the Carmichael function λ(n) is the least positive integer d such that

ud ≡ 1(modn)

for every u ∈ (ℤ∕nℤ)×. Equivalently, λ(n) is the exponent of the group (ℤ∕nℤ)×.

Remark 3.3.9 (The jump function). With respect to the divisibility order, the Carmichael function has a right adjoint λ∗: ℕ∗→ ℕ∗, whose notation λ∗ we borrow from [CP09, Theorem 5.2(b)]. Thus

λ(b′)∣b⟺b′∣λ∗(b).

Therefore, λ∗(b) records the possible next periods b′ for which exponentiation raises the precision.

Lemma 3.3.10 (A metric for an inverse sequence). Let

X = lim (⋯ ↠ X2 ↠ X1 ↠ X0)

be an inverse limit of non-empty finite sets and surjective maps, where X0 has one element. Write πn: X → Xn for the projections. For distinct x,y ∈ X, let

ν(x,y) = max{n ∈ ℕ∣πi(x) = πi(y) for every i ≤ n},

and put

d(x,y) = 2−ν(x,y),d(x,x) = 0.

Then (X,d) is a complete ultrametric space.

Suppose that a map F : X → X satisfies

πn(x) = πn(y)⟹πn+1(F(x)) = πn+1(F(y))

for every n ≥ 0. Then F is a contraction with Lipschitz constant 1 2.

Proof. The ultrametric inequality follows because two elements that agree with a third element through level n also agree with each other through level n. A Cauchy sequence stabilizes at every finite level. These stable coordinates form an element of X, so the metric is complete.

If x and y agree through level n, then F(x) and F(y) agree through level n + 1. Hence

d(F(x),F(y)) ≤ 1 2d(x,y).
□

We now use another inverse sequence for the same topological rieg:

Diagram 7 from the Rieg theory manuscript

All the displayed objects are quotient riegs by Theorem 3.2.2. Adding the last four quotients does not change the limit in Definition 3.3.6. The periods in this sequence are

1∣2∣4∣20∣100∣1000∣104∣⋯.

The standard formula for the Carmichael function [Car10] gives

λ(2) = 1,λ(4) = 2,λ(20) = 4,λ(100) = 20,

and

λ(10k+1)∣10k(k ≥ 2).

The modular parts of this inverse sequence have limit

ℤ10 = ℤ2 × ℤ5.

We equip this limit with the metric in Lemma 3.3.10. Since ℕ∞,10∞ is a topological rieg, its total exponentiation defines a map

T : ℕ∞,10∞→ℕ∞,10∞,T(x) = 3x.

This map preserves the modular part: at each finite quotient, a modular exponent can be represented by an integer m above the tail, and 3m is again above the tail. If two elements of the modular part agree modulo a period b, then their images agree modulo the next period b′: indeed, λ(b′)∣b and 3 is invertible modulo b′. Therefore, T raises the agreement by one level. By Lemma 3.3.10, the restriction

T : ℤ2 × ℤ5→ℤ2 × ℤ5

is a contraction with Lipschitz constant 1 2.

Lemma 3.3.11 (Stabilization of the last two digits). For every element ξ of the modular part ℤ∕102ℤ ⊂ ℕ2,102, we have T4(ξ) = [87].

Proof. The first four transitions of the above sequence give

T(ξ) ≡ 1(mod2),T2(ξ) ≡ 3(mod4),

and then

T3(ξ) ≡ 33 ≡ 7(mod20),T4(ξ) ≡ 37 ≡ 87(mod100).
□

Proposition 3.3.12 (The unique fixed point of 3x). The function T(x) = 3x on ℕ∞,10∞ has a unique fixed point α, which belongs to ℤ2 × ℤ5, and every iteration sequence x,T(x),T2(x),… starting from x ∈ ℤ2 × ℤ5 converges to α. Furthermore, for each k ≥ 2, the reduction of α is the unique solution of 3x = x in ℤ∕10kℤ ⊂ ℕk,10k (cf. Figure 1).

Proof. There is no fixed point in the natural-number part, since 3n > n for every n ∈ ℕ. On the modular part, the map T is a contraction on a non-empty complete metric space. The Banach fixed-point theorem gives a unique fixed point α ∈ ℤ2 × ℤ5, and every iteration sequence in the modular part converges to α.

It remains to prove the statement at each finite level. If r ∈ ℤ∕102ℤ satisfies 3r = r, then

r = T4(r) = [87]

by Lemma 3.3.11. Suppose that the solution is unique modulo 10k, and let r and r′ be solutions modulo 10k+1. Their reductions modulo 10k coincide. Since λ(10k+1)∣10k, we obtain

r = 3r ≡ 3r′ = r′(mod10k+1).

Hence the solution is unique at every level. The reduction of α gives its existence. □

We close this subsection by determining the solutions of another exponential equation, xx = x, in the same topological rieg.

Proposition 3.3.13 (The solutions of xx = x). In ℕ∞,10∞ = ℕ ⊔ (ℤ2 × ℤ5), the equation xx = x has exactly 11 solutions: the element 1 of the summand ℕ, and the ten elements

(0,0),(0,1),(1,0),(1,1),(1,i),(1,−1),(1,−i),(−1,0),(−1,1),(−1,−1)

of ℤ2 × ℤ5, where μ4 = {1,i,−1,−i} denotes the group of fourth roots of unity in ℤ5×.

Proof. In the summand ℕ, we have 00 = 1≠0, 11 = 1, and nn > n for n ≥ 2, so 1 is the only solution there.

Let x belong to the modular part ℤ2 × ℤ5, let k ≥ 2, and choose an integer n ≥ k with n ≡ x(mod10k). Since ℕk,10k is a rieg (Example 3.2.8), the reduction of xx to ℤ∕10kℤ is [nn]. Therefore,

(∗) xx = x⟺nn ≡ n(mod10k) for every k ≥ 2.

Below, vp denotes the p-adic valuation and p ranges over {2,5}.

First, suppose 1 ≤ vp(x) < ∞ and put a := vp(x). Taking k > a in (∗), we have vp(n) = a, and nn − n = n(nn−1 − 1) with p∣n and n − 1 ≥ 1, so p ∤ nn−1 − 1 and

vp(nn − n) = a < k,

contradicting (∗). Hence each of the two coordinates of x ∈ ℤ2 × ℤ5 is either 0 or a unit.

Next, suppose that the p-coordinate u of x is a unit. Then (∗) modulo pk reads nn−1 ≡ 1(modpk), i.e., the order of u in (ℤ∕pkℤ)× divides n − 1. If u = 1, it is a torsion element of ℤp×; assume u≠1 and put s := vp(u − 1) < ∞. Once k > s, we have vp(n − 1) = s. For p = 2, the order of u is a power of 2 dividing n − 1, hence divides 2s. For p = 5, the order divides 4 ⋅ 5k−1; its 5-part divides n − 1, hence divides 5s, while its prime-to-5 part divides 4. In both cases the orders are bounded independently of k, so u is a torsion element of ℤp×. The torsion subgroup of ℤ2× is {±1}, and that of ℤ5× is μ4.

Finally, we determine which of the remaining 3 × 5 combinations satisfy (∗); note that the same representative n serves both coordinates. If the 2-coordinate is 0, then 4∣n, so n − 1 is odd, and the order of the 5-coordinate must be odd: the 5-coordinate is 0 or 1. If the 2-coordinate is 1, then 2k∣n − 1, so every order in {1,2,4} divides n − 1: the 5-coordinate can be any element of {0}∪ μ4. If the 2-coordinate is − 1, then n ≡ 3(mod4), so v2(n − 1) = 1: the order 2 divides n − 1 but 4 does not, and the 5-coordinate lies in {0,1,−1}. This yields the 2 + 5 + 3 = 10 listed elements. Conversely, each listed element satisfies (∗): a coordinate 0 gives vp(n) ≥ k, hence nn ≡ 0 ≡ n(modpk), and the case analysis above shows precisely that the order of each torsion coordinate divides n − 1. □

Remark 3.3.14 (The levelwise counts diverge). Contrast Proposition 3.3.13 with Proposition 3.3.12, where every level has a unique solution. The numbers of solutions of xx = x in ℤ∕10kℤ ⊂ ℕk,10k for k = 2,3,…,7 are

22,38,196,196,1412,1412,

and they are unbounded, even though the limit has only 11 points: for a unit u ≡ 1(mod4) with s := v2(u − 1), one has v2(u2s − 1) = 2s, so u solves the 2-adic part of the equation modulo 2k precisely when k ≤ 2s. Pairing such u with the 5-coordinate 0 produces, for k ≥ 4, at least 2⌊k∕2⌋ solutions at level k, almost all of which die at a later level. Note also that 0 is a solution in the summand ℤ2 × ℤ5 but not in the summand ℕ: modulo 10k, the class [0] is represented by 10k, and (10k)10k ≡ 0. This illustrates the splitting ℕ ⊔ ℤ10 of Definition 3.3.6.

4. Example 2: Riegs in category theory and combinatorics

In this section, we consider the rieg structures obtained through decategorification. Since many of these are defined via universality, the verification of the rieg structure follows from abstract nonsense. However, when one tries to write them down explicitly, it looks non-trivial that all fourteen axioms of a rieg are satisfied.

For example, from abstract nonsense, it follows immediately that the set of all isomorphism classes of finite directed graphs admits a natural rieg structure. But can you imagine what the exponentiation looks like? For instance, consider a single edge →, and then consider taking the exponential of that edge with itself, denoted →→. The resulting graph is depicted in Figure 6: its four vertices are the four functions from the vertex set {0,1} of → to itself (the two constant functions c0,c1, the identity id, and the transposition swap), and it has exactly four edges, one of which is a loop. Notice that the vertex swap is isolated.

Diagram 8 from the Rieg theory manuscript

Figure 6. The exponential graph →→

Regarding the decategorified semiring structure, numerous studies have been conducted under the name Burnside rig. However, to the author’s knowledge, research on its exponential structure has been very limited. Despite the exponential structure satisfying numerous equations in the definition of rieg, these have been left unused. See [Sch00b]

4.1. Burnside rieg of CCC with finite coproducts (and a topos).

4.1.1. The Burnside rieg functor 𝔅.

Definition 4.1.1. The (1-)category 𝐂𝐂𝐂𝐜 is defined as follows:

  • Its objects are small categories that are cartesian closed and have finite coproducts.
  • Its morphisms are functors between these categories that preserve finite products, finite coproducts, and exponentials, where we consider functors up to natural isomorphism3 .

Proposition 4.1.2. For a cartesian closed category 𝒞 with finite coproducts, the set of isomorphism classes of 𝒞 forms a rieg with the categorical initial object, the binary coproducts, the terminal object, the binary products, and the exponentials.

Proof. All of the fourteen axioms are implied by the universalities. □

Remark 4.1.3 (Symmetric monoidal closed categories). This construction does not work for a symmetric monoidal closed category with finite coproducts, since we use the fact that the product A × B is a categorical product to prove some axioms including (x × y)z = xz × yz.

We borrow the following notation from [Men17] (and from [Sch90]).

Definition 4.1.4. The Burnside rieg functor 𝔅: 𝐂𝐂𝐂𝐜 →𝐑𝐢𝐞𝐠𝐬 sends an object 𝒞 of the category 𝐂𝐂𝐂𝐜 to the rieg induced by 𝒞 (Proposition 4.1.2), and a functor in 𝐂𝐂𝐂𝐜 to the induced rieg homomorphisms.

Proposition 4.1.5 (Topos to rieg functor). For a locally cartesian closed small category 𝒞 with finite coproduct (in particular, an elementary topos 𝒞), the correspondence

ob(𝒞) ∋ c↦𝔅(𝒞∕c) ∈𝐑𝐢𝐞𝐠𝐬

defines a contravariant functor

𝔅(𝒞∕−): 𝒞op →𝐑𝐢𝐞𝐠𝐬.

Example 4.1.6 (Finite sets to natural numbers). The Burnside rieg of finite sets is isomorphic to the rieg of natural numbers

𝔅(𝐅𝐢𝐧𝐒𝐞𝐭)≅ℕ.

Example 4.1.7 (Infinite sets to cardinal arithmetic). Let κ be an infinite cardinal and 𝐒𝐞𝐭<κ be the (essentially small) category of sets with cardinality less than κ. The category 𝐒𝐞𝐭<κ is a topos if and only if it is cartesian closed if and only if κ is a strong limit cardinal. Therefore, if κ is a strong limit cardinal, then we obtain the induced rieg 𝔅(𝐒𝐞𝐭<κ), which is the (small) rieg of cardinal arithmetic.

Obviously, the structure of this rieg is interconnected with set theory, especially continuum hypothesis. With the generalized continuum hypothesis, 𝔅(𝐒𝐞𝐭<ℶω) is a countable rieg with 2ℵ0 = ℵ1. However, without the GCH, there are many other possibilities of this rieg structure.

Example 4.1.8 (Finite presheaf topos). For any small category 𝒞, 𝐅𝐢𝐧𝐒𝐞𝐭𝒞 has all finite limits and colimits. Therefore, if 𝐅𝐢𝐧𝐒𝐞𝐭𝒞 is cartesian closed, we obtain a rieg 𝔅(𝐅𝐢𝐧𝐒𝐞𝐭𝒞). For example, if 𝒞 is a finite category or a (possibly infinite) group, then 𝐅𝐢𝐧𝐒𝐞𝐭𝒞 is known to be cartesian closed (and a topos).

  • The category of finite functions [→,𝐅𝐢𝐧𝐒𝐞𝐭] forms a rieg of multisets of natural numbers (or Dirichlet polynomials [SM23]).
  • More generally, for any finite category 𝒞, its finite (co)presheaf category [𝒞,𝐅𝐢𝐧𝐒𝐞𝐭] forms a rieg. Examples include the category of finite graphs [⇉,𝐅𝐢𝐧𝐒𝐞𝐭] and the category of rooted forests [ωop,𝐅𝐢𝐧𝐒𝐞𝐭].
  • For any (possibly infinite) group G, the category of its finite actions [G,𝐅𝐢𝐧𝐒𝐞𝐭] forms the Burnside rig of the group G. Examples include the category of sets with involution [ℤ∕2ℤ,𝐅𝐢𝐧𝐒𝐞𝐭] and the category of finite loops [ℤ,𝐅𝐢𝐧𝐒𝐞𝐭].

On the other hand, the category of finite discrete dynamical systems [ℕ,𝐅𝐢𝐧𝐒𝐞𝐭] is not cartesian closed.

Question 4.1.9. When is a finite presheaf category [𝒞op,𝐅𝐢𝐧𝐒𝐞𝐭] cartesian closed? The author is also interested in the closely related question: when is a finite presheaf category a topos?

Example 4.1.10.

HeyAlg Every Heyting algebra H is a cartesian closed poset with finite coproducts, and thus a rieg.

4.1.2. Enumerations. In many cases, to utilize the Burnside rieg 𝔅(𝒞) to enumerate something, we consider a enumeration function |−| : 𝔅(𝒞) → ℕ, which is typically a rig homomorphism. There are several ways to choose such enumeration function.

Example 4.1.11 (Global sections for a connected topos). For example, for a locally finite topos 𝒞 we have the global section functor Γ = 𝒞(1,−): 𝒞 →𝐅𝐢𝐧𝐒𝐞𝐭, which preserves products, but not necessarily coproducts. If the topos 𝒞 is connected (i.e., if Γ preserves finite coproduct), then the functor Γ: 𝒞 →𝐅𝐢𝐧𝐒𝐞𝐭 induces the rig homomorphism |−|Γ: 𝔅(𝒞) → ℕ that counts the number of global sections.

Example 4.1.12 (Connected components for a stably locally connected topos). If a topos 𝒞 admits an adjoint triple π0 ⊣Δ ⊣Γ: 𝒞 →𝐅𝐢𝐧𝐒𝐞𝐭 such that π0 preserves finite products (see stable local connectedness in [Joh11]), then the functor π0 induces a rig homomorphism |−|π0: 𝔅(𝒞) → ℕ.

Example 4.1.13 (Enumeration with a torsion-free atom.). Let E be a small topos. Due to Proposition 4.1.5, every object X ∈ob(E) induces a rieg homomorphism 𝔅(E) →𝔅(X → 1)𝔅(E∕X). Therefore, if E∕X ≃𝐅𝐢𝐧𝐒𝐞𝐭, then we obtain a rieg homomorphism 𝔅(E) → ℕ. According to the terminology in [KM91], we call such an object X a torsion-free atom4 .

4.2. Riegs of functions 𝔅(𝐅𝐢𝐧𝐒𝐞𝐭→): List of numbers and Liouville’s divisor theorem.

4.2.1. The Burnside rieg of functions 𝔅(𝐅𝐢𝐧𝐒𝐞𝐭→). Our first example is the Burnside rieg of the topos 𝐅𝐢𝐧𝐒𝐞𝐭→, which is called the topos of (finite) fucntions or the (finite) Sierpinski topos. An object of the topos is a function p: A → B.

Since a function f : A → B can be regarded as a family of sets {f−1(b)}b∈B, the isomorphism classes of the topos 𝐅𝐢𝐧𝐒𝐞𝐭→ are in bijective correspondence with the finite multiset of natural numbers. Let mult(ℕ) denote the set of all multisets of natural numbers. Each multiset will be denoted as a list of natural numbers (),(3,1,4),(1,2,3,1,1,0,0) ∈mult(ℕ), and rearranged lists are identified (1,2,3,1,1,0,0) = (0,0,1,1,1,2,3) ∈mult(ℕ).

The addition is given by a concatenation of lists, and the product is given by the list of all ccomponent-wise products.

(3,1,4) + (1,2,7,0) = (3,1,4,1,2,7,0) (3,1,4) × (1,2,7,0) = (3,6,21,0,1,2,7,0,4,8,28,0)

[SM23] explains that this rig structure is isomorphic to the rig of (formal) Dirichlet polynomials

mult(ℕ)≅ℕ [0−x,1−x,2−x,…] in 𝐑𝐢𝐠𝐬.

However, exponentials are different:

(n1,…,nk)(m) = (n 1m,…,n km).

There is a unique extension of this exponential to all exponentials. For example,

(3,1,4)(2,0) = (3,1,4)(2)(3,1,4)(0) = (9,1,16)(1,1,1) = (9,9,9,1,1,1,16,16,16).

Remark 4.2.1 (mult(R) and Artin gluing).

4.2.2. Liouville’s theorem. The equation

(1 + 2 + 3 + ⋯ + n)2 = 13 + 23 + ⋯ + n3

is well-known. As studied in [BS13], there are so many similar equations including

  • (1 + 2 + 2 + 4)2 = 13 + 23 + 23 + 43,
  • (1 + 2 + 2 + 3 + 5)2 = 169 = 13 + 23 + 23 + 33 + 53, and
  • (1 + 2 + 2 + 3 + 4 + 6)2 = 13 + 23 + 23 + 33 + 43 + 63.

So the question is: for a given finite list of natural numbers (a1,…an), does the equation

(∑ i=1na 1) 2 = ∑ i=1na i3. (1)

holds?

Writing d(n) for the number of divisors of a positive integer n, Liouville’s theorem states that

(∑ m|nd(m))2 = ∑ m|nd(m)3.

For example, in the case of n = 10, we obtain the equation

(d(1) + d(2) + d(5) + d(10))2 = d(1)3 + d(2)3 + d(5)3 + d(10)3.

The proof of this theorem is not difficult, and is just a highschool level problem. Given the equation (1 + 2 + 3 + ⋯ + n)2 = 13 + 23 + ⋯ + n3, it suffices to prove that the problem can be reduced to the case of m = pe for some prime number p. In other words, the only non-trivial part is to prove that if two sequence (a1,…,an) and (b1,…,bm) satisfies the equation, so does their “product sequence” (a1 ⋅ b1,a1 ⋅ b2,…,a1 ⋅ bm,a2 ⋅ b1,…,an ⋅ bm).

Here is a categorified (or Burnside) point of view. Consider the Burnside rieg 𝔅(𝐅𝐢𝐧𝐒𝐞𝐭→) of the topos of functions between finite sets. We obtain the equivalence

a list C = (a1,…an) satisfies Equation (1) ⟺#Hom(A,C) = #Hom(B,C),

where A,B ∈𝐅𝐢𝐧𝐒𝐞𝐭→ are given in Figure 7.

Diagram 9 from the Rieg theory manuscript

Figure 7. The representing object of sum-of-cubes and square-of-sum in 𝐅𝐢𝐧𝐒𝐞𝐭→

4.3. Rieg of finite groups.

Counting problems. We will give a rieg-theoretic method to answer the following undergraduate-level counting problems.

Question 4.3.1. For a finite group G, let nG denote the number of subgroups whose index (= the number of cosets) is 2. Write nG×H using nG and nH.

Question 4.3.2. Count the number of group homomorphisms from D6 to itself. (Hint: D6 is the dihedral group of order 12, and D6≅S3 × C2.)

Notice that all operations 0,1,+,×,↑ of the cartesian closed category of finite groupoids Groupoidfin are well-defined up to equivalence. Therefore, (as a 2-categorical analogy of Proposition 4.1.2), the equivalence classes of finite groupoids, which are the formal sums of finite groups, form a rieg! The aim of this section is to concretely describe the rieg structure.

Definition 4.3.3. We definite the rig of finite groups 𝒢 as follows:

  • The underlying additive commutative monoid is the free commutative monoid of the set of isomorphism classes of finite groups, which is given by the finite formal sum.

    𝒢 := ⊕ [G]: iso.classℕ
  • The multiplication is given by the unique extension of

    [G] × [H] := [G × H].

Hereafter, the isomorphism class of a group G will be simply referred to as G.

Notation 4.3.4. The (isomorphism class of) the cyclic group ℤ∕nℤ is denoted by Cn.

Example 4.3.5. Some examples of calculations include:

  • C12 = C3 × C4
  • D6 = S3 × C2
  • (S3 + C2)2 = S32 + 2(S3 × C2) + C22 = S32 + 2D6 + C22

Definition 4.3.6 (Conjgacy classes of homomorphism). Let G and H be finite groups.

  • Two homomorphisms ϕ,ψ: G → H are conjugate if there exists h ∈ H such that ϕ = hψh−1 holds.
  • The automorphism group (or the stabilizer) of a homomorphism ϕ: G → H is the subgroup of H, defined by Aut(ϕ) := {h ∈ H∣hϕh−1 = ϕ}.

Theorem 4.3.7. The rig of finite groups 𝒢 admits the following rieg structure:

  • For two finite groups G and H, its exponential HG is defined by

    HG := ∑ [ϕ]:conj.classAut(ϕ)
  • In general, the exponential is defined by

    (∑ jHj)(∑ iGi ) := ∏ i ∑ j (HjGi ) ,

    for any two families of finite groups (Gi)i and (Hj)j.

Proof. This is the rieg of equivalent classes of finite groupoids, explained at the beginning of this section. □

Example 4.3.8. Let’s try to calculate D6D6. In order to utilize the exponential rules, we first calculate smaller parts:

  • C2C2 = 2 ⋅ C2
  • C2S3 = 2 ⋅ C2
  • S3C2 = C2 + S3
  • S3S3 = 1 + C2 + S3

Then we obtain

D6D6 = (S3S3 )C2 ⋅(C2S3 )C2 = (1 + C2 + S3)C2 ⋅(2 ⋅ C2)C2 = (1 + 3 ⋅ C2 + S3) ⋅(4 ⋅ C2) = 4 ⋅ C2 + 12 ⋅ C22 + 4 ⋅ D 6.

Some properties of finite groups, like being abelian, are characterized in terms of this rieg structure.

Proposition 4.3.9. For a finite group A,

  • if A is abelian, then we have AG = #Hom(G,A) ⋅ A.
  • conversely, if AG is a sum of A for any G, then A is abelian.

Proof. The former statement is immediate from the definition. If AG is a sum of A for any G, then AA is a sum of A. Especially, we have Aut(idA) = A and A is abelian. □

Now we can answer the first problem, Question 4.3.1!

Answer (to Question 4.3.1). Since C2G = (nG + 1) ⋅ C2, we have

(nG×H + 1) ⋅ C2 = C2G×H = (C 2G)H = (n G + 1) ⋅ C2H = (n G + 1) ⋅ (nH + 1) ⋅ C2.

This proves nG×H = nG ⋅ nH + nG + nH.

Remark 4.3.10. For a limit cardinal κ, one can consider a rieg of larger groups, in which we can similarly consider representations of product groups, like ℂ𝐕𝐞𝐜𝐭f.d.iso.G×H = (∑ k=0∞GLk(ℂ))G×H.

4.3.1. Groupoid cardinality. In order to utilize riegs for counting something, the basic rieg theoretic method is to consider a rig homomorphism from a given rieg to the rig of numbers, like ℕ, ℤ, ℚ. For the rieg of finite groups 𝒢, the notion of groupoid cardinality, introduced in [BD01], is useful. We can utilize it to obtain rational numbers from elements of 𝒢.

Definition 4.3.11. For x = ∑ iGi ∈𝒢, its groupoid cardinality |x| ∈ ℚ is defined by

|x| := ∑ i 1 #Gi.

Proposition 4.3.12. The groupoid cardinality function |−| : 𝒢 → ℚ is a rig homomorphism. Furthermore, for finite groups G and H, we have

|GH| = #Hom(H,G) #G .

Proof. For sums and products, the proof is easy. Regarding exponentials, one can use the orbit-stabilizer theorem for the conjugate action of G on Hom(H,G). □

Exercise 4.3.13. Check that Proposition 4.3.12 is compatible with Proposition 4.3.9.

Let us finish this introductory note by giving the answer to Question 4.3.2!

Answer (to Question 4.3.2). In Example 4.3.8, we obtained D6D6 = 4 ⋅ C2 + 12 ⋅ C22 + 4 ⋅ D6. By taking the groupoid cardinality, we obtain

#Hom(D6,D6) 12 = 4 2 + 12 4 + 4 12.

This proves that #Hom(D6,D6) = 64.

5. Example 3: Riegs in logic

In this subsection, we investigate the riegs that consist of “propositions.” The motivating examples are the rieg of truth values and the rieg of subsets:

Example 5.0.1 (Rieg of truth values). The set of truth values 2 = {⊥,⊤} has a natural rieg structure, as (2,⊥,⊤,∨,∧,←). This is isomorphic to the overflow rieg with threshold 1 and ℕ1,1.

Example 5.0.2 (Powerset). For any set X, its powerset 𝒫(X) has a rieg structure with (𝒫(X),∅,X,∪,∩,←), where A ← B is defined to be A ∪ Bc. More generally, every Heyting algebra5 is a rieg.

When X = ∅, its powerset rieg is the degenerate rieg. When X is a singleton, its power rieg is the rieg of truth values.

The above riegs have special properties:

  • 1 + 1 = 1
  • The underlying rig is a lattice
  • They are Heyting algebras.

For each property, we have a class of riegs. We will study them one by one.

5.1. Riegs in which 1 + 1 = 1.

Proposition 5.1.1. For a rieg R, the followings are equivalent:

(1)
1 + 1 = 1
(2)
charR = (0,1) or (1,1)
(3)
There is a rieg homomorphism from the rieg of truth values 2 = {⊥,⊤}
(4)
The addition is idempotent. (i.e., for any x ∈ R, x + x = x)
(5)
Both addition and multiplication are idempotent. (i.e, for any x ∈ R, x + x = x × x = x)

Proof. The equivalence between 1, 2 and 3 follows from the definition of characteristics. By multiplying x, we have 1 ⟹ 4. To prove 5 from 1, we can use

x × x = x1+1 = x1 = x.

The remaining implications are trivial. □

Example 5.1.2 (Max-plus is not a rieg). The max-plus algebra (ℝ ∪{−∞},−∞,⊕,0,⊗) admits no rieg structures. If it admits a rieg structure, since 1 ⊕ 1 = max(1,1) = 1, the multiplication ⊗, (which is the addition in the usual sense) should also be idempotent. However, 1 ⊗ 1 = 1 + 1 = 2≠1, contradiction.

Definition 5.1.3. A rieg R is said to be idempotent, if it satisfies the conditions of Proposition 5.1.1.

Remark 5.1.4. (Multiplicatively idempotent riegs) Even if a rieg R is multiplicatively idempotent, it might not be additively idempotent. For example, the parity rieg Example 3.2.4 is multiplicatively idempotent but not additively idempotent.

However, there is a strong restriction on the characteristics. For a multiplicatively idempotent rieg R, its characteristics charR should be less than (2,2) since 2 = 2 × 2 = 4. Therefore, there are only finitely many possible characteristics, namely

charR = (0,1),(1,1),(1,2),(2,1),or (2,2).

And all of them are possible, since ℕ0,1, ℕ1,1, ℕ1,2, ℕ2,1, and ℕ2,2 are multiplicatively idempotent.

5.2. Riegs that are lattices. Next, we will study when (the underlying rig of) a rieg is a lattice. The necessary condition is being idempotent (Definition 5.1.3). However, this is not sufficient:

Example 5.2.1 (Idempotent rieg that is not a lattice). In Example B.1.9, we observed that ℕ[x] with exponential

f(x)g(x) = f(x)g(0) = f(x) ×⋯ × f(x) g(0) times

is a rieg.

Dividing ℕ[x] into 4 groups

ZC = {f(x) ∈ ℕ[x]∣f(0) = 0 and f: constant} = {0} ZI = {f(x) ∈ ℕ[x]∣f(0) = 0 and f: increasing} PC = {f(x) ∈ ℕ[x]∣f(0) > 0 and f: constant} PI = {f(x) ∈ ℕ[x]∣f(0) > 0 and f: increasing},

we obtain a 4-element rieg {ZC,ZI,PC,PI} as a quotient rieg. Indeed, the group to which f belongs is determined by the pair of data (whether f(0) = 0 or f(0) > 0; whether f is constant or increasing), and this pair for f + g and for f × g is determined by the pairs for f and g (for the multiplication, note that ZC = {0} consists of the single element 0). For the exponential, observe that fg = fg(0) is equal to 1 ∈PC if g(0) = 0, and belongs to the same group as f if g(0) > 0; in both cases, the group of fg depends only on the groups of f and g.

The underlying rig of this quotient is the rig presented by a single generator x and two equations 1 + 1 = 1 and x2 = x. (The additive idempotency follows from 1 + 1 = 1 via f + f = f × (1 + 1) = f; hence the presented rig consists of the four elements 0,1,x, and 1 + x, corresponding to ZC,PC,ZI, and PI, respectively.) This rieg satisfies 1 + 1 = 1, but the underlying rig is not a lattice since 1 + x≠1.

Proposition 5.2.2 (Rieg and Lattice). For a rieg R, the following conditions are equivalent:

(1)
The underlying rig is a (bounded) lattice.
(2)
For any x ∈ R, 1 + x = 1.

Proof. The implication 1⟹2 follows from the fact that 1 is the maximum element in a lattice.

We prove 1 from 2. Substituting x with 1, we obtain 1 + 1 = 1, i.e., this rieg is idempotent (Definition 5.1.3). Now it is enough to prove the absorption laws

x + xy = x x(x + y) = x.

It is proven as x(x + y) = x2 + xy = x + xy = x(1 + y) = x1 = x. □

5.3. Heyting algebras are riegs. A Heyting algebra is not just related but is a rieg satisfying additional equations.

Proposition 5.3.1. Every Heyting algebra (H,0,1,∨,∧,←) is a rieg.

Proof. We can directly prove this. Alternatively, this is also an immediate corollary of Proposition 4.1.2, since a Heyting algebra is a bicartesian closed category whose underlying category is a poset. □

Example 5.3.2 (A rieg that is a lattice but not a Heyting algebra). Let 𝒫(X) be the power set of a non-empty set X, equipped with a rig structure given by ∅,X,∪ and ∩. We consider a rieg structure on 𝒫(X) that is different from Example 5.0.2.

First, fix an arbitrary element x ∈ X and define the exponential as

AB := { A(x ∈ B) X(x∉B).

This rieg structure is given by the composition of two rig homomorphisms. The first one is a rig homomorphism to the rig of truth value 2 = {⊥,⊤} defined as

𝒫(X) →2: A↦“x ∈ X”.

The other is the unique rig homomorphism

2 →End(𝒫(X),X,∩),

which sends 0 ∈2 to the constant funciton constX: 𝒫(X) →𝒫(X) and 1 ∈2 to the identity function idX: 𝒫(X) →𝒫(X).

Remark 5.3.3 (Riegs from ultrafilters). Rig homomorphisms 𝒫(X) →2 correspond exactly to ultrafilters on X. For each ultrafilter 𝒰 on X, the same construction equips the rig 𝒫(X) with a rieg structure: the exponentiation is given by AB = A if B ∈𝒰, and AB = X otherwise. The example above is the special case where 𝒰 is the principal ultrafilter generated by x ∈ X.

If X has at least two elements, this rieg is not a Heyting algebra, because

{x}{x} = {x}≠X.

Proposition 5.3.4. A rieg R is a Heyting algebra, if and only if it satisfies the following additional equations:

  • 1 + x = 1
  • xx = 1
  • xyx = xy

Corollary 5.3.5. The category of Heyting algebras HeyAlg is a reflective full subcategory of the category of riegs 𝐑𝐢𝐞𝐠𝐬.

Diagram 10 from the Rieg theory manuscript

The construction of the left adjoint is the usual one: quotient by new equations. For example, the associated Heyting algebra of ℕ is the Heyting algebra of truth values {⊥,⊤}.6

The relation between Heyting algebras and riegs is not an addition of structures, but an addition of properties. In this respect, it is closer to the relationship between abelian groups and groups rather than to the relationship between rings and groups.

Appendix A. Preliminaries on equational theory and profinite topology

In this section, let 𝕋 be a (finite arity single-sorted) algebraic theory. We are especially interested in the case T = (the theory of riegs). The category of 𝕋-algebras is denoted by Alg𝕋.

A.1. Categorical properties of riegs (and general 𝕋-algebras).

Remark A.1.1 (Categorical properties of riegs). Several category-theoretic properties of 𝐑𝐢𝐞𝐠𝐬 immediately follow from the mere existence of an equational definition by the theory of categorical universal algebra. Some of them are noted below.

  • The category of riegs 𝐑𝐢𝐞𝐠𝐬 has all small limits and colimits.
  • Furthermore, the forgetful functor U : 𝐑𝐢𝐞𝐠𝐬 →𝐒𝐞𝐭 (preserves and) creates all limits.
  • The forgetful functor U : 𝐑𝐢𝐞𝐠𝐬 →𝐒𝐞𝐭 has a left adjoint F : 𝐒𝐞𝐭 →𝐑𝐢𝐞𝐠𝐬
  • Every rieg homomorphism f : R → S is uniquely (up to the canonical isomorphism) decomposed into the composition of a surjective homomorphism followed by an injective homomorphism R ↠Im(f) ↣ S.

This means we can construct arbitrary small limits using limits in 𝐒𝐞𝐭. For a given family of riegs {Rλ}λ∈Λ, its categorical product is given by the product set ∏ λ∈λRλ with the index-wise operations.

An element of the free rieg F({x}) looks like

(3x + xxx )x2x+5x+0x+1 + 0xx.

A.2. Topological 𝕋-algebras and their profinite completion. A topological 𝕋-algebra is a model of 𝕋 in the category Top, that is, a 𝕋-algebra equipped with a topology such that all operations of 𝕋 are continuous. Morphisms of topological 𝕋-algebras are continuous 𝕋-algebra homomorphisms. The category of topological 𝕋-algebras is denoted by TopAlg𝕋.

It is straightforward to verify the next lemma.

Lemma A.2.1 (Limits in the topological 𝕋-algebras). TopAlg𝕋 admits all small limits, which are preserved by the two forgetful functors TopAlg𝕋 →Alg𝕋 and TopAlg𝕋 →Top.

In otherwords, a small limit in TopAlg𝕋 is just the limit at the underlying set level equipped with the canonical 𝕋-algebra structure and the initial topology.

For a given 𝕋-algebra R, a quotient 𝕋-algebra is a surjective 𝕋-algebra homomorphism R ↠ Q, up to the canonical isomorphisms. The quotients form a complete lattice Quot𝕋(R). We put Q ≤ Q′ if there exists a homomorphism Q′↠ Q that commutes with the quotient maps from R. Let FinQuot𝕋(R) denote the subposet of finite quotient 𝕋-algebras.

Definition A.2.2 (Profinite completion of a 𝕋-algebra). For a 𝕋-algebra R, its profinite completion R^ is the limit of all finite quotients of R. In other words, R^ is the limit of the following diagram:

FinQuot𝕋(R)op →Alg 𝕋 →discreteTopAlg𝕋.

Remark A.2.3 (As a codensity monad). It is much more natural to define the profinite completion as an endofunctor on the category TopAlg𝕋 although we will not need it in the present paper. The category of finite (and discrete) 𝕋-algebras finAlg𝕋 is a full subcategory of the category TopAlg𝕋, and the codensity monad of the embedding finAlg𝕋↪TopAlg𝕋 is the profinite completion.

Appendix B. Other (counter)examples

B.1. Basic examples.

Example B.1.1 (Non-negative real numbers). The set of non-negative real numbers [0,∞) has the natural rieg structure, defining

0x = { 1 (x = 0) 0 (x > 0).

The rieg of natural numbers is a subrieg of this rieg. However, the subset of all non-negative rational numbers is not subrieg, since it is not closed under exponentials.

Proposition B.1.2 (J. Koizumi, personal communication). Let f : (0,∞) → (0,∞) preserve the exponentiation, i.e. satisfy

f(xy) = f(x)f(y)(x,y > 0).

Then f is either the constant function 1 or the identity function.

Proof. Putting x = y = 1, the value a := f(1) satisfies a = aa, hence (a − 1)log⁡a = 0 and a = 1. Applying f to xyz = (xy)z gives

f(x)f(yz) = f(xy)f(z) = f(x)f(y)f(z)(x,y,z > 0).

Assume f is not constantly 1, and fix x0 with f(x0)≠1; then f(yz) = f(y)f(z), i.e. f is multiplicative. Put φ(u) := log⁡f(eu) for u ∈ ℝ. Multiplicativity gives φ(u + v) = φ(u) + φ(v), while the original equation with x = eu gives φ(yu) = f(y)φ(u) for all u ∈ ℝ and y > 0. Fixing u0 with φ(u0)≠0, we obtain f(y) = φ(yu0)∕φ(u0), so additivity of φ makes f additive: f(y + y′) = f(y) + f(y′). An additive function with positive values is strictly increasing, and f(1) = 1 forces f(q) = q for every positive rational q; being increasing and agreeing with the identity on a dense subset, f = id. Conversely, both functions do satisfy the equation. □

Remark B.1.3. The hypothesis has to be f(xy) = f(x)f(y), that is, f is an endomorphism of the binary operation ↑. Under the weaker reading f(xy) = f(x)y the statement is false, since f(x) = xc satisfies (xy)c = (xc)y for every c > 0. Note also that (0,∞) is not a rieg, having no 0; it is a model of the eleven high school identities, so the correct reading of Proposition B.1.2 is a rigidity statement about ↑-endomorphisms.

Remark B.1.4 (The rig [0,∞] is not a rieg). It is natural to ask whether this rieg structure can be extended to the rig [0,∞]. The answer is no. The rig [0,∞] cannot have a rieg structure. Suppose that there is an extension to a rieg structure. First, since 2∞ = 2∞+1 = 2∞× 2, we have 2∞∈{0,∞}. And, since 2∞(0.5)∞ = 1∞ = 1, we have 2∞∈ (0,∞), contradiction.

B.1.1. Numbers with Infinites.

Example B.1.5 (Complemented natural numbers). ℕ ∪{∞} has a canonical rieg structure, which is obtained as the limit of the overflow riegs

Diagram 11 from the Rieg theory manuscript

In this sense, this rieg is the overflow rieg with threshold ∞. Notice that this construction is similar to the construction of p-adic integers.

B.1.2. Rig with many rieg structures.

Example B.1.6 (Rieg of dual numbers constructed by Yuhi Kamio). Recall that a dual number is a number plus an “infinitesimal number”

a + b𝜀(a,b ∈ ℝ)

The ring of infinitesimal numbers is denoted by ℝ[𝜀] and algebraically defined as ℝ[𝜀] = ℝ[x]∕(x2).

There is a similar rieg, which we call the dual number rieg.

Definition B.1.7 (Dual numbers, concrete definition). The dual number rieg ℕ[𝜀] has ℕ × ℤ as the underlying set and its element (n,m) is denoted by n + m𝜖. Three operations are defined as follows:

(1)
(n + m𝜀) + (n′ + m′𝜀) = (n + n′) + (m + m′)𝜀
(2)
(n + m𝜀) × (n′ + m′𝜀) = (nn′) + (nm′ + n′m)𝜀
(3)
The exponential depends only on the natural-number part n′ of the exponent:
(n+m𝜀)↑(n′+m′𝜀) = (n + m𝜀) ×⋯ × (n + m𝜀) n′times = { 1 (n′ = 0), nn′ + n′nn′−1 m𝜀(n′ > 0).

This definition has a more conceptual paraphrase, with which we can easily prove that it’s actually a rieg. First notice that the projection

ℕ[𝜀] → ℕ: n + m𝜀↦n

is a rig homomorphism to the initial rig ℕ. Thus we have the composite rig homomorphism

ℕ[𝜀] → ℕ →End(ℕ[𝜀],×,1),

and this gives the exponential structure of the dual number rieg.

More generally, we have the following proposition.

Proposition B.1.8 (Rieg structure by a natural number evaluation). For a commutative rig R and a rig homomorphism (or say, “natural number evaluation”)

v: R → ℕ,

R admits an induced rieg structure

R → ℕ →End(R,×,1).

In other words, the exponential is given by

xy = x ×⋯ × x v(y) times.

For example, we can replace the additive group ℤ𝜀 in the definition of the dual number rieg with an arbitrary commutative monoid.

Example B.1.9. In Proposition B.1.8, letting v: R → ℕ be the evaluation function

ev0: ℕ[x] → ℕ: x↦0,

we have a rieg structure on ℕ[x] with

f(x)g(x) = f(x)g(0) = f(x) ×⋯ × f(x) g(0) times.

Example B.1.10 (Rieg of polynomials). Is it possible to have more than one rieg structure on a single rig? The answer is (not at all surprisingly) yes. A typical example is the rig of polynomials ℕ[x]. Since ℕ[x] is the free rig with a generator x, rieg structures on ℕ[x]

ℕ[x] →End(ℕ[x],×,1)

bijectively correspond to the elements of End(ℕ[x],×,1).

B.2. Riegs of three elements. The degenerate rieg 1 is the unique 1-element rieg. The rieg of truth values 2 is the unique 2-element rieg. This subsection aims to provide all 3-element riegs.

Proposition B.2.1 (3-element riegs). There are 8 isomorphism classes of 3-element riegs.

Let R be a 3-element rieg. There are three possible characteristics char(R) = (1,1),(1,2),(2,1).

Case 1: char(R) = (1,2),(2,1). If char(R) = (1,2),(2,1), then R is the parity rieg (Example 3.2.4) or the overflow rieg with the threshold 2 (Example 3.2.5).

Case 2: char(R) = (1,1). So we need to classify all 3-element riegs with characteristics char(R) = (1,1). Let R = {0,1,a}. We distinguish cases based on the value of 1 + a.

- Case 2.1: char(R) = (1,1) and 1 + a = 1. If 1 + a = 1, then the rieg R satisfies the equation 1 + x = 1, which means that the underlying rig of R is the lattice ({0 < a < 1},0,∨,1,∧) (Proposition 5.2.2). This rig 0 < a < 1 is isomorphic to the rig

R≅⟨a∣1 + 1 = 1,1 + a = 1,a2 = a⟩ in 𝐑𝐢𝐠𝐬

Therefore, the rieg structures R →End(R,1,∧) on the rig R are in bijective correspondence with the monoid homomorphisms ϕ: (R,1,∧) → (R,1,∧), which correspond to the exponentiation function −a: R → R, that satisfy

  • x ∧ x = x, which always holds,
  • x ∧ ϕ(x) = x, and
  • ϕ2(x) = ϕ(x).

In other words, the rieg structures on the rig ({0 < a < 1},0,∨,1,∧) are in bijective correspondence with the idempotent ∧-semilattice homomorphisms −a = ϕ: (R,1,∧) → (R,1,∧) with ϕ(x) ≥ x. One can show that there are exactly 4 rieg structures on the rig ({0 < a < 1},0,∨,1,∧), which are visualized in Figure 8.

Diagram 12 from the Rieg theory manuscript

Figure 8. The four possible exponentiations x↦xa

- Case 2.2: char(R) = (1,1) and 1 + a = a. The only remaining case is 1 + a = a, since 1 + a = 0 is impossible (Claim 2.3.1). Since R is idempotent (Definition 5.1.3), the underlying rig structure is determined:

R≅⟨a∣1 + 1 = 1,1 + a = a,a2 = a⟩ in 𝐑𝐢𝐠𝐬

Therefore, the rieg structures on R are in bijective correspondence with monoid homomorphisms ϕ: (R,1,×) → (R,1,×) with

  • x × x = x, which always holds
  • x × ϕ(x) = ϕ(x), and
  • ϕ2(x) = ϕ(x).

There are two such morphisms visualized in Figure 9.

Diagram 13 from the Rieg theory manuscript

Figure 9. The two possible exponentiations x↦xa

References

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ZEN University, Tokyo, Japan

Email address: ryuya_hora@zen.ac.jp

Footnotes

Results and notes

Connected compact Hausdorff riegs§

Question Does a nontrivial connected compact Hausdorff topological rieg exist? All three binary operations—addition, multiplication, and exponentiation—are required to be continuous. Current result Compact rieg no-go proves that every compact Hausdorff topological rieg with …

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