Grothendieck Ring of the Pairing Function Without Cycles

Grothendieck ring of the pairing function without cycles

Esther Elbaz February 12, 2020

Abstract A bijection (l, r) between M 2 and M is said to be a pairing function with no cycles, if any composition of its coordinate functions has no ﬁxed point. We compute here the Grothendieck ring of the pairing function without cycles to be isomorphic to 2 2 Z ' Z[X]/(X − X ). In a previous article, we illustrated ideas to construct, given a quotient R of a polynomial ring over Z, a structure whose Grothendieck ring is Z. Following these very ideas, the 2 structure we should consider to obtain a structure with Grothendieck ring Z[X]/(X − X ) is precisely the pairing function without cycles. The method we use here to compute its Grothendieck ring has already allowed us to treat the case of a bijection, without cycles, between a set and itself deprived of N point, for any N ∈ N (the Grothendieck ring is Z[X]/(N)), and will allow us to treat in forthcoming papers the bijection without cycles between a set and the union of two disjoint copies of itself (the Grothendieck ring is Z), and for every integer N, an enrichment of this structure whose Grothendieck ring is Z/NZ.

Contents

1 Introduction1 1.1 Method...... 3 1.2 Pairing function without cycles...... 5

2 Representation of formulas and deﬁnable sets by binary trees6

3 Closed tree and simple formulas9

4 Form of the deﬁnable injections 11

5 Simple sets 13

6 Decomposition of deﬁnable sets 14

7 Deﬁnable injection on simple set 16

8 Representation of a decomposition: tree of a decomposition 17

9 Computation of the Grothendieck ring 20

10 Example of pairing function without cycles on N 22

1 1 Introduction

The notion of Grothendieck ring of a theory has been introduced in the early 2000. F. Loeser an J. Denef [LD] on the one hand introduced the notion of the Grothendieck ring of a theory through motivic integration. T. Scanlon and J. Krajicek [SK] on the other hand considered the Grothendieck ring of a structure and built up a dictionary between the geometric properties of the Grothendieck ring of a structure, and the combinatorial properties of this structure.

Definition 1.1. Let M be a non empty structure and let Def(M) be the set of definable (with parameters) subsets of some Cartesian product of M. We define an equivalence relation on Def(M) by: A, B ∈ Def(M) are equivalent if and only if, there exists a definable bijection between them. We consider the quotient set, denoted Def(g M), and denote by [A] the class of any subset A of Def(M). On this quotient set, we define two laws:

• An additive law deﬁned by: [A] + [B] = [A ∩ B] + [A t B].

• A multiplicative deﬁned by: [A] ∗ [B] = [A × B].

The additive law corresponds on the level of sets to the disjoint union and admits the class of the empty set as neutral element. The multiplicative law corresponds on the level of sets to the Cartesian product and admits the class of a singleton as neutral element. The additive law is not cancellative: a + b = a + c doesn’t imply that b = c. In order to make it cancellative, we quotient Def(g M), by the equivalent relation deﬁned by: a, b ∈ Def(g M) are equivalent if and only if there exists c ∈ Def(g M) such that a + c = b + c. We then obtain a cancellative monoide for the additive law. There exists a unique (up to isomorphism) ring that embeds this last quotient. This ring is called the Grothendieck ring of M and is denoted K0(M). The Grothendieck ring satisﬁes the following universal property: let f be a {+, ×}-homomorphism from Def(g M) to a ring. Then f factors through K0(M).

As an example, one can easily checked that the Grothendieck ring of any finite structure is Z: the class of a definable set in the Grothendieck ring will correspond to the class of its cardinality. The real closed field, R also admits Z as its quotient. It has been demonstrated by J. Krajicek and T. Scanlon [SK] and relies on the fact that two sets that have the same dimension and the same geometric Euler characteristic constructed on the definable subsets of R through triangulation (see for example [VD]) are in definable bijection. The geometric Euler characteristic is s a result the quotient map that associates to any definable subset its class in the Grothendieck ring.In particular, it does not account for the dimension of the definable <0 F F >0 subsets. For example R is equal to the disjoint union R {0} R and, in the language of the fields, both R>0, R<0 are in definable bijection with R. Noting [R] the class of R in its Grothendieck ring, we obtain that 2[R] + 1 = [R] which implies that [R] = −1 : the class of R is equal to the opposite of the class of a singleton. Thanks to the second quotient, we can easily prove a property, called the onto-pigeonhole principle and abbreviated onto-PHP, which states that the Grothendieck ring of a structure is non-zero if and only if none of its definable sets is in definable bijection with itself deprived of a point.

2 This property demonstrated by J. Krajicek and T. Scanlon in [SK] is an example of the par- allel between the combinatorial properties of a structure and the algebraic properties of its Grothendieck ring.It has been used to show the triviality of the Grothendieck ring of several rings and fields, like for example, of Laurent series fields [?], or more generally of certain valued fields in Z [?, ?]. In this article, we compute the Grothendieck ring of the pairing function without cycles to be isomorphic to Z2 ≡ Z[X]/(X − X2). In [?], we illustrated a method to construct given R, a fixed polynomial ring over Z, a structure that admits R as its Grothendieck ring. To do so we consider a structure with exactly the functions that are require to obtain this Grothendieck ring . Indeed, noting X the class of a structure M in its Grothendieck ring, the relation X = X2 precisely means that M is in definable bijection with M 2. The property of being without cycles is, in ad-equation with the general ideas explained below, required to get quantifier elimination. In the next subsection, we give an idea of what this method consists of. We then apply it to the pairing function without cycles and prove our theorem.

Theorem 1.2. Let L := {l; r} be a language with two unary function. Let M be a L-structure such that the 2-ary function (l, r) is a bijection between M and M 2. Assume furthermore, that this pairing function has no cycles: any function f obtained as a composition of the function {l, r} has no ﬁxed point. 2 Then K0(M) is isomorphic to Z .

1.1 Method

We use the convention N = Z≥0. Let R be any quotient of Z[X]. We have explained in [?] how one can construct a somehow minimalistic theory such that any model of this theory will admit R as its Grothendieck ring.We recall in this section, the basic ideas. In order to construct a structure admitting a prechosen ring, we proceed in two steps. First, we design the structure. Second, we compute its Grothendieck ring and thus show that it is indeed the one we are expecting.

Remark 1.3. We should highlight that, in the general case, the Grothendieck ring of a structure is not determined by its theory. Two elementary equivalent structures do not necessarily have elementary equivalent Grothendieck ring (this is due to the fact that we consider set deﬁnable with parameters). It has been proven [SK] that if M ≡ N, then K0(M) ≡∃1 K0(N) and, later, that if N is an elementary extension of M, we have an injection of its Grothendieck ring into K0(M). In the case of the theory of modules, S. Perera [?] proved that all of their elementary equivalent models have elementary isomorphic Grothendieck rings. But this does not hold in general. The theories of our construction have the particularity that any of their model has the same Grothendieck ring.

The ring R being a quotient of Z[X], there exists an ideal I of Z[X] such that R = Z[X]/I. It is well known that any ideal of Z[X] admits a generating subset consisting of a polynomial P (X) (possibly a constant polynomial) and an integer (possibly zero). We can show that if we

3 can construct a structure whose Grothendieck ring is R for any R generated by a polynomial over N and possibly a constant, then we can treat the general case. This results from the following lemma:

Lemma 1.4. Let Q(X) ∈ Z[X] whose leading coefficient is positive. Then there exists P (X) ∈ N[X] and k ∈ N such that P (X − k) = Q(X). Let M be a structure whose Grothendieck ring is Z[X]/(P (X)) (respectively Z[X]/(N,P (X)) where N ∈ N). Then there exists a structure N whose Grothendieck ring is Z[X]/(Q(X)) (respectively Z[X]/(N,Q(X)) where N ∈ N). Proof. We won’t prove this lemma in detail here but just give the general idea. It is easy to prove the first part. Let us assume that there exists a structure M whose class in K0(M) annihilates P (X). Then by noting N the disjoint union of M and of a finite set of cardinality k, we have that P ([N] − k) = Q([N]) = 0. It can be proved that if we obtain, through the general ideas explained below, a structure M whose Grothendieck ring is Z[X]/(P (X)) (respectively Z[X]/(N,P (X)) where N ∈ N)hen we can endow the disjoint union of M and of a finite set of cardinality k with a structure such that the Grothendieck ring of M is Z[X]/(Q(X)) (respectively Z[X]/(N,Q(X)) where N ∈ N). In the following, we will thus assume that R := Z[X]/I where I is generated by Pd i • P (X) := i=0 aiX , where for any 0 ≤ i ≤ d, ai ≤ 0; • N, a constant polynomial, possibly 0.

We first design the natural theory we have to place ourselves in. This theory will be such that all its models have the desired Grothendieck ring.We prove the consistency of the theory by exhibiting a model. Finally we compute the Grothendieck ring of any of this model and prove that it is equal to the desired Grothendieck ring. Step 1: Construction of the theory Let M be the candidate structure. Let X be the class of M in K0(M). In order to obtain that in the Grothendieck ring, P (X) = 0, we want our structure to be in definable bijection with the disjoint union: i 0 0 (ti|i6=1M × Ai) t (M × A1) where for every i 6= 1, Ai is a set of cardinality ai, and A1 a set of cardinality a1 + 1. Thus we exhibit a structure M endowed with a bijection α between M and a subset of M d+1 in definable bijection with this disjoint union. In the case where the set of generator of I contains a constant polynomial N,we want to have a bijection, β, between M and M \ F0 where F0 is a finite set of cardinality N.

A key ingredient of the demonstrations is to be able to define specific definable sets, called "simple sets", that have two properties:

• they allow to rebuild any deﬁnable set by Boolean combination,

• they are in deﬁnable bijection with a Cartesian product of M,

4 • injections whose graph is deﬁned by a conjunction of equalities between terms, transform a simple set in another simple set.

We place ourselves in a minimal theory containing axioms defining the bijection and some others ensuring the existence of this class of definable sets. Namely, we will ask that any composition of the coordinate functions of the bijection has no fixed point and the two bijections α and β to be as independent as possible. In order to obtain quantifiers elimination, we place ourselves in a minimal language containing a symbol for each of this coordinate function. We prove the consistency of the theory by exhibiting a model. If we would like to apply these ideas to construct a structure whose Grothendieck ring is Z[X]/(X − X2), the structure we would have to consider is precisely the one of the pairing function without cycles. Indeed, following these ideas, to obtain Z[X]/(X − X2) as a Grothendieck ring, we should consider a set M such that there is a bijection, p, from M to M 2. The function p defines two functions l and r such that p(x) := (l(x), r(x)). In order to obtain quantifier elimination, we place ourselves in the language {l, r}. We will also require p to be without cycles: any composition of functions l and r has no fixed point. Step 2 Computation of the Grothendieck ring : The following ideas can be applied generally, independently of the first steps, provided we have exhibited a subclass of definable sets that satisfy the properties of the simple sets mentioned above. Here we will apply them to compute the Grothendieck ring of the pairing functions without cycles. We first prove the quantifiers elimination and the existence of the simple sets mentioned above. Their properties imply that K0(M) is a quotient of Z[X]. Because of the bijections α and β, we know that K0(M) is a quotient of Z[X]/I. We want to prove that it is actually equal to Z[X]/I. To do so, we follow the following steps:

• We deﬁne a subclass of the class of simple sets that are the irreducible closed sets of the Noetherian topology they generate. This allows us to refer to the Krull dimension of these sets considered as closed set of this Noetherian topology.

• For every deﬁnable set A and any deﬁnable injection h from A to A, there exists two such subclasses of simple sets, S1 and S2, such that:

1. A = ti(Bi \ (jBi,j)) is a partition of A in simple sets of S1,

2. For every i, h restricted to Bi \ (∪jBi,j) can be extended to Bi on an injection that transforms any simple set of S1 in a simple set of S2.

• Finally, relying on these properties and using the property of the Noetherian topology and some speciﬁc to the structure , we prove that I = J.

Remark 1.5. The subclasses of simple sets S1 and S2 mentioned above are in fact the simple sets with com- plexity bounded by a ﬁxed integer.

In [?], we gave some example of application of these ideas.

5 1.2 Pairing function without cycles Let M be a set. We call pairing function any bijection Θ from M to M 2. If a set M is endowed with such a bijection, Θ, then to each element x ∈ M, we can associate a binary tree whose root is x, and such that for every node, the right child corresponds to the right coordinate of Θ(x) and the left child corresponds to its left coordinate. The pairing function is said to be without cycles if for every x ∈ M, the branches of the tree Tx never contains two identical nodes. E. Bouscaren and B. Poizat have studied this theory in [?]. As they explain the pairing functions have interested numerous authors. One of their interesting feature is the fact that they are "the simplest example of a theory stable which is not limiter of superstable theory". It is indeed a single statement, the existence of a bijection between M 2 and M, that prevents the superstability". A pairing function leads to the deﬁnition of two other functions:

• the function l called "left member" which at all x ∈ M associates the element y such that there exists z ∈ M with Θ(x) = (y, z)

• the function d called "right member" which at all x ∈ M associates the element z such that there exists y ∈ M with Θ(x) = (y, z).

In other words, Θ(x) = (l(x), r(x)). If x ∈ M, we will note xl := l(x) and xr := r(x). E. Bouscaren and B. Poizat proved that in the language consisting of a function symbol for the left member and another for the right member, the theory of the pairing functions without cycles admits the elimination of quantiﬁers and is complete ([8]). It is in this language, L = {l, r}, that we will be placing ourselves in the following.

2 Representation of formulas and deﬁnable sets by binary trees

A natural way to represent an element is to use an infinite binary tree: the left child of a node will correspond to its image by the function l and the right child will correspond to its image by the function r. We will also use binary trees to represent certain definable set and definable functions. We explain how in this section.

Definition 2.1. Let T be an infinite and complete binary tree. We define the depth of a node recursively:

• by convention, the root is of depth 0;

• if a node is of depth p, then its children are of depth p + 1.

There is thus one node at depth 0, two at depth 1,..., and 2p at depth p. So we count 2p+1 − 1 nodes at depth less than or equal to p.

6 Definition 2.2. Labelling of the nodes Let T be an infinite binary tree. We label the nodes of T by associating them with a finite sequence of elements of {l, r} as follows: • the root is labelled by the empty sequence;

• if a node is labelled by a sequence {u1, . . . , un}, then his left child is labelled by {u1, . . . , un, l} and his right child by {u1, . . . , un, r}. The nodes of depth p are thus labelled by a sequence of p elements. Remark 2.3. We’ll note Tl (respectively Tr) the left subtree (respectively right subtree) of T , that is the full subtree of T of root l(x) (respectively of root r(x)) where x is the root of T . In the following, all trees considered will be binary inﬁnite trees. Deﬁnition 2.4. Tree associated with an element Let x be an element of M. We associate to x a binary tree labelled by elements of M as follows: • its root is labeled by x; • if a node is labeled by y, then its left child is labeled by l(y) and its right child by r(y).

We will denote this tree Tx. The nodes of Tx are enumerated by finite sequences of elements of {l, r} as explained above. If i is a finite sequence of {l, r}, we will note xi the term that labels the i-th node of Tx. This notation is consistent with the one proposed above: l(x) which is denoted xl corresponds to the left child of the tree of root x; and r(x) noted xr corresponds to its right child. The nodes of Tx correspond exactly to the set of one variable-terms in x. Remark 2.5. The fact that the pairing function has no cycles is equivalent to saying that for every x ∈ M there is no branch of Tx where an element of M appears more than once. Definition 2.6. Depth of a formula, a set, a definable function Let ψ be a formula. We call depth in x of ψ the largest integer n such that ψ involves a term in x corresponding to a node of depth n in the tree Tx. Let A be a definable set. We call depth in x of A the smallest integer n such that there is a formula defining A whose depth in x is n. Let h be a definable application. We call depth in x of h the smallest integer n such that there is a formula defining the graph of h whose depth in x is n. Thanks to quantifiers elimination, any formula is equivalent to a Boolean combination of the most simple formulas: equalities and inequalities on term (possibly constant term). We will say that a disjunction of such formulas is a simple formula. We will now explain how this "primitive formulas" can be represented by a binary tree. Definition 2.7. Primitive formula Let us consider a formula φ with possibly several free variables. If φ is a conjunction of equalities and inequalities between terms, then we say that φ is primitive. If φ is a conjunction of equalities between terms, then we say that φ is positive primitive.

7 Deﬁnition 2.8. Primitive formula associated with a tree Let T be a binary tree whose nodes are labelled with formulas. Let us suppose two things:

1. For every i, ﬁnite sequence of {l, r}, the i-th node of T is labelled with a ﬁnite list of equalities or inequalities between xi and other terms in x (possibly constant), that is by a list of formulas φj of the form:

• t(x) = c • t(x) = t0(x) • t(x) 6= c • t(x) 6= t0(x)

0 where t(x) is the term corresponding to xi, t (x) is a term having for single variable x, and c is a constant.

2. Only a ﬁnite number of nodes of T is labelled by a non-empty list.

We can associate with T the formula ∧jφj obtained by taking the conjunction of all the formulas that appear in the labels of T . We denote φ(T )(x) this formula.

Reciprocally to a primitive formula with only one variable can be associated a tree.

Definition 2.9. Tree associated with a single variable primitive formula Let ψ(x) = ∧jφj(x) a primitive formula with a single free variable x. We can associate to it a binary tree, noted T (ψ), such that for every i, finite sequence of {l, r} the i-th node of T is labelled by the list of formulas φj(x) that involve the term corresponding to xi. By abuse, we can also say that this tree is associated with ψ(M) or that it defines ψ(M).

Remark 2.10. If F (x, y) is a primitive formula with 2 variables, we can associate to F the tree in x constructed as before considering y as a parameter. The association of a tree to such formulas is a bijection. But two equivalent formulas can have distinct trees.

Example 2.11. 1. Let c ∈ M. Consider the formula ψ(x) := (r(x) = c) ∧ (l(x) = r(x)). The tree associated with ψ(x) is the tree, T , whose root has a left child labelled by {l(x) = r(x)} and a right child labelled by {(r(x) = c), (l(x) = r(x))}. The other nodes are labelled by the empty list. The formula associated with the tree T is the formula φ(T )(x) = (r(x) = c) ∧ (l(x) = r(x)) ∧ (l(x) = c). The formulas ψ(x) and φ(T )(x) are equivalent.

l(x) = r(x) l(x) = r(x), r(x) = c

8 2. Let ψ(x, y) be the formula (l(x) = c) ∧ (r(x) = l(y)). The tree in x associated with this formula is the tree whose root has a left child labelled by {l(x) = c}, and a right child labelled by {(r(x) = l(y))}.

l(x) = c r(x) = l(y)

Remark 2.12. Let T be a tree labelled as in 2.8. Suppose the root of T is labelled by the empty list, then φ(T )(x) is equivalent to Φ(Tl)(l(x)) ∧ φ(Tr)(r(x)). 0 Since Θ is bijective, for every y ∈ φ(Tl)(M) and every z ∈ φ(Tr)(M), the tree T such that 0 0 Tl = Ty, Tr = Tz, and whose root is labelled by the empty list, is the tree of an element of φ(T )(M).

3 Closed tree and simple formulas

Deﬁnition 3.1. Let ψ be a primitive formula and T be its associated tree. We say that T (or by abuse ψ) is closed if the nodes of T , f1, . . . , fn, with a non-empty label satisfy the following conditions:

• every node of T is either a descendant or an antecedent of one of the node f1, . . . , fn;

• for every i 6= j, fi is not a descendant of fj.

Let T be a closed tree and f1, . . . , fn be the nodes of T satisfying the previous properties. Let 0 T be the tree obtained from T by removing the label of all its nodes except f1, . . . , fn. We say that T 0 is a skeleton tree of T . Let T be a tree associated with a primitive formula ψ. We say that T 0 is a closed subtree of T , if T 0 is closed and if T 0 is obtained from T by removing the labels of some of the nodes of T .

Remark 3.2. Let T be a closed subtree and f1, . . . , fn be nodes of T with a non-empty label that satisfy the condition of the deﬁnition. Let T 0 be a skeleton tree of T . Then φ(T 0) is a formula equivalent to φ(T ).

Lemma 3.3. Let ψ(x) be a primitive formula obtained as a conjunction of formulas of the form t(x) = c where t is a non-constant term and c is a constant. Suppose moreover that its tree is closed. Then ψ deﬁnes a singleton.

Proof. This is an immediate consequence of the fact that Θ is a bijection and can be proved by recurrence on the depth of ψ, q. For q = 1, since Θ is a bijection, the property is obvious.

9 Suppose that the property is satisfied for q ∈ N∗. Let us show that it is true for q + 1. Let T be the tree associated with ψ. Since q + 1 is strictly greater than 0, and T is closed, its root is labelled by the empty list, and ψ(x) is equivalent to φ(Tl)(l(x)) ∧ φ(Tr)(r(x)). The formulas φ(Tl) and φ(Tr) are also closed and by hypothesis of recurrence define a singleton. The same is thus true of φ(T ). Lemma 3.4. Let ψ(x) be a primitive formula that defines a finite and non-empty set. Then the tree associated to ψ admits a closed subtree. Furthermore, ψ(x) is equivalent to a primitive positive formula and ψ defines in fact a singleton. More precisely, let T be the skeleton of T (ψ) and T 0 be the tree obtained from T by removing of the labels of T every formula that is an inequality. Then ψ(x) is equivalent to φ(T )(x). Proof. Let suppose that, T , the tree associated to ψ does not admit a closed subtree. Then there exists a node of T that is neither a descendant nor a ascendent of any node of T that is labelled by the empty list. Therefore this node is totally independent of ψ. Let i be such that this node corresponds to the term xi. Let x ∈ ψ(M) and z ∈ M. Consider the element y of M such that for any term yk that has not a filiation link with xi, yk = xk and such that yi = z. This element satisfies the formula ψ since x does. As this is true for any element z, ψ(M) is actually infinite. This contradiction proves that T admits a closed subtree. Let T 0 be a skeleton tree of T . Let us suppose that the formula associated with T 0 is not positive. If one of the label does not contain an equality then it is easy to see that the formula defines an infinite set by proceeding for example exactly the same way we proved that T admits a closed subtree. Let T 00 be the tree obtained from T 0 by removing the formulas of the labels that are an inequality. The formula associated to T 00 is obviously primitive positive. It is obvious that φ(T ) and φ(T 00) are equivalent and that φ(T 0) defines a singleton. Definition 3.5. Consider a formula of the form

l n 1 m 2 3 k 4 5 ψ(x) := (∧j=1tj(x) = cj) ∧ (∧j=1tj (x) 6= cj) ∧ (∧j=1tj (x) = tj (x)) ∧ (∧j=1tj (x) 6= tj (x))

1 2 3 4 5 0 where for every j, tj(x), tj (x), tj (x), tj (x), tj (x), tj (x) are terms in a single variable and cj, cj are constants. Suppose furthermore that

2 3 • for every j there is no ﬁliation links between tj (x) and tj (x),

l m 2 3 • for every 1 ≤ j ≤ m, the conjunction (∧j=1tj(x) = cj) ∧ (∧j=1tj (x) = tj (x)) does not 2 3 imply any equality between tj (x) (respectively tj (x)) and a constant. We will then say that ψ is simple. If ψ(x, y1, . . . , yn) is a primitive formula, then we say it is simple in x, if for any constants c1, . . . , cn, the formula in one variable, ψ(x, c1, . . . , cn), is simple. Remark 3.6. l m 2 3 Any formula of the form (∧j=1tj(x) = cj)) ∧ (∧j=1tj (x) = tj (x)) is obviously equivalent to a simple formula.

10 Lemma 3.7. 0 0 Let ψ(x, y1, . . . , yl, y1, . . . , yn), a formula that is simple in x:

l n 1 0 m 2 3 k 4 5 ψ(x) := (∧j=1tj(x) = yj) ∧ (∧j=1tj (x) 6= yj) ∧ (∧j=1tj (x) = tj (x)) ∧ (∧j=1tj (x) 6= tj (x))

1 2 3 4 5 where for every j, tj(x), tj (x), tj (x), tj (x), tj (x), tj (x) are terms in x. 0 0 0 0 If there exists c1, . . . , cl, c1, . . . , cn such that ψ(x, c1, . . . , cl, c1, . . . , cn) defines a finite set, then 0 0 l ψ(x, y1, . . . , yl, y1, . . . , yn) is equivalent to ∧j=1tj(x) = yj. 0 0 0 0 Moreover, for every c1, . . . , cl, c1, . . . , cn, elements of M, ψ(x, c1, . . . , cl, c1, . . . , cn) defines either a singleton or the empty set.

Remark 3.8. The formula (xg = c) ∧ (xg = xd) deﬁnes a singleton and is not a conjunction of equalities between terms in y and constants. However, it is equivalent to (yg = c) ∧ (yd = c). It is in order to take such equivalences into account that we must assume that ψ is simple.

Proof. 0 0 0 0 Let c1, . . . , cl, c1, . . . , cn such that ψ(x, c1, . . . , cl, c1, . . . , cn) defines a finite set. By lemma 3.4, 0 0 the tree associated to ψ(x, c1, . . . , cl, c1, . . . , cn) admits a closed subtree. Moreover, let T be the skeleton of T (ψ) and T 00 be the tree obtained from T by removing off the labels of T 0 every formula that is an inequality. Then ψ(x) is equivalent to φ(T )(x). This formula is obviously simple. It is easy to check that if this formula isn’t a conjunction between 0 0 terms and constants, it then defines an infinite set. That implies that ψ(x, c1, . . . , cl, c1, . . . , cn) l is in fact equivalent to ∧j=1tj(y) = cj. 0 0 Since the fact that ψ(x, c1, . . . , cl, c1, . . . , cn) defines a finite set does only depends on the 0 0 shape of its associated tree, it is obvious that for every c1, . . . , cl, c1, . . . , cn, elements of M, 0 0 l ψ(x, c1, . . . , cl, c1, . . . , cn) is equivalent to ∧j=1tj(x) = yj and defines either a singleton or the empty set.

4 Form of the deﬁnable injections

Lemma 4.1. Form of the definable injections Let A be a definable subset of M and h be an injective definable function from A to A. The 0 graph of h can be defined by a formula of the form ∨j(φj(x) ∧ φj(x, y)) where:

1. every φj(x) is primitive in x,

2. the sets φj(M) are all disjoint,

0 3. for every j, φj(x, y) is a disjunction of positive primitives formulas. Proof. By elimination of the quantifiers, we can assume that we have partitioned A into a disjoint union of sets Aj defined by formulas φj such that points 1 and 2 are satisfied. 0 Let φj(x, y) be a formula that defines the graph of h restricted to Aj. It is enough to show that these formulas are equivalent to simple formulas in y: as for every x, the formula in y, φ0(x, y) defines a singleton, we will then apply lemma 3.7. The next lemma thus prove the theorem.

11 Lemma 4.2. Let F (x, y) be a primitive formula in y. Then there exists G(x, y) such that G(x, y) is simple in x and G(x, y) is equivalent to F (x, y). Proof. It is obvious that we can ﬁnd a formula F 0(x, y) such that F 0(x, y) is equivalent to F (x, y) and

l n 1 m 2 3 k 4 5 F (x, y) = (∧j=1tj(y) = cj) ∧ (∧j=1tj (y) 6= cj) ∧ (∧j=1tj (y) = tj (y)) ∧ (∧j=1tj (y) 6= tj (y))∧

n 6 7 m 8 9 (∧j=1tj (y) 6= tj (x)) ∧ (∧j=1tj (y) = tj (x)) 1 2 3 4 5 6 8 7 9 where tj(y), tj (y), tj (y), tj (y), tj (y), tj (y), tj (y), tj (y) are terms in y and tj (x), tj (x) are terms in k x satisfying for every i, j and every 1 ≤ k ≤ 8, tj is not a descendant of ti. Let us show that we can even suppose that there is no ﬁliation link between them. k Suppose, there exists i, j and 1 ≤ k ≤ 8 such that tj (x) is an ascendant of ti, then the condition k on tj (x) is equivalent to a conjunction of conditions on its children. We can go down this way until we reach the depth of ti. For any term intervening in this formula, we can replace an equality on term by an equality with a constant if such is implied by the formula ti(y) = ci. Reiterating this operation, we obtain a simple formula in y equivalent to F (x, y).

Definition 4.3. Let h be a definable injection from a definable set of M into M. We say that a formula, φ(x, y) defining the graph of h is a normal formula if it is a conjunction of a formula F (x) defining the set of definition of h and a formula ψ(x, y) simple in y, conjunction of formulas of the form

• t(x) = t0(y) where t0(y) is a term in y and t(x) a term in x,

• t(y) = c where t(y) is a term in y and c ∈ M.

Remark 4.4. Le G(x, y) be a conjunction of formulas of the form

• t(x) = t0(y) where t0(y) is a term in y and t(x) a term in x,

• t(y) = c where t(y) is a term in y and c ∈ M.

Then G(x, y) is obviously equivalent to a formula ψ(x, y) positive and simple in y.

Lemma 4.5. Let h be an injection definable by a normal formula on a set A \ B where A and B are definable subsets of M and B ( A. Then h can be extended on A by an injection h˜ definable by a normal formula.

Proof. Let φ(x, y) be a normal formula defining h. We can write φ(x, y) as a conjunction of a formula F (x) defining the set of definition of h and, ψ(x, y), a simple formula, conjunction of formulas of the form:

• t(x) = t0(y) where t0(y) is a term in y and t(x) a term in x,

12 • t(y) = c where t(y) is a term in y and c ∈ M.

Furthermore, according to lemma 3.7, if there exists x0 ∈ M such that ψ(x0, y) is a singleton, then for any element x ∈ M, ψ(x, y) defines a singleton. Thus this formula can indeed define the function which to x ∈ M associates the only element y such that ψ(x, y) is true. Let G(x) be a formula defining A. The last point is to prove that G(x) ∧ ψ(x, y) defines the graph of an injection. Since the formula, ψ(x, y) is simple in y it is also in x. There exists an element x0 (any x0 ∈ A \ B) such that the formula in y, ψ(x0, y) defines a singleton. According to lemma 3.7, for any x, ψ(x, y) defines a singleton: this precisely means that the function ψ(x, y) is injective.

5 Simple sets

We define here sets that we will call "simple" and which, at fixed depth, p, will be an analogous of the closed irreducible sets of a Noetherian topology: these simple sets will be such that they can not be written as non-trivial union of other simple sets. Furthermore, they allow to obtain by Boolean combination any definable set of depth p. p In order to define them, we’ll set the equalities between terms xi where i ∈ {l, r} and, eventu- ally, the equality of some of these terms xi with constants. To set equalities between the terms p p n xi where i ∈ {l, r} is equivalent to partitioning {l, r} in a disjoint finite union, tk=1Ik, where p for every i, j ∈ {l, r} , xi = xj, if and only if there is k with i, j ∈ Ik. We add conditions fixing equalities between some coordinates and constants from M.

Deﬁnition 5.1. ∗ n p Let p ∈ N Let tj=1Ij be a partition of {l, r} and let C be a set of couples (i, c) where i ∈ {g, d}p and c ∈ M. n We associate tj=1Ij and C, the A subset of M, x elements such as:

• For every 1 ≤ j ≤ n, and all j1, j2 ∈ Ij, xj1 = xj2 ,

• for every (i, c) ∈ C, xi = c,

• for every i, j such that there exists k with i, j ∈ Ik, if (i, c) ∈ C, then (j, c) ∈ C .

The set A is called simple set of depth p (the depth is not specified when there is no n p ambiguity). We call tj=1Ij the partition of {l, r} adapted to A and C its set of constants. Remark 5.2. 1. The depth of a set is not intrinsic to this set: any simple set of depth p is also a simple set of q for every q > p. The same set admits different presentations as simple sets of different depths. 2. The intersection of two simple set of depth p is a simple set of depth p. 3. If B1 is a simple set and B2 ⊆ B1 is a set strictly included in B1 and defined by a primitive positive formula of the same depth as B1, then B2 is also a simple set whose partition refines the one adapted to B1, and whose set of constants contains the one of B1. 4. If B1,B,2 ,B3 are simple sets, not all identical, and of the same depth, then B1 = B2 ∪ B3 implies that B2 = ∅ or B3 = ∅.

13 Deﬁnition 5.3. ∗ m q Let p ∈ N and ti=1Ji be a disjoint union of subsets of {l, r} . m Consider the formula ∧i=1(∧s,t∈Ji xs = xt) which for every i, imposes equality between all xk for k ∈ Ji. This formula possibly implies equalities between other terms. Consider the maximal q subsets, Ii, of {l, r} such that for every k, l ∈ Ii, the equality xk = xl is implied by the formula m ∧i=1(∧s,t∈Ji xs = xt). q q We call partition of {l, r} generated by tiJi, the partition of {l, r} made up of Ii.

q q Let tiKi be a partition of {l, r} and C a set of pairs (j, c) where j ∈ {l, r} and c ∈ M. The m 0 formula [∧i=1(∧s,t∈Ji xs = xt)] ∧ [∧(j,c)∈C xj = c] deﬁnes a simple set whose set of constants, C , is called set of constants generated by C and tiKi.

Lemma 5.4. Let p be an integer. Consider the topology on M whose closed sets are generated by the simple sets of depth p. Then this topology is Noetherian and the simple sets of depth p are its irreducible closed sets.

Proof. Let F be the set of finite unions of simple set of depth p. Let’s check that the axioms defining closed sets are satisfied by the elements of F . F is clearly stable by finite union. The set M is the simple set of depth p whose set of constants is empty and whose partition only includes singletons. It is thus an element of F . It is obvious that F is stable by intersection. The fact that the simple sets are irreducibility closed sets of this topology is exactly the point 4 of the remark 5.2. Noetheriennity is immediate to verify: two simple sets are included one in the other if, and only if, the partition of the smallest refines the one of the largest or the set of constants of the smallest contains the one of the largest. Since a set of constants has at most 2p elements and since there is a finite number of possible partitions, it cannot exist infinite chains of simple sets.

Lemma 5.5. Every non-empty simple set is in deﬁnable bijection either with M or a singleton.

Proof. Let A be a non-empty simple set of M of depth p, of partition tjIj, and of set of constants C. k 0 Let us consider the maximal subpartition ti=1Ii of tjIj such that for every i, the value of the 0 terms corresponding to Ik are not determined by the constant set. If this maximal subpartition is empty then A is reduced to a singleton. Otherwise, it is obvious that for every a1, . . . , ak ∈ M, there is a unique element x of M such that

0 • for every 1 ≤ j ≤ k and every i ∈ Ik, xj = aj,

• all the other terms of depth less or equal to p have their value implied by C.

The set A is hence in definable bijection with M k. Since M 2 is in definable bijection with M, any Cartesian power of M is in definable bijection with M and so is A.

14 6 Decomposition of deﬁnable sets

It is obvious by quantifier elimination that these simple sets of depth p permit to construct by Boolean combination all definable sets of depth less than or equal to p. Lemma 6.1. Let A be a set of depth less than p. It is possible to write A as a Boolean combination of simple sets of depth p. Proof. This is clear by the elimination of quantifiers. Definition 6.2. Let p ∈ N. Let A be a definable set and ti(Bi \ Ci) a partition of A such that:

• for every i, Bi is a simple set of depth p,

• for every i, Ci ( Bi is a union of simple sets of depth p.

We say that ti(Bi \ Ci) is a decomposition of A into simple sets of depth p. The sets Bi are called positive sets, and the sets Ci, negative sets. The sets Bi \ Ci are called elementary sets. Lemma 6.3. n n Let A ⊆ M be a deﬁnable set and A = ti=1(Bi \ Ci) a decomposition in simple sets of depth p. Let T be the topology generated by the simple sets of depth p. It is possible to modify this decomposition in order to get a decomposition that satisﬁes that for every 1 ≤ i ≤ n there are k i1, . . . , ik such that the closure of Ci ∩ A in T admits tm=1Bim as a decomposition in simple sets of depth p. Proof. The simple sets of depth p are the irreducible closed sets of the topology that they generate. We place ourselves in this topology and will refer to the simple and irreducible closed sets of this topology. Let 1 ≤ j ≤ n. 0 0 We consider B1,...,Br, the components of the decomposition in irreducible of the closure of A ∩ Cj. Let Ac be the complementary set of A in M n. 0 c We consider C := Bj ∩ A . 0 0 0 We add, for every 1 ≤ j ≤ r, the elementary set Bj \ (C ∩ Bj). r 0 0 0 By construction, it is clear that tj=1Bj \ (C ∩ Bj) ⊆ A.

It is equally clear, since the topology is Noetherian, that we cannot repeat this operation indefinitely. By iterating this operation as many times as necessary, we obtain a decomposition satisfying the condition of the lemma. Definition 6.4. n A decomposition A = tj=1(Bj \ Cj) which satisfies all the conditions of the lemma 6.3 is said elementary. We say that n is the cardinality of the partition.

15 Lemma 6.5. n Let tj=1(Bj \ Cj) be an elementary decomposition of depth p.

Then, for every 1 ≤ j ≤ n, Cj ∩ A = t{i|Bi⊆Cj }(Bi \ Ci) and

Bj ∩ A = (Bj \ Cj) t ∪{i|Bi⊆Cj }(Bi \ Ci). Proof. For this lemma, we will use the following deﬁnition: Deﬁnition 6.6. n Let tj=1(Bj \ Cj) be a decomposition. Let 1 ≤ j ≤ n We say that Bj is of height h if h is the bigger number such that there exists

h elements of [|1; h|], i1, . . . , ip with Bi1 ( ... ( Bih = B.

Let 1 ≤ j ≤ n and let h be the height of Bj. Let’s prove the lemma by induction on h. Suppose that h = 1. Since the decomposition is elementary, the decomposition in irreducible of Cj ∩ A is an union of positive sets. Since h = 1, this means that Cj ∩ A = ∅. The set Bj \ Cj is equal to Bj ∩ A

and Cj ∩ A = ∅ = t{i | Bi⊆Ci}(Bi \ Ci) since this last union is on the void. Let h ∈ N∗. Suppose the result shown for h. Let’s show it for h + 1. k Since the decomposition is elementary, there exists i1, . . . , ik such that Cj ∩ A admits tm=1Bim as decomposition in irreducible closed sets (in the topology generated by the simple sets of depth p). These sets are obviously of height less than Bj.

By hypothesis of induction, for every 1 ≤ m ≤ k, Cim ∩ A = t{i | Bi⊆Cim }(Bi \ Ci) and

Bim ∩ A = (Bim \ Cim ) t{i | Bi⊆Cim } (Bi \ Ci). k k k Since Cj ∩ A = ∪m=1Bim , we have Cj ∩A = ∪m=1(Bim ∩A) and therefore Cj ∩A = [tm=1(Bim \

Cim )] t ∪{i | Bi⊆Cim }(Bi \ Ci).

This last union is obviously equal to t{i | Bi⊆Cim }(Bi \ Ci). Finally, since Bj ∩ A = (Bj \ Cj) t (Cj ∩ A), we have the second equality sought. This completes the induction.

7 Deﬁnable injection on simple set

Lemma 7.1. Let A be a simple set of depth p. Let h be an injection deﬁned on A whose graph is deﬁned by a simple formula. Then, there exists an integer q depending only on h and p such that for any simple set X ⊆ A, of depth p the image of X by h is a simple set of depth q. We say that the integer q is adapted to A and h.

Proof. 0 00 Let F (x, y) := (∧jtj(y) = tj(x)) ∧ (∧jtj (y) = cj) a normal formula deﬁning the graph of h on 0 00 A, and let p the maximum of depth in y of all the terms tj(y) and tj (y). We can always replace a formula of depth p by an equivalent formula of depth q for any q ≥ p. Up to considering a formula equivalent to F and of depth at most p0 + p in y, we can assume 0 that all the terms tj(x) are of depth p. Let q be the depth in y of such a formula. k Let ti=1Ii be the partition associated with A. For every i let Ji be the set of elements, u, of

16 q {l, r} such that ∃k ∈ Ii yu = xk is implied by F (x, y). q Let C be the set of couples; (i, c), with c ∈ M and i ∈ {l, r} such that yi = c is implied by F (x, y). ˜ q We note tiJi the partition of {l, r} generated by tiJi and C. ˜ ˜ Let C be the set of constants generated by C and tiJi. ˜ ˜ The partition tiJi and the set C define a simple set of depth q which is the image of A by h. Lemma 7.2. Let A be a definable set, h a definable injection from A to itself. Then there exists two integers p, q such that

• A = ti(Bi \ Ci) is a decomposition of A into simple sets of depth p,

• for every j, the restriction of h on Bj \ Cj is deﬁnable by a normal formula,

• for every j, h restricted to Bj \ Cj can be extended on Bj by a deﬁnable injection hj,

0 0 • for every j, Bj := hj(Bj) is a simple set of depth q and Cj := hj(Cj) is a ﬁnite union of simple sets of depth q,

• for every j, the decomposition in irreducible of Cj in the topology generated by the simple 0 sets of depth p has the same number of elements as the decomposition in irreducible of Ci in the topology generated by the simple sets of depth q. Furthermore, the number of irreducible components of Ci reduced to a singleton is equal 0 to the number of components irreducible of Ci reduced to a singleton. Proof. By elimination of the quantifiers, there exists an integer p and a partition of A := tiAi in definable sets such that for every i, Ai is a simple set of depth at most p and h restricted to Ai is a normal function. As every definable set of depth p is a Boolean combination of simple sets of depth p, by decomposing the sets Ai, we get a finite partition of A, A = ti(Bi \ Ci), where

• for every j, the set Bj is a simple sets of depth p and the set Cj is a ﬁnite union of simple sets of depth p;

• for every j, the function h restricted to Bj \ Cj is a normal function.

According to lemma 4.5, for every j, h restricted to Bj \ Cj can be extended on Bj by a normal injection that we will denote hj. According to the previous lemmas, for every j, there exists qj such that for any simple set X of depth p and included in Bj, hj(X) is a simple set of depth qj. Let q := maxj(qj). As any simple set of given depth q0 can be considered as a simple set of depth greater than q0, by taking q, 0 the maximum of all the qj, we get that for every j, Bj := hj(Bj) is a simple set of depth q 0 and Cj := hj(Cj) is a ﬁnite union of simple sets of depth q. Let us write, for any integer i, the decomposition of Ci in the topology generated by the simple sets of depth p: Ci = ∪jBi,j. Since hi is injective, for every j, the sets hi(Bj,i) are distinct and are the components of the 0 decomposition in irreducible of Ci = hi(Ci) in the topology generated by the simple sets of depth q. It is obvious, that the number of irreducible components of Ci reduced to a singleton is equal 0 to the number of irreducible components of Ci reduced to a singleton.

17 Definition 7.3. Let A be a definable set, and h a definable injection from A to itself. Let p, q be two integers. We say that A = ti(Bi \ Ci), a finite decomposition of A (which is not necessarily a partition) is adapted to h if it satisfies the properties of lemma 7.2.

8 Representation of a decomposition: tree of a decomposition

One way to represent a deﬁnable set is to associate a tree with one of its decompositions. A branch corresponds to a chain of simple sets that are included one in the other.

Lemma 8.1. h Let A be a deﬁnable set and A = tj=1(Bj \ Cj) a decomposition. Then, there is a tree T and φ a surjection from T to the set {A} t {B1 \ C1,...,Bh \ Ch} such that:

• if r is the root of T , then φ(r) = A,

• if r is the root and B1,...,Bm are the maximal positive sets in the sense of inclusion, then r has exactly m children n1, . . . , nm, and furthermore, B1 \ C1 = φ(n1),...,Bm \ Cm = φ(nm),

• let n be a node of T not equal to its root and let 1 ≤ j ≤ h be such that Bj \Cj = φ(n). Let

Bj1 ,...,Bjs be the positive maximal sets (in the sense of inclusion) among those included

in Cj; then n has s children, n1 . . . , ns, and furthermore, Bj1 \Cj1 = φ(n1),...,Bjs \Cjs = φ(ns).

h We say that T is the tree associated to the decomposition tj=1(Bj \ Cj) = A.

Remark 8.2. a. We want to distinguish the root A from the other nodes: even if the decomposition has only one element, B \ C, the tree has a root whose image by φ is A and a leaf whose image by φ is B \ C. b. The function φ is not a bijection: some elementary sets can be associated with several nodes of the tree. All the nodes corresponding to the same set B \ C, are roots of the same subtree. c. In the construction proposed in the previous demonstration, the order in which trees T1,...,Ts are planted on the root is arbitrary. Nevertheless, this is the only phenomenon that could contra- dicts the uniqueness of a tree associated with a given decomposition: its graphical representation is obviously not unique, but there is a single tree associated with a given decomposition.

Proof. Let p be the length of a maximum chain formed by positive sets. Let’s prove the result by induction on p. h Suppose that p = 1. Then A = tj=1(Bj \ Cj) where no negative set contains positive set. (Positive sets are the closed sets ones of the irreducible decomposition of A.) The associated tree is the tree of root A of depth 1 with h leaves f1, . . . , fl corresponding to elementary sets. To conclude we just have to set φ as φ(r) = A, φ(fi) = Bi \ Ci.

18 Let p ∈ N∗. Suppose the result shown for every 1 ≤ i ≤ h, show it for p + 1.

Up to renumbering the sets Bi, we can assume that B1,...,Bs are the maximum positive sets (in the sense of inclusion) of A. We consider T0 the tree of root A and whose s leaves f1, . . . , fs, correspond to B1\C1,...,Bs\Cs. h We can apply the hypothesis of induction to tj=s+1(Bj \ Cj) to get the tree T1. Each node of level 1 of T1 corresponds to a set Bj \ Cj where j ranges between s + 1 and h, such that, according to the lemma 6.5, Bj is included in one of the sets B1,...,Bs. Up to renumbering the sets, we can assume that the children of the root of T1 are Bs+1,...,Bs+t where s + t ≤ h. We consider the tree T of root A whose s child f1, . . . , fs, correspond to B1 \ C1,...,Bs \ Cs and such that for every 1 ≤ i ≤ s and all j between s + 1 and s + t, we plant on the the node corresponding to Bi \ Ci the subtree of T1 of root Bj \ Cj if and only if, Bi ( Bj. It is obvious that T satisﬁes the conditions of the statement.

Here are some examples of trees associated with a decomposition: Example 8.3. Let B1,B2,B3 be simple sets and C1 a union of simple sets, such as:

• B1 and B2 are not included in each other

• C1 ( B1 ∩ B2

• B3 ( C1.

Let’s put C2 := B1 ∩ B2. Let A = (B1 \ C1) t (B2 \ C2) t B3. A tree associated with this decomposition of A is:

(B1 \ C1) (B2 \ C2)

B3 B3

Example 8.4. Let B1,...,B3 simple sets such as:

• B3 ( B2 ( B1.

Let A = (B1 \ B2) t (B2 \ B3) t (B3). A tree associated with this decomposition of A is:

19 (B1 \ B2)

(B2 \ B3)

Lemma 8.5. Let p ∈ N. One places oneself in T , the topology generated by the simple sets of depth p. Let A be a closed sets and ti(Bi \Ci) = A an elementary decomposition of A into simple sets of depth p. Let T0 be a tree associated with this decomposition. Then any node of T0 corresponding to Bj \ Cj has as many children as Cj has irreducible components in T .

Proof. Since A is a closed set, for every j, Cj ∩ A = ∪iXi where ∪iXi is the irreducible decomposition of Cj. According to the lemma 6.5, Cj ∩ A is also equal to ti|Bi(Cj (Bi \ Ci). By taking the closure of this last union, it follows that every set Xi is equal to one of the set Bi : these correspond to the children of the node associated with Bj \ Cj which therefore has as many children as Cj has irreducible components (in T ).

9 Computation of the Grothendieck ring

Lemma 9.1. Let A be a deﬁnable set and h a deﬁnable injection from A to A. Let A = ti(Bi \ Ci) a decomposition adapted to h, and p an integer such that all the elements of this decomposition are of depth p. Let T be the the topology whose closed sets are generated by the simple sets of depth p. Then A admits an elementary decomposition in T adapted to h.

Proof. Let j be an index and X an irreducible component of Cj ∩ A. Let us assume that X is not one of the positives sets of the decomposition A = ti(Bi \ Ci). Let us show that there exists C a closed set of T such that A = (ti(Bi \ Ci)) t (X \ C) is a decomposition adapted to h. Since the set X is irreducible, there exists an index i such that X ⊆ Bi and Bi is minimal for this property. The minimality of Bi and the fact that X is included in the closure of A imply that X is not included in Ci. Let C := X ∩ Ci. The set X \ C is then included in Bi \ Ci. It is enough to check that:

• h is deﬁned on X \ C by a normal formula,

• h restricted to X \ C can be extended on X by an injection h˜ deﬁnable by the same normal formula than h on X \ C,

• X0 := h˜(X) is a simple set of depth q and C0 := h˜(C) is an union of simple sets of depth q.

20 ˜ Since X \C is included in Bi \Ci, the two first properties are obvious. Let h be the prolongation of h on X that satisfies the second property. This prolongation is defined by a normal formula, X0 := h˜(X) is a simple sets of depth q and C0 := h˜(C) is an union of simple sets of depth q.

The decomposition A = (ti(Bi \ Ci)) t (X \ C) is thus adapted to h. By iterating this construction, we obtain an elementary decomposition adapted to h. Corollaire 9.2. Let A a definable set, and h a definable injection from A into A. Then A admits a decomposition adapted to h (in the meaning of the previous demonstration) and elementary in the topology generated by the positive and negative sets of this decomposition. Theorem 9.3. Any model of the theory of the pairing function without cycles admits a Grothendieck ring isomorphic to Z[X]/(X − X2), itself isomorphic to Z2. Proof. Let M be a model of the theory of the pairing functions without cycles. Let X be the class of M in the Grothendieck ring. It is obvious that X = X2. Let A be a non-empty definable subset of M. Since A is a Boolean combination of simple sets, and since every simple set is in definable bijection with M or with a singleton, the class of A in the Grothendieck ring of M is an element of Z[X]/(X − X2). Let us now consider Z[X]/(X − X2) where X is an undetermined variable. Let us fix a family, F, of finite sets containing for each n ∈ N exactly one set of cardinality n. To each elements aX + b ∈ Z[X]/(X − X2), with a and b positives, we associate the definable set (M × F ) t G where F is the element of F of cardinality a and G is the element of F of cardinality b. Let us check that two distinct elements of this ring are associated with definable sets that are not in definable bijection. In other words, let us show that for every finite sets F, G, F 0 and G0, if there exists a definable bijection between (M × F ) t G and (M × F 0) t G0, then |F | = |F 0| and |G| = |G0|. Let us assume that there is such a bijection, that we will denote by h, between (M ×F )tG and 0 0 N (M × F ) t G . We consider ti=1(Bi \ Ci) a decomposition in simple sets adapted to h. We can furthermore assume that this decomposition is elementary. We consider the tree T associated to this decomposition as explained in 8.1.

To each node ni of T (distinct of the root), we associate di the number of its children. The P total number of the nodes of T is N = i di + I where I is the number of children of the root. N 0 0 Since the decomposition is adapted to h, h(M) = ti=1(Bi \ Ci) where for every i, h(Bi \ Ci) = 0 0 0 0 0 Bi \Ci. Let T be a tree associated with this decomposition. To each node ni of T , we associate 0 di the number of its children. Let us show that the number of children of the root of T is equal to the number of children of the root of T 0. Since the total number of nodes of T is obviously equal to the one of T 0, P P 0 0 0 0 i di + I = i di + I where I is the number of children of the root of T ans it is suﬃcient to P 0 P verify that i di = i di. Since (M ×F )tG is closed, and since the decomposition is elementary, according to the lemma 8.5, every node n of T corresponding to a set Bi \ Ci has as much children as Ci has irreducible

21 components. According to the lemma 7.2 and the fact that the decomposition is adapted to h, for every i, the 0 number of irreducible components of Ci is equal to the number of irreducible components of Ci. As the set (M × F 0) t G0 is closed, the lemma 8.5 implies that every node n of T 0 corresponding 0 0 0 to a set Bi \ Ci has as much children as Ci has irreducible components. P P 0 0 The combination of these two points lead to i di = i di and thus to I = I . But the sets corresponding to the children of the root of T (respectively T 0) are the elements of the decomposition in irreducible of (M × F ) t G (respectively h(M)). As (M × F ) t G has a decomposition in irreducible of cardinality |F | + |G| (respectively |F 0| + |G0|), it thus follows that |F | + |G| = |F 0| + |G0|. Let tN (B \ C ). This set admits M × F as closure. i=1,|Bi|>1 i i N 0 0 0 Let t 0 (B \ C ). This set admits M × F as closure. i=1,|Bi|>1 i i By proceeding as precedently, and by noticing that if two sets are in definable bijection, then so is their closure, we show that |F | = |F 0|. It then follows that |G| = |G0|: two distinct elements of N[X]/(X − X2) correspond to two definable sets of M which are not in definable bijection. As announced, the Grothendieck ring of M is Z[X]/(X − X2). It remains to show that Z[X]/(X − X2) is isomorphic to Z2. Let a := 1 − X and b := X. Then a2 = a, b2 = b, ab = 0. This allows to define an isomorphism between Z[X]/(X − X2) and Z2 by putting a 7→ (1, 0) and b 7→ (0, 1).

10 Example of pairing function without cycles on N

There exists numerous examples of pairing functions on N. For instance, the pairing function 1 of Cantor: (m, n) 7→ 2 (m + n)(m + n + 1) + m. But this function is not without cycles since 0 7→ 0. The pairing function that we think of the most naturally are often defined algebraically and verify a certain property of decreasingness: if a ∈ N, then its image (b, c) is such that b < a or c < a ... So they have a cycle in 0. Here we will give the example of a function of pair without cycles. Let us consider a table whose rows and columns are indexed by N. To each box of the table, we associate an integer in the following way: we start by associating 0 in the box (0, 0). If n ≥ 0 and all boxes of the square [|0; n|]2 have already been filled up, then we consider the square of side n + 1, [|0; n + 1|]2, and we start by filling up the boxes of the line n + 1 before going up along the last column, using each time the minimal integer that has not already been associated to a box. Here’s what it gives for the first three squares:

0 3 8 1 2 7 4 5 6 This deﬁnes a bijection of N in N2. This bijection corresponds to the function deﬁned as follows: for n = k2 + t where t ≤ k, we set Θ(n) = (t, k) for n = k2 + t where k ≤ t < 2k + 1, we set Θ(n) = (k, k − t).

22 Θ is not without cycles: indeed, 0 is sent on (0, 0). We are going to modify Θ so that it becomes without cycles. The problem with Θ is that it actually checks a property of "decreasingness": For every n ∈ N its image (a, b) is such as a ≤ n and b ≤ n. To remedy to this, we will make point 0, a point that makes "diverge to infinity". We proceed as follows: to define the images of integers, we will again use the table whose lines and the columns are indexed by N. On all the boxes of coordinates of the form (2k, 2k) where k > 1, we enter the number 2k−1. To (2, 2), we associate 0. Then we enter the remaining digits (which are not powers of 2) one after the other by browsing the table of the same way that we had when defining Θ. Here are the first three lines and rows of this table 1 6 11 3 5 10 7 9 0

It is obvious that p0 is without cycles. Indeed, p0 is strictly "decreasing" on the set of strictly positive integers that are not a power of 2 (in the sense that p0(n) = (a, b) with a < n and b < n). k 0 And on 2 |k ∈ N t {0}, p is strictly increasing.

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[E2] E. Elbaz, A Grothendieck ring of ﬁnite characteristic, not submitted yet.

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