Forms of Choice in Ring Theory

We investigate the relationship between various choice principles and nth\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\hbox {th}$$\end{document}-root functions in rings. For example, we show that the Axiom of Choice is equivalent to the statement that every ring has a square-root function. Furthermore, we introduce a choice principle which implies that every integral domain has an nth\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\hbox {th}$$\end{document}-root function (for odd integers n), and introduce another choice principle which is equivalent to the Prime Ideal Theorem restricted to certain ideals. Finally, we investigate the dependencies between the two new choice principles and a choice principle for families of n-element sets.


Introduction Some Forms of Choice Related to Algebra
The investigation of consequences of the Axiom of Choice in algebra has a long tradition. Below we list a few choice principles in the context of rings and vector spaces. For more choice principles specifically related to rings we refer the reader to Howard and Rubin [5, pp. 71-75].
• Krull's Theorem (FORM 1 CD in [5]): Every proper ideal in a commutative ring can be extended to a maximal ideal. Krull proved in [7] that every non-zero ring has a maximal ideal. Since he used explicitly the Well-Ordering Principle (FORM 1 CD in [5]) in his proof (see [7, p. 735 f]), one may ask how much of the Axiom of Choice we get a non-square is a non-square. If the characteristic is even, it is equal to 2 and then all elements of a finite field are squares. If R is an integral domain and −1 = 1, then x → x 2 is injective, and therefore a square root function exists. If R is an ordered field, e.g., R, one may always choose the larger of two possible solutions of x 2 = a 2 as square root of a 2 . In the case F = C one can use its structure of a Riemann surface to define a root by identifying an analytical principal branch of the root function. However, in general we need some form of the Axiom of Choice to define a square root function, and therefore it is not surprising that the existence of root functions in rings is related to some forms of choice.
In the present work we investigate the relationship between the following choice principles: Classical Choice principles.
• Axiom of Choice AC: Every family of non-empty sets has a choice function.
• Prime Ideal Theorem: Every ideal in a Boolean algebra can be extended to a prime ideal. • Axiom of Choice for Families of n-element Sets C n : Every family of nelement sets has a choice function.

New Choice Principles for Rings.
• nRR: Every ring has an nth-root function.
• nRID: Every integral domain has an nth-root function.
• nRF: Every field has an nth-root function.

Choice Principles for Families of n-Element Sets.
• Bounded Multiple Choice for Families of n-element Sets kC n (see Zuckerman [11]): If F = {Y λ : λ ∈ Λ} is a family of n-element sets and k is a positive integer, then from each Y λ ∈ F we can choose a non-empty set with at most k elements. • Cycle Choice for Families of n-element Sets cC n (new): If F = {Y λ : λ ∈ Λ} is a family of n-element sets, then on each set Y λ ∈ F we can choose a cyclic order.
In particular, we show the following relations: 1. For all n > 1, nRR ⇔ AC (Theorem 3.1) 2. 10. For two different primes p and q, qC p+q cC p+q ∨C p+q (Theorem 7.7). In particular, for two different primes p and q, qC p+q implies neither cC p+q nor C p+q (Corollary 7.8). In Sect. 2, we first give a construction of polynomial rings in arbitrarily many variables which does not use any non-trivial form of the Axiom of Choice. Then we give the definition of three choice principles for families of n-element sets, which are related to nth-root function in rings. In Sect. 3, we show that the Axiom of Choice is equivalent to the existence of square-root functions in rings. In Sect. 4 we investigate the relationship between nth-root functions in integral domains and some choice principles for families of n-element sets, and in Sect. 6 we show that one of these choice principles is equivalent to a weak form of the Prime Ideal Theorem. In Sect. 5 we investigate the relationship between two choice principles for families of n-element sets, and in Sect. 7 we prove an independence result. Sections 2-6 are self-contained, whereas Sect. 7 requires some basic knowledge in the construction of Fraenkel-Mostowski type permutation models of set theory with atoms.

Polynomial Rings
For the sake of completeness, we show that we do not need any non-trivial form of the Axiom of Choice in order to construct polynomial rings in arbitrarily many variables.
Let R be a ring and Λ a set. In the literature, polynomial rings in arbitrarily many variables {X λ : λ ∈ Λ} are usually defined very explicitly by first defining the set of monomials M as finitely supported functions Λ → N and then the set of polynomials as finitely supported functions M → R. In this work we choose a different approach. We assume the polynomial ring in one variable as given and define arbitrary polynomial rings as direct limits.

Direct Limit of Rings
We first need to establish the notions of a directed set and a direct system.

Definition.
A directed set is a set I together with a binary relation satisfying the three conditions 1. reflexivity: ∀i ∈ I : i i 2. transitivity: ∀i, j, k ∈ I : (i j) ∧ (j k) ⇒ (i k) 3. existence of an upper bound: ∀i, j ∈ I ∃k ∈ I : i, j k.
We will first define the direct limit of a direct system via its universal property and then give the explicit construction.
Definition. Let {R i : i ∈ I} be a direct system of rings over a directed set (I, ). A direct limit, denoted colim i∈I R i , of the direct system is a ring R together with, for every i ∈ I, a ring homomorphism f i : R i → R with the following properties.
For all j ∈ I with j i holds f i • f ji = f j . Moreover, R together with {f i : i ∈ I} is universal with respect to this property. This means that for every ring S and for all ring homomorphisms g i : It is straight forward to show that this actually defines a ring structure on colim i∈I R i . We omit the proof that this is indeed the direct limit of the direct system of rings.

Polynomial Rings in Arbitrarily Many Variables
Now we apply the direct limit of rings to define polynomial rings. Let R be a ring and Λ a set. Denote by X = {X λ : λ ∈ Λ} the set of variables.
The directed set. Let fin(X ) be the set of all finite subsets of X . For every S ∈ fin(X ) let Enum(S) be the set of all bijections {1, . . . , |S|} → S. Let so the elements of I are ordered pairs S, f with S ∈ fin(X ) and f ∈ Enum(S). It is not difficult to see that I, together with the relation is a directed set.  (1), which is also an element of S since S ⊆ S . In particular the extension is independent of f . Inductively, there exists a unique ring homomorphism mapping for all n ∈ {1, . . . , |S|} the element f (n) to f (n). It is obvious that this defines a direct system.

Definition.
The polynomial ring R[X ] is defined as the direct limit of the above direct system.
The polynomial ring satisfies the following universal property.
Universal property of the polynomial ring in arbitrarily many variables. Let S be a ring and let ϕ : R → S be a ring homomorphism. Then, for every map

Choice Principles for Families of n-Element Sets
Let n be a positive integer. If F = {Y λ : λ ∈ Λ} is such that for each λ ∈ Λ we have |Y λ | = n (i.e., Y is an n-element set), then F is called a family of n-element sets.
The Axiom of Choice for Families of n-element Sets, denoted C n , states that every family F = {Y λ : λ ∈ Λ} of n-element sets has a choice function, i.e., there is a function A weaker choice principle than C n we obtain by requiring that f (Y λ ) ⊆ Y λ is a non-empty set with at most k elements.
Definition. For positive integers k, n with k ≤ n, k-Bounded Multiple Choice for Families of n-element Sets, denoted kC n , states that if F = {Y λ : λ ∈ Λ} is a family of n-element sets, then there exists a function which chooses from each Y λ ∈ F a non-empty set with at most k elements.
Below, we shall only consider the case when k = 2, denoted 2C n . The fact summarizes two results which are used later. The first part is well-known in the case k = 1, but we see that the generalization to k ≥ 1 is almost immediate.
λ is an n-element set and the family F n = {Y k λ : λ ∈ Λ} is a family of n-element sets. By C n , F n has a choice function choosing subsets of F of size at most k, say f . Let g : F m → F m be defined by stipulating , i.e. we "forget" the second coordinate of the elements chosen by f . Since f is a choice function for F n , g is a choice function for F m .
In order to define cycle choice for families of n-element sets, we recall first the notion of a cyclic order: The triples (x, y, z) occurring in the following definition can be read and thought of as "after x comes y and then z".
One source of cyclic orders is the following. Assume that (S, <) is a totally ordered set. Then, we get a cyclic order by stipulating [x, y, z] if x < y < z and "closing" this relation under cyclicity.
There are two important concepts in the notion of cyclic orders, namely immediate successors and intervals.
Definition. Let S be a cyclically ordered set and let s ∈ S. The immediate successor of s is the unique element s + not equal to s such that there does not exist any element t with [x, t, x + ].
Immediate successors might not always exist, but they do if S is a finite set. They also can be used to define cyclic orders, because a cyclic order is uniquely determined by the immediate successors of every element.

Roots in Rings and the Axiom of Choice
In this section we consider the most classical of all choice axioms: the Axiom of Choice itself. The strategy we use to relate it to root functions in rings is the following. When constructing a choice function for a family of sets, we will use these sets as well as their elements as indefinite variables in a polynomial ring over some convenient ring, and then we will divide out an appropriate ideal to set a set and its elements in relation as well as getting rid of algebraic difficulties. Proof.
(1 ⇒ 2) Let R be a ring, let n ∈ N and denote as in the introduction

Denote the equivalence class of x by [x] and let
be the set of equivalence classes. Clearly F is a partition of R into pairwise disjoint non-empty sets. Hence, by AC there is a choice function f for F . Define now the nth root function n √ · : R (n) → R by stipulating For some index-set Λ, let F = {Y ι : ι ∈ Λ} be a family of pairwise disjoint non-empty sets. Denote by A = F the union over all sets in F .
It is easy to see that as a Z-module R is freely spanned by {1} ∪ A (1) ∪ A (2) ∪ · · · ∪ A (n−1) ∪ F , and thus, as a Z-module, where A (k) and F are the free Z-modules over A (k) and F , respectively. For Y ∈ F denote by π ∪Y the projection and by π 1 the projection By nRR, there is an nth-root function n √ · : R (n) → R. Let us identify the elements of F and F with the corresponding elements in F ⊆ R (n) and A (1) respectively. To define the function g : First observe that since π 1 is a ring homomorphism (identifying Z1 with Z) and π 1 (Y ) = 0, π 1 (r) n = π 1 (r n ) = 0, and since π 1 (r) ∈ Z, we obtain π 1 (r) = 0. We can write r = s +t, where s = π ∪Y (r) andt = r − s ∈ ker(π ∪Y ) ∩ ker(π 1 ). In particular, An easy computation shows that as all terms of the form s ktn−k belong to the ideal I, except when k = n. Writing Thus m i=1 α n i = 1. We consider two cases: Case 2 : n is odd. Then either m = 1 and α 1 = 1 or there exist some i, j such that α i > 0 and α j < 0. In this case, set It is clear that in both cases g(Y ) satisfies the requirements of AC'.
(4 ⇒ 1) Let F be a collection of non-empty sets. Consider the collection of all their subsets F = Y ∈F P(Y ). By AC', there is a function g : F → Note that for every Y ∈ F the intersection n∈N g n (Y ) is a singleton since g(Y ) is finite and the sequence {g n (Y )} n∈N is strictly decreasing until it stabilizes to a singleton. It is clear that the function f : As we can see in the proof, we do not need root functions on all rings to get AC. It is enough to have root functions on all rings of characteristic 0. Note however that the ring R we constructed has an abundance of zero divisors. This is unavoidable, as we will see later: If we allow only integral domains for defining square root functions, we get much weaker choice principles than AC. Proof. Let F be an arbitrary field and denote as before F (n) := {y ∈ F : ∃x(x n = y)}. Recall that for any y in F, the polynomial X n − y has at most n zeros in F. More precisely, for all y ∈ F (n) with y = 0 the cardinality of the set W y := {x ∈ F : x n = y} equals the number of nth roots of unity in F. Recall as well that if we write n = p r · k, where p is the characteristic of F and p r the highest power of p dividing n, then the number of nth roots of unity in a splitting field of F is precisely k. In particular, it divides n. It is clear that the nth roots of unity of F form a subgroup of the nth roots of unity in any splitting field. So, by Lagrange's Theorem, this number and therefore also the cardinality |W y | divides n.

Relationships Between n-Element
However, by Fact 2.1.(a), for any k | n, C n ⇒ C k . Hence, define and apply C k to F to get an nth root function on F (n) \{0}. By choosing 0 as nth root of 0, we get an nth root function on F (n) .

Proposition 4.2. Every field has an nth-root function if and only if every in-
tegral domain has an nth root function. In short nRF ⇔ nRID.
In particular, C n ⇒ nRID.
Proof. (⇒) Let R be an arbitrary integral domain. Consider F := Quot(R), the quotient field of R. Let y = 0 be an element of Recall that k := |W y | divides n and is independent of y. Let ζ be a primitive kth root of unity in F. Note that such a ζ exists, since for every y ∈ F (n) the set { x x : x, x ∈ W y } is a cyclic group of roots of unity of order k which is independent of y. Now we explain how ζ induces a cyclic order on W y . If n = 2 then it is the empty cyclic order. Otherwise, for every x ∈ W y the element ζx is the immediate successor of x. This means that for every With Proposition 4.1, we have an nth-root function n √ · on F (n) . Now, we can define the nth root of y in R (n) as the first element after n √ y that is

Small Values of n
In this section we'll see that for small values of n the converse of the implications above are true as well.
((ii) ⇒ (iii)) This implication is clear since every field is also an integral domain.
((iii) ⇒ (i)) Let F be a family of pairwise disjoint 2-element sets and denote the union by X := F . Furthermore, let K be a field of characteristic not equal to 2 and let F := Quot(K[X ]) be the field of rational functions with variables in X and coefficients in K. As in the introduction denote F (2) Notice that S ⊆ F (2) . Also notice that the square roots of x 2 − 2xx +x 2 are precisely x −x andx − x, and these two elements are different because the characteristic of F is not equal to 2. By assumption there exists a square root function sq : F (2) → F for F. Now, we define a choice function f for F by stipulating is a primitive third root of unity. For every Y ι ∈ F consider the set The set S ι has 6 elements and is closed under multiplication by ζ. Note that for every p ∈ R holds p 3 = (ζp) 3 . Consequently the set {p 3 : p ∈ S ι } has only two elements. Denote them by p ι and q ι . With 3RID we can choose a third root 3 √ p ι of p ι and a third root 3 √ q ι of q ι . Note that 3 √ p ι and 3 √ q ι are elements of S ι . Let x, y ∈ Y ι be the variables of 3 √ p ι and 3 √ q ι which occur with coefficient 1. Now we get a choice function for F as follows. If x = y we can choose x ∈ Y ι . Otherwise we choose the unique element in Y ι \{x, y}. Proof. We prove 2C m ⇒ mRF and then use Proposition 4.2.

Root Functions in Integral Domains, Bounded Choice and Cyclic Choice
Let F be an arbitrary field. Denote as always F (m) := {y ∈ F : ∃x ∈ F(x m = y)}. For y ∈ F (m) denote W y := {x ∈ F : x m = y}, i.e., the set of mth roots of y. Then there is l | m such that for every non-zero element y ∈ F (m) we have |W y | = l. By Fact 2.1 the principle 2C l holds, and we get a function f from F := {W y ⊆ F : y ∈ F (m) ∧ y = 0} to the power set of F with the following properties: If, for a given y in F (m) , the set f (W y ) has only one element, choose that element as the mth root of y. If f (W y ) has two elements x 1 and x 2 , let k := m−1 2 and observe that Hence this element is invariant under permutation of x 1 and x 2 . In addition so we can choose this element as a root of y. Proof. This follows immediately from Proposition 4.5 and Corollary 7.8. Consider the set Note that S ι has cardinality n and is closed under multiplication with ζ. Moreover for every element p ∈ S ι holds (ζp) n = p n , so all elements of S ι have the same nth power p. With nRID we can choose an nth-root n √ p of p. Our choice function chooses the unique element x ∈ Y ι that in n √ p occurs with coefficient 1.

Relations Between Cycle Choice and the Axiom of Choice for Families of n-Element Sets
In this section we leave rings aside and investigate the relation between two weak choice principles, Cycle Choice and the Axiom of Choice for Families of n-element sets.
be a family of k-element sets. Let A = {a 0 , a 1 , . . . , a n−k } be a (n − k + 1)element set with A ∩ Y λ = ∅ for all λ ∈ Λ. Since cC n+1 holds, for every λ ∈ Λ we can choose a cyclic order on Y λ ∪ A. We define now a choice function by choosing the first element in the cyclic order on Y λ ∪ A that comes after the element a 0 , i.e., the unique element y ∈ Y λ such that there does not exist any y ∈ Y λ with [a 0 , y , y].
(⇐) Let n ∈ N and assume that C k holds for every 1 ≤ k ≤ n. Let F = {Y λ : λ ∈ Λ} be a family of (n + 1)-element sets. For every 1 ≤ k ≤ n let be a choice function. Let λ ∈ Λ. We define a directed graph on Y λ by putting a directed edge from x ∈ Y λ to y ∈ Y λ if and only if f n (Y λ \ {x}) = y. Note that every vertex has precisely one outgoing edge, and in total there are as many edges as vertices. Now there are three cases to investigate.
Case 0 The graph on Y λ is a cycle graph.
In this case we are done, we simply define the cyclic order by stipulating that y is the immediate successor of x if there is a directed edge from x to y.
In the other two cases, we will see that we can actually define a choice function on F . We will first do that and then treat the cases together.
Case 1 There is a vertex which is not the endpoint of any edge.
Since the set of such vertices A without incoming edge cannot be all of Y λ , we can choose x 0,λ := f |A| (A).

Case 2 Every vertex has an incoming edge, but the graph is no cycle graph.
In this case the graph on Y λ is a union of disjoint cycles C 0 , C 1 , . . . , C l for an l ≥ 1. Now because |C i | ≤ k we can choose a vertex c i from every cycle via c i := f |Ci| (C i ), and then we can choose out of those {0, 1, . . . , l}}) .
In Cases 1 and 2 we can now recursively define a total order on Y λ by declaring the immediate successor of , x 1,λ , . . . , x i,λ }). Recall that every total order induces a cyclic order, and we are done.

Cyclic Choice for Primes and a Weak Form of the Prime Ideal Theorem
Corollary 5.2 means that for composite numbers n, the axiom cC n is much stronger than it seems on first sight. The following proposition stands a bit in contrast to that. For prime numbers p we show that cC p is equivalent to being able to extend just certain ideals in certain rings to prime ideals, which looks like a very weak assertion.
Proof. Assume first the weak version of the Prime Ideal Theorem. Let K = C. Let F = {Y λ : λ ∈ Λ} be a collection of sets each of which has precisely p elements. Let p be a prime ideal in the ring Let R := K[ F ∪ F ]/p and denote by π : K[ F ∪ F ] → R the usual projection. Since p is a prime ideal, the ring R is an integral domain. Let Y ∈ F . The equation T p = π(Y ) in R by construction has p pairwise distinct solutions, namely π(x) for every x ∈ Y . Recall that the number of zeroes of a polynomial in an integral domain is at most its degree. Let ζ ∈ C be a primitive pth root of unity. Observe that for every x ∈ Y the polynomial T p = π(Y ) also has the pairwise distinct solutions ζ k π(x) with k ∈ {0, . . . , p − 1}. Therefore, for every x, x ∈ Y there exists a k such that π(x) = ζ k π(x ). Now we define a cyclic order on Y by saying that x is the immediate successor of For the other direction, let K be a field with characteristic not equal to p. Let F = {Y λ : λ ∈ Λ} be a family of pairwise disjoint sets with p elements. Assume that each Y ∈ F is cyclically ordered.
Step 1: Let L be a splitting field of the polynomial T p − 1 over K. It suffices to consider L instead of K.
There is an obvious injection ι : Then p := ι −1 (p L ) will be a prime ideal in K[ F ∪ F ], because under ring homomorphisms it is always true that the inverse image of a prime ideal is a prime ideal. In addition, observe Step 2: Assume K = L.
Let ζ ∈ L be a primitive pth root of unity. For x ∈ F let x + be the immediate successor of x, i.e., the unique element not equal to x such that there does not exist any element x with [x, x , x + ]. We claim that has the desired properties. First we show that p is a prime ideal. Let a, b ∈ L[ F ∪ F ] be such that ab ∈ p. If we can show that either a or b is in p, then we are done by the definition of a prime ideal. The idea here is the following: The polynomial ring L[ F ∪ F ] is the direct limit of its polynomial subrings with finitely many variables. Likewise, both a and b only involve finitely many variables. In this way we reduce the situation to finitely many variables. Let F ⊂ F be a finite subset as follows. First we require that a, b ∈ L[ F ∪ F ]. Next, denote by S :  that α 1 , . . . , α n , c 1 , . . . , c l ∈ S ∈ L[ F ∪ F ]. Since F is finite, for every Y ∈ F we can choose an element x Y,1 . By considering it "minimal" the cyclic order yields a total order on Y .
. By the universal property of the polynomial ring there exists a unique homomorphism ϕ :

Also by considering L[ F ∪ F ] = L[F ]
[ F ] the map ϕ is just an evaluation homomorphism followed by the injection Y → Y p . From this we see that Note that by construction ab ∈ ker(ϕ). Also note that ϕ surjects on an integral domain, hence its kernel is a prime ideal. Therefore, a or b is in ker(ϕ), which is a subset of p, and we are done.
It remains to prove that for all x, x ∈ F with x = x the difference x − x is not in p. We argue by contradiction, i.e., we assume that Obviously p ⊂ ker(ψ), but still x / ∈ ker(ψ). If x / ∈ Y then we are already done since clearly x ∈ ker(ψ). If x ∈ Y , observe that since there exists a k ∈ {1, . . . , p−1} with x−ζ k x ∈ p we also have that x −ζ k x = (1−ζ k )x ∈ p. This implies that x ∈ p. But then x, x lie in ker(φ), which is a contradiction. Now we want to say a word about the case p = 2. Recall that cC 2 is true since for 2-element sets, the empty order is a cyclic order. Let Y ∈ F and denote Y = {x,x}. Since Y − x 2 and Y −x 2 are in the ideal, so is x 2 −x 2 , which is equal to (x−x)(x+x). If these elements are supposed to lie in a prime ideal not containing x −x, then it must contain x +x. Now we can check that can be extended to a prime ideal.

A Consistency Result
In this section, we show that for two different primes p and q, qC p+q does neither imply cC p+q nor C p+q . We will do this by showing that the statement qC p+q ∧ ¬cC p+q ∧ ¬C p+q is consistent with the axioms of Zermelo-Fraenkel Set Theory (denoted ZF). In fact, we will show that there exists a model of ZF in which qC p+q holds, but both statements cC p+q and C p+q fail. To show this we construct a permutation model V p+q in which we have ¬cC p+q ∧¬C p+q ∧qC p+q for all primes p = q. Recall that in order to establish the corresponding independence result in ZF, by Pincus [9] it suffices to construct a permutation model of ZFA in which qC p+q holds but cC p+q and C p+q fail -where ZFA is a model of Zermelo-Fraenkel set theory with atoms (see Halbeisen [2,Chapter 8]). For the sake of completeness, we give a short introduction to permutation models, which are models ZF with a set of atoms A, denoted ZFA. An atom is a set which does not contain any element, but which is distinct from the empty set ∅. The development of the theory ZFA is very much the same as that of ZF. Similar to the cumulative hierarchy of sets in ZF, we define by induction on the class of ordinals Ω the sets Now, the underlying idea of permutation models, which are models of ZFA, is the fact that the axioms of ZFA do not distinguish between the atoms, and so a permutation of the set of atoms induces an automorphism of the universe.
Let A be a set of atoms and let M = α∈Ω M α be a model of ZFA. Furthermore, in M, let G be a group of permutations (or automorphisms) of A. We say that a set F of subgroups of G is a normal filter on G if for all subgroups H, K of G we have: For every set x ∈ M there is a least ordinal α (in fact a successor ordinal) such that x ∈ P α (A). So, by induction on the ordinals, for every φ ∈ G and for every set x ∈ M we can define φ x by stipulating Notice that for all x, y ∈ M and every φ ∈ G we have φ x = y ⇐⇒ x = φ −1 y and x ∈ y ⇐⇒ φ x ∈ φ y. For x ∈ M, the symmetry group of x, denoted sym G (x), is the group of all permutations in G which map x to x, in other words A set x is said to be symmetric (with respect to F) if the symmetry group of x belongs to F, i.e., sym G (x) ∈ F. By (E) we have that every atom a ∈ A is symmetric. A set x is called hereditarily symmetric if x as well as each element of its transitive closure (i.e., the smallest transitive set which contains x) is symmetric. Notice that for all x ∈ M and every φ ∈ G, x is hereditarily symmetric if and only if φ x is hereditarily symmetric. This follows from (D).
Let V ⊆ M be the class of all hereditarily symmetric sets. Then V is a transitive model of ZFA and we call V a permutation model. Because A, as well as every a ∈ A, is symmetric, we get that the set of atoms A belongs to V.
Because ∅ is hereditarily symmetric and for all ordinals α the set P α (∅) is hereditarily symmetric too, the kernel V is a subclass of V. In particular, by induction on α one easily verifies the following Fact 7.1. For any set x ∈ V and any φ ∈ G we have φ x = x.
Since the atoms a ∈ A do not contain any elements, but are distinct from the empty set, the permutation models are not models of ZF. However, one can embed arbitrarily large fragments of a permutation model into a well-founded model of ZF.
For each set S ⊆ A, let fix G (S) = {φ ∈ G : φa = a for all a ∈ S} and let F be the filter on G generated by the subgroups fix G (E) : E ∈ fin(A) , where fin(A) is the family of all subsets of A which have finitely many elements. Then F is a normal filter. Furthermore, a set x is symmetric if and only if there exists a set of atoms E x ∈ fin(A) such that where E x is called a support of x. Notice that if E x is a support of x and E x ⊆ F x ∈ fin(A), then F x is a support of x as well. Now, fix an arbitrary prime p and an arbitrary prime q = p. We start with a ground model M p+q of ZFA + AC with a set of atoms where N is the set of natural numbers, and the sets P i and Q j are called blocks. Blocks are pairwise disjoint finite sets with |P i | = p and |Q j | = q. So we have that 1. for all i, j ∈ N, The group G of permutations of A is defined as follows: 3. For every i ∈ N,σ i is the cyclic permutation of P i with a i,0 → a i,1 → · · · → a i,p−1 → a i,0 . In other words,σ i is the permutation of P i defined by stipulatingσ If we replace the elements of P i with the elements of Z/pZ, thenσ i becomes addition with 1 (modulo p) andσ t i becomes addition with t (modulo p). Now, for each i ∈ N we extend the permutationσ i of P i to a permutation σ i of A by stipulating Similarly, for every j ∈ N, letτ j be the permutation of Q j defined by stipulatingτ and like forσ i , for each j ∈ N we extend the permutationτ j of Q j to a permutation τ j of A by stipulating otherwise. Now, we define the group G of permutations of A by requiring Let F be the normal filter on G generated by the subgroups where E ∈ fin(A), and let V p+q be the class of all hereditarily symmetric sets. Then, as mentioned above, V p+q is a permutation model. Below we shall prove that V p+q |= ¬cC p+q ∧ ¬C p+q ∧ qC p+q . For this, we first prove that cC p+q and C p+q fail in V p+q .
Then F q is an infinite family of q-element sets, F p is an infinite family of p-element sets, and F p+q is an infinite family of (p + q)element sets.
First of all note that F q , F p , F p+q are sets in V p+q , because ∅ is a support for all these sets. Now we show that in V p+q , neither of F q , F p , F p+q has a choice function (i.e., C q , C p , C p+q fail in V p+q ), and finally we show that also cC p+q fails in V p+q .
Assume towards a contradiction that there exists a function f : Since E f is finite, such a j 0 ∈ N exists. Without loss of generality assume that f (Q j0 ) = b j0,0 and consider the permutation τ j0 , which belongs to fix G (E f ).
Since τ j0 ∈ fix G (E f ), we must have τ j0 (f ) = f . On the one hand, we have This shows that there is no support of a choice function f : F q → F q , or in other word, there is no choice function for F q in V p+q , and since F q is a family of q-element sets, this implies that C q fails in V p+q .
With similar arguments, using the families F p and F p+q , we can show that also C p and C p+q fail in V p+q . Now, let us assume towards a contradiction that V p+q |= cC p+q . So there exists a function ζ in V p+q which assigns to every Y ∈ F p+q a cyclic order ζ Y such that for all x ∈ Y , Let E ζ ∈ fin(A) be a support of ζ. Let j 0 ∈ N be such that Q j0 ∩ E ζ = ∅ and let Y 0 ∈ F p+q be such that Q j0 ⊆ Y 0 . Since E ζ is a support of ζ and therefore Let a := ζ Y0 (b j0,0 ). If a ∈ Q j0 we have that by (2) This is a contradiction and therefore we have that a / ∈ Q j0 . Let 2 ≤ k 0 ≤ p + 1 be minimal such that . So, the exponent k 1 is not minimal, which is a contradiction. Hence, there is no function in V p+q which defines a cyclic order on each Y ∈ F p+q , which shows that cC p+q fails in V p+q .
Before we can show that V p+q |= qC p+q , we have to prove a few facts. Proof. Notice first that for any x, x ∈ S, the relation is an equivalence relation. Now, take an arbitrary x ∈ S. Since E is a support of S, for each φ ∈ fix G (E) we have φ x ∈ S. In particular, In order to show that there is a well-ordering of x ∈ S}. Notice also that since α is in the kernel V, by Fact 7.1, for every β ∈ α and for each φ ∈ G we have φ β = β. Now, since for all φ ∈ fix G (E) and every This shows that E is a support of w, and hence, w ∈ V p+q . 14 Page 24 of 33 L. Halbeisen et al.

Results Math
Below we show that every x ∈ V p+q has a unique least closed support. For this, we introduce first the notion of closed supports.
For a finite set E ∈ fin(A) we say that E is closed, if for all i, j ∈ N, In other words, E ∈ fin(A) is closed if and only if E is the union of finitely many blocks P i and Q j . Notice that the filter generated by the groups fix G (E * ), where E * is closed, is the same as F . Therefore, every set x ∈ V p+q has a closed support. We can even prove that every set has a smallest closed support. This smallest closed support will be called canonical support.
Furthermore, let σ 2 , τ 2 ∈ fix G (E x ) be such that and τ 2 | Ex\E x = τ | Ex\E x and τ 2 is the identity on A\(E x \E x ) .
Then στ = σ 2 τ 2 σ 1 τ 1 . So we can write φ as a product of permutations of fix G (E x ) ∪ fix G (E x ). Hence, since φ ∈ fix G (E x ∩ E x ) was arbitrary, this shows that E x ∩ E x is a support of x. Now, let S ∈ V p+q be an arbitrary set and let E S be a support of S. Look at the function which assigns to every set x ∈ S the set E ∈ fin(A) : E is a closed support of x .
Then E S is a support of this function and therefore, the function is in V p+q . In order to see this, we have to show that if E is a support of x, φ(E) is a support of φ(x) for every φ ∈ fix G (E S ). Let ψ ∈ fix G (φ(E)). For every e ∈ E we have that ψ • φ(e) = φ(e). So, . Since E is a support of x, we have that Therefore, ψ ∈ sym G (φ(x)). So, φ(E) is indeed a support of φ(x).

and since
x and x have the same canonical support E, Since G is a commutative group, we have that , and since φ| A\E ∈ fix G (E), this shows that which completes the proof.
In order to prove the main result of this section, we have to define an order on the set of closed supports. For this, we define first a total order on the set of blocks {P i : i ∈ N} ∪ {Q j : j ∈ N}.
Let A and B be two distinct elements of {P i : i ∈ N} ∪ {Q j : j ∈ N}. We define If A, B ∈ {P i : i ∈ N} ∪ {Q j : j ∈ N}, then, by the definition of the group G, for any φ ∈ G we have A < B ⇐⇒ φA < φB .
Hence, the relation "<" belongs to V p+q . For all non-empty setsỹ ⊆ y −1 , all 0 ≤ r < u + v and all natural numbers n define λ r,n (ỹ) as follows: Let λ r,0 (ỹ) := ∅ and for every n ∈ N \ {0} let In other words, λ r,1 (ỹ) is the set of all x ∈ỹ such that χ r (x,ỹ) is -minimal, λ r,2 (ỹ) is the set of all x ∈ỹ such that χ r (x,ỹ) is the second smallest set with respect to and so on. Note that the union of all λ r,n (ỹ) is equal toỹ. Note that n∈ω λ r,n (ỹ) =ỹ and only finitely many λ r,n (ỹ) are non-empty.
In the last step, we choose at most q elements from Y , which is done by induction on r. For each 0 ≤ r < u + v we define a non-empty set y r ⊆ y −1 with q |y r | and show that y u+v−1 contains at most q elements. Notice that q |y −1 |.
In this case, let l := |y r−1 |. Notice that q l and that by definition of χ r (x, y r−1 ) we have q < |y r−1 |. So, there exists a unique natural number m ≤ q and pairwise disjoint sets Z 1 , . . . , Z m ⊆ y r−1 with m i=1 |Z i | = l, such that for every 1 ≤ i ≤ m, all x, x ∈ Z i the ≤G-smallest function φ ∈G with φ(x) = x satisfies φ| r = 0.
If there is a set Z i (for some 1 ≤ i ≤ m) such that |Z i | = 1, we define Then, since m ≤ q < |y r−1 |, W 0 is a proper subset of the set of all Z i 's. Otherwise let Then W 0 is a proper subset of the set of all Z i 's. If |W 0 | = 1, let y r be the unique set which belongs to W 0 . Now, if |W 0 | > 1, then for each 1 ≤ i, j ≤ m such that Z i , Z j ∈ W 0 define d i := min{φ| r : there is a j ∈ {1, . . . , m} \ {i} such that for all z i ∈ Z i there is a z j ∈ Z j such that φ is ≤G -minimal with φ(z i ) = z j }.
Define d := min d i : 1 ≤ i ≤ m ∧ Z i ∈ W 0 . Then the set is a proper non-empty subset of W 0 . If |W 1 | > 1, then we repeat this process, starting with the set W 1 instead of W 0 , and obtain a non-empty set W 2 with |W 2 | < |W 1 |. After repeating this process at most m times, we finally obtain a set W = {Z i0 } which contains a unique set Z i0 (for some i 0 ∈ {1, . . . , m}) with q Z i0 and we define y r := Z i0 . Case 3. 0 ≤ r < u and for all x ∈ y r−1 , |χ r (x, y r−1 )| = p.
In this case, first notice that since |χ r (x, y r−1 )| = p we have p ≤ |y r−1 |. Now, if p ≤ |y r−1 | < p + q, then the set y −1 \ y r−1 contains at most q elements and we are done. So, we can assume that |y r−1 | = p + q, and we can argue similar as in Case 2. Finally, we consider y u+v−1 , which is a subset of Y , and show that y u+v−1 contains at most q elements. For this, let {x 0 , x 1 , . . . , x q } ⊆ y u+v−1 .
We show that at least two of the x i 's are equal, i.e., y u+v−1 does not contain a (q + 1)-element subset. Let φ ∈G be ≤G-minimal such that there is a k ∈ {1, . . . , q} with φ(x 0 ) = x k .
By construction of y u+v−1 , for each r with 0 ≤ r < u + v we have φ| r = 0. Therefore, φ = τ l k jv for an l k ∈ {1, . . . , q − 1}. Moreover, for each i ∈ {1, . . . , q} there is an l i ∈ {1, . . . , q − 1} with τ li jv (x 0 ) = x i . Since there are more i's than l i 's, there must be distinct i, i ∈ {1, . . . , q} with The following theorem summarizes the preceding results. which shows that the statement ¬cC p+q ∧¬C p+q ∧¬C p ∧¬C q ∧qC p+q is relatively consistent with ZF.
As an immediate consequence we get Corollary 7.8. If p and q are two different primes, then qC p+q implies neither cC p+q nor C p+q .
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