DEFINABLE TYPES IN PRESBURGER ARITHMETIC

DEFINABLE TYPES IN PRESBURGER ARITHMETIC GABRIEL CONANT Abstract. We consider the first order theory of (Z, +, <), also known as Presburger arithmetic. We prove a characterization of definable types in terms of prime models over realizations, which has a similar flavor to the Marker-Steinhorn Theorem of o-minimality. We also prove that a type over a model is definable if and only if it has a unique coheir to any elementary extension, which is a characterization of definable types that Presburger arithmetic shares with stable theories and o-minimal theories. 1. Presburger Arithmetic Let L = {+,, 0, 1, <} and let T P r = Th(Z, +,, 0, 1, <). Note that all integers in Z are definable from + and <. We specify the following definable relations: (1) For n Z +, a unary predicate P n defining the elements divisible by n, i.e., T P r = x(p n (x) y ny = x). (2) For n Z +, a binary relation n defining congruence mod n, i.e. T P r = x y(x n y P n (x y)). Note that nx is shorthand for x +... + x (n times). We will also use n x to mean P n (x). Theorem 1.1. 2] T P r has quantifier elimination in the language L {P n : n Z + }. See 2, Section 3.1] for more details on T P r, including a full axiomatization in L {P n : n Z + }. An important property of T P r, which is shown in 1] and follows easily from quantifier elimination, is that it is quasi-o-minimal, i.e., for any M = T P r, any definable subset of M (in one dimension) is a Boolean combination of -definable sets and intervals with endpoints in M. For the rest of this section, let M = T P r. Definition 1.2. Let a, b M with a < b. For n Z +, set b a = n if M =! n 1 x a < x < b. If there is no such n, set b a =. Set b a = 0 for a = b; and let a b = b a. Date: March 26, 2013. 1

2 GABRIEL CONANT Note that if a, b M then a b < if and only if b a Z. Moreover, if a, b M such that a < b and b a =, then a + k < b k for all k N. Definition 1.3. Given A M, let A be the subgroup of M generated by A. Let cl(a) be the divisible hull of A {1}, i.e., x cl(a) if and only if there are n Z +, k 1,..., k m, l Z and a 1,..., a m A such that nx = k 1 a 1 +... + k n a n + l. Theorem 1.4. Let A M. Then dcl(a) = acl(a) = cl(a). Proof. Let A 1 = A {1} and suppose w acl(a). By quantifier elimination, w is one of only finitely many solutions to a formula ϕ(x) of the form i I 0 m i x = a i i I 1 b i < m i x < c i k j (n j x + d j ) k j (n j x + d j ), j J 0 j J 1 } {{ } θ(x) where a i, d j A 1, b i A 1 {- }, c i A 1 { }, m i, n j Z, and k j Z +. If I 0 then w cl(a), so assume I 0 =. Set k = j J 0 J 1 k j. We have k j (n j w + b j ) for all j J 0 and k j (n j w + b j ) for all j J 1. It follows that for all m Z, k j (n j (w + mk) + b j ) for all j J 0 and k j (n j (w + mk) + b j ) for all j J 1. Therefore {w + mk : m Z} is an infinite set of solutions to θ(x). For any i I 1, if b i m i w < then m i w = b i + r for some r Z and so m i w A 1, which implies w cl(a). Similarly, if c i m i w < then x cl(a). So assume, towards a contradiction, that b i m i x = = c i m i x for all i I 1, and so b i < m i (w + rk) < c i for all r Z. Then {w + rk : k Z} is an infinite set of solutions to ϕ(x), which is a contradiction. Now suppose a cl(a). Then we have ma A 1 for some m Z +, and so ma is an L(A)-term. It follows that a is the unique solution to the L(A)-formula mx = ma, so a dcl(a). Altogether, we have shown acl(a) cl(a) dcl(a), which proves the result. Proposition 1.5. Any nonempty definable subset of M with a lower bound has a least element. Proof. Any nonempty subset of Z with a lower bound has a least element. So for any ϕ(x, ȳ) L, we can express this property with a sentence in T P r. In particular, Z = ȳ ( xϕ(x, ȳ) t x(ϕ(x, ȳ) t x) ) t ( ϕ(t, ȳ) x(ϕ(x, ȳ) t x) )]. Therefore, for any b M, we have the result for ϕ(m, b) by elementarity. Corollary 1.6. T P r has definable Skolem functions.

Proof. Suppose ϕ(x, ȳ) is an L-formula. DEFINABLE TYPES IN PRESBURGER ARITHMETIC 3 For any b, if ϕ(m, b) is bounded above then it has a greatest element, and otherwise ϕ(m, b) has a least positive element. So we can define the following Skolem function: f(ȳ) = x ϕ(x, ȳ) z > x ϕ(z, ȳ) ( z t > z ϕ(t, ȳ) x > 0 z ( 0 < z < x ϕ(z, ȳ) ))] From the last corollary, it follows that the definable closure of any set is a model of T P r. In particular, if M N and ā N, then we use the notation M(ā) for cl(mā). 2. Irreducible Types and Cuts Definition 2.1. A type p S n (M) is reducible if there is some 1 i n and ψ( x) L(M) such that p is irreducible if it is not reducible. p ψ( x)!xψ(x 1,..., x i 1, x, x i+1,..., x n ). Note that p S n (M) is reducible if and only if for some (equivalently all) realizations ā = p, there is some 1 i n such that a i M(a 1... a i 1 a i+1... a n ). Definition 2.2. (1) Let Z n := Z n \{ 0}. Given variables x 1,..., x n and k = (k 1,..., k n ) Z n we define the L-term s k( x) := k 1 x 1 +... + k n x n. (2) Given C M, C is downwards closed if c C and a < c implies a C, for all a, c M. C is finitely upwards closed if c C implies c + m C, for all m Z + and c M. C is a cut if it is downwards closed and finitely upwards closed. Note that if C is a cut then C has no maximal element and M\C has no minimal element. (3) Fix p S n (M). Given 1 i n, let σ i : Z + N such that for all m Z +, we have σ i (m) < m and p x i m σ i (m). Set σ p = (σ 1,..., σ n ). Next, given k Z n, define C k p := {c M : p s k( x) > c}.

4 GABRIEL CONANT Lemma 2.3. Suppose p S n (M) is irreducible. Then for all k Z n, C k p is a cut. Proof. Fix k Z n. Then C k p is clearly downwards closed. If C k p is not finitely upwards closed then it follows that p s k( x) = c for some c M. By assumption, there is some k i 0. Therefore if ā = p then a i M(a 1... a i 1 a i+1... a n ), contradicting that p is irreducible. Proposition 2.4. Suppose p, q S n (M) are irreducible such that σ p = σ q and C k p = C k q for all k Z n. Then p = q. Proof. By quantifier elimination, it suffices to show that p and q agree on atomic L(M)-formulas in variables x 1,..., x n. These are of the form (i) s k( x) = c, (ii) s k( x) > c, (iii) P m (s k( x) + c), where k Z n, c M, and m Z +. Since p and q are irreducible, it follows by Lemma 2.3 that both p and q must contain the negation of all formulas in (i). Moreover, p and q agree on all formulas in (ii) since C k p = C k q for all k Z n. Let ā = p and b = q in some elementary extension of M. Let σ p = (σ 1,..., σ n ) and σ q = (τ 1,..., τ n ). Then σ i = τ i for all 1 i n by assumption. Therefore, for all 1 i n and for all m Z +, we have a i m σ i (m) = τ i (m) m b i. By properties of modular arithmetic, it follows that for all k Z n, c M, and m Z +, we have s k(ā)+c m s k( b) + c. Therefore p P m (s k( x) + c) if and only if q P m (s k( x) + c). 3. Definable Types Definition 3.1. Let p S n (A) and B A. Then p is definable over B if for all L-formulas ϕ( x, ȳ), there is an L(B)-formula d p ϕ](ȳ) such that for all ā A, ϕ( x, ā) p = d p ϕ](ā). A type p S n (A) is definable if it is definable over B for some B A. Definition 3.2. Suppose p S n (M) is reducible. Pick i minimal such that a i M(a 1... a i 1 a i+1... a n ) for any ā = p, and define p red := p x1...x i 1x i+1...x n S n 1 (M).

DEFINABLE TYPES IN PRESBURGER ARITHMETIC 5 Proposition 3.3. Let p S n (M) be a reducible type (here T can be any complete theory). (a) p is definable if and only if p red is definable. (b) For all N M and q S n 1 (N ) extending p red, there is a unique q S n (N ) such that q p q. Proof. Part (a). If p is definable then p red is definable by definition. Conversely, suppose p red is definable. Without loss of generality, suppose ψ( x, z) L and c M are such that ψ( x, c)!xψ(x 1,..., x n 1, x, c) p, and so p red = p x1...x n 1. Fix ϕ( x, ȳ) L. Then for b M, we have ϕ( x, b) p x(ϕ(x 1,..., x n 1, x, b) ψ(x 1,..., x n 1, x, c)) p red M = d pred x(ϕ(x1,..., x n 1, x, ȳ) ψ(x 1,..., x n 1, x, z)) ] ( b, c). Therefore p is definable via d p ϕ( x, ȳ)](ȳ) := d pred x(ϕ(x1,..., x n 1, x, ȳ) ψ(x 1,..., x n 1, x, z)) ] (ȳ, c) L(M). Part (b). Fix N M and q S n 1 (N ) extending p red. Suppose q 1, q 2 S n (N ) are such that q p q 1 and q p q 2. Let ψ( x) = ψ( x, c) L(M) be as in part (a). Let ā i = q i, for i = 1, 2, in some elementary extension M of N. Then ā i := (a i 1,..., a i n 1) = q for i = 1, 2, so there is some σ Aut(M/N ) such that σ(ā 1 ) = ā 2. By assumption, q i ψ( x)!xψ(x 1,..., x n 1, x) for i = 1, 2. Therefore ψ(a 1 1,..., a 1 n 1, M) = {a 1 n} and ψ(a 2 1,..., a 2 n 1, M) = {a 2 n} But M = ψ(a 2 1,..., a 2 n 1, σ(a 1 n)), and so σ(a 1 n) = a 2 n. Therefore σ(ā 1 ) = ā 2, which implies q 1 = tp(ā 1 /N ) = tp(ā 2 /N ) = q 2. Definition 3.4. Let M = T P r and N M. Then N is an end extension of M if for all x N \M, either x > c for all c M, or x < c for all c M. A type p S n (M) is an end type if M(ā) is a proper end extension of M for some (i.e., all) ā = p. Proposition 3.5. Any irreducible end type is definable. Proof. By quantifier elimination it suffices to give definitions for the atomic formulas s k( x) = t(ȳ), s k( x) > t(ȳ) and P l (s k( x) + t(ȳ)), where t(ȳ) is an L-term, k Z n, and l Z +. Fix ā = p. Since p is irreducible, we have by Lemma 2.3 that k Z n implies s k(ā) M. But M(ā) is a proper end extension of M, which implies that for all k Z n, we either have s k(ā) < M or s k(ā) > M. Therefore we can partition Z n into S + S where S + := { k : s k(ā) > M} and S := { k : s k(ā) < M}.

6 GABRIEL CONANT Given k Z n and l Z +, let m( k, l) {0,..., l 1} such that s k(ā) l m( k, l). It follows that for any k Z n, any l Z +, and any ā = p, we have (1) s k(ā ) > M if k S + ; (2) s k(ā ) < M if k S ; (3) s k(ā ) l m( k, l). Therefore we can use the following definitions for p: d p s k( x) = t(ȳ) ] (ȳ) := y 1 y 1 d p s k( x) > t(ȳ) ] y 1 = y 1 if k S + (ȳ) := y 1 y 1 if k S d p Pl (s k( x) + t(ȳ)) ] (ȳ) := P l (m( k, l) + t(ȳ)) We now have all the pieces necessary to prove, for Presburger Arithmetic, a version of the Marker- Steinhorn Theorem in o-minimality (see 3]). Given a M and m Z +, we define a m = a ā m M, where ā is the unique integer in {0,..., m 1} such that a m ā. Then a m cl(a) and ( x ( x )) M = x m x y < y x < my. m m Theorem 3.6. Let p S n (M). The following are equivalent: (i) p is definable; (ii) for all ā = p, M(ā) is an end extension of M (i.e., either p is realized or an end type). Proof. First suppose ā = p, with M(ā) not an end extension of M. Then there is some b M(ā)\M and m 1, m 2 M such that m 1 < b < m 2. By definition, there is some r Z + such that rb M, ā. So let m 3 M and k Z n such that rb = m 3 + s k(ā). Note that s k(ā) M, so, with d 1 = m1 m3 r M and d 2 = m2 m3 r M, we have d 1 < s k(ā) < d 2. Let X = {m M : p s k( x) < m}. If p is definable, then X is a definable subset of M. Note that X is bounded below by d 1, and X since d 2 X. Therefore X has a least element d, by Proposition 1.5. It follows that d 1 < c < d, which is a contradiction. Therefore p is not definable. Conversely, suppose that for all ā = p, M(ā) is an end extension of M. If p is realized in M then it is clearly definable, so we may assume p is an end type. We prove p is definable by induction on n. The base case, as well as the induction step for irreducible p, follows from Proposition 3.5. So suppose p is reducible.

DEFINABLE TYPES IN PRESBURGER ARITHMETIC 7 Then p red is still either realized in M or an end type, and therefore definable by induction. So p is definable by Proposition 3.3. Corollary 3.7. Suppose M = T P r and N M. Then every type over M realized in N is definable if and only if every 1-type over M realized in N is definable. Proof. If every 1-type over M realized in N is definable, then N is an end extension of M. 4. Coheirs Definition 4.1. Let T be a theory and M = T. Given p S n (M), N M, and M A N, a type q S n (A) is a coheir of p to A if p q and q is finitely satisfiable in M, i.e., for every ϕ( x, ā) q there is some c M such that N = ϕ( c, ā). A type q S n (A) is an heir of p to A if p q and every formula represented in q is represented in p, i.e., for any ϕ( x, ā) q there is some c M such that ϕ( x, c) p. The next two results hold in any complete theory T. Theorem 4.2. 4] Suppose M = T and p S n (M). (a) p is definable if and only if p has a unique heir to any N M. (b) Suppose T is stable. Then p is definable. Moreover, if N M and M A N, then the unique heir of p to A is also the unique coheir of p to A. Proposition 4.3. Let M = T. Fix p S n (M), N M, and M A N. Suppose q is a partial n-type over A, which is finitely satisfiable in M and extends p. Then there is a complete type q S n (A), which extends q and is a coheir of p. Proof. Let Σ = {ϕ( x) L(A) : c M, N = ϕ( c)}. We first show that Σ q is consistent. Since Σ is closed under conjunctions, it suffices to show that Σ 0 := {ϕ( x)} {ψ 1 ( x),..., ψ k ( x)} is consistent, where ϕ( x) Σ and ψ i ( x) q, for 1 i k. But by assumption, there is c M such that N = k i=1 ψ i( x). By definition of Σ, it follows that c realizes Σ 0. Let q S n (A) be a complete type extending Σ q. For any ϕ( x) L(A), if ϕ( x) has no solution in M then ϕ( x) Σ q. Therefore every formula in q has a solution in M, and so q is a coheir of p. We now return to M = T P r. Lemma 4.4. If p S n (M) is an irreducible end type, and N M, then p has a unique coheir to N.

8 GABRIEL CONANT Proof. Let S + = { k Z n : s k( x) > 0 p} and S = { k Z n : s k( x) < 0 p}. By Lemma 2.3, and since p is an end type, it follows that for all k Z n, C k if k S + p = M if k S Fix N M. We define two cuts in N : C + M := {c N : b M, c < b} and C M := {c N : b M, c < b}. In particular, C + M is the downwards closure of M in N, and C M is the cut of elements in N that are less than every element of M. We partition N = D D D +, where D = C M, D = C+ M (N \C M ), and D+ = N \C + M. Note that D < D < D +, and also that D and D D are both cuts in N. Moreover, we have M D, and for all x in D there are c 1, c 2 M with c 1 < x < c 2. Let q := p {c < x i < d : 1 i n, c D, d D + }. If ϕ(x, b) p, with b M, then there is some ā M such that M = ϕ(ā, b). Clearly then, c < a i < d for any 1 i n, c D, and d D +. Therefore q is finitely satisfiable in M, and so can be extended to a coheir q S n (N ) of p by Proposition 4.3. Claim 1 : σ q = σ p and, for k Z n, C k C + M q = if k S + if k S C M Proof : We clearly have σ p = σ q since p q. Let ā = q in some elementary extension of N. First suppose k S +. We want to show D < s k(ā) < D +. But s k(ā) > D since k S + implies s k(ā) > M. Suppose, towards a contradiction, that s k(ā) > c for some c D +. Let k = (k 1,..., k n ) and set k j a j := max{k i a i : 1 i n}. Then c < s k(ā) nk j a j. Case 1 : k j > 0. Then a j > M and a j > c nk j, so there is some b M such that c nk j < b. It follows that c < nk j b, which is a contradiction since nk j b M and c D +. Case 2 : k j < 0. Then a j < M and a j < c nk j, so there is some b M such that b < c nk j. Again, this implies nk j b < c, which is a contradiction. By a similar proof, we have D < s k(ā) < D for all k S. Suppose r S n (N ) is a coheir of p to N. Claim 2 : r is irreducible. Proof : Suppose not. Without loss of generality, there is a realization ā = r such that a n N (a 1,..., a n 1). Then there is m Z +, d N, and k 1,..., k n 1 Z such that ma n = d + k 1 a 1 +... + k n 1 a n 1. With k = (-k 1,..., -k n 1, m), we have r s k( x) = d. Since r is a coheir, there is some b M such that s k( b) = d, and so d M. But this implies s k( x) = d p, contradicting that k 0 and p is irreducible.

DEFINABLE TYPES IN PRESBURGER ARITHMETIC 9 Now suppose, towards a contradiction, that q r. By Claim 1, Claim 2, and Proposition 2.4, it follows that there is some k Z n such that k S + and C k r C + M, or k S and C k r C M. Case 1 : k S + and C k r C + M. Since p r and C k p = M, it follows that C + M C k r. Then there is some d D + such that r s k( x) > d. But this formula cannot be satisfied in M, contradicting that r is a coheir. Case 2 : k S and C k r C M. Then C k r C M and we obtain d D such that r s k( x) < d, which again contradicts that r is a coheir. From these contradictions, we have shown that q = r, and so q is the unique coheir of p to N. We can now prove, for Presburger arithmetic, another characterization of definable types, which is true in any stable theory by Theorem 4.2, and also true in any o-minimal theory (see 3]). Theorem 4.5. Let M = T P r and p S n (M). Then p is definable if and only if p has a unique coheir to any N M. Proof. Suppose p is definable. By Theorem 3.6, p is either realized in M or is an end type. Note that any realized type will have a unique extension to any N M, which must be a coheir since it is fully satisfied in M. Also, if p is irreducible then the result follows by Lemma 4.4. So we may assume p is reducible. We prove, by induction on n, that p has a unique coheir to any N M. The base case is trivial since reducible 1-types are realized. So assume the result for types in less than n variables, and fix N M. Then p red is still either realized or an end type and so has a unique coheir to N by induction. If q is a coheir of p to N, then q red is the unique coheir of p red. Therefore p has a unique coheir to N by Proposition 3.3. Conversely, suppose p is not definable. Let ā be a realization of p in some elementary extension N M. By Theorem 3.6, M(ā) is not an end extension of M. Therefore there is some k Z n such that C k p M. Let C = C k p. Define X := {b N : c < b < d for all c C, d M\C}. Note that X since s k(ā) X. We define two partial types over N : q 1 := p {c < s k( x) < b : c C, b X} and q 2 := p {b < s k( x) < d : b X, d M\C}. Claim: For i {1, 2}, q i is finitely satisfiable in M. Proof : First consider q 1. Fix ψ( x) p and c C. Let A = ψ(m). Considering s k : M n M as a definable function, we have that B := s k(a) is definable (over M). Note that N = x B, x > c and so M = x B, x > c. Then {x B : x > c} is definable, nonempty, and bounded below, and so has a least element c by Proposition 1.5. Then M = x((x B x > c) x > c ), and so N satisfies this sentence. In particular c < s k(ā), and so c C. But c s k(a), and so we have shown that for all c C there is some d M such that M = ψ( d) and c < s k( d) < M\C. It follows that q 1 is finitely satisfiable in M. The

10 GABRIEL CONANT proof for q 2 is similar. By the claim and Proposition 4.3, q 1 and q 2 extend to two distinct coheirs of p to N. References 1] O. Belegradek, Y. Peterzil & F. Wagner, Quasi-o-minimal structures, Journal of Symbolic Logic 65 (2000) 1115-1132. 2] D. Marker, Model Theory: An Introduction, Springer, 2002. 3] D. Marker & C. Steinhorn, Definable types in o-minimal theories, The Journal of Symbolic Logic 59 (1994), no. 1, 185-198. 4] A. Pillay, An Introduction to Stability Theory, Dover, 2008. Department of Mathematics, Statistics and Computer Science, University of Illinois at Chicago, 851 S Morgan St, 322 SEO, Chicago, IL, 60607, USA; E-mail address: gconan2@uic.edu