We prove that for every Scott set $S$ there are $S$-saturated real closed
fields and models of Presburger arithmetic

Dolich, Alf

Knight, Julia

Lange, Karen

Marker, David

English

arXiv

We prove that for every Scott set S there are S-saturated real closed fields and S-saturated models of Presburger arithmetic

Knight, Julia F.

Wellesley College

Wellesley College
Wellesley College Digital Scholarship and Archive
Faculty Research and Scholarship
8-2015
Representing Scott Sets in Algebraic Settings
Alf Dolich
Julia F. Knight
Karen Lange
klange2@wellesley.edu
David Marker
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Recommended Citation
Representing Scott sets in algebraic settings, A. Dolich, J. Knight, K. Lange, and D. Marker, Archive for Mathematical Logic, 54(5-6)
(2015), 631-637.
Representing Scott Sets in Algebraic Settings∗
Alf Dolich
Kingsborough Community College
Julia F. Knight
University of Notre Dame
Karen Lange †
Wellesley College
David Marker
University of Illinois at Chicago
1 Introduction
Recall that S ⊆ 2ω is called a Scott set if and only if:
i) S is a Turing ideal, i.e., if x, y ∈ S and z ≤T x⊕ y, then z ∈ S, where
x⊕ y is the disjoint union of x and y;
ii) If T ⊆ 2<ω is an infinite tree computable in some element of S, then
there is f ∈ S an infinite path through T .
Scott sets first arose in the study of completions of Peano arithmetic
(PA) and models of PA. Scott [9] shows that the countable Scott sets are
exactly the families of sets “representable” in a completion of PA. If M is a
nonstandard model of Peano arithmetic and a ∈M, let
r(a) = {n ∈ ω :M |= pn|a}
where p0, p1, . . . is an increasing enumeration of the standard primes. The
standard system of M
SS(M) = {r(a) : a ∈M}
∗This work was begun at a workshop on computable stability theory held at the Amer-
ican Institute of Mathematics in August 2013.
†Partially supported by National Science Foundation grant DMS-1100604
1
is a Scott set. A longstanding and vexing problem in the study of models
of arithmetic is whether every Scott set arises as the standard system of a
model of Peano arithmetic. The best result is from Knight and Nadel [6].
Proposition 1.1 If S is a Scott set and |S| ≤ ℵ1, then there is a model of
Peano Arithmetic with standard system S.
Thus the Scott set problem has a positive solution if the Continuum Hypo-
thesis is true, but the question remains open without additional assumptions.
Later in this section, we sketch a proof of Proposition 1.1.
Scott sets also are important when studying recursively saturated struc-
tures. We assume that we are working in a computable language L. We fix a
Go¨del coding of L and say that a set of L-formulas is in S if the corresponding
set of Go¨del codes is in S.
Let T be a complete L-theory. The following ideas were introduced in
[5], [10] and [7].
Definition 1.2 Let S ⊆ 2ω. We say that a modelM of T is S-saturated if:
i) every type p ∈ Sn(∅) realized in M is computable in some element of
S;
ii) if p(x, y) ∈ Sn+1(∅) is computable in some element of S , a ∈Mn and
p(x, a) is finitely satisfiable in M, then p(x, a) is realized in M.
If a model is S-saturated for some S ⊆ 2ω, then the model is certainly
recursively saturated.
Proposition 1.3 If M |= T is recursively saturated, then M is S-saturated
for some Scott set S.
If the theory T has limited coding power, then we can say little about S.
For example, an algebraically closed field of infinite transcendence degree will
be S-saturated for every Scott set S. On the other hand, the associated Scott
set is unique for many natural examples, such as Peano arithmetic, divisible
ordered abelian groups, real closed fields, Z-groups (models of Th(Z,+)) and
Presburger arithmetic (models of Th(Z,+, <)).
Definition 1.4 We say that a theory T is effectively perfect if there is a tree
(φσ : σ ∈ 2<ω) of formulas in n-free variables computable in T such that:
i) T + ∃v φσ(v) is consistent for all σ;
ii) if σ ⊂ τ , then T |= φτ (v)→ φσ(v);
iii) φσ ̂ 0(v) ∧ φσ ̂ 1(v) is inconsistent with T for all σ.
2
Proposition 1.5 If T is effectively perfect, then every recursively saturated
model of T is S-saturated for a unique Scott set S.
The theories we will be considering are all effectively perfect. For Peano
arithmetic, Presburger arithmetic and Z-groups we can use the formulas pn|v
to find such a tree. For real closed fields we can use q < v for q ∈ Q and for
ordered divisible abelian groups we can use the binary formulas mv < nw
for m,n ∈ Z.
We now sketch a proof of Proposition 1.1. We first note that for models
M of Peano arithmetic, M is recursively saturated if and only if M is S-
saturated where S is the standard system of M. (For more details see [4]).
Lemma 1.6 If S is a countable Scott set and T ∈ S is a completion of Peano
arithmetic, then there is an S-saturated model of T .
Proof Sketch Build M by a Henkin construction. At any stage, we will
have a finite tuple a and will be committed to tp(a) the complete type of a,
where T ⊆ tp(a) ∈ S. At alternating stages, we either witness an existential
quantifier or realize a type p(v, a) ∈ S, using the join property of Scott sets
to compute p(v, x) ∪ tp(a), and using the tree property to find completions.

Lemma 1.7 Suppose S0 ⊂ S1 are countable Scott sets, T ∈ S0 is a comple-
tion of Peano arithmetic, and M0 and M1 are countable recursively satu-
rated models of T , where Si is the standard system of Mi. Then there is an
elementary embedding of M0 into M1.
Proof Sketch Let a0, a1, . . . be a list of the elements ofM0. Suppose we have
a partial elementary map (a0, . . . , an) 7→ (b0, . . . , bn). If tp(an+1, a0, . . . , an) =
p(v, a0, . . . , an), there is b ∈M1 realizing p(v, b0, . . . , bn), and we can extend
the embedding. Box
We can now prove Proposition 1.1. Suppose |S| = ℵ1 and S is the union
of an ω1-chain of countable Scott sets
S0 ⊆ S1 ⊆ . . . ⊆ Sα ⊆ . . .
where Sα =
⋃
β<α Sβ when α is a limit ordinal. We can build an elementary
chain (Mα : α < ω1) where Mα is recursively saturated with standard
3
system Sα. Then
⋃
α<ω1
Mα is recursively saturated with standard system
S.
While we have nothing new to say about the Scott set problem for Peano
arithmetic, we show that the analgous problem for recursively saturated mod-
els has a positive solution in some related algebraic settings.
For divisible ordered abelian groups, this follows easily from an unpub-
lished result of Harnik and Ressayre. Let (G,+, <) be a divisible ordered
abelian group. Define an equivalence relation on G by g ≡ h if and only
if there is a natural number n such that |g| < n|h| and |h| < n|g|. Let
Γ = {|g|/ ≡ : g ∈ G}, the set of equivalence classes of positive elements. The
ordering of G induces an ordering on Γ. Suppose S is a Scott set and kS is
the set of real numbers computable in some element of S (where we identify
a real with its cut in the rationals). It is easy to see that kS is a real closed
field.
Theorem 1.8 (Harnik–Ressayre) A divisible ordered abelian group G is
S-saturated if and only if Γ is a dense linear order without endpoints and
each equivalence class under ≡ is isomorphic to the ordered additive group
of kS.
A complete proof is given in [3].
Corollary 1.9 For any Scott set S, there is an S-saturated divisible ordered
abelian group.
Proof Let G be the set of functions f : Q→ kS such that {q ∈ Q : f(q) 6= 0}
is finite. We add elements of G coordinatewise and order G lexicographically.
By the Harnik–Ressayre Theorem, G is S-saturated. 
2 Real Closed Fields
In [2] D’Aquino, Knight and Starchenko show that if M is a nonstandard
model of Peano arithmetic with standard system S, then the real closure of
the fraction field ofM is an S-saturated real closed field. Thus, it is natural
to ask whether we can find an S-saturated real closed field for every Scott
set S.
Theorem 2.1 For any Scott set S, there is an S-saturated real closed field.
4
The value group of an S-saturated real closed field will be an S-saturated
divisible ordered abelian group. Thus, Corollary 1.9 will also follow from
Theorem 2.1.
Theorem 2.1 is a simple induction using the following Lemma.
Lemma 2.2 Let S be a Scott set. Let K be a real closed field such that
every type realized in K is in S. Suppose p(v, w) is a set of formulas in S,
a ∈ K, and p(v, a) is finitely satisfiable in K. Then we can realize p(v, a) by
a (possibly new) element b such that every type realized in K(b)rcl is in S.
Proof The set of formulas p(v, a) ∪ tp(a) is a consistent partial type in S,
and, hence, has a completion in S. Thus, without loss of generality, we may
assume p(v, a) is a complete type. If p(v, a) is realized in K, then there is
nothing to do. If p(v, a) is not realized in K then it determines a cut in the
ordering of Q(a)rcl that is not realized in K, and, hence, by o-minimality, it
determines a unique type over K. Let b realize p(v, a) and let c ∈ K. We
need to show that tp(b, c) is in S.
How do we decide whether K(b)rcl |= φ(b, c)? By o-minimality, φ(v, c)
defines a finite union of points and intervals with endpoints in Q(c)rcl. Since
b 6∈ K, b is neither one of the distinguished points nor an end point of of one
of the intervals. There are ∅-definable Skolem functions f and g such that:
i) f(a) < v < g(a) ∈ tp(b/a);
ii) f(a) < v < g(a)→ φ(v, c) or f(a) < v < g(a)→ ¬φ(v, c);
Given tp(b/a) and tp(a, c) we can computably search and find the decom-
position of φ(v, c) and f and g as above. We can then decide whether φ(b, c)
holds. Since S is closed under join and Turing reducibility, tp(b, c) is in S.

The above argument works for any o-minimal theory T ∈ S.
Every real closed field K has a natural valuation for which the valuation
ring is
O = {x : |x| < n for some n ∈ N}.
If K is recursively saturated, then the value group is a recursively saturated
divisible ordered abelian group. It is natural to ask if every recursively sat-
urated divisible ordered abelian group arises this way.
D’Aquino, Kuhlmann and Lange [3] gave a valuation-theoretic character-
ization of recursively saturated real closed fields. In the following argument
we assume familiarity with their results.1
1This is essentially our original proof of Theorem 2.1.
5
Proposition 2.3 Let G be a recursively saturated divisible ordered abelian
group. There is a recursively saturated real closed field with value group G.
Proof Sketch Let S be the Scott set of G. Start with the field
kS(t
g : g ∈ G)rcl.
This is a real closed field with residue field kS, value group G and all types
recursive in S. Given K a real closed field K with value group G and all
types recursive in S and suppose we have a ∈ K and (f0, f1, . . .) a sequence of
Skolem functions recursive in S such that (f0(a), f1(a), . . .) is pseudo-Cauchy.
If the sequence has no pseudo-limit in K it determines a unique type over K.
Adding a realization b does not change the value group or residue field. As
above, every type realized in K(b)rcl is in S. We can iterate this construction
to build the desired real closed field. 
3 Presburger Arithmetic
In [6] Knight and Nadel proved that for every Scott set S there is an S-
saturated Z-group, i.e., an S-saturated model of Th(Z,+). They asked
whether the same is true for the theory of (Z,+, <). This is Presburger
arithmetic, which we denote Pr. We answer this question in the affirmative.
Theorem 3.1 For every Scott set S, there is an S-saturated model of Pres-
burger arithmetic.
We will consider Presburger arithmetic in the language that includes con-
stants for 0 and 1 and unary predicates Pn(v) for n = 2, 3, . . . that hold if
n divides v. We can eliminate quantifiers in this language and the resulting
structure is quasi-o-minimal; i.e., any formula φ(v, a) defines a finite Boolean
combination of ∅-definable sets and intervals with endpoints in dcl(a)∪{±∞}.
We will use this in the following form. (See, for example [8] §3.1 for quantifier
elimination and [1] for quasi-o-minimality.)
Lemma 3.2 i) Any formula φ(v, a) is equivalent over tp(a) to a Boolean
combination of formulas of the form v ≡ m mod n, v = α, v < β where α, β
are in the definable closure of a and m,n ∈ N .
ii) tp(b, a) is determined by:
6
• tp(a);
• the sequence b mod 2, b mod 3, b mod 4, . . .;
• the cut of b in the definable closure of a.
We obtain Theorem 3.1 by an iterated construction using the following
lemma.
Lemma 3.3 Let S be a Scott set. Let G |= Pr such that every type realized
in G is computable in S. Suppose a ∈ G and p(v, w) is a complete type in
S such that p(v, a) is finitely satisfiable. Then there is H ⊇ G such that
H |= Pr, such that p(v, a) is realized in G and every type realized in H is in
S.
Proof If p(v, a) is realized in G, then there is nothing to do, so we assume
p(v, a) is not realized in G. Let p−(v, a) be the partial type describing the
cut of v over the definable closure of a, i.e., p− consists of all formulas of the
form mv <
∑
niai or mv >
∑
niai that are in p where m,ni ∈ Z.
Case 1: Suppose p− is omitted in G.
Let b be any realization of p, and let H be the definable closure of G∪{b}.
It is enough to show that if c ∈ G, then tp(b, c) is in S. We will show
that tp(b, c) is recursive in tp(b, a) and tp(a, c). Using only tp(b, a), we can
determine b mod n for all n. Thus, we only need to consider formulas of the
form α < v < β where α, β ∈ dcl(c). Since p−(v, a) is omitted, we can, as in
the case of real closed fields, search to find γ, δ ∈ dcl(a) such that γ < b < δ
and either α ≤ γ < δ ≤ β, δ < α or γ > β. This can be done recursively in
tp(b, a) and tp(a, c). Thus, every type realized in H is in S.
Case 2: Suppose b ∈ G realizes p−.
Let b̂ be a realization of p(v, a) and let q0(v) be the divisibility type of
b̂− b, i.e., if b̂ ≡ m mod n and b ≡ l mod n then “v ≡ m− l mod n” ∈ q0 for
l,m ∈ Z and n > 1.
Let q(v) ∈ S1(G) be the unique type containing
• q0(v);
• n < v for all n ∈ Z;
• v < g for all g ∈ G such that Z < g.
7
Let  realize q and let H be the definable closure of G ∪ {}. Suppose
α ∈ dcl(a) and b < α. Since G |= Pr, we have b+ n < α for all n ∈ Z. Thus,
 < α − b and b +  < α. Similarly, if α < b, then α < b +  and, thus, b + 
realizes p−(v, a). By the choice of q0, b+  realizes p(v, a).
Suppose c ∈ G. It suffices to show that tp(, c) is in S. Without loss
of generality, we may assume that c = (c1, . . . , cn) where all of the ci are
positive infinite and 1, c1, . . . , cn are linearly independent over Q. We need
to decide the signs of expressions of the form
r + s+
n∑
i=1
tici
where r, s, ti ∈ Z. Such an expression is positive if and only if
• ∑ tici > 0, or
• ∑ tici = 0 and q > 0, or
• ∑ tici = s = 0 and r > 0
This can be computed using tp(c). Thus tp(, c) is recursive in q0 and tp(c).
Hence, every type realized in H is in S. 
References
[1] O. Belegradek, Y. Peterzil and F. Wagner, Quasi-o-minimal structures,
J. Symbolic Logic 65 (2000), no. 3, 1115–1132.
[2] P. D’Aquino, J. Knight and S. Starchenko, Real closed fields and models
of Peano arithmetic, J. Symbolic Logic 75 (2010), no. 1, 111.
[3] P. D’Aquino, S. Kuhlmann and K. Lange, A valuation theoretic charac-
terization of recursively saturated real closed fields, J. Symbolic Logic,
to appear.
[4] R. Kaye, Models of Peano arithmetic, Oxford Logic Guides, 15. Oxford
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8
[6] J. Knight and M. Nadel, Models of arithmetic and closed ideals, J.
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[7] A. Macintyre and D. Marker, Degrees of recursively saturated models.
Trans. Amer. Math. Soc. 282 (1984), no. 2, 539–554.
[8] D. Marker, Model theory: an introduction, Graduate Texts in Mathe-
matics, 217. Springer-Verlag, New York, 2002.
[9] D. Scott, Algebras of sets binumerable in complete extensions of arith-
metic. 1962 Proc. Sympos. Pure Math., Vol. V pp. 117–121 American
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Representing Scott Sets in Algebraic Settings

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