Model Theory

Stephen G. Simpson

May 2, 1998

Department of Mathematics The Pennsylvania State University University Park, State College PA 16802

[email protected] www.math.psu.edu/simpson/courses/math563/

Note: Chapters 12 and 13 are not ﬁnished. 2 Contents

1 Sentences and models 7 1.1Symbols...... 7 1.2Formulas...... 8 1.3Structures...... 9 1.4Truth...... 10 1.5Modelsandtheories...... 10

2 Complete theories 13 2.1Deﬁnitionsandexamples...... 13 2.2Vaught’stest...... 15 2.3ApplicationsofVaught’stest...... 16

3 The compactness theorem 21 3.1Proofofthecompactnesstheorem...... 21 3.2Someapplicationstoﬁeldtheory...... 23 3.3 The L¨owenheim-Skolem-Tarskitheorem...... 24

4 Decidability 27 4.1Recursivelyaxiomatizabletheories...... 27 4.2Decidabletheories...... 30 4.3Decidablemodels...... 31

5Elementaryextensions 35 5.1Deﬁnitionandexamples...... 35 5.2Existenceofelementaryextensions...... 38 5.3Elementarymonomorphisms...... 40

3 4 CONTENTS

6 Algebraically closed ﬁelds 43 6.1Simpleﬁeldextensions...... 43 6.2Algebraicclosure...... 47 6.3Completenessandmodelcompleteness...... 49 6.4Hilbert’sNullstellensatz...... 52

7 Saturated models 55 7.1Elementtypes...... 55 7.2Saturatedmodels...... 58 7.3Existenceofsaturatedmodels...... 60 7.4Preservationtheorems...... 63

8 Elimination of quantiﬁers 71 8.1Themodelcompletionofatheory...... 71 8.2 Substructure completeness ...... 73 8.3Theroleofsimpleextensions...... 76

9 Real closed ordered fields 79 9.1Orderedfields...... 79 9.2Uniquenessofrealclosure...... 83 9.3 Quantifier elimination for RCOF ...... 86 9.4ThesolutionofHilbert’s17thproblem...... 89

10 Prime models (countable case) 93 10.1Theomittingtypestheorem...... 93 10.2Primemodels...... 96 10.3Thenumberofcountablemodels...... 101 10.4Decidableprimemodels...... 104

11 Differentially closed fields of characteristic 0 109 11.1Simpleextensions...... 109 11.2Differentiallyclosedfields...... 115 11.3Differentialclosure(countablecase)...... 117 11.4Ritt’sNullstellensatz...... 120

12 Totally transcendental theories 125 12.1 Stability ...... 125 12.2Rankofanelementtype...... 125 12.3Indiscernibles...... 125 CONTENTS 5

12.4Existenceofsaturatedmodels...... 125

13 Prime models (uncountable case) 127 13.1Stronglyatomicmodels...... 127 13.2Normalsets...... 127 13.3Uniquenessandcharacterizationofprimemodels...... 127 6 CONTENTS Chapter 1

Sentences and models

1.1 Symbols

1. We assume the availability of a large supply of nonlogical symbols of the following kinds: 1. n-ary relation symbols R( ,..., ), n 1; ≥ 2. n-ary operation symbols o( ,..., ), n 1; ≥ 3. constant symbols c. These collections of symbols are assumed to be disjoint.

2. We make use of the following logical symbols : 1. propositional connectives (negation), , (conjunction, disjunction), , (implication, biimplication);¬ ∧ ∨ → ↔ 2. quantiﬁers , (universal, existential); ∀ ∃ 3. equality =;

4. variables v0,v1,...,vn,.... Note that = is a logical symbol although syntactically it behaves as a binary relation symbol.

7 8 CHAPTER 1. SENTENCES AND MODELS 1.2 Formulas

1. The notion of a term is deﬁned inductively as follows. A constant symbol is a term. A variable is a term. If t1,...,tn are terms and o is an n-ary operation symbol, then o(t1,...,tn) is a term.

The notion of atomic formula is deﬁned as follows. If t1 and t2 are terms, then t1 = t2 is an atomic formula. If t1,...,tn are terms and R is an n-ary relation symbol, then R(t1,...,tn)isanatomicformula.

3. The notion of a formula is deﬁned inductively as follows. An atomic formula is a formula. If ϕ and ψ are formulas then so are ϕ, ϕ ψ, ϕ ψ, ϕ ψ, ϕ ψ.Ifϕ is a formula and v is a variable, then ¬vϕ and∧ vϕ are∨ formulas.→ ↔We assume familiarity with the concept of a∀free variable∃ , i.e. one not bound by a quantiﬁer. We assume unique readability of formulas.

4. If S is a set of formulas and/or terms, the signature of S is the set of all nonlogical symbols occurring in it. This is sometimes called in the literature the similarity type of S. Note that = never bolongs to the signature since it is a logical symbol. We write sig(S) = signature of S.

5. A sentence is a formula with no free variables.

Examples: The formula x y(x+y = 0) is a sentence. Here + is a binary operation symbol, 0 is a constant∀ ∃ symbol, and = is a logical symbol. The formula x + y = y + x is not a sentence. If we write x,y,... in the same formula, we tacitly assume that x,y,... are distinct variables. 1.3. STRUCTURES 9

Examples, continued: The formula x( y(y y = x) z(z z = x)) is a sentence. It is “logically equivalent” to∀ the∃ sentence· x∨∃y(y y· = x− y y = x) but these two sentences are not identical. We asume∀ ∃ that· the∨ student· −has some previous acquaintance with the syntactical and semantical notions of logical equivalence. These notions will be deﬁned later.

1.3 Structures

A structure is an ordered pair =( , Φ) where is a nonempty set, called the universe of , and ΦA is a function|A| whose domain|A| is a set of nonlogical symbols. The domainA of Φ is called the signature of .Toeachn-ary relation symbol R sig( ) we assume that Φ assigns an nA-ary relation ∈ A R n = . ⊆|A| |A|×···×|A| n times | {z } To each n-ary operation symbol o sig( ) we assume that Φ assigns an n-ary operation ∈ A

o : n . |A| →|A| To each constant symbol c sig( ) we assume that Φ assigns an individual constant c . ∈ A ∈|A|

Example: the structure of the reals

=( , +, , , 0, 1,<) R |R| − · where = R. We cannot include because it is not an operation on |R| ÷ |R| (because not everywhere deﬁned). Here the universe is = R =( , ); +, are binary operations; is a unary operation; 0, 1|R| are constants;−∞< ∞is a binary· relation. − 10 CHAPTER 1. SENTENCES AND MODELS 1.4 Truth

1. Given a structure and a sentence σ such that sig(σ) sig( ), we assume known the meaningA of ⊆ A

= σ ( satisﬁes σ, σ is true in ) . A| A A

For example, = x y(y y = x y y = x) expresses the fact that every real numberR| or its∀ negative∃ · is a square.∨ · Note− that the structure

=(Z, +, , , 0, 1,<) , Z · − where = Z = ..., 2, 1, 0, 1, 2,... , has the same signature as but satisﬁes|Z| the negation{ of− the− above sentence.} R In general, = σ means that σ is true in when the variables are interpreted as rangingA| over , the other symbolsA in σ being given their obvious interpretation. |A| Another example:

2 2 2 2 = x(x>0 y1 y2 y3 y4(x = y + y + y + y )) Z| ∀ →∃ ∃ ∃ ∃ 1 2 3 4 and this expresses the fact that every positive integer is the sum of four squares.

1.5 Models and theories

1. Let S be a set of sentences. A model of S is a structure such that = σ for all σ S,andsig( )=sig(S). M M| ∈ M 1 For example, a group can be described as a model =( , ,− , 1) of the axioms of group theory: G |G| · x y z((x y) z = x (y z)) ∀ ∀ ∀ · · · · x(x 1=1 x = x) ∀ · 1 · 1 x(x x− = x− x =1) ∀ · · 1.5. MODELS AND THEORIES 11

2. The class of all models of S is denoted Mod(S). A sentence τ is said to be a logical consequence1 of S (written S = τ)ifsig(τ) sig(S), and = τ for all Mod(S). | ⊆ M| M∈A theory is a set T of sentences which is consistent and closed under logical consequence; in other words, T has at least one model, and τ T whenever τ is a sentence such that sig(τ) sig(T )and = τ for all ∈ Mod(T ). For example, the theory of groups⊆ is theM| set of all logicalM∈ consequences of the axioms of group theory. These axioms have many nonobvious logical consequences, e.g. the Jacobi identity

((x, y),zx) ((y,z),xy) ((z, x),yz)=1 · · 1 1 y 1 whereweuseabbreviations(x, y)=x− y− x y and x = y− x y. · · · · · 3. A model class is a nonempty class of structures all having the same signature. An elementary model class is a model class of the form Mod(S)whereS is a consistent set of sentences. There are lots of nonelementary model classes, e.g. the class of ﬁnite groups. If K is a model class, we write Th(K)forthetheory of K, i.e. the set of sentences σ such that sig(σ) sig(K)and = σ for all K.Notethat Th(K) is a theory and for any⊆ theory T weM| have T = Th(Mod(M∈T )). There is a natural 1-1 correspondence between theories and elementary model classes.

4. If T is a theory and S T ,wesaythatS is a set of axioms for T if T =Th(Mod(S)). If there⊆ exists a finite set of axioms for T ,wesaythatT is finitely axiomatizable. For example, the theory of groups is finitely axiomatizable (a finite set of axioms for it is displayed above). We shall see later that the theory of fields of characteristic 0 is not finitely axiomatizable.

1Note: This deﬁnition is somewhat unusual because, for instance, x(x<0 x = 0 x>0) is not a logical consequence of x y(xy∀), owing∨ to the restriction∨ on signature. ∀ ∀ ∨ ∨ 12 CHAPTER 1. SENTENCES AND MODELS Chapter 2

Complete theories

2.1 Deﬁnitions and examples

1. AtheoryT is complete if for all sentences σ,eitherσ T or σ T , provided sig(σ) sig(T ). ∈ ¬ ∈ ⊆ Examples: The theory of groups is not complete. The theory of ﬁelds of characteristic 0 is not complete (e.g. x(x x = 1 + 1) is true in R,false ∃ · in Q). We shall see later that the theory of algebraically closed ﬁelds of characteristic 0 is complete.

2. Two structures and are elementarily equivalent (written )ifthey have the sameA signatureB and satisfy the same sentences. InA≡B other words, Th( )=Th( ). A B 3. Proposition. For a theory T the following are equivalent.

1. T is complete;

2. all models of T are elementarily equivalent;

13 14 CHAPTER 2. COMPLETE THEORIES

3. T =Th( ) for some structure . A A Proof. Trivial.

Exercise. Prove that the theory of dense linear ordreings without end points is complete. Use the method of elimination of quantiﬁers, described below.

axioms for linear orderings: x y z(x

We use the following notational convention: ϕ(x1,...,xn) denotes a formula ϕ whose free variables are among x1,...,xn. AtheoryT is said to admit elimination of quantiﬁers if for all formulas ϕ(x1,...,xn), n 1, there exists a quantiﬁer free formula ϕ∗(x1,...,xn) such that ≥

T = x1 ... xn(ϕ(x1,...,xn) ϕ (x1,...,xn)) . | ∀ ∀ ↔ ∗ For the above exercise, show that the theory of dense linear orderings without end points admits elimination of quantifiers. (Do this by induction on the number of quantifiers, working from the inside out.) Once this has been done, completeness follows easily. Historically, the method of elimination of quantifiers came very early. It is also the most versatile method for showing that algebraic theories are complete and/or decidable. The drawback is that the method is syntactical. We shall develop various model-theoretic methods for proving the same results. 2.2. VAUGHT’S TEST 15 2.2 Vaught’s test

Vaught’s test is another method for establishing completeness of theories. It is less versatile than quantiﬁer elimination, but much easier to use.

Two structures and are said to be isomorphic (written ∼= )if sig( )=sig()A and thereB exists an isomorphic map of ontoA B, i.e. i : A suchB that A B |A| → |B| 1. i is one-one and onto;

2. RA(a1,...,an) if and only if RB(a1,...,an);

3. i(oA(a1, ..., an)) = oB(i(a1),...,i(an));

4. i(cA)=cB.

Note that isomorphic structures are “essentially identical”. In particular they satisfy the same sentences, i.e. = implies . A ∼ B A≡B

The cardinality or power of a structure is the cardinality of its universe. For instance, we say that is countable Aif and only if is countable, etc. The power of is denotedA or card( ). |A| A kAk A

AtheoryT is said to be κ-categorical if

(i) T has at least one model of power κ;

(ii) any two such models are isomorphic. 16 CHAPTER 2. COMPLETE THEORIES

4. Theorem (Vaught’s test for completeness). Let T be a countable theory such that

(i) T has no ﬁnite models;

(ii) for some inﬁnite cardinal κ, T is κ-categorical.

Then T is complete.

We shall prove this later as an application of the L¨owenheim-Skolem- Tarski theorem. For now we content ourselves with giving various examples of how the theorem is used to prove completeness of speciﬁc theories.

2.3 Applications of Vaught’s test

1. We can use Vaught’s test to show that the theory of dense linear orderings without end points is complete. Clearly the theory has no ﬁnite models. We shall show that the theory is 0-categorical. Let =( ,

2. We use Vaught’s test to show that the theory of nontrivial torsion free divisible Abelian groups is complete. 2.3. APPLICATIONS OF VAUGHT’S TEST 17

Abelian groups: x y z((x + y)+z = x +(y + z)) ∀ ∀ ∀ x y(x + y = y + x) ∀ ∀ x(x +0=x) ∀ x(x +( x)=0) ∀ − An Abelian group is said to be divisible if it satisﬁes

x y(ny = x) ∀ ∃ for n =1, 2, 3,.... This is an inﬁnite set of axioms. Note that ny is an abbreviation for

y + + y . ··· n times An Abelian group is said to| be torsion{z } free if it satisfies x(nx =0 x =0) ∀ → for n =1, 2,.... This is again an infinite set of axioms. An Abelian group is nontrivial if it contains a nonzero element. Clearly any nontrivial torsion free Abelian group is infinite. Thus our theory has no finite models. To prove completeness, we shall apply Vaught’s test by showing that our theory is κ-categorical for any uncountable cardinal κ. An example of a torsion free divisible Abelian group is the additive group of the rationals (Q, +, , 0). More generally, if V is any vector space over the field of rational numbers,− then the additive group of V is torsion free and divisible. Fact: Any torsion free divisible Abelian group is of this form. Proof. Let =( , +, , 0) be a torsion free divisible Abelian group. A |A| − For a and r Q we want to define ra.Letr = m/n, m, n Z, n>0. Define∈|A| ∈ ∈

ra =someb with ma = nb .

We need to show that ra is well deﬁned. So suppose ma = nb1 = nb2.Then n(b1 b2) = 0. hence b1 b2 = 0 by torsion freeness of . The vector space − − A 18 CHAPTER 2. COMPLETE THEORIES axioms

(r1 + r2)a = r1a + r2a

r(a1 + a2)=ra1 + ra2 1a = a ( r)a = ra − − are easily veriﬁed. Thus is the additive group of a vector space over Q. A We shall need the following facts from the theory of vector spaces. (1) Any vector space V over a ﬁeld F has a basis, i.e. a set U V such that every element of V is uniquely expressible in the form ⊆

r1u1 + + rnun ,ri F, ··· ∈ where the ui are distinct elements of U,1 i n. (Proof: use Zorn’s lemma.) (2) Any two bases of V over F have≤ ≤ the same cardinality, the dimension of V .(3)IfV1 and V2 are vector spaces over F of the same dimension, then they are isomorphic over F . Let and be torsion free divisible Abelian groups of power κ where A B κ is uncountable. As vector spaces over Q they have bases U and U respectively. Put λ =card(U ). Since every element of is ofA the formB A |A| r1u1 + + rnun, ri Q, ui U ,wehave λ 0.Fromthiswe easily deduce··· card(U ∈)=λ =∈κ.A Similarly card(|A|U ≤)=·ℵκ. So by fact (3), A B ∼= . This shows that our theory is κ-categorical. Hence by Vaught’s test itA is complete.B

A corollary of the result we have just proved is that the Abelian groups (Q, +, , 0) and (R, +, , 0) are elementarily equivalent. Here Q = rationals, − − R =reals.

3. Remark. The theory of nontrivial torsion free divisible Abelian groups is not 0 categorical, since there are countable vector spaces over Q of dimensions ℵ1, 2, 3,.... The theory of dense linear orderings without end points is not κ-categorical, κ> 0. ℵ 2.3. APPLICATIONS OF VAUGHT’S TEST 19

Remark. Let Tp be the theory of algebraically closed ﬁelds of characterstic p,wherep is a prime or p = 0. We shall see later that Vaught’s test can be used to show that Tp is complete. Namely, Tp is κ-categorical for all κ> 0. ℵ The ﬁeld axioms are as follows:

axioms for a commutative ring: x y z((x + y)+z = x +(y + z)) ∀ ∀ ∀ x y(x + y = y + x) ∀ ∀ x(x +0=x) ∀ x(x +( x)=0) ∀ − x y z((x y) z = x (y z)) ∀ ∀ ∀ · · · · x y(x y = y x) ∀ ∀ · · x y z(x (y + z)=x y + x z) ∀ ∀ ∀ · · · 0 =1 6 ﬁeld axiom: x(x =0 y(x y =1) ∀ 6 →∃ · If we weaken the last axiom to

x y((x =0 y =0) x y =0) ∀ ∀ 6 ∧ 6 → · 6 we get the axioms for a domain (usually called integral domain). A ﬁeld (or domain) is said to be of characteristic p if p is the least n such that

n =1+ +1=0. ··· n times

Here p will have to be a prime| number{z (since} m n = 0 implies m =0or n = 0). If there is no such p, then we say that the· field is of characteristic 0. For example Q, R, C, Z are of characteristic 0. Afield =( , +, , 0, , 1) is said to be algebraically closed if it satisfies F |F| − · n x0 x1 ... xn(xn =0 y(xny + + x1y + x0 =0)) ∀ ∀ ∀ 6 →∃ ··· for n =1, 2, 3,.... For example, the fundamental theorem of algebra asserts that the complex field =(C, +, , 0, , 1) is algebraically closed. C − · 20 CHAPTER 2. COMPLETE THEORIES

5. Remark. A diﬃcult theorem of Morley says that if a theory T is countable and κ-categorical for some uncountable cardinal κ,thenitisκ-categorical for all uncountable cardinals κ. This shows that the class of theories to which Vaught’s test applies is rather limited. Many important complete theories are not κ-categorical for any κ. This limitation of Vaught’s test will be partly overcome by means of saturated models. (See theorem 7.3.6 below.) Chapter 3

The compactness theorem

In this chapter and in chapter 4 it will be convenient to assume that formulae have been deﬁned in terms of , , only; then , , , can be introduced by deﬁnition as usual. ¬ ∨ ∃ ∧ → ↔ ∀

3.1 Proof of the compactness theorem

1. AsetS of sentences is said to be consistent if it has a model. The following theorem is due to G¨odel 1929 in the countable case and Malcev 1936 in the uncountable case.

2. Theorem (compactness theorem). If every ﬁnite subset of S is consistent, then S is consistent.

Proof. We give a proof by transfinite induction in the style of Henkin 1949. Assume that S is finitely consistent, i.e. every finite subset of S has a model. Let κ =max( 0, card(S)). Thus κ is a cardinal; we identify cardinals with initial ordinals.ℵ Let c : γ<κ be new constant symbols, and put W = { γ } sig(S) c : γ<κ .Let σγ : γ<κ be an enumeration of all sentences σ ∪{ γ } { } with sig(σ) W . Fix a variable x and let ϕγ(x):γ<κ be an enumeration of all formulas⊆ ϕ with only free variable {x and sig(ϕ) } W . By induction ⊆ 21 22 CHAPTER 3. THE COMPACTNESS THEOREM

on γ define finitely consistent sets of sentences Sγ , γ κ, and a function h : κ κ as follows. ≤ → Stage 0: S0 = S. Stage 2γ +1:Leth(γ)betheleastβ such that cβ does not occur in S2γ or in ϕγ(x). Define

S2γ+1 = S2γ ( xϕγ (x)) ϕγ(c ) . ∪{ ∃ → h(γ) }

The sentence ( xϕγ (x)) ϕγ (c ) is called a Henkin sentence. We claim ∃ → h(γ) that S2γ+1 is finitely consistent (assuming S2γ was). Let be a model of M some finite subset of S2γ+1. We want to expand to a model of those sentences plus the Henkin sentence. First, interpretM in arbitrarily the |M| nonlogical symbols in ϕγ(x) which are not interpreted in .Calltheresult- M ing structure 0.Let 00 be the result of interpreting the Henkin constant M M ch(γ) so as to make the Henkin sentence true. Stage 2γ +2: Let S2γ+2 = S2γ+1 σγ if this is finitely consistent; ∪{ } otherwise S2γ+2 = S2γ+1 σγ . If the former is not fintely consistent then the latter is. ∪{¬ }

Stage δ κ, δ limit ordinal: Let Sδ = γ<δ Sγ. This is still ﬁnitely consistent. ≤ S We now build a model directly from Sκ.LetT be the set of all variable M free terms in Sκ. Deﬁne an equivalence relation on T by t1 t2 if and ≈ ≈ only if t1 = t2 Sκ.Put = T/ .If[t1],...,[tn] T/ and R is an n-ary relation symbol,∈ put |M| ≈ ∈ ≈

R([t1],...,[tn]) if and only if R(t1,...,tn) Sκ . ∈ If o is an n-ary operation symbol deﬁne

o([t1],...,[tn]) = [o(t1,...,tn)] .

For c a constant symbol put c =[c]. Our model =( , Φ) is defined by = T/ ,Φ(R)=R,Φ(o)=o,Φ(c)=c. M |M| |M|We claim≈ that, for all sentences σ with sig(σ) W , = σ if and ⊆ M| only if σ Sκ. We prove this by induction on the number of propositional ∈ connectives in σ. For atomic σ this holds by definition of .Forσ = σ1 σ2 M ∨ or σ = σ1 it is easy to check. Suppose σ = yϕ(y). If = σ then ¬ ∃ M| = ϕ(t)forsomet T . Hence by induction ϕ(t) Sκ. Therefore M| ∈ ∈ yϕ(y) Sκ by finite consistency. Conversely, suppose yϕ(y) Sκ.By ∃ ∈ ∃ ∈ 3.2. SOME APPLICATIONS TO FIELD THEORY 23

relabeling variables we may assume that y = x.Alsoϕ(x)=ϕγ(x)forsome γ<κ. Hence ( xϕ(x)) ϕ(c ) Sκ where β = h(γ). Hence ϕ(c ) Sκ ∃ → β ∈ β ∈ by ﬁnite consistency. Hence by induction = ϕ(cβ), hence = yϕ(y). This completes the proof. M| M| ∃

3.2 Some applications to ﬁeld theory

1. Theorem (A. Robinson). Suppose σ is a sentence which is true in all fields of characteristic 0. Then σ is true in all fields of characteristic p, for all but finitely many primes p (the “exceptional primes”).

Proof. If the conclusion fails, then σ is consistent with the ﬁeld axioms plus 1 + +1 = 0, for inﬁnitely¬ many primes p. But, for primes p, ··· p 1+ +1 = 0 implies 1 + +1 =0foralln

An aﬃne variety is a set X Cn such that ⊆

X = (z1,...,zn):fi(z1,...,zn)=0, 1 i k { ≤ ≤ } where fi,1 i k are polynomials in n variables with coeﬃcients from C. ≤ ≤ A mapping G : X Cn is said to be polynomial if there exist polynomials → gj,1 j n in n variables with coeﬃcients from C, such that for all ≤ ≤ (z1,...,zn) X, ∈

G(z1,...,zn)=(g1(z1,...,zn),...,gn(z1,...,zn)) . 24 CHAPTER 3. THE COMPACTNESS THEOREM

Theorem (Ax). Let G : X X be a polynomial map of an affine variety → in Cn into itself. If G is one-one, then G is onto. Proof. More generally, we shall prove that the theorem holds for an arbitrary algebraically closed field in place of C. Besides the compactness theorem, we shall use completeness of the theory Tp of algebraically closed fields of characteristic p where p is 0 or a prime. For prime p let Fp be the algebraic closure of the prime field Fp = Z/pZ. Note that any finite set of elements in Fp is contained in a finite subfield of Fp. Hence the theorem holds for Fp. Observe that the theorem holds for a particular field if and only if satisfies a certain collection of sentences. If one of these sentencesF σ is falseF in C, then by completeness of T0 we have T0 = σ. Hence Tp = σ for all | ¬ | ¬ sufficiently large primes p. Hence Fp = σ. This is a contradiction. | ¬ For more applications of the compactness theorem in field theory, see Robinson’s book and more recent literature. For applications to “local properties” in group theory, see Malcev’s book.

3.3 The L¨owenheim-Skolem-Tarski theorem

1. Note that the proof of the compactness theorem also established the following result: If S is (finitely) consistent then S has a model of power max( 0, card(S)). ≤ ℵ 2. Theorem (Löwenheim-Skolem-Tarski). Let S be a set of sentences which has an infinite model. Let κ be a cardinal max( 0, card(S)). Then S has amodelofpowerκ. ≥ ℵ

Proof. Let c : γ<κ be new constant symbols. Put { γ }

S∗ = S c = c : β<γ<κ . ∪{ β 6 γ }

Since S has an infinite model, S∗ is finitely consistent. Hence by the compactness theorem S∗ is consistent. Applying the remark above we find that 3.3. THE LOWENHEIM-SKOLEM-TARSKI¨ THEOREM 25

S∗ has a model of power max( 0, card(S∗)) = κ. This model must have power exactly κ. ≤ ℵ

3. Corollary (Vaught’s test). Let T be a theory which

1. has no ﬁnite models

2. is κ-categorical for some κ card(T ). ≥ Then T is complete.

Proof. Suppose not. Let , Mod(T ) such that .Bythe A B∈ A6≡B L¨owenheim-Skolem-Tarski theorem ﬁnd 0, 0 of power κ such that 0, A B A≡A 0.Then 0 0. Hence 0 = 0 so T is not κ-categorical. B≡B A 6≡ B A 6∼ B 26 CHAPTER 3. THE COMPACTNESS THEOREM Chapter 4

Decidability

4.1 Recursively axiomatizable theories

Let V be a fixed recursive1 signature. The formulas ϕ with sig(ϕ) V can be Gödel numbered in the usual way. This permits notions from recursion⊆ theory to be introduced. For instance, a set F of formulas is said to be recursive if its set of Gödel numbers ](ϕ):ϕ F is recursive, etc. { ∈ }

Deﬁnition. AtheoryT is said to be recursively axiomatizable if (i) sig(T ) is recursive, and (ii) T has a recursive set of axioms.

Remark. The notion of recursive axiomatizability is an important generalization of that of finite axiomatizability. Most theories which arise in prac- tice have finite signature and are recursively axiomatizable, e.g. the theory of groups, the theory of fields of characteristic 0, first order Peano arithmetic, ZF set theory. Examples of theories which are not recursively axiomatizable are: the theory of finite groups; Th( )where =(Z, +, , , 0, 1,<). Z Z − ·

1i.e. V is a recursive set, and we can recursively identify the elements of V as n-ary relation or operation symbols or constant symbols. This is the case if V is ﬁnite.

27 28 CHAPTER 4. DECIDABILITY

3. A very important general result about recursively axiomatizable theories is the following, which is a reformulation of Gödel’s completeness theorem. Theorem (essentially Gödel 1929). Let T be a recursively axiomatizable theory. Then T is recursively enumerable. This theorem would usually be proved, following Gödel, by showing that T can be generated from its axioms and a certain explicit recursive set of “logical axioms” by means of a certain explicit recursive set of “logical rules”. One would then apply Church’s thesis to conclude that T is recursively enumerable. Such a proof would give more or less information depending on the choice of logical axioms and rules. We shall give a “bare bones” proof which avoids these concepts entirely.

4. Deﬁnition. Asentenceτ is logically valid if = τ for all structures with sig( ) sig(τ). A| A A ⊆ 5. Lemma. Let V be a ﬁxed recursive signature. The set of all logically valid sentences τ with sig(τ) V is recursively enumerable. ⊆ Proof. WemaysafelyassumethatV contains no operation symbols. We may also assume that the only logical symbols are , , .(Thusweare dispensing with =.) ¬ ∨ ∃ Let c : n ω be a recursive set of new constant symbols and put { n ∈ } W = V c : n ω .Let ϕn(vi ):n ω be a recursive enumeration of ∪{ n ∈ } { n ∈ } all formulas ϕ(vi) with exactly one free variable and with sig(ϕ) W .We can choose2 our enumeration so that ⊆

sig(ϕn(vin )) V c0,...,c2n 1 . ⊆ ∪{ − }

We shall work with Henkin sentences ( vin ϕn(vin )) ϕ(c2n). Let S = σn : n ω be a recursive enumeration of all∃ sentences σ→with sig(σ) W . { ∈ } ⊆ 2This amounts to arranging things so that, in the proof of the compactness theorem 3.1, the function h is given by h(n)=2n. 4.1. RECURSIVELY AXIOMATIZABLE THEORIES 29

Given a sentence τ with sig(τ) V ,wehave ⊆ τ is logically valid τ has no model ⇔¬ τ has no countable model (L¨owenheim-Skolem theorem) ⇔¬ there is no structure such that = c : n ω and sig( )=W ⇔ A |A| { n ∈ } A and = τ and, for all n, =( vi ϕn(vi )) ϕn(c ) A| ¬ A| ∃ n n → 2n there is no function f : S 0, 1 such that for all n, f( σn)= ⇔ →{ } ¬ 1 f(σn), and for all m and n, f(σm σm)=max(f(σm),f(σn)) and − ∨ f( vi ϕn(vi )) = f(ϕn(c ) f(ϕn(c )). ∃ n n 2n ≥ m f : S 0, 1 nR(f Sn,τ)whereSn = σ0,...,σn 1 and R is a ﬁxed⇔∀ primitive→{ recursive}∃ relation { − }

N f : S 0, 1 ( n N) R(f Sn,τ)(byK¨onig’s lemma) ⇔∃ ∀ →{ } ∃ ≤ NP(N,τ)whereP is a ﬁxed primitive recursive relation. ⇔∃ This proves the lemma.

Remark. The set of logically valid sentences is not recursively enumerable (“Church’s theorem”).

6. Proofofthetheorem. Given a recursively axiomatizable theory T ,letA = τn : n ω be a recursive set of axioms for T .PutAn = τ0,...,τn 1 .If {σ is a sentence∈ } with sig(σ) sig(T ), we have { − } ⊆ σ T ∈ A = σ ⇔ | nAn = σ (compactness theorem) ⇔∃ | n 1 n − τi σ is logically valid . ⇔∃ i=0 → By the lemma, V the expression in square brackets is recursively enumerable. Hence so is T . 30 CHAPTER 4. DECIDABILITY

7. Exercise (Craig). Prove the converse of the above theorem: If a theory T with sig(T ) recursive is recursively enumerable, then it is recursively axiomatizable.

4.2 Decidable theories

1. Deﬁnition. AtheoryT is decidable if (i) sig(T ) is recursive, (ii) T is recursive.

Remark. Every decidable theory is recursively axiomatizable, but the converse fails, e.g. first order Peano arithmetic. there even exist finitely axiomatizable undecidable theories, e.g. Robinson’s Q, the theory of groups, the theory of fields.

2. Remark. The ﬁrst order theories which are important in algebra tend to be decidable. We make a short table.

Decidable Undecidable Abelian groups (finite) groups Boolean algebras (finite) distributive lattices linear orderings ∗algebraically closed fields fields of characteristic p ∗Th(R) = real closed fields fields of characteristic 0 finite fields ordered fields ∗differentially closed fields of char. 0 ordered Abelian groups Th(Qp), Qp = p-adic rationals

Note: A ∗ indicates that these decidability results will be proved later.

A useful suﬃcient condition for decidability is the following: 4.3. DECIDABLE MODELS 31

3. Theorem. Let T be a theory which is recursively aximatizable and complete. Then T is decidable.

Proof. By the previous theorem, T is recursively enumerable. Also its com- plement T = σ : σ is a sentence, sig(σ) sig(T ),σ/T { ⊆ ∈ } = σ : σ T (by completeness) { ¬ ∈ } is recursively enumerable. Hence T is recursive.

4. Examples. We have seen in exercise 2.1.4 that the theory of dense linear orderings without end points is complete. It is also recursively (in fact ﬁnitely) axiomatizable. Hence by the previous theorem, it is complete.

Remark. Each of the following theories will be proved complete later: algebraically closed fields of characteristic 0 or a prime p, real closed fields, differentially closed fields of characteristic 0. Decidability then follows by the previous theorem.

5. Remark. The most versatile method for proving that an algebraic theory T is decidable is quantiﬁer elimination. In order to decide whether σ T ∈ one constructs an equivalent quantiﬁer-free sentence σ∗. It should be easy to decide whether σ∗ T . ∈ 4.3 Decidable models

1. Deﬁnition. A countable structure is decidable if (i) sig( ) is ﬁnite; (ii) A A there exists an enumeration of the universe of , = ak : k ω ,such that A |A| { ∈ }

(](ϕ(v1,...,vn)), k1,...,kn ): = ϕ(ak ,...,ak ) { h i A| 1 n } 32 CHAPTER 4. DECIDABILITY is recursive.

2. Deﬁnition. A countable structure is computable if (i) sig( ) is ﬁnite, and (ii) as above but restricted to atomicA formulas ϕ. A

For example, the structure =(Z, +, , , 0, 1,<) is computable but not Z − · decidable. The structure (Q,<) is computable and hence decidable in view of the following proposition.

3. Proposition. Let T be a recursively axiomatizable theory which admits elimination of quantiﬁers. If Mod(T ) is computable, then it is decidable. M∈

Proof. To decide whether = ϕ(a1,...,an) find a quantifier free ϕ∗ such M| that T = x1 ... xn(ϕ ϕ∗). Since T is recursively enumerable, we can | ∀ ∀ ↔ find ϕ∗ recursively. So = ϕ(a1,...,an) if and only if = ϕ∗(a1,...,an). M| M| Since ϕ∗ is a Boolean combination of atomic formulas, we can recursively decide whether = ϕ∗(a1,...,an). M|

Example. It is fairly easy to see that the field Q of algebraic numbers is recursive. However, it is also a model of the theory of algebraically closed fields of characteristic 0. We shall see later that this theory admits elimination of quantifiers. Hence Q is decidable. Similarly, all countable algebraically closed fields are computable and hence decidable.

4. Theorem. Let T be a decidable theory such that sig(T ) is ﬁnite. Then T has a decidable model.

Proof. We imitate the proof of the compactness theorem 3.1.2. Put V = sig(T )andlet c : m ω be a recursive list of new constant symbols. { m ∈ } Put W = V cm : m ω . Fix a variable x and let ϕn(x):n ω be a recursive enumeration∪{ ∈ of} all formulas ϕ(x) with only{ one free variable∈ } x 4.3. DECIDABLE MODELS 33

andwithsig(ϕ) W .Let σn : n ω be a recursive enumeration of all sentences σ with⊆ sig(σ) W{. Perform∈ the} following recursive construction. ⊆ Stage 0: Let S0 = T . Stage 2n +1:Leth(n)betheleastm such that cm does not occur in S2n ϕn(x) .PutS2n+1 = S2n ( xϕn(x)) ϕn(c ) . ∪{ } ∪{ ∃ → h(n) } State 2n+2: Put S2n+2 = S2n+1 σn if this is consistent; S2n+2 = S2n+1 ∪{ } ∪ σn otherwise. To see that we can make this decision recursively, note that {¬ } S2n+1 = T τ0,...,τ2n for some finitely many sentences τ0,...,τ2n.Let ∪{ } τ0,...,τ2n, σn be the result of replacing the new constant symbols cm by new variables zm.Then e S2n+1e e σn is consistent ∪{ } 2n T z0 ...zj τi σn is consistent ⇔ ∪∃ i=0 ∧ 2Vn z0 ...zj τi σn / T , ⇔¬∃ i=0 ∧e e ∈ and we can decideV this recursively. e e At the end of the construction, Sω = n ω Sn is a complete recursive theory. As in the proof of 3.1.2 we can build a∈ model =( , Φ) where S M |M| = T/ , T = Gödel numbers of variable free terms , t1 t2 if and |M| ≈ { } ≈ only if t1 t2 Sω,etc. is decidable because we can identify with the set of≈ least∈ elementsM of equivalence classes under ,andthen|M| = ≈ M| ϕ(t1,...,tn) ϕ(t1,...,tn) Sω. ⇔ ∈ 5. Remark. There exist examples of recursively (even finitely) axiomatizable theories with no decidable (even computable) model. For example, we may 0 take as axioms a finite fragment of Peano arithmetic together with a false Σ1 sentence. 34 CHAPTER 4. DECIDABILITY Chapter 5

Elementary extensions

5.1 Deﬁnition and examples

1. Deﬁnition. Let , be structures with sig( )=sig( )and . A B A B |A| ⊆ |B| We say that e if for all formulas ϕ(x1,...,xn) with only the free A⊆ B variables shown, and all a1,...,an , = ϕ(a1,...,an) if and only if ∈|A| A| = ϕ(a1,...,an). We then say that is an elementary substructure of , orB| is an elementary extension of . A B B A 2. Let , be as above. We say that ( is a substructure of , is an extensionA B of ) if the above conditionA⊆B holdsA for atomic formulas ϕB. B A 3. Example. Algebra is full of examples of substructures. E.g. if and are groups, if and only if is a subgroup of . Similarly forA rings, ﬁelds,B linear orderings,A⊆B etc. A B

Example. Let =(Q, +, , , 0, 1,<)and =(R, +, , , 0, 1,<). Then Q − · R − · clearly , but e since, for example, = x(x x = 1 + 1) while Q⊆R Q6⊆ R R| ∃ · 35 36 CHAPTER 5. ELEMENTARY EXTENSIONS

does not. This example actually shows that , i.e. and are not elementarilyQ equivalent (deﬁnition 2.1.2). Q6≡R Q R

Proposition. If e then . A⊆ B A≡B Proof. Obvious.

The converse does not hold, e.g.

Example. Let N = 0, 1, 2,... , P = 1, 2,... .Then(P,<) (N,<), and { } { } ⊆ (P,<) (N,<) since they are isomorphic. However (P,<) = x(1 = x 1 < ≡ | ∀ ∨ x)and(N,<) = x(x<1) so (P,<) e (N,<). | ∃ 6⊆ However, this phenomenon does not occur in theories which admit elimination of quantiﬁers:

7. Proposition. If , Mod(T )whereT admits elimination of quantiﬁers, A B∈ then implies e . A⊆B A⊆ B Proof. Obvious.

8. Examples. Let X be any densely ordered subset of the real numbers such that X hasnoﬁrstorlastelement.E.g.X =(0, 1) (1, 2) or X =(0, 1) Q. ∪ ∪ Then (X, <) e (R,<) because the theory of dense linear ordering without endpoints admits⊆ elimination of quantiﬁers.

9. Example. The fact that the theory of algebraically closed ﬁelds admits elimination of quantiﬁers yields the following result: 5.1. DEFINITION AND EXAMPLES 37

Let Q be the ﬁeld of algebraic numbers. Let

f1(x1,...,xn)= = fk(x1,...,xn)=0= g(x1,...,xn) ··· 6 be a finite system of equations and inequations with coefficients in Q. If the system has a solution in some extension field of Q, then it has one in Q.

Proof. We may assume that the extension ﬁeld is algebraically closed. Hence it is an extension of Q, and by quantiﬁer elimination the extension is elementary. This gives the result immediately if we look at the formula

x1 ... xn fi(x1,...,xn)=0 g(x1,...,xn) =0 . ∃ ∃ " i=1 ∧ 6 # ^ In other words, if a variety is nonempty, then it has a point with coordinates in Q. This is closely related to Hilbert’s Nullstellensatz. (See chapter 6.)

9. Definition (A. Robinson). A theory T is said to be model complete if for all , Mod(T ), implies e . A B∈ A⊆B A⊆ B The previous proposition says that if T admints elimination of quantifiers, then T is model complete. The converse is not true, e.g. T = Th(Z, +, , 0, 1,<) = Presburger arithmetic. This theory is model complete but− does not admint elimination of quantifiers. However, it admits elimination of quantifiers if we add defined relations m, m 2. ≡ ≥

10. Example. We shall show later that the theory of algebraically closed fields is model complete and admits elimination of quantifiers. These concepts have algebraic meaning: model completeness expresses the Hilbert Nullstel- lensatz, while quantifier elimination is an abstract formulation of the theory of resultants. 38 CHAPTER 5. ELEMENTARY EXTENSIONS 5.2 Existence of elementary extensions

Theorem. Any infinite structure has a proper elementary extension ∗. A A Proof. Let =( , Φ). Introduce new constant symbols a for each a andA form the|A| elementary diagram of , i.e. the set of all sen- ∈|A| A 1 tences ϕ(a ,...,a ), sig(ϕ(x1,...,xn)) sig( ), a1,...,an ,such 1 n ⊆ A ∈|A| that = ϕ(a1,...,an). Note that an elementary extension of is virtually the sameA| thing as a model of the elementary diagram of . A Let c be a new constant symbol and let S = (elementaryA diagram of ) c = a : a .Since is infinite, S is finitely consistent. [ Details: A ∪{ 6 ∈|A|} |A| Let S0 be a finite subset of S.Letc0 , c0 = a for all a mentioned in S0. ∈|A| 6 Then = S0 if we interpret a as a, c as c0. ] Therefore by the compactness A| theorem, S is consistent, i.e. has a model =( , Ψ). Put ∗ =( ∗ , Φ∗) B |B| A |A | where ∗ = ,Φ∗ =Ψ sig( ). Clearly e ∗ since =elementary |A | |B| A A⊆ A B| diagram of .Also = ∗ since c =Ψ(c) ∗ . A |A| 6 |A | ∈|A|\|A| 2.

Example. Let =(R, +, , , 0, 1,<, )where represents other struc- R − · ··· ··· ture on R.Let ∗ be a proper elementary extension of .Then ∗ will R R R contain inﬁnitesimals, i.e. quantities δ such that 0 <δ0. Elements of ∗ are sometimes called hyperreal numbers.Theycanbe used to give a rigorousR development of calculus based on inﬁnitesimals. This is the beginning of a subject known as nonstandard analysis (A. Robinson).

3. Exercise. Prove the upward Löwenheim-Skolem-Tarski theorem: Let be an infinite structure. Let κ be a cardinal max( , sig( ) ). Then Ahas a proper elementary extension of power κ≥. kAk | A | A (Hint: Combine the elementary diagram method (used in the proof of the previous theorem) with the proof of the Löwenheim-Skolem-Tarski theorem in 3.3.) § 1 To be precise, instead of = ϕ(a1,...,an) we should write = ϕ(a1,...,an) where =( , Φ (a,a):A|a ). A|A| | A|A| |A| ∪{ ∈|A|} 5.2. EXISTENCE OF ELEMENTARY EXTENSIONS 39

In addition to the compactness theorem, another useful method for con- structing elementary extensions is the method of elementary chains (Tarski/Vaught):

A chain is a collection of structures ( α)α<δ where δ is a limit ordinal, such A that α β for all α<β<δ. A ⊆A 5. Proposition. Given a chain as above, there is one and only one structure δ = α with the following properties: A α<δ A

1. S δ = α ; |A | α<δ |A |

2. α Sδ for all α<δ. A ⊆A

δ α δ α δ 0 Proof. Obvious. Note that RA = α<δ RA , oA = α<δ oA ,andcA = cA . S S 6. Theorem (Tarski/Vaught elementary chain principle). Given an elementary chain, i.e. a chain ( α)α<δ such that α e β for all α<β<δ, put A A ⊆ A δ = α.Then α e δ for all α<δ. A α<δ A A ⊆ A S Proof. Show by induction on the length of a formula ϕ(x1,...,xn)that, for all α<δand a1,...,an α , α = ϕ(a1,...,an) if and only if ∈|A| A | δ = ϕ(a1,...,an). This is trivial if ϕ is atomic or a Boolean combination of A | shorter formulas. Assume that ϕ(x1,...,xn)= yψ(x1,...,xn,y). If α = ∃ A | yψ(a1,...,an,y), let an+1 α be such that α = ψ(a1,...,an,an+1). ∃ ∈|A| A | Then δ = ψ(a1,...,an,an+1) by induction, so δ = yψ(a1,...,an,y). A | A | ∃ Conversely, suppose δ = yψ(a1,...,an,y). Then δ = ψ(a1,...,an,b) A | ∃ A | for some b δ = α .Letβ<δbe such that b β .Then ∈|A| α<δ |A | ∈|A| β = ψ(a1,...,an,b) by induction. Hence β = yψ(a1,...,an,y). Hence A | S A | ∃ α = yψ(a1,...,an,y)since α e β. A | ∃ A ⊆ A Later we shall use this theorem plus compactness to construct saturated models. 40 CHAPTER 5. ELEMENTARY EXTENSIONS 5.3 Elementary monomorphisms

1. Let , be structures with sig( )=sig( ). We say that f : is an elementaryA B embedding or elementaryA monomorphismB if f is one-one|A| → |B| and, for all formulae ϕ(x1,...,xn)andalla1,...,an , = ϕ(a1,...,an) ∈|A| A| ⇔ = ϕ(f(a1),...,f(an)). B| 2. If the above equivalence holds for atomic ϕ,wesaythatf is an embedding or monomorphism of into ,oranisomorphism of into (not necessarily onto) . For example,A whatB is usually called a monomorphismA of groups, rings, etc.B is a monomorphism in this sense. The concepts of monomorphism and elementary monomorphism are essentially just trivial variants of the concepts of extension and elementary extension.

3. Theorem. If then there exists a structure such that both and are elementarilyA≡B embeddable into . (The converseC is obvious.) A B C Proof. Let S = (elementary diagram of ) (elementary diagram of ). [ We assume that a : a b :Ab ∪ = .]ClearlyanyB model of S yields a{ as desired.∈|A|}∩{ So by the∈|B|} compactness∅ theorem it suf- C fices to show that S is finitely consistent. Let S0 be a finite subset of S. Say S0 = ϕ(a ,...,a ),ψ(b ,...,b ) where sig(ϕ ψ) sig( )=sig( ), { 1 m 1 n } ∧ ⊆ A B a1,...,am , b1,...,bn .Since = x1 ...xmϕ(x1,...,xm)and ∈|A| ∈|B| A| ∃ = y1 ...ynψ(y1,...,yn)and ,weseethat x1 ...xmϕ(x1,...,xm) B| ∃ A≡B ∃ ∧ y1 ...ynψ(y1,...,yn) is consistent. Hence S0 is consistent, Q.E.D. ∃ 4. From the above we can derive a criterion for model completeness:

Theorem (Robinson’s test). A theory T is model complete for any , Mod(T ) and monomorphism , we can ﬁnd an elementary⇔ extensionA B∈ A→B 5.3. ELEMENTARY MONOMORPHISMS 41

∗ of and a monomorphism ∗ so that the diagram A A B→A

OB B BB BB BB B e ∗ A ⊆ A commutes.

Proof. :SinceT is model complete, the monomorphism is elemen- ⇒ A→B tary, so we may take ∗ = . : Given a monomorphismA B where , Mod(T ). Use the hypothesis⇐ repeatedly to constructA→B a pair of elementaryA B∈ chains

e e e e = / / / / 0 1 B n BO B BB BOBB ···B BOBB ··· BB B BB B BB BB BB BB BB BB BB BB e ! e B! e ! e B! = 0 / 1 / / n / A A A ··· A ··· where all triangles commute. Hence = = so we ω n ω n ∼= n ω n ω have by Tarski/Vaught A ∈ A ∈ B B S S

ω = ω AO ∼ BO e e

/ A B whence is elementary. So T is model complete. A→B 42 CHAPTER 5. ELEMENTARY EXTENSIONS Chapter 6

Algebraically closed ﬁelds

6.1 Simple ﬁeld extensions

1. Let be any structure. For each a introduce a new constant symbol A ∈|A| a.Thediagram of is the set of atomic sentences ϕ(a1,...,an) which hold in . Recall that anA extension of is any structure .Notethatan extensionA of is virtually the sameA thing as a modelB⊇A of the diagram of . A A

2. Let be a field. A field extension of is an extension such that is aA field. A field extension of is thusA virtually the sameB⊇A thing as a modelB of (field axioms) (diagram ofA ). ∪ A

3. If , are fields, ,andb ,let [b] be the smallest substructure of A containingB A⊆Bb , i.e. the∈|B| subringA generated by b .Let (b) beB the smallest subfield|A| ∪ { } of containing b . A field|A| extension ∪ { } of A is said to be simple if it is ofB the form (b).|A| ∪ { } A A We want to survey all possible simple field extensions of . This survey is the main piece of algebraic information that we need for ourA model-theoretic analysis.

43 44 CHAPTER 6. ALGEBRAICALLY CLOSED FIELDS

4. Let be a ﬁeld and consider terms t(x) such that sig(t(x)) sig(diagram of A ⊆ )) and x is the only free variable in t(x). Two such terms t1(x)andt2(x) Aare said to be equivalent if

(field axioms) (diagram of ) = x(t1(x)=t2(x)) . ∪ A | ∀ The set of equivalence classes is naturally a commutative ring and is denoted [x]. We shall see that [x]isadomain. A A 5. A polynomial is a term f(x) as above with either f(x) 0orf(x)= n ≡ anx + + a1x + a0, ai . Clearly each equivalence class in [x] contains··· a polynomial. We∈|A| claim that each equivalence class containsA only one polynomial. Thus the polynomials are a system of representatives for the equivalence classes in [x]. A m Obviously the polynomials over form a domain, since if f(x)=amx + n A + a1x + a0, g(x)=bnx + + b1x + b0, am =0= bn,thenf(x) g(x)= ··· m+n ··· 6 6 · ambnx + = 0. We now use the following fact: any domain is embeddable in a field,···6 its field of quotients. In particular, the set of polynomials over is embedded in a field extension of . Hence distinct polynomials representA distinct equivalence classes in [x],A as claimed above. A 6. It follows that [x] is a domain. The quotient field of [x] is denoted (x). This is in agreementA with our earlier notation for simpleA extensions. A(x)is our first example of a simple extension of . A A 7. The degree of a nonzero polynomial f(x) [x]isdeg(f)=n where f(x)= n ∈A anx + + a1x + a0, an = 0. The degree of the zero polynomial, deg(0), is undefined.··· 6

Lemma (division algorithm). If f(x),g(x) [x], g =0,thereexist q(x),r(x) [x] such that f(x)=g(x) q(x)+∈Ar(x) and6 either r 0or deg(r) < deg(∈Ag). · ≡ 6.1. SIMPLE FIELD EXTENSIONS 45

Proof. As usual.

In the above lemma, if r 0wesaythatg divides f. ≡ 8. Deﬁnition. For f(x) [x]wesaythatf(x)isnonconstant if deg(f) > 0. We say that f(x)isirreducible∈A if f is nonconstant and there is no nonconstant g(x) [x] of lower degree than f(x) such that g(x) divides f(x). ∈A 9.

Let f(x) [x] be nonconstant. For g1,g2 [x]sayg1 = g2 mod f if ∈A ∈A f divides g1 g2.Let [x]/(f(x)) be the set of equivalence classes mod f. Clearly −[x]/(f(x))A has the structure of a commutative ring with as a subfield.A Clearly each equivalence class mod f contains one and onlyA one polynomial g(x) [x] such that deg(g) < deg(f)org 0. Thus [x]/(f(x)) may be viewed∈A as a vector space of dimension deg(f≡)over . A A 10. Lemma. If f(x) is irreducible then [x]/(f(x)) is a field. A Proof. Given g [x], deg(g) < deg(f), g =0 modf, to find an inverse mod f.Put ∈A 6

I = sf + tg : s, t [x] . { ∈A } Let h I, h 0 be of least possible degree. Apply the division algorithm to get ∈f(x)=6≡g(x) q(x)+r(x). Note that r I, hence deg(r) < deg(h)is impossible. Hence r· 0, i.e. h divides f.Alsodeg(∈ h) deg(g) < deg(f). Since f is irreducible≡ it follows that h is a constant. h =0since≤ f =0.We 1 6 1 6 now have sf + tg = h ,soh− tg =1 modf,soh− t is the desired inverse, Q.E.D. ∈|A|

11. In the above lemma, let b be the equivalence class of x in [x]/(f(x)). Then (b)= [x]/(f(x)) is a simple ﬁeld extension of .NotealsothatA [b]= A A A A 46 CHAPTER 6. ALGEBRAICALLY CLOSED FIELDS

(b) in this case, so the dimension of (b)over as a vector space is ﬁnite, Abeing just deg(f). A A

We now have two kinds of examples of simple ﬁeld extensions of .The next theorem says that there are no others. A

12. Theorem. Let (b) be any simple ﬁeld extension of . Then either A A 1. (b) = [x]/(f(x)) for some irreducible f [x], or A ∼ A ∈A 2. (b) = (x). A ∼ A In both cases the isomorphism is the identity on and sends b to x. A Proof. Case 1: b is algebraic over , i.e. f(b) = 0 for some nonzero f [x]. Let f be of least degree such thatA this holds. Clearly f is irreducible.∈A We claim that g(b) = 0 if and only if f divides g. One direction is obvious. For the other direction, suppose g(b) = 0. The division algorithm gives g(x)=f(x) q(x)+r(x). Hence r(b)=f(b) q(b)+r(b)=g(b)=0. Hencewecannothavedeg(· r) < deg(f). Hence r ·0, i.e. f divides g.This proves the claim. ≡

It follows that g1(b)=g2(b) if and only if f divides g1 g2.Thismatches − the definition of [x]/f,sowehave [b] ∼= [x]/f. But by lemma 10 the latter is a field, henceA so is the former,A i.e. A[b]= (b). A A Case 2: not case 1, i.e. b is transcendental over .Thusg1(b)=g2(b) A if and only if g1(x) g2(x). Thus [b] = [x] as rings. Hence the quotient ≡ A ∼ A fields are also isomorphic, i.e. (b) = (x). A ∼ A

13.

Corollary. Let (b1)and (b2) be two simple extensions of .Iff(b1)= A A A f(b2) = 0 for some ireducible f [x], then (b1) = (b2). ∈A A ∼ A

14. Corollary. b is algebraic over if and only if (b) is ﬁnite dimensional over . A A A 6.2. ALGEBRAIC CLOSURE 47 6.2 Algebraic closure

1. A ﬁeld extension is algebraic over if each b is algebraic over . B⊇A A ∈|B| A

/ (c) A finite A 2. Definition. Afield is algebraically closed if every nonconstant g(x) [x] has a root in , i.e. Bg(b)=0forsomeb . For any field ,analgebraic∈B closure of isB a field extension such∈|B| that (i) is algebraicA over , and (ii) Ais algebraically closed.B⊇A B A B 3. Proposition. Every field has an algebraic closure. A Proof. Note first that for any nonconstant g(x) [x] we can adjoin a root, i.e. we can find a simple algebraic field extension∈A (b) such that g(b)=0. (Just factor g(x) into irreducibles and apply the constructionA of Lemma 6.1.10 to one of these irreducible factors.) Now let gα(x), α<βbe an enumeration of all the nonconstant g(x) ∈ [x]. Put 0 = , α+1 = α(bα)wheregα(bα) = 0, and δ = α for A A A A A A α<δ A limit ordinals δ β. Finally put 1 = β. By Lemma 6.2.1 1 is algebraic ≤ B A B S over , and by construction every nonconstant g(x) [x]hasarootin 1. A ∈A B 48 CHAPTER 6. ALGEBRAICALLY CLOSED FIELDS

Repeating this construction ω times, we get

= 0 alg 1 alg alg n alg , A B ⊆ B ⊆ ···⊆ B ⊆ ··· n ω, such that each nonconstant g(x) n[x]hasarootin n+1.Thus ∈ ∈B B = n ω n is an algebraic closure of . B ∈ B A S 4.

Lemma. If is algebraically closed and alg ,then = . A A⊆ B A B Proof. Since is algebraically closed, the only irreducible polynomials in [x]arelinearA, i.e. of degree 1. Hence by Theorem 6.1.12 every element of anA algebraic extension of actually belongs to . A A

5. Proposition. Any two algebraic closures of a ﬁeld areisomorphicover . A A

Proof. Let 1 and 2 be two algebraic closures of . B B A

1 2 B `AA }>B AA }} AA }} A }} A

If is algebraically closed, then 1 = 2 = by the previous lemma. A B B A By Zorn’s lemma or transfinite induction, it suffices to find bi i , ∈|B|\|A| i =1, 2, such that (b1) ∼= (b2)over .Since is not algebraically closed, let f [x] be nonconstantA A with noA root in A. We may assume that f is ∈A A irreducible. Take bi i so that f(bi)=0,i =1, 2. Then (b1) ∼= (b2) over by Corollary∈|B 6.1.13.| This completes the proof. A A A By the above proposition we are justified in writing = the algebraic closure of (unique up to isomorphism over ). By the proofA of Proposition A A 6.2.3 we have =max( , 0). kAk kAk ℵ 6.3. COMPLETENESS AND MODEL COMPLETENESS 49

Corollary. If is algebraically closed, then is isomorphic over to a subﬁeld of .B⊇A A A B

|= | | | A ⊇ A Proof. Let be the subﬁeld of with = b : b is algebraic over . By LemmaC 6.2.4 is algebraicallyB closed.|C| Hence{ ∈|B|is an algebraic closureA} of . Hence by PropositionC 6.2.5 we have that isC isomorphic to over . A A C A

6.3 Completeness and model completeness

In this section we show that the theory Tp of algebraically closed ﬁelds of characteristic p (p = 0 or prime) is complete and model complete.

Lemma. Let be a ﬁeld and let 1, 2 be algebraically closed extensions A B B of such that 1 = 2 = κ>max( 0, ). Then 1 and 2 are isomorphicA over kB. k kB k ℵ kAk B B A

1 2 B `AA }>B AA }} AA }} A }} A

γ Proof. We use a back-and-forth argument. Let i = bi : γ<κ , i =1, 2. |B | { } γ At stage γ of our construction we will have a partial isomorphism iγ : = A1 ∼ 50 CHAPTER 6. ALGEBRAICALLY CLOSED FIELDS

γ γ γ where i, <κ, i =1, 2. A2 A⊆Ai ⊆B kAi k

1 2 BO BO

i : γ γ γ 1 ∼= 2 A `AA }>A AA }} AA }} A }} A 0 0 Stage 0: 1 = 2 = , i0 =identity. A A γ A 2γ 2γ Stage 2γ +1:If b1 is algebraic over 1 ,letf 1 [x] be irreducible γ A ∈A γ such that f(b1 ) = 0, and take b 2 such that i2γ (f)(b)=0.Ifb1 is 2γ ∈|B| 2γ 2γ transcendental over ,takeb 2 ;thisispossiblesince < A1 ∈|B|\ A2 A2 κ,andb is then transcendental over 2γ by Lemma 6.2.4. In either case, by 2 Theorem 6.1.12 and Corollary 6.1.13A we can deﬁne an isomorphism

2γ+1 2γ γ 2γ 2γ+1 i2γ+1 : = (b ) = (b)= A1 A1 1 ∼ A2 A2 γ by i2γ+1 i2γ , i2γ+1(b )=b. ⊇ 1 Stage 2γ + 2: Same as stage 2γ + 1 except reverse the roles of 1 and 2. δ γ B B Stage limit δ κ: i = γ<δ i , i =1, 2, iδ = γ<δ iγ . Finally we have κ ≤κ A A iκ : 1 = = = 2, Q.E.D. B A1 ∼ A2 B S S 2.

Theorem. Tp is κ-categorical for all κ> 0. ℵ Proof. Let be the smallest field of characteristic p, i.e. = =(Q, +, , , 0, 1) A A Q − · if p =0, = Z/p if p is prime. Note that is a subfield of any field of A A characteristic p.Let 1, 2 Mod(Tp), 1 = 2 = κ.Thenwehave B B ∈ kB k kB k that 1 = 2 over by Lemma 6.3.1. B ∼ B A 3.

Corollary. Tp is complete.

Proof. Immediate by Vaught’s test 2.2.4. 6.3. COMPLETENESS AND MODEL COMPLETENESS 51

Corollary. Tp is decidable.

Proof. Immediate from theorem 4.2.3.

5. Exercise. Prove that the theory of algebraically closed ﬁelds (of all charac- teristics) is decidable.

Theorem (A. Robinson). Tp is model complete.

Proof. Given a monomorphism where , are algebraically closed, we want to show that this monomorphismA→B is elementary.A B Let κ be a cardinal > . By Exercise 5.2 we can get elementary extensions ∗ of and ∗ of kBksuch that = = κ.Then = over byA LemmaA 6.3.1.B B kA∗k kB∗k A∗ ∼ B∗ A = AO∗ ∼ BO∗ e e

/ A B Since the diagram commutes, the monomorphism is elementary, Q.E.D. A→B

7. Corollary (Hilbert). Let be a ﬁeld and let be its algebraic closure. Let A A

g1(x1,...,xn)= = gm(x1,...,xn)=0 ··· ( ) f(x1,...,xn) =0 ∗ 6 ) be a finite system of polynomial equations and inequations with coefficients in .If() has a solution in some field extension of , then it has a solution in A. ∗ A A (Algebraic reformulation: The algebraic points on a variety are dense with respect to the Zariski topology on that variety.) 52 CHAPTER 6. ALGEBRAICALLY CLOSED FIELDS

Proof. This is immediate from model completeness of the theory of algebraically closed ﬁelds. Consider the sentence

σ = x1 xn gi =0 f =0 . ∃ ··· i=0 ∧ 6 ! ^ If = σ then = σ. By Corollary 6.2.6 we have . Hence A⊆B|= σ by model completeness.B| A⊆A⊆B A| 6.4 Hilbert’s Nullstellensatz

Let be a ﬁeld and let = [x1,...,xn] be the ring of polynomials in n A R A variables x1,...,xn over .Azero of f is an n-tuple b1,...,bn , A ∈R ∈|B| where is a ﬁeld extension of , such that f(b1,...,bn)=0.Analgebraic B A zero of f is a zero whose coordinates b1,...,bn lie in , the algebraic closure of . A A 2.

Theorem (Nullstellensatz). Suppose f,g1,...,gm such that every com- ∈R mon algebraic zero of g1,...,gm is a zero of f. Then there exists a nonnegative integer k such that

m k f = pigi i=1 X where p1,...,pm . (The converse holds trivially.) ∈R Proof. Let I be the set of f for which the conclusion holds. Suppose the conclusion fails, i.e. f/I.∈R then we claim that I is an ideal in , i.e. ∈ R (i) g,h I g + h I; ∈ ⇒ ∈ (ii) g I,h g h I; ∈ ∈R⇒ · ∈ (iii) 0 I and 1 / I. ∈ ∈ 6.4. HILBERT’S NULLSTELLENSATZ 53

k l To prove (i), suppose g,h I,sayg = pigi and h = qigi.Then ∈ P k + l P (g + h)k+l = gihj . i i+j=k+l X Since i + j = k + l,wemusthaveeitheri k or j l. Therefore each term on the right hand side contains gk or hl. Therefore≥ ≥ each term is of the form k+l rigi.So(g + h) is also of this form. So g + h I. To prove (ii) just k k k ∈ note that if g = pigi then (g h) = h pigi so g h I. Part (iii) is Pobvious since f/I. · · ∈ Thus I is an∈ idealP containing no powerP of f.ByZorn’slemmaletJ I be an ideal containing no power of f and maximal with this property. ⊇ We claim that J is a prime ideal1, i.e. g/J, h / J g h/J.Ifg/J then the ideal generated by J g is properly∈ larger∈ than⇒ J·,sosomepower∈ ∈ of f belongs to it, say f k = sg∪{+ u} where s , u J. Similarly, if h/J then f l = th + v where t , v J. Hence ∈R ∈ ∈ ∈R ∈ f k+l =(sg + u) (th + v)=stgh + sgv + thu + uv . · J ∈ If gh J then f k+l J, contradiction. This proves| the{z claim.} ∈ ∈ Put 1 = /J = the quotient ring of by J.SinceJ is prime, 1 is a R R R R domain. Let be the quotient field of 1. We claim that is canonically isomorphic toB a subfield of . This isR clear since AJ = 0 (since J = 0 implies 1 J).B |A| ∩ { } |A| ∩ 6 { } ∈ Let b1,...,bn correspond to x1,...,xn , i.e. bi =[xi]J .Then ∈|B| ∈R for any g , g(b1,...,bn) = 0 if and only if g J. ∈R ∈ In particular, g1(b1,...,bn)= = gm(b1,...,bn)=0andf(b1,...,bn) = ··· 6 0. Hence by model completeness (Corollary 6.3.7) we can find a1,...,an ∈ such that g1(a1,...,an)= = gm(a1,...,an)=0andf(a1,...,an) = 0.|A| Q.E.D. ··· 6

3. Given the statement of Hilbert’s Nullstellensatz, it is natural to ask whether a bound on the exponent k can be computed. We can use G¨odel’s completeness theorem 4.1.3 to show that this is the case: 1The Nullstellensatz actually implies that J is a maximal ideal, i.e. /J is a ﬁeld. R 54 CHAPTER 6. ALGEBRAICALLY CLOSED FIELDS

Theorem (eﬀective bounds for Hilbert’s Nullstellensatz). We can ﬁnd a recursive function K : ω3 ω with the following property: →

Let be a ﬁeld and let f,g1,...,gm = [x1,...,xn]be A ∈R A such that every common algebraic zero of g1,...,gm is a zero of k f.Iff,g1,...,gm are of degree d,thenf = pigi for some ≤ k K(m, n, d)andsomepi of degree K(m, n, d). ≤ ∈R ≤ P

Proof. Let T be the theory of algebraically closed fields with a distinguished subfield. Let σmndk be a sentence in the language of T asserting that, for all polynomials f,g1,...,gm of degree d in n variables x1,...,xn with ≤ coefficients in the distinguished subfield, if all common zeros of g1,...,gm are zeros of f, then there exists polynomials p1,...,pm of degree k in ≤ variables x1,...,xn with coefficients in the distinguished subfield, such that l f = pigi for some l k. ≤ We claim that m, n, d kσmndk T . To see this, assume the contrary for P ∀ ∃ ∈ some fixed m, n, d and consider a theory T 0 T with new constant symbols ⊇ intended to denote the coefficients of f,g1,...,gm.LetT 0 say that for all k, σmndk fails for this choice of f,g1,...,gm. By the compactness theorem, T 0 is consistent. But any model of T 0 would give a counterexample to Hilbert’s Nullstellensatz. Let R(m, n, d, k) σmndk T .ByGödel’s completeness theorem, T is recursively enumerable.⇔ Hence∈ so is R. Hence we can find a recursive function K : ω3 ω such that R(m, n, d, K(m, n, d)) for all m, n, d.This completes the proof.→

4. One could actually show that the bounding function K can be taken to be primitive recursive. This would be done by giving an explicit quantiﬁer elimination procedure for the theory of algebraically closed ﬁelds. Chapter 7

Saturated models

7.1 Element types

Given a theory T and a nonnegative integer n,letFn(T )bethesetofall formulas ϕ(v1,...,vn) with no free variables other than v1,...,vn, such that sig(ϕ) sig(T ). We say that Y Fn(T )isconsistent over T if there exist ⊆ ⊆ Mod(T )anda1,...,an such that = ϕ(a1,...,an) for all ϕ Y . NoteA∈ that by compactness, if∈|A| each ﬁnite subsetA| of Y is consistent over T ∈then so is Y .

A realization of Y Fn(T )isann-tuple a1,...,an , Mod(T ), such ⊆ ∈|A| A∈ that = ϕ(a1,...,an) for all ϕ Y .WethensaythatY is realized in the modelA|. ∈ A We say that ψ Fn(T )isalogical consequence of Y Fn(T ) over T if every realization of∈Y is a realization of ψ. ⊆

Deﬁnition. An n-type over T is a set p Fn(T ) which is consistent over T and closed under logical consequence over⊆T .

55 56 CHAPTER 7. SATURATED MODELS

Deﬁnition. An n-type p over T is said to be complete if, for all ϕ Fn(T ), either ϕ p or ϕ p. The set of all complete n-types over T is∈ denoted ∈ ¬ ∈ Sn(T ).

Remark. Let us say that ϕ, ψ Fn(T ) are equivalent over T if ∈

T = v1 vn(ϕ ψ) . | ∀ ···∀ ↔

The equivalence classes form a Boolean algebra Bn(T ) where the Boolean operations are given by

[ϕ] [ψ]=[ϕ ψ] · ∧ [ϕ]+[ψ]=[ϕ ψ] ∨ [ϕ]=[ ϕ] − ¬

1=[v1 = v1], 0=[v1 = v1] 6 (For n = 0 this Boolean algebra is sometimes known as the Lindenbaum algebra of T .) A complete n-type over T is essentially the same thing as an ultraﬁlter on Bn(T ). Thus Sn(T ) is just the Stone space of Bn(T ).

6. Example. Suppose T admits elimination of quantifiers. Then an n-type over T is determined by the quantifier-free formulas in it (at least for n 1). For example, let T be the theory of dense linear order without end points.≥ A complete n-type over T is determined by specifying the order relations among n n v1,...,vn.ThusSn(T ) = k . For example S1(T ) =1, S2(T ) =3, | | k=1 | | | | S3(T ) = 13, etc. | | P We shall see later that a theory T is 0-categorical if and only if n(Sn(T ) is finite). Here T is assumed to be countableℵ and complete and to∀ have an infinite model. 7.1. ELEMENT TYPES 57

7. Remark. Let T be a complete theory and let p be a complete n-type over T , n 1. It is easy to see that every finite subset of p is realized in every ≥ model of T . (Consider sentences of the form v1 vn(ϕ1 ϕk)where ∃ ···∃ ∧···∧ ϕ1,...,ϕk p.) However, p itself need not be realized in a given model of ∈ T . For example, let T = T0 = the theory of algebraically closed fields of characteristic zero. Let p S1(T ) be a 1-type saying that v1 is transcenden- n ∈ tal, i.e. anv1 + + a1v1 + a0 =0forai Z, an = 0. Clearly every finite subset of p is realized··· in every6 algebraically∈ closed6 field of characteristic 0. But p itself is not realized in = the algebraic closure of the rational field Q =(Q, +, , , 0, 1). Q − ·

8. Proposition. Let be an inﬁnite model of a complete theory T and assume T . A | |≤kAk

(i) Given p Sn(T ) we can ﬁnd an elementary extension of of the same power as∈ in which p is realized. A A

(ii) has an elementary extension of power max( , Sn(T ) )inwhich A kAk | | every p Sn(T ) is realized. ∈

Proof. (i) Let c1,...,cn,a(a ) be new constant symbols. Since each finite subset of p is realized n ∈|A|,wehavethat A (elementary diagram of ) ϕ(c ,...,c ):ϕ p A ∪{ 1 n ∈ } is finitely consistent. Hence by the compactness and Löwenheim-Skolem theorems, this has a model of cardinality max( , p )= .Thisisan elementary extension of and the interpretationkAk of c| ,...,c| kAkrealizes p. A 1 n (ii) We use an elementary chain argument. Let Sn(T )= pα : α<κ { } where κ = Sn(T ) .Put 0 = , α+1 = an elementary extension of α | | A A A A of the same power as α in which pα is realized; δ = α for limit A A α<δ A δ κ; finally = κ. We can show by induction on α κ that α max(≤ , α ).B HenceA max( ,κ). If need be,≤S use thekA upwardk≤ Löwenheim-SkolemkAk | | theoremkBk to ≤ raise thekAk cardinality of to max( ,κ). B kAk 58 CHAPTER 7. SATURATED MODELS 7.2 Saturated models

1. Given a structure =( , Φ) and a set X , introduce new constant B |B| ⊆|B| symbols a (a X)andlet X =( , ΦX )whereΦX =Φ (a,a):a ∈ B |B| ∪{ ∈ X .Thus X is just like but with its signature expanded to include } B B constant symbols denoting the elements of X.Th(X )isjustthesetof B all sentences ϕ(a ,...,a )witha1,...,an X,sig(ϕ) sig( ), such that 1 n ∈ ⊆ B = ϕ(a1,...,an). B| 2. Definition. Let κ be an infinite cardinal. We say that is κ-saturated if for all X of cardinality less than κ, every completeB 1-type over X is realized in⊆|B|. B (To be precise we should say: every complete 1-type over Th( X )is B realized in X .) B 3. Exercise. Show that if is κ-saturated, then for all X of cardinality <κand all n-types p overB X (n 1), p is realized in .⊆|B| ≥ B 4. Example. Let =( ,<) be a linear ordering, and let κ be an infinite B |B| cardinal. We say that is κ-dense if for every pair of sets X0,X1 B ⊆|B| of cardinality <κ,ifX0

5. We note that a finite structure is κ-saturated for all infinite κ. If an infinite structure is κ-saturated, then κ. B kBk ≥

6. Deﬁnition. A structure is saturated if it is -saturated. B kBk

For example, (Q,<)is 0-saturated, hence saturated. ℵ

7. Proposition. (i) every uncountable algebraically closed ﬁeld is saturated. (ii) Q and Fp are not saturated. (iii) There exist countable saturated algebraically closed ﬁelds of every characteristic.

Proof. Let X where is an algebraically closed field. A complete 1- type over X is⊆|B| virtually theB same thing as a simple extension of the subfield generated by X. We may therefore apply our classification of simple field Aextensions (see 6.1). The details are left to the reader. §

8. Theorem (uniqueness of saturated models). Let and be saturated models of the same power. If and are elementarilyA equivalent,B then they are isomorphic. A B

Proof. We use a back-and-forth argument. Let = = κ and fix well orderings of and of order type κ. WekAk shall definekBk enumerations |A| |B| = aγ : γ<κ and = bγ : γ<κ in such a way that ( ,aγ)γ<κ |A| { } |B| { } A ≡ ( ,bγ )γ<κ. B Stage γ<κ, γ even: We have inductively ( ,aα)α<γ ( ,bα)α<γ.Letaγ be the least element of (with respect to theA fixed well≡ ordering)B different |A| from aα, α<γ.Let

pγ S1(Th(( ,aα)α<γ )) = S1(Th(( ,bα)α<γ)) ∈ A B 60 CHAPTER 7. SATURATED MODELS

bethecomplete1-typeover aα : α<γ realized by aγ .Since is κ- { } B saturated we can ﬁnd bγ realizing pγ over bα : α<γ .Thus ∈|B| { }

( ,aα)α γ ( ,bα)α γ . A ≤ ≡ B ≤ Stage γ<κ, γ odd: Reverse the roles of , . A B Finally ( ,aγ)γ<κ ( ,bγ )γ<κ and these enumerations exhaust the uni- verses of Aand respectively≡ B since the elements were chosen by means of ﬁxed well|A| orderings|B| of order type κ. Thus aγ bγ is an isomorphism of onto . 7→ A B 9. Theorem (universality of saturated models). Let be a κ-saturated model and , κ. Then there exists an elementaryB monomorphism of into A≡B. kAk ≤ A B Proof. Similar to the previous proof.

10. Example. For algebraically closed fields, theorems 8 and 9 have the following significance. Let p = 0 or a prime. There is exactly one countable saturated algebraically closed field. Every countable algebraically closed field of characteristic p is embeddable into this one.

7.3 Existence of saturated models

1. We begin with an existence theorem for countable saturated models.

Theorem (Vaught). Let T be a complete countable theory. The following are equivalent.

(1) T has a countable saturated model;

(2) Sn(T ) 0 for all n. | |≤ℵ 7.3. EXISTENCE OF SATURATED MODELS 61

Proof. (1) (2): Easy. Let be the countable saturated model of T .Then ⇒ A each p Sn(T ) is realized in . Hence Sn(T ) is countable. (2) ∈ (1): Note ﬁrst thatA (2) implies ⇒

(3) for all finite X , Mod(T ), one has S1(X) X0. ⊆|B| B∈ | |≤ We shall prove (3) (1) by an elementary chain argument. ⇒ Stage 0. Let 0 be any countable model of T . A i Stage n +1.Wehaveacountable n Mod(T ). Let Xn : i ω A ∈ 0 { i∈+1 } be an enumeration of the finite subsets of n .Put n = n, n = i |A | A A A a countable elementary extension of n in which all complete 1-types over i i+1 A i Xn are realized. ( n exists because S1(Xn) is countable.) Finally put i A n+1 = i ω n. This completes stage n. A ∈ A Finally put = n ω n. We claim that is 0-saturated. Let X S B ∈ A B ℵ i ⊆B be finite. Then X n for some n. Hence X = Xn for some i. Hence ⊆|AS | i+1 every 1-type over X is realized in n . This is an elementary submodel of , so these types are realized in ,A Q.E.D. B B 2. Example. We give an example of a countable complete theory T with no countable saturated model. Let T =Th( Q)where =(Q,<). Then 0 Q Q S1(T ) =2ℵ because each Dedekind cut in (Q,<) gives rise to a different complete| | 1-type over T . (There are other complete 1-types over T which do not correspond to Dedekind cuts.)

3. The basic theorem on the existence of uncountable saturated models is the following:

Theorem (Morley and Vaught). Let be an inﬁnite model and let κ be an inﬁnite cardinal such that κ sig( ).A Then has a κ+-saturated elementary extension of power κ. ≥ A A kAk Proof. We use an elementary chain argument as in the proof of the previous theorem. Put λ = κ and note that λκ = λ. We construct an elementary chain in which eachkAk model has power λ. Stage 0: Let e 0 where 0 = λ. A⊆ A kA k 62 CHAPTER 7. SATURATED MODELS

+ α Stage γ +1,γ<κ :Wehave γ of power λ.Let X : α<λ be an A { γ } enumeration of all X γ of cardinality κ. Stage γ +1 has λ substages: 0 α+1 ⊆|A | ≤ α Put γ = γ; γ an elementary extension of γ realizing all 1-types over α A A A α κ A β α Xγ (we can do this because S1(Xγ ) 2 λ); and γ = α<β γ for | λ |≤ ≤ A A limit β λ. Finally put γ+1 = γ . ≤ + A A S Stage δ κ , δ limit: Put δ = γ. ≤ A γ<δ A Finally put = κ+ . Clearly e and = λ. We claim is κ+-saturated. SupposeB A X , XA⊆S κB.SincekBkκ+ is a regular cardinal,B + ⊆|B| | |≤α X γ for some γ<κ . Hence X = Xγ for some α<λ.Soevery1-type ⊆A α+1 over X is realized in e . This completes the proof. Aγ ⊆ B 4. Corollary. Let T be a complete theory of cardinality κ, and suppose that T has an inﬁnite model. Then T has a κ+-saturated model≤ of power 2κ. Proof. By L¨owenheim-Skolem let be a model of T of power κ. Apply the previous theorem to get a κ+-saturatedA elementary extension of power κκ =2κ.

5. Corollary. Assume the Generalized Continuum Hypothesis. Let T be a complete theory of cardinality κ.ThenT has a saturated model of power κ+. This model is unique up to≤ isomorphism. Proof. The G.C.H. tells us that 2κ = κ+. Apply the previous corollary and the uniqueness result for saturated models ( 7.2). § 6. The previous corollary may be applied to give the following useful variant of Vaught’s test. Unlike Vaught’s test itself, the following provides a necessary and suﬃcient condition for completeness.

0 Theorem. Assume 2ℵ = 1.LetT be a countable complete theory with no ﬁnite models. T is completeℵ any two saturated models of T of power ⇔ 1 are isomorphic. ℵ Proof. Immediate from the previous corollary. 7.4. PRESERVATION THEOREMS 63

7. Remark. The G.C.H. is needed for the above results. For example, let T be the complete theory of dense linear ordering without end points. Then 0 any 1-saturated model of T has cardinality 2ℵ . Thus we do not get ℵ 0 ≥ satisfactory results unless we assume 2ℵ = 1. However, the G.C.H. can be eliminatedℵ from most applications of saturated models, by observing that the conclusions are usually absolute. Thus, if we are trying to prove that a certain theory T is complete, we may assume G.C.H., apply the variant of Vaught’s test referred to above, and then eliminate G.C.H. by noting that completeness is an absolute property of T . Actually, there are several ways to avoid the need to assume the G.C.H. (1) Assume G.C.H. and then eliminate it by absoluteness arguments. (2) Assume the existence of inaccessible cardinals (some people might consider this assumption more reasonable than G.C.H.). (3) Use “special” models (Morley-Vaught). (4) Use “recursively saturated” models (Barwise-Schlipf). We shall simply assume the G.C.H. when needed.

7.4 Preservation theorems

In this section we present a typical application of saturated models, namely to the proofs of “preservation theorems”. Preservation theorems relate model theoretic properties of T to syntactic properties of axioms for T .

1. The following syntactical classiﬁcation of formulas is occasionally useful.

Deﬁnition. Aformulaψ(y)isΣk if it is of the form

ψ(y)= x1 x2 xk ϕ(x1, x2,...,xk, y) ∃ ∀ ··· where ϕ is quantifier free and xi = xi1 ...xini .ThusaΣk formula consists of k alternating blocks of quantifiers followed by a quantifier free matrix, and the first block is existential. The class of Πk formulas is defined similarly except that the first block is universal. 64 CHAPTER 7. SATURATED MODELS

Let Γ be a class of formulas (e.g. Γ = Σk or Πk). A Γ-theory is a theory T Γ such that T ΓisasetofaxiomsforT . We write +3 to mean that sig( )=sig(∩ )andTh( ) Γ Th( ) Γ. A B A B A ∩ ⊆ B ∩ 3. Lemma (localization lemma). Let Γ be a class of formulas which is closed under disjunction. Let T be a theory such that Mod(T ) is closed under Γ Γ +3, i.e. Mod(T ), +3 implies Mod(T ). Then T is a Γ-theory. A∈ A B B∈

Proof. Let = T Γ, sig( )=sig(T ). We must show that = T . Suppose not. Let ∆B| = σ ∩ Γ: B= σ .Thenforeach Mod(B|T ) there exists { ∈ B| ¬ } A∈ k σ ∆ such that = σ. Hence by compactness T = i=1 σi for some ∈ A| k k | σ1,...,σk ∆. Hence σi T Γ. Hence = σi, a contradiction. ∈ i=1 ∈ ∩ B| i=1 W W W 4. Γ If sig( )=sig( ), we write f : / (f is a Γ-morphism of into ) to meanA that f B: and fApreservesB satisfaction of Γ formulas,A i.e. forB |A| → |B| all Γ formulas ϕ(x1,...,xn)anda1,...,an ,if = ϕ(a1,...,an)then ∈|A| A| = ϕ(f(a1),...,f(an)). B| Examples:

Γ (1) If Γ = Π0 =Σ0 = quantifier free formulas ,thenf : / means that f is a monomorphism{ of into . } A B A B Γ (2) If Γ = atomic formulas ,thenf : / means that f is a homomorphism{ of into .} (We have notA previouslyB defined the concept of homomorphism,A soB this may be taken as the definition. If and are groups or rings, this coincides with the usual definition ofA homo-B morphism.)

Γ (3) If Γ = all formulas then f : / means of course that f is an elementary{ embedding} of intoA . B A B 7.4. PRESERVATION THEOREMS 65

5. We now present our ﬁrst preservation theorem.

Theorem (Tarski). Suppose that T is preserved under substructures, i.e. every substructure of a model of T is a model of T .ThenT is a universal (i.e. Π1-) theory. Proof. By the localization lemma, it suﬃces to prove that if Mod(T ), B∈ Π1 +3 ,then Mod(T ). Let where is κ-saturated, κ = B A A∈ B≡B∗ B∗ Σ1 Σ1 .Wehave +3 , hence +3 ∗ . kAkClaim: ThereA existsB a monomorphismA B of into . To see this, let A B∗ = aγ : γ<κ. As in the proof of Theorem 7.2.9 we choose bγ , |A| { } ∈|B∗| γ<κ, so that the embedding aγ bγ is a Σ1-morphism. Stage γ<κ: Let 7→ Yγ be the set of Σ1 formulas ψ(a ,...,a ,v1), α1 < <αn <γ,such α1 αn ··· that = ψ(a1,...,an,aγ). By inductive hypothesis we have A|

Σ1 ( ,aα)α<γ +3( ,bα)α<γ . A B∗

Hence = v1ψ(bα ,...,bα ,v1)foreachψ(a ,...,a ,v1) Yγ. Hence B∗ | ∃ 1 n α1 αn ∈ by κ-saturation there exists bγ such that = ψ(bα ,...,bα ,bγ )for ∈|B∗| B∗ | 1 n all ψ(a ,...,a ,v1) Yγ. α1 αn ∈ [ Details: To show that Yγ is realized in ( ∗,bα)α<γ it suﬃces to show B k that each ﬁnite subset is, i.e. ( ,bα)α<γ = v ψi(bα ,...,bα ,v) for all B∗ | ∃ i=1 1 n ψ1,...,ψk Yγ. But this is a Σ1 sentence, true in ( ,aα)α<γ , hence true in ∈ V A ( ∗,bα)α<γ .] B Now clearly

Σ1 ( ,aα)α γ +3( ,bα)α γ A ≤ B∗ ≤ so the induction hypothesis is preserved. Finally

Σ1 ( ,aγ)γ<κ +3( ,bγ )γ<κ A B∗ so in particular aγ bγ is a monomorphism of into ∗. This proves the claim. 7→ A B Now ∗ Mod(T ), hence Mod(T )sinceMod(T ) is closed under substructure.B ∈ This completes theA∈ proof. 66 CHAPTER 7. SATURATED MODELS

6. The above argument actually established the following result. Lemma (embedding lemma). Let Γ be a class of formulas which is closed Γ under and (e.g. Γ = Σk, k 1). If +3 and is κ-saturated, ∧ ∃ Γ ≥ A B B κ ,then f : / . ≥kAk ∃ A B We can use this lemma for Γ = Σ2 to prove the following preservation theorem.

7. Theorem (Chang-Los-Susko). Let T be a theory such that Mod(T )isclosed under unions of ω-chains, i.e. 0 n (n ω), n Mod(T ) A ⊆ ··· ⊆ A ⊆ ··· ∈ A ∈ implies n ω n Mod(T ). Then T is a Π2-theory. (The converse is obvious.) ∈ A ∈ S Proof. By the localization lemma, in order to show that T is a Π2-theory, it Π2 suﬃces to show that if Mod(T )and +3 then Mod(T ). A∈ A B B∈ Σ2 We have +3 Mod(T ). Let 1 where 1 is -saturated. B A∈ A≡A A kBk Σ2 Then by the embedding lemma we can ﬁnd a Σ2 embedding f : / 1 . B A Let e 1, 1 1 -saturated. Then B⊆ B B kA k Π2 ( 1,f(b))b +3( 1,b)b A ∈|B| B ∈|B|

Σ1 so by the embedding lemma again we can ﬁnd an embedding g : 1 / 1 A B such that the following diagram commutes:

1 ?A Σ2 ~~ ~~ Σ1 ~~ ~~ e 1 B ⊆ B Repeating ω times we get

1 2 n+1 >A >A A< Σ2 }} Σ2 || Σ2 yy }} Σ1 || Σ1 yy Σ1 }} || yy }} || yy = 0 e 1 e 2 e e n e n+1 e . B B ⊆ B ⊆ B ⊆ ··· ⊆ B ⊆ B ⊆ ··· 7.4. PRESERVATION THEOREMS 67

So ﬁnally ω = n ω n = n ω n.Since n Mod(T )wehaveby B ∈ B ∈ A A ∈ assumption n ω n Mod(T ). Since e ω, it follows that Mod(T ). This completes∈ theAS proof.∈ S B⊆ B B∈ S

Corollary (Robinson). If T is model complete then T is a Π2-theory.

Proof. If T is model complete, then any chain of models of T is an elementary chain. Hence Mod(T ) is closed under unions of chains, so we apply the theorem.

For example, the theory of algebraically closed ﬁelds is model complete, and the given axioms for this theory are Π2.

9. We now prove Lyndon’s preservation theorem for positive sentences. A formula is said to be positive if it is built up from atomic formulas using only , , , (and not using , , ). A positive theory is a theory with a set ∀of positive∃ ∧ ∨ sentences as axioms.¬ → ↔

Theorem (Lyndon). Let T be a theory which is preserved under homomorphic images, i.e. any homomorphic image of a model of T is a model of T . Then T is a positive theory. (The converse is easy.)

For example, the theory of groups is positive, and each homomorphic image of a group is a group. The theory of commutative rings is not positive because of the presence of the axiom 0 = 1. If we drop this axiom, the theory becomes positive, and every homomorphic6 image of a commutative ring is a commutative ring.

Proof. By the localization lemma, it suﬃces to show that if Mod(T ), pos A∈ +3 ,then Mod(T ). Let , where = = A B B∈ A≡A∗ B≡B∗ kA∗k kB∗k κ and ∗, ∗ are κ-saturated. (We use the Generalized Continuum Hypoth- esis toA obtainB saturated models. By absoluteness, the use of G.C.H. can be eliminated.) 68 CHAPTER 7. SATURATED MODELS

pos We have ∗ +3 ∗ . We claim that ∗ is a homomorphic image of ∗. The proof is similarA toB that of the embeddingB lemma (7.4.6), except we needA a back-and-forth argument. [ Details: Choose well orderings of and of order type κ.We |A∗| |B∗| construct enumerations ∗ = aγ : γ<κ , ∗ = bγ : γ<κ . Stage γ, γ even: We|A have| { } |B | { }

pos ( ,aα)α<γ +3( ,bα)α<γ . A∗ B∗

Let aγ be the least element of not among aα : α<γ .LetYγ be the set |A∗| { } of positive formulas ϕ(a ,...,a ,v) such that = ϕ(aα ,...,aα ,aγ ). As α1 αn A∗ | 1 n in the proof of the embedding lemma, we have = v k ϕ (a ,...,a ,v), ∗ i=1 i α1 αn k A | ∃ hence = v ϕi(bα ,...,bα ,v) for each ﬁnite set of formulas ϕ1,...,ϕk B∗ | ∃ i=1 1 n V ∈ Yγ. Hence by saturation of it follows that Yγ is realized in ( ,bα)α<γ, V B∗ B∗ i.e. there exists bγ such that ∈|B∗| pos ( ,aα)α γ +3( ,bα)α γ . A∗ ≤ B∗ ≤ Stage γ, γ odd: We have

pos ( ,aα)α<γ +3( ,bα)α<γ , A∗ B∗ hence

neg ( ,bα)α<γ +3( ,aα)α<γ , B∗ A∗ where a negative formula is deﬁned to be the negation of a positive one. Let bγ be the least element of ∗ not among bα : α<γ . According to the DeMorgan laws, the class of|B negative| formulas{ is closed} under and . ∃ ∧ Hence we can apply the same argument as before to ﬁnd aγ such that ∈|A∗| neg ( ,bα)α γ +3( ,aα)α γ , B∗ ≤ A∗ ≤ i.e.

pos ( ,aα)α γ +3( ,bα)α γ . A∗ ≤ B∗ ≤

Note that aγ = aα, α<γ,sincev = a is a negative formula. 6 6 α 7.4. PRESERVATION THEOREMS 69

Finally we get enumerations = aγ : γ<κ and = bγ : γ<κ |A∗| { } |B∗| { } such that α<γ<κ aα = aγ and ⇒ 6 pos ( ,aγ)γ<κ +3( ,bγ)γ<κ . A∗ B∗

Hence in particular the mapping aγ bγ is a homomorphism of ∗ onto .] 7→ A B∗ Since ∗ Mod(T ), it follows by hypothesis that ∗ Mod(T ). Hence A≡AMod(T ),∈ Q.E.D. B ∈ B∈ For more on preservation theorems, see Chang and Keisler, Model Theory. 70 CHAPTER 7. SATURATED MODELS Chapter 8

Elimination of quantiﬁers

8.1 The model completion of a theory

Deﬁnition (Robinson). Let T and T ∗ be theories with the same signature. We say that T ∗ is a model completion of T if (i) T T ; ⊆ ∗ (ii) every model of T can be extended to a model of T ∗; (iii) for any Mod(T ), T (diagram of ) is complete. A∈ ∗ ∪ A (Recall that the diagram of is the set of all quantiﬁer free sentences A ϕ(a ,...,a ), sig(ϕ) sig( ), a1,...,an , = ϕ(a ,...,a ).) 1 n ⊆ A ∈|A| A| 1 n Condition (iii) is equivalent to

(iii) if Mod(T ), i Mod(T ∗), i =1, 2, then 1 and 2 are “elementarilyA∈ equivalentA⊆B over∈ .” B B A

1 2 B `AA ≡ }>B AA }} AA }} A }} A 71 72 CHAPTER 8. ELIMINATION OF QUANTIFIERS

2. Example. The theory of algebraically closed ﬁelds is a model completion of the theory of ﬁelds.

Proof. (i) is trivial, and (ii) holds since every field can be extended to an algebraically closed field (Proposition 6.2.3). For (iii), let be a field and A let 1 and 2 be two algebraically closed fields extending .Wewantto B B A show that 1 2 over . By the upward Löwenheim-Skolem theorem let B ≡B A i e , i =1, 2, where = = κ and κ>max( 0, ). By B ⊆ Bi∗ kB1∗k kB2∗k ℵ kAk Lemma 6.3.1 we have that = over . Hence 1 2 over . B1∗ ∼ B2∗ A B ≡B A

1 = 2 BO∗ ∼ BO∗ e e

1 2 `A > B AA }}B AA }} AA }} A }} A 3. Remarks. The following facts about model completions are obvious.

(a) If T ∗ is a model completion of T then T ∗ is model complete. (b) T is model complete if and only if T is its own model completion.

Theorem (Robinson). Let T0, T1 be model completions of T .ThenT0 = T1.

Proof. Given Mod(T0), we shall show that Mod(T1). We use an A∈ A∈ elementary chain argument. Let = 0 Mod(T0). By (i) and (ii) of the A A ∈ deﬁnition, we can ﬁnd 1 Mod(T1), 0 1. More generally, given 2n A ∈ A ⊆A A ∈ Mod(T0)let 2n 2n+1 Mod(T1)and 2n+1 2n+2 Mod(T0). Put A ⊆A ∈ A ⊆A ∈ = n ω n.SinceT0 is model complete, we have 2n e 2n+2 so by the B ∈ A A ⊆ A elementary chain principle, = 0 e . But since T1 is model complete, S A A ⊆ B we also have 2n+1 e 2n+3 so Mod(T1). Hence Mod(T1), Q.E.D. A ⊆ A B∈ A∈ Thus the model completion of T is unique if it exists. 8.2. SUBSTRUCTURE COMPLETENESS 73

4. Another example. The theory of dense linear order without end points is the model completion of the theory of linear order. (This is an easy consequence of quantiﬁer elimination.)

8.2 Substructure completeness

1. Deﬁnition. AtheoryT is substructure complete if T (diagram of )is complete whenever is a substructure of a model of T .∪ A A

2. Definition. T admits elimination of quantifiers if for all n 1andall ≥ ϕ Fn(T ), there exists a quantifier free formula ϕ Fn(T ) such that ∈ ∗ ∈ T = v1 ... vn[ϕ(v1,...,vn) ϕ (v1,...,vn)]. | ∀ ∀ ↔ ∗

3. Theorem. Let T be a theory. The following are equivalent.

(i) T is the model completion of a universal (i.e. Π1)theory.

(ii) T is substructure complete.

(iii) T admits elimination of quantiﬁers.

Proof. (i) (ii). Let T be the model completion of a universal theory U. Let ⇒ Mod(T ). Then Mod(U)sinceU is universal. Hence T (diagramA⊆B∈ of ) is complete,A∈ by the definition of model completion. ∪ A (ii) (iii). Assume that T is substructure complete. Given ϕ Fn(T ), ⇒ ∈ to find a quantifier free ϕ∗ Fn(T ) such that T ∗ = ϕ ϕ∗. We proceed as in the proof of the localization∈ lemma 7.4.3. Let|Y be↔ the set of quantifier free ψ Fn(T ) such that T = ϕ ψ.Ifϕ does not exist, then by ∈ | → ∗ compactness we can find Mod(T )anda1,...,an so that = B∈ ∈|B| B| 74 CHAPTER 8. ELIMINATION OF QUANTIFIERS

ϕ(a1,...,an) ψ(a1,...,an) for all ψ Y .Let be the substructure of ¬ ∧ ∈ A B generated by a1,...,an. By substructure completeness we have

T (diagram of ) = ϕ(a ,...,a ) . ∪ A | ¬ 1 n

Hence there exists a quantiﬁer free formula θ Fn(T ) such that θ(a1,...,an) diagram of and ∈ ∈ A

T = v1 ...vn [θ(v1,...,vn) ϕ(v1,...,vn)] , | ∀ →¬ i.e. T = v1 ...vn[ϕ θ]. Hence θ Y . Hence = θ(a1,...,an). This is a contradiction.| ∀ →¬ ¬ ∈ B| ¬ (iii) (i). Suppose T admits elimination of quantifiers. Let U be the theory whose⇒ axioms are the universal sentences of T . We claim that T is a the model completion of U.If Mod(U) then an easy compactness argument shows that T (diagramA∈ of ) is consistent. Hence is extendible toamodelofT . The∪ last part ofA the definition of modelA completion is immediate from elimination of quantifiers. (Note: Mod(U) is just the class of substructures of models of T .)

4. Example. The theory of dense linear ordering without end points is the model completion of the theory of linear orderings. the latter theory is universal so we conclude:

Theorem. The theory of dense linear ordering without endpoints admits elimination of quantiﬁers.

(This can also be proved by a direct syntactical argument. See Exercise 2.1.4.)

5. Example. We have seen that the theory T of algebraically closed fields is the model completion of the theory of fields. Unfortunately, the theory of fields is not universal. (The key axiom is x(x =0 y(x y = 1)).) However, T is also the model completion of the theory∀ 6 of→∃ domains,· and the latter theory is universal. (The key axiom is x y((x =0 y =0) x y =0).) ∀ ∀ 6 ∧ 6 → · 6 8.2. SUBSTRUCTURE COMPLETENESS 75

(To see that T is the model completion of the theory of domains, just note that any ﬁeld containing a domain also contains its fraction ﬁeld, and apply Example 8.1.2. See also Example 8.3.3 below.) Hencewehave:

Theorem (Tarski). The theory of algebraically closed ﬁelds admits elimination of quantiﬁers. (In the language of +, , , 0, 1.) · −

6. This theorm has algebraic applications; e.g.

Corollary (elimination theory). Let S be a finite system of equations and inequations in n unknowns (variables) x1,...,xn with coefficients (constants) c1,...,cN . Then we can effectively find a finite set S1,...,Sk of systems of equations and inequations in c1,...,cN alone, such that for any c1,...,cN in an algebraically closed field, S is solvable if and only if one of S1,...,Sk holds.

Proof. Regard c1,...,cN as variables and effectively eliminate quantifiers from the formula x1 xn S. Put the resulting quantifier free formula into disjunctive normal∃ form.···∃

Special cases of the above result are known classically. For example, in van der Waerden (vol. 1, 27) one ﬁnds the following result. Let S be a system of two equations §

m m 1 a0x + a1x − + + am =0 n n 1 ··· b0x + b1x − + + bn =0. ··· These two polynomials have a common root if and only if the determinant

a0 a1 ... am a a ... a 0 1 m a0 a1 ... am

b0 b1 ... bn

76 CHAPTER 8. ELIMINATION OF QUANTIFIERS vanishes. This (m + n) (m + n) determinant is called the resultant of the system S. In general, the× theory of “resultants” or “elimination theory” is a branch of classical algebra concerned with explicit determination of S1,...,Sk given S as in the corollary.

Recall that a variety X Cn is the solution set of a system of equations in n unknowns with complex⊆ coeﬃcients.

Corollary (Tarski). Let X Cn be a variety. Let Y Ck be the image of ⊆ ⊆ X under the projection (x1,...,xn) (x1,...,xk), 1 k n.ThenY is a Boolean combination of varieties. 7→ ≤ ≤ (This is already nontrivial for k =2,n =3.)

8.3 The role of simple extensions

In this section we discuss the role of simple extensions in quantiﬁer elimination.

1. Deﬁnition. Let U be a universal theory. Let Mod(U). Given b , we write [b] for the substructure of generatedA⊆B∈ by b .This is∈|B| called a simpleA extension of . B |A| ∪ { } A

2. Theorem (L. Blum). Let U be a universal theory and let T be a theory with sig(T )=sig(U), U T , such that every model of U is extendible to a model of T . The following⊆ are equivalent.

(i) T is the model completion of U.

+ (ii) Suppose Mod(T ), κ -saturated, where κ =max( 0, ). Then upA⊆B∈ to isomorphism overB , contains every simple extensionℵ kAk of . A B A 8.3. THE ROLE OF SIMPLE EXTENSIONS 77

(iii) Suppose i Mod(T ), i =1, 2. If 2 is κ-saturated, κ 1 , A⊆B ∈ B ≥kBk then there exists an embedding of 1 into 2 such that the following diagram commutes. B B

1 ______/ 2 B A`A }>B AA }} AA }} A }} A

Proof. (i) (ii). Immediate from the deﬁnitions. ⇒ (ii) (iii). Let 1 = bα : α<κ and deﬁne a tower of simple ⇒ |B | { } extensions 0 = , α+1 = α[bα], β = α for limit β κ. Apply A A A A A α<β A ≤ (ii) κ times to get an embedding of 1 into 2. B SB (iii) (i). Given i Mod(T ), i =1, 2. To show 1 2 over . We use⇒ an elementaryA⊆B chain argument.∈ B ≡B A

e e e 1 / 10 / 100 / ?B 0 BF 0 BF ··· ~~ 00 0 ~~ 0 00 ~~ 0 0 ~~ 00 0 0 00 00 0 A @@ 0 00 @@ 0 0 @@ 00 00 @@ 0 0 e e e 2 / 0 / 00 / B B2 B2 ···

First choose 0 e 2 suﬃciently saturated, and apply (ii) to get 1 0 B2 ⊇ B B →B2 over . Then choose 0 e 1 suﬃciently saturated and apply (ii) to get A B1 ⊇ B 2 0 over . Etc. Finally we get B →B1 A e (n) 1 / = ?B B1∗ n ω 1 ÐÐ ∈ B ÐÐ ÐÐ S ÐÐ = A >> ∼ >> >> >> e / = (n) 2 2∗ n ω 2 B B ∈ B S and the diagram gives an isomorphism of 1∗ onto 2∗ over . Hence 1 2 over . B B A B ≡B A 78 CHAPTER 8. ELIMINATION OF QUANTIFIERS

3. Example. We can use the above criterion to prove (using a minimum of algebra) that the theory of algebraically closed fields is the model completion of the theory of domains. Suppose , algebraically closed and κ+- A⊆B B saturated, where κ max( 0, ). We must show that contains every simple extension [≥b] of theℵ domainkAk . All such extensionsB are contained in simple field extensionsA of the quotientA field of . Using our classification of simple field extensions ( 6.1) we see immediatelyA that contains all such extensions. § B In the next chapter we shall use this criterion to show that the theory of real closed ordered fields is the model completion of the theory of ordered fields. Chapter 9

Real closed ordered ﬁelds

9.1 Ordered ﬁelds

1. The axioms for ordered ﬁelds are as follows:

(a) the ﬁeld axioms (as in example 2.3.4)

(b) the axioms for linear order:

x y z((x

x y z(x

2. Examples. =(Q, +, , , 0, 1,<) = the ordered ﬁeld of rationals. Q − · 79 80 CHAPTER 9. REAL CLOSED ORDERED FIELDS

=(R, +, , , 0, 1,<) = the ordered ﬁeld of real numbers. R − · =(Z, +, , , 0, 1,<) = the ordered ﬁeld of rationals. Z − ·

3. The following facts are easily proved.

(a) In an ordered domain, we have

x0 ↔ − x>0 x<0 ↔− x =0 x2 > 0 6 → (x0) x z 0.) ··· n

(c)| The ordering{z } of an ordered ﬁeld is dense (because x

(d) Any ordered ﬁeld contains as an ordered subﬁeld. Any ordered domain contains as an orderedQ subdomain. Z

Proposition. Let be an ordered domain and let 1 be its ﬁeld of quo- A A tients. Then there is one and only one way to order 1 so that it becomes an ordered ﬁeld with as an ordered subdomain. A A

Proof. We are forced to order 1 by A x x > 0 x y = y2 > 0 . y ⇔ · y ·

It is easy to check that this makes 1 an ordered field, etc. For example, there is only oneA way to order the rational field so as to make it an ordered field. 9.1. ORDERED FIELDS 81

5. Deﬁnition. Let be an ordered ﬁeld and let f [x]. We say that has the intermediateF value property (IVP) for f ∈Fif a, b , a

9. Lemma. Let be an ordered ﬁeld which is not real closed. Then there is F an ordered ﬁeld 1 % such that 1 is algebraic over . F F F F Proof. Pick f [x] such that the IVP fails for f but holds for all g [x]of smaller degree.∈F Let a, b [x] be such that a

B =[a, b] A. Obviously A, B = ,[a, b]=A B,anda0 0ifg>0onsomeintervalF [a0,b0], a0 A, b0 B; g(α) < 0ifg<0onsomeinterval[a0,b0], a0 A, ∈ ∈ ∈ b0 B; g(α)=0ifg 0. ∈Claim: this makes≡ (α) an ordered ﬁeld. The only nontrivial axiom to be checked is the productF rule: the product of positive elements is positive. Suppose gi(α) > 0, deg(gi) < deg(f), i =1, 2. Let g1(α) g2(α)=g3(α), · deg(g3) < deg(f). We must show g3(α) > 0. We have

g1(x) g2(x) g3(x)=f(x) h(x) . · − · Put m =deg(f). The left side has degree 2m 2. Hence so does the right ≤ − side, so deg(h) m 2. Let a0 A, b0 B be such that g1,g2,g3,h all have ≤ − ∈ ∈ constant sign on [a0,b0]. Case 1: h>0on[a0,b0]. Let c [a0,b0] be such that f(c) < 0. then ∈ g3(c) > 0since

g1(c) g2(c) = g3(c)+f(c) h(c) , · · pos neg hence g3 > 0on[a0,b|0]. Hence{z g}3(α) > 0. | {z } Case 2: h<0on[a0,b0]. Take c [a0,b0]sothatf(c) > 0. Then by an ∈ argument similar to the above, g3(c) > 0, so g3(α) > 0. Case 3: h =0on[a0,b0]. Hence h 0sog3 = g1 g2 > 0on[a0,b0], hence ≡ · g3(α) > 0. This completes the proof. 9.2. UNIQUENESS OF REAL CLOSURE 83

10. Definition. Let be an ordered field. A real closure of is an ordered field so thatF (1) is algebraic over ;(2) is real closed.F G⊇F G F G 11. Proposition. Any ordered field has a real closure. F Proof. This is an easy consequence of the previous lemma plus Zorn’s lemma (or transfinite induction). Let be the algebraic closure of regarded as B F a field. Form a transfinite sequence of extensions = 0, α+1 =proper F F F ordered extension of α, β = α<β β for limit β, all within . The process must eventually stop,F soF by lemma 9.1.9F we then have a realB closure of . S F 9.2 Uniqueness of real closure

The purpose of this section is to prove that the real closure of an ordered ﬁeld is unique up to isomorphism over . The usual proofs of this result use Feither Galois theory or Sturm’s test.F We give a more elementary argument based on Rolle’s theorem.

1. Lemma (Rolle’s theorem). Let be a real closed ordered ﬁeld, g [x]. G ∈G Between any two distinct roots of g liesatleastonerootofg0. n (Here g0 is the formal derivative of g.Ifg(x)=anx + + a1x + a0 then n 1 ··· g0(x)=nanx − + + a1.) ··· Proof. Given a

k 1 l 1 g(x)=(x a) − (x b) − q(x) − − where

q(x)=[k(x b)+l(x a)]p(x)+(x a)(x b)p0(x) . − − − − 84 CHAPTER 9. REAL CLOSED ORDERED FIELDS

We have q(a)=k(a b)p(a), q(b)=l(b a)p(b)soq(a)andq(b)haveopposite − − sign, hence q(c)=0forsomea

2. Lemma (maximum principle). Let be a real closed ordered ﬁeld, g [x], a, b , a

Proof. If g is a linear function there is nothing to prove. Otherwise g0 has only ﬁnitely many roots in the interval [a, b]. Let

a = c0 g(ci) for all i.Thenci

3. Lemma. Let be an ordered field. Let g [x] be a nonconstant polynomial such thatF satisfies the IVP for all polynomials∈F of degree deg(g). Suppose g(α)=0whereF α lies in some ordered field extension of ≤ .Then actually α . F ∈|F| Proof. We prove the lemma by induction on n =deg(g). Assume that the conclusion fails, i.e. α/ . It follows that g is irreducible. (Otherwise we could apply the induction∈|F| hypothesis to the factors and get a contradiction.) In particular g has no roots in ,sobytheIVPg has constant sign on . Without loss of generality, assume|F| that g<0on . |F| |F| Since g(α) = 0, an easy calculation shows that α < 1+ a1 + + an = n n 1 | | | | ··· | | M where g(x)=x + a1x − + + an.Thuswehaveg( M) < 0, g(α)=0, g(M) < 0, and M<α

Proposition. If is a RCOF and 1 is an ordered ﬁeld extension of which G G G is algebraic over ,then 1 = . G G G

Proof. Immediate from the previous lemma.

Proposition. Let be an ordered ﬁeld. Then any two real closures of areisomorphicoverF . F F

Proof. Let 1 and 2 be real closures of .If is real closed, then 1 = 2 = by the previousG G proposition. Therefore,F byF Zorn’s lemma or transfiniteG G induction,F it suffices to show that if is not real closed, then we can find F αi i , i =1, 2, such that (α1) = (α2)over as ordered fields. ∈|G|\|F| F ∼ F F In order to find α1,α2 we imitate the proof of lemma 9.1.9. Let f [x] be such that the IVP fails for f but holds for all g of lower degree.∈F Let a, b, A, B be as in the proof of lemma 9.1.9. Since 1 is real closed, 1 G |G |\|F| contains α1 such that f(α1)=0anda0 <α1 0onsomeinterval[a0,b0]in where a0 A, b0 B.By F ∈ ∈ the previous lemma we know that any root of g lying in 1 already lies in |G | . Hence g>0on[a0,b0]in 1. Hence in particular g(α1) > 0. Since every |F| G element of (α1)isoftheformg(α1), g [x], deg(g) < deg(f), our claim F ∈F is proved. Similarly we can find α2 2 such that f(α2)=0and ∈|G|\|F| a0 <α2

Deﬁnition. If is an ordered ﬁeld, we denote its real closure by .(The previous result showsF that is unique up to isomorphism over .)F F F 86 CHAPTER 9. REAL CLOSED ORDERED FIELDS

7. Corollary. Let be an ordered ﬁeld and let be a RCOF extending . Then containsF (an isomorphic copy of) . G F G F

Proof. Let 1 be the algebraic closure of within . Clearly 1 is a real F F G F closure of . By the previous result it follows that is isomorphic to 1 over . F F F F

9.3 Quantiﬁer elimination for RCOF

1. Lemma. Let be an RCOF and let (α)beasimpleorderedﬁeldextension of , α/ .ThenG α is transcendentalG over , and the ordering of is completelyG ∈|G| determined by the sets A = a G: a<α and B = b G : α

(ii) If A = = B,theng(α) > 0 if and only if there exist a A, b B such6 that∅6g(c) > 0 for all c , a

2. Theorem. The theory of real closed ordered ﬁelds is the model completion of the theory of ordered ﬁelds. 9.3. QUANTIFIER ELIMINATION FOR RCOF 87

Proof. It suffices to show that if 1 are ordered fields and if where F⊆F F⊆G is a real closed ordered field and is κ-saturated, κ = 1 ,then 1 is Gembeddable into over . (Cf. theoremG 8.3.2.) By transfinitekF k inductionF we G F may safely assume that either (1) 1 is algebraic over ,or(2) 1 = (α), α transcendental over , α/ ,andF is real closed.F F F F ∈|F| F In case (1), we have 1 and by corollary 9.2.7 is embeddable into over . In case (2), putF A⊆=F a : a<α and BF= b : α

Proof. Any RCOF has an ordered subﬁeld isomorphic to

=(Q, +, , , 0, 1,<) . Q − · By the previous theorem, any two RCOFs are elementary equivalent over . In particular any two RCOFs are elementarily equivalent, so the theory RCOFQ is complete. Decidability follows by theorem 4.2.3. (Tarski actually gave a decision procedure based on a speciﬁc explicit quantiﬁer elimination procedure.)

4. Corollary (Tarski’s transfer principle). Any first order sentence which is true in the real field =(R, +, , , 0, 1,<) is true in any real closed ordered field. R − ·

5. Examples. (1) The fact that the complex field is algebraically closed (i.e. the fundamental theorem of algebra) is expressibleC as a set of first order 88 CHAPTER 9. REAL CLOSED ORDERED FIELDS sentences in the theory of RCOF.(Weidentifya b√ 1 C with an ordered ± − ∈ pair (a, b) R2.) It follows by Tarski’s principle that, if is any RCOF,then (√ 1) =∈ [x]/(x2 + 1) is algebraically closed. G G (2)− BottG and Milnor have used homotopy theory to prove that the only finite dimensional nonassociative algebras over R are those of dimensions 1, 2, 4, and 8 respectively (i.e. R, C, quaternions, octonians). By Tarski’s transfer principle, the same holds for any RCOF.

6. Corollary (Tarski). The ﬁrst order theory of the real ﬁeld

=(R, +, , , 0, 1,<) R − · is decidable. (In other words, there is a decision procedure for “high school mathematics”, including plane and solid geometry since the latter can be interpreted into the first order theory of using Cartesian coordinates.) R 7. Corollary (Tarksi). Given a finite system S of equations and inequalities in n variables with rational coefficients, we can decide recursively whether S has a solution in reals.

8. Theorem (Tarksi). The theory of real closed ordered ﬁelds admits elimination of quantiﬁers.

Proof. From proposition 9.1.4 and theorem 9.3.3 it follows that RCOF is the model completion of the theory of ordered domains. The latter theory is universal so by theorem 8.2.3 it follows that RCOF admits elimination of quantiﬁers.

9. Corollary (Tarski). The theory of real closed ordered ﬁelds is model complete. 9.4. THE SOLUTION OF HILBERT’S 17TH PROBLEM 89

10. Corollary (Artin). Let S be a finite system of equations and inequalities in n unknowns with coefficients in an ordered field .IfS has a solution in some ordered field extension of ,thenS has aF solution in ,thereal closure of . F F F Proof. Immediate from model completeness plus corollary 9.2.7.

In the next section we shall apply this result to obtain Artin’s solution of Hilbert’s 17th problem.

9.4 The solution of Hilbert’s 17th problem

Let be an ordered ﬁeld, its real closure. Put = (x1,...,xn)=the F F B F ﬁeld of rational functions over in n variables x1,...,xn. Hilbert’s 17th problem is concerned with the representationF of rational functions as sums of squares.

Theorem (Artin). Let f , f 0. Suppose that f(a1,...,an) 0for ∈|B| 6≡ ≥ all a1,...,an such that f(a1,...,an) is deﬁned. Then there exists a positive integer∈|kF|such that

k 2 f = cigi i=1 X where ci , ci > 0, gi , gi 0. ∈|F| ∈|B| 6≡ Proof. Let I be the set of all f of this form. We have ⊂|B| (i) g2 I for all g , g 0 ∈ ∈|B| 6≡ (ii) g,h I implies g + h, g h I ∈ · ∈ (iii) 0 / I. ∈ 90 CHAPTER 9. REAL CLOSED ORDERED FIELDS

AsetI satisfying (i), (ii) and (iii) is called an order ideal. The⊂|B| following lemma holds for any order ideal in any ﬁeld . B Lemma. Let I be an order ideal, f/I, f 0. Then there exists an order ideal J I such that f J. ∈ 6≡ ⊇ − ∈ Proofoflemma. Assume f =0.LetJ be the set of elements of of the form 6 |B|

m1 ml p( f)=(f) g1 + +( f) gl − − · ··· − · where l 1, gi I, mi 0. Clearly f J and J satisﬁes (i) and (ii). Suppose≥ (iii) fails∈ for J,say≥ p( f)=0.Wethenhave− ∈ − p( f)=q(f 2) f r(f 2)=0 − − · where q(f 2)andr(f 2) I or = 0. If r(f 2)=0thenq(f 2) = 0 so both are 0. This is impossible since∈l 1. Hence r(f 2) = 0. Hence ≥ 6 q(f 2) 1 2 f = = q(f 2) r(f 2) I r(f 2) · · r(f 2) ∈ since f =0.Thisprovesthelemma. 6 Now to prove the theorem, suppose f/I, f = 0. By the lemma, get an order ideal J I containing f. By Zorn’s∈ lemma6 let K J be a maximal order ideal. Then⊇ K satisﬁes− (i), (ii), (iii) and also ⊇

(iv) for all g = 0 either g K or g K. 6 ∈ − ∈ This is immediate from the lemma, by maximality of K. For g,h deﬁne g

2. Corollary (Artin’s solution of Hilbert’s 17th problem). Let be either the F real field R or the rational field Q. Suppose f(x1,...,xn) (x1,...,xn) ∈F is definite over , i.e. f(a1,...,an) 0 for all a1,...,an such that F ≥ ∈|F| f(a1,...,an) is defined. Then there exist g1,...,gk in (x1,...,xn)such that F

k 2 f = gi . i=1 X

Proof. For the reals this is immediate from the previous theorem. For the rationals, it is immediate from the theorem plus the following two observa- tions: (1) Q is dense in R; (2) every positive element of Q is the sum of 4 squares in Q.

3. Remark. An ordered field is said to be Hilbertian if (1) is dense in , (2) every positive elementF of is a sum of squares in .F For example, F =reals, = rationals, =anyF RCOF are Hilbertian. ClearlyF the above corollaryF holdsF for any HilbertianF ordered field. McKenna1 has proved the converse: If the corollary holds for ,then is Hilbertian. F F 4. We now derive a corollary concerning effective bounds, analogous to theorem 6.4.3.

Corollary. We can ﬁnd a recursive function K : ω2 ω with the following property: →

Let be an ordered ﬁeld, and let f (x1,...,xn) be such that F ∈F f(a1,...,an) 0 for all a1,...,an such that this is deﬁned. ≥ k∈ F 2 If f is of degree d then f = cig for some k K(n, d), ≤ i=1 i ≤ ci , ci 0, gi (x1,...,xn)ofdegree K(n, d). ∈F ≥ ∈F P ≤ 1SLNM 498, pp. 220-230. 92 CHAPTER 9. REAL CLOSED ORDERED FIELDS

Proof. Let T be the theory of real closed ordered ﬁelds with a distinguished subﬁeld, .Letσndm be the sentence asserting that the desired conclusion F holds with K(n, d) replaced by m.Wehave n d mσndm T (otherwise we could use compactness to construct a counterexample∀ ∀ ∃ to∈ theorem 9.4.1). So put K(n, d)=leastm such that σndm T . This is recursive since T is decidable. ∈

5. Remark. Ax (unpublished) and Pﬁster (1967)2 have shown that for real closed we can always get k 2n (but apparently with no bound on the degrees).F There are some open≤ problems in this area.

6. Exercise. Prove the following result which is known as the Nullstellensatz for ordered ﬁelds.

3 Theorem (Dubois ). Let f,h1,...,hm [x1,...,xn] be polynomials in n variables over an ordered ﬁeld . Suppose∈F that every common zero of F h1,...,hm in , the real closure of , is a zero of f. Then there exist nonnegative integersF r and k such thatF

k m 2r 2 f + cigi = pjhj i=1 j=1 X X where ci , ci 0, gi,pj [x1,...,xn]. ∈|F| ≥ ∈F Deduce a version with eﬀective bounds.

2Proc. Symp. Pure Math. 28, pp. 483-491. 3Ark. Mat. 8 (1969), pp. 111-114. See also Prestel’s monograph. Chapter 10

Prime models (countable case)

10.1 The omitting types theorem

1. Definition. Let p be an n-type over a theory T .Amodel of T is said to A omit p if there is no n-tuple a1,...,an which realizes p. ∈|A| We have already seen how to use the compactness theorem to construct a model which realizes p. It is somewhat more difficult to construct a model which omits p. Indeed, such a model may not even exist. The omitting types theorem gives a sufficient condition for the existence of a model of T which omits p.(IfT is complete, this sufficient condition is also necessary.)

2. Deﬁnition. We say that p is principal over T if it is generated by a single formula, i.e. there exists ϕ Fn(T ) such that p = ψ Fn(T ):T = ∈ { ∈ | v1 vn(ϕ(v1,...,vn) ψ(v1,...,vn)). Such a ϕ is called a generator of p. ∀ ··· → We say that p is essentially nonprincipal over T if p is not included in any principal n-type over T .

3. Theorem (the omitting types theorem). Let T be a countable theory and let p be an n-type over T . Suppose that p is essentially nonprincipal over T .

93 94 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

Then there exists a model of T which omits p.

Proof. We go back to the Henkin proof of the completeness theorem. Let

C = ci : i ω be a countable set of new constant symbols. Let σs : s ω{ be an∈ enumeration} of all sentences σ with sig(σ) sig(T ) C.Let{ ∈ } ⊆ ∪ ϕS(v):s ω be an enumeration of all formulas ϕ(v) with no free variables { ∈ } s s other than v, such that sig(ϕ(v)) sig(T ) C.Let c1,...,cn : s ω be an enumeration of all n-tuples of⊆ constant∪ symbols from{h C. i ∈ } Stage 0: T0 = T . Stage 3s +1: Put T3s+1 = T3s σs if this is consistent, otherwise ∪{ } T3s+1 = T3s σs . ∪{¬ } Stage 3s +2:Let h(s)betheleasti such that ci does not appear in T3s+1 ϕs(v) .PutT3s+2 = T3s+1 ( vϕs(v)) ϕs(c ) . ∪{ } ∪{ ∃ → h(s) } Stage 3s +3: Let q be the set of all ψ Fn(T ) such that T3s+2 ψ(cs,...,cs ) is inconsistent. Clearly q is∈ a principal n-type over T∪. {¬ 1 n } Hence q does not include p.Pickaformulaϕ p q and put T3s+3 = s s ∈ \ T3s+2 ϕ(c ,...,c ) . ∪{¬ 1 n } Finally put T = s ω Ts. Deﬁne a Henkin model as in the proof of theorem 3.1.2. Each∞ n-tuple∈ of elements of is denotedM by an n-tuple of Henkin constants cs,...,cS s .Thus omits|M|p. 1 n M 4. Remark. The hypothesis that T is countable cannot be omitted from the omitting types theorem.

5. We digress to present a typical application of the omitting types theorem.

Theorem. Let =( , A) be a countable model of ZF set theory. Then A |A| ∈ has a proper elementary extension such that ωA = ωB, i.e. and Ahave the same natural numbers. B A B

Proof. Let T be the theory generated by (elementary diagram of ) rank(c) > rank(a):a A ∪{ ∈|A|} where c is a new constant symbol. T is consistent by the compacttness theorem, and every model of T gives rise to a proper elementary extension 10.1. THE OMITTING TYPES THEOREM 95

of .Letp be the 1-type over T generated by v1 ω v1 = n : n ωA . ToA prove the theorem, it suﬃces to ﬁnd a model{ of∈T }∪{which omits6 p.∈ } Suppose this were not possible. By the omitting types theorem, p is included in a principal 1-type over T .Letϕ Fn(T ) be a generator of ∈ such a principal 1-type. In particular T = ϕ(n) for all n ωA.Let ϕ(v)=ψ(c,v)wheresig(ψ) sig(elementary| diagram¬ of ). Then∈ for each ⊆ A n ωA we have T = ψ(c,n). Hence by compactness, for each n ωA there exists∈ a such| that¬ the elementary diagram of contains ∈ ∈|A| A x (rank(x) > rank(a) ψ(x, n)) . ∀ →¬ Hence

= n ω y x (rank(x) > rank(y) ψ(x, n)) . A| ∀ ∈ ∃ ∀ →¬ By the replacement axiom in it follows that A = y n ω x (rank(x) > rank(y) ψ(x, n)) . A| ∃ ∀ ∈ ∀ →¬ Let a be such that ∈|A| = n ω x (rank(x) > rank(a) ψ(x, n)) . A| ∀ ∈ ∀ →¬ Since T =rank(c) > rank(a)itfollowsthatT = n ω ψ(c,n). In other | | ∀ ∈ ¬ words, T = v1 (ϕ(v1) v1 / ω). This contradicts the assumption that the principal| 1-type∀ generated→ by∈ϕ includes p.

6. The omitting types theorem can be generalized as follows:

Theorem. Let T be a countable theory. For each i ω,letpi be an ∈ essentially nonprincipal ni-type over T .ThenT has a model which omits pi for each i ω. ∈ The proof is a straightforward generalization of the proof of the omitting types theorem. 96 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

7. Exercise. Use the above generalization of the omitting types theorem to prove the following result, which generalizes theorem 10.1.4 above.

Theorem (Keisler and Morley). Let be a countable model of ZF set theory. Then has a proper elementary Aend extension, i.e. a proper elementary extension Asuch that rank(a) < rank(b) for all a , b . B ∈|A| ∈|B|\|A| In the next section, the omitting types theorem will be used to study prime models.

10.2 Prime models

1. Deﬁnition. Let T be a complete theory. A model of T is said to be prime if every model of T has an elementary submodel whichA is isomorphic to . A 2. Examples.

1. Let T = ACF(0). This theory has a prime model = the algebraic Q closure of the rational field =(Q, +, , , 0, 1). Q − · 2. Similarly for T = ACF(p), the theory of algebraically closed fields of prime characteristic p. The prime model is the algebraic closure of p =(Fp, +, , , 0, 1). F − · 3. Generalizing examples 1 and 2, let be any field and let T = ACF (diagram of ). We know that TAis complete because ACF admits∪ elimination ofA quantifiers. The prime model of T is , the algebraic closure of . A A 4. T = RCOF. The prime model is = the real closure of the ordered Q field =(Q, +, , , 0, 1,<). Q − · 5. T = RCOF (diagram of )where is any ordered field. The prime model is ,∪ the real closureF of . F F F 10.2. PRIME MODELS 97

6. T = theory of dense linear ordering without end points. The prime model is (Q,<). 7. We give an example of a complete countable theory with no prime model. Let T have 1-place relations Si(x), i ω, and axioms saying ∈ that the Si’s are independent, i.e. any nontrivial Boolean combination of them is nonempty. The completeness of T can be proved by quantiﬁer elimination. It is easy to see that no model of T is prime.

3. Remark. It is clear from the deﬁnition of prime model that any elementary submodel of a prime model is prime. Example 6 shows that a prime model may have proper elementary submodels.

4. The purpose of this section is to establish necessary and suﬃcient conditions for a complete theory T to have a prime model, and to establish the uniqueness of prime models when they exist. Throughout this chapter we deal with countable theories T ; for the uncountable case, see chapter 13.

5. The following lemma is just a restatement of the omitting types theorem in the special case when T and p are complete.

Lemma. Let T be a countable complete theory, and let p Sn(T )bea complete n-type over T . The following are equivalent. ∈ (i) p is principal over T .

(ii) p is realized in every model of T .

Proof. Assume that p is principal, say generated by ϕ Fn(T ). In particular ∈ T v1 vn ϕ(v1,...,vn) is consistent. Since we are assuming that T is ∪{∃ ··· } complete, it follows that T = v1 vn ϕ(v1,...,vn). Let be any model of | ∃ ··· A T .Since = v1 vn ϕ(v1,...,vn), there exist a1,...,an such that A| ∃ ··· ∈|A| = ϕ(a1,...,an). Then clearly the n-tuple a1,...,an realizes p. A| 98 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

Conversely, suppose that p is nonprincipal over T .Sincep is complete, it follows that p is essentially nonprincipal over T . Hence by the omitting types theorem, there exists a model of T which omits p.

Deﬁnition. A structure is atomic if for every a1,...,an the com- A ∈|A| plete n-type realized by a1,...,an is principal over Th( ). A 7.

Theorem (Vaught). Let T be a countable complete theory. The following are equivalent for Mod(T ). A∈ (i) is prime. A (ii) is countable and atomic. A

Proof. (i) (ii). Assume is prime. Since T has a countable model and is isomorphic⇒ to an elementaryA submodel of , it follows that is countable.B A B A To show that is atomic, let a1,...,an and let p be the complete A ∈|A| n-type over T realized by a1,...,an.Since is prime, p is realized in every model of T . Hence by the previous lemma itA follows that p is principal. Thus is atomic. A (ii) (i). Assume that is countable and atomic. Let = an : n ω ⇒be an enumeration ofA .Let be a model of T .Wewantto|A| { construct∈ } an elementary embedding|A| of Binto . Assume inductively that A B we have chosen b0,...,bn 1 so that ( ,ai)i

8. Theorem (Vaught). Let T be a countable complete theory. If T has a prime model, it is unique up to isomorphism.

Proof. By the previous theorem it suﬃces to show that any two countable atomic models of T are isomorphic. We use a back-and-forth argument. The inductive step is as in the proof of the previous theorem. The details are routine.

9. Theorem (Vaught). Let T be a countable complete theory. The following are equivalent. (i) T has a prime model.

(ii) T has an atomic model.

(iii) Every principal n-type over T is included in a complete principal n-type over T .

10. Remark. A Boolean algebra B is said to be atomic if for all b B, b =0, ∈ 6 there exists a b such that a is an atom, i.e. a = 0 and there is no a1 ≤ 6 such that 0

s s theorem 10.1.3. Let c1,...,cns : s ω be an enumeration of all ﬁnite sequences of Henkin{h constants. Stagesi ∈ 0,} 3s +1,and3s +2areasinthe proof of theorem 10.1.3.

Stage 3s +3:Putn = ns and let q be the set of ψ Fn(T ) such that s s ∈ T3s+2 ψ(c ,...,c ) is inconsistent. Let p q be a principal complete ∪{¬ 1 n } ⊇ n-type over T .Letϕ Fn(T ) be a generator of p.PutT3s+3 = T3s+2 s s ∈ ∪ ϕ(c ,...,c ) . Clearly T3s+3 is consistent. { 1 n } At the end of the construction we obtain a countable atomic model of T . This is prime by theorem 10.2.7.

12.

Corollary (Vaught). Let T be a countable complete theory. Suppose Sn(T ) is countable for each n ω.ThenT has a prime model. ∈

Proof. Assuming that Sn(T ) is countable, we shall show that Bn(T )isatomic, i.e. every principal n-type q over T is included in a complete principal n-type over T . Suppose not, i.e. any principal n-type which includes q is incomplete. We use a splitting argument. For each ﬁnite sequence s of 0’s and 1’s we deﬁne a principal n-type qs which includes q.Beginwithq = q.Givenqs, ∅ we know that qs is incomplete, so let ϕs Fn(T ) be a formula such that ∈ qs ϕs and qs ϕs are both consistent. Let qs1 be the n-type generated ∪{ } ∪{¬ } by qs ϕs and let qs0 be the n-type generated by qs ϕs . ∪{ } ω ∪{¬ } For each f 2 let qf = qf n : n ω .Thenqf is an n-type over ∈ { ∈ } T ,andf = g implies qf qg is inconsistent. Let pf be a complete n-type 6 ∪ S 0 over T which includes qf .Thenf = g implies pf = pg. Hence Sn(T ) =2ℵ 6 6 | | contradicting our assumption that Sn(T ) is countable.

13.

Remark. In terms of Boolean algebras, the previous argument shows that if B is a countable Boolean algebra with S(B) countable, then B is atomic. 10.3. THE NUMBER OF COUNTABLE MODELS 101 10.3 The number of countable models

1. Let T be a complete theory and let κ be a cardinal. We write ν(T,κ)=the number of nonisomorphic models of T of cardinality κ. This function is a central object of study in pure model theory. In this section, we study ν(T, 0)whereT is complete and countable. We ℵ beginwiththecasewhenν(T, 0) = 1, i.e. T is 0-categorical. ℵ ℵ 2. Theorem (Ryll-Nardzewski). Let T be a complete countable theory with no ﬁnite models. The following are equivalent:

(1) T is 0-categorical. ℵ

(2) Sn(T ) is ﬁnite for each n ω. ∈ (3) Every countable model of T is prime.

(4) Every countable model of T is saturated.

(The equivalence of (1) and (2) is due to Ryll-Nardzewski.)

Proof. (1) (2). Suppose that Sn(T ) is infinite. We claim that there ⇒ exists a nonprincipal q Sn(T ). To see this, for each principal p Sn(T )let ∈ ∈ ϕp Fn(T ) be a generator of p.PutY = ϕp : p Sn(T ),pnon-principal . ∈ {¬ ∈ } Since Sn(T ) is infinite, it follows by compactness that Y is consistent over T . Hence Y can be extended to a complete n-type q Sn(T ). Clearly q is nonprincipal. ∈ By lemma 10.2.5 there exists a countable model Mod(T ) such that omits q. Also there exists a countable Mod(T )A∈ such that realizes q. A B∈ B and are nonisomorphic, so T is not 0-categorical. A B ℵ (2) (3). Since Sn(T ) is finite, every p Sn(T ) is principal. Hence every model⇒ of T is atomic. Hence by theorem∈ 10.2.7 it follows that every countable model of T is prime. (2) (4). Let be a countable model of T ., and let X be finite. ⇒ A ⊆|A| From (2) it follows that S1(X) is finite. hence each p S1(X)isprincipal, ∈ hence realized in X . This shows that is saturated. A A 102 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

(3) (4) (1). This is because of uniqueness of prime models (10.2.8) or saturated∨ models⇒ (7.2.8) respectively.

Remark. In terms of Boolean algebras, the argument for (1) (2) above shows that if B is a Boolean algebra with S(B) infinite (equivalently⇒ B is infinite), then B has a nonprincipal ultrafilter.

Next we consider the situation where 1 <ν(T, 0) < 0.Wesaythata ℵ ℵ countable model of T is weakly saturated if every p Sn(T ), n ω is realized in . A ∈ ∈ A

Theorem (Rosenstein). Suppose 1 <ν(T, 0) < 0 where T is a countable complete theory. Then T has a weakly saturatedℵ ℵ countable model which is not saturated.

Proof. Suppose not. Let 1,..., k be the countable models of T which are not saturated. By assumptionA theseA models are not weakly saturated, so let pi Sn (T ) be omitted in i,1 i k. ∈ i A ≤ ≤ Introduce new constant symbols ci ,...,ci ,1 i k,andletT be a 1 ni ∗ ≤ i ≤ i countable complete theory which includes T ϕ(c1,...,cni ):ϕ pi, 1 i k .Let =( , Φ ) be any countable∪{ model of T .Then∈ ≤= ≤ } A∗ |A∗| ∗ ∗ A ( ∗ , Φ∗sig(T )) is a countable model of T . Since each pi is realized in |A,1| i k, it follows that is saturated. Hence is saturated. We A ≤ ≤ A A∗ have now shown that any countable model of T ∗ is saturated. Hence T ∗ is 0-categorical. Hence by Ryll-Nardzewski’s theorem 10.3.2, it follows that ℵ Sn(T )isﬁniteforalln ω. Hence Sn(T ) is ﬁnite for all n ω. Hence by ∗ ∈ ∈ 10.3.2 again it follows that T is 0-categorical, i.e. ν(T, 0) = 1, contradicting the hypothesis of the theorem. ℵ ℵ 10.3. THE NUMBER OF COUNTABLE MODELS 103

Theorem (Vaught). Let T be a countable complete theory. Then ν(T, 0) = 2. ℵ 6

Proof. Suppose 1 <ν(T, 0) < 0.Fromν(T, 0) 0 it follows that Sn(T ) is countable for each n ℵω. Henceℵ among theℵ countable≤ℵ models of T are a prime one, a saturated one,∈ and a weakly saturated one (corollary 10.2.12 and theorems 7.3.1 and 10.3.5). These three countable models are nonisomorphic (theorems 10.2.7 and 10.3.5).

Examples. (1) We give examples of countable complete theories Tk with ν(T, 0)=k,1 k< 0.WemaytakeT1 to be the theory of dense ℵ ≤ ℵ linear ordering without endpoints ( 0-categorical by 2.3.1). By the previous ℵ theorem, T2 does not exist. We may take T3 to be the theory whose axioms are T1 c

T3 P (c ):n ω ∪{0 n ∈ } x (P (x) for exactly one i, 0 i m) ∪{∀ i ≤ ≤ } x y (x

Thus a model of T3+m is a model of T3 in which the points are colored with m + 1 densely shuﬄed colors Pi,0 i m, and the distinguished points ≤ ≤ cn have color P0. In a given model of T3+m we may have cn’s unbounded, bounded with no limit, or bounded with a limit in Pi,0 i m.This gives m + 3 distinct countable models. (All except the prime≤ one≤ are weakly saturated.) (2) There are plenty of examples of countable complete theories T with 0 ν(T, 0)= 0 or 2ℵ . For instance ν(ACF(0), 0)= 0 and ν(RCOF, 0)= 0 ℵ ℵ ℵ ℵ ℵ 2ℵ . 104 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

8. We omit the proof of the following theorem.

Theorem (Morley). Let T be a countable complete theory. If ν(T, 0)is 0 ℵ inﬁnite, then it is equal to either 0, 1,or2ℵ . ℵ ℵ 9. Remark. Vaught has conjectured that for all countable complete theories 0 T ,ifν(T, 0) > 0 then ν(T, 0)=2ℵ . No counterexample to Vaught’s conjectureℵ is known.ℵ Steel1 hasℵ shown that Vaught’s conjecture holds for theories T whose models are trees. (A tree is a partially ordered set such that the predecessors of any element are linearly ordered.) Shelah has shown that Vaught’s conjecture holds for superstable theories. Also, Lachlan2 has shown that for superstable theories, ν(T, 0) < 0 implies ν(T, 0)=1. ℵ ℵ ℵ 10.4 Decidable prime models

1. Let be a countable model such that sig( ) is ﬁnite (or recursive). Recall A A that is said to be decidable if there exists an enumeration = an :1 n<ωA of the elements of , such that |A| { ≤ } A

ϕ Fn(T ): = ϕ(a1,...,an) ( ∈ n ω A| ) [∈ is recursive. (This is easily seen to be equivalent to deﬁnition 4.3.1.)

2. Recall theorem 4.3.4 which said that if T is a decidable theory, then T has a decidable model. However, it is not in general true that if T is a complete decidable theory which has a prime model, then T has a decidable prime model. We now present a theorem which gives a necessary and suﬃcient condition for T to have a decidable prime model. 1SLNM 689 2FM 81 10.4. DECIDABLE PRIME MODELS 105

3. Theorem (Harrington). Let T be a complete decidable theory. The following are equivalent.

(a) T has a decidable prime model.

(b) Given n 1andϕ Fn(T ) such that ϕ is consistent with T ,wecan ≥ ∈ ϕ ϕ recursively enumerate a complete n-type p Sn(T ), such that ϕ p and pϕ is principal. ∈ ∈

Note. We do not assume in (b) that it is possible to pass recursively from ϕ to a generator of pϕ.

Proof. (a) (b). Let be a decidable prime model of T .Givenϕ Fn(T ) ⇒ A ∈ consistent with T ,sinceT is complete we have T = v1 vn ϕ so we can | ∃ ··· recursively find an n-tuple a1,...,an such that = ϕ(a1,...,an). We ∈|A| A| ϕ can then enumerate the complete n-type realized by a1,...,an and define p to be this type. Clearly ϕ pϕ,andpϕ is principal by theorem 10.2.7. (b) (a). Let T be a∈ complete decidable theory such that (b) holds. Introduce⇒ a recursive set of new constant symbols C = c :1 n< { n ≤ ω and let T0 = T Hn(c ,...,c ):1 n<ω where Hn(c ,...,c ) } ∪{ 1 n ≤ } 1 n is the nth Henkin sentence ( xψn(x)) ψn(c ). Here we assume that ∃ → n ψn(x):1 n<ω is an enumeration of all formulas ψ(x)withonefree { ≤ } variable x such that sig(ψ(x)) sig(T ) C, and we assume that sig(ψn(x)) sig(T ) c ,...,c . ⊆ ∪ ⊆ ∪{ 1 n} We shall construct a complete decidable theory T extending T0 such ∞ that, for each m 1, there will exist a formula ϕm Fm(T ) such that ≥ϕm ∈ c1,...,cm realizes p . This clearly implies that the Henkin model associated with T (as in the proof of theorem 4.3.4) is decidable and prime. If it were possible to pass recursively from ϕ to a generator of pϕ,we could construct T by straightforwardly combining the proofs of theorem ∞ 4.3.4 and 10.2.9. It would then be the case that ϕm is a recursive function of m. Unfortunately this is not possible. We shall have to define ϕm as the s limit of a recursive sequence of approximations ϕm. We shall use a priority argument to show that the approximations converge. We shall construct T recursively in stages T0 T1 Ts . ∞ ⊆ ⊆ ··· ⊆ ⊆ ··· At each stage s 0, Ts will consist of T0 plus finitely many sentences. ≥ 106 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

s Let τ (c1,...,cNs ) be the conjunction of this finite set of sentences, where s τ FN (T ). Put ∈ s s s τ (v1,...,vm) vm+1 vn [ H1(v1) Hn(v1,...,vn) τ (v1,...,vN )] m ≡∃ ··· ∧···∧ ∧ s s where n =max(m, Ns). Thus τm expresses all the assertions about c1,...,cm s to which we are committed at the end of stage s.Notethatτm Fm(T ). ∈ s At each stage s, there will be a recursive sequence of formulas ϕm s ∈ Fm(T ), 1 m<ω.Hereϕ is the stage s approximation of the desired ≤ m formula ϕm. We now present the construction. 0 0 Stage 0. Let T0 be as above. Put ϕm(v1,...,vm)=τm(v1,...,vm). Stage s +1. Let s =(n 1,k,t) in some fixed recursive enumeration − of ω ω ω.Letθ(v1,...,vn)bethekth formula in some fixed recur- × × sive enumeration of Fn(T ). At this stage we shall consider whether to add θ(c1,...,cn) to our theory. s ϕn s s s If θ p and T = τn θ and T = τm vm+1 vn [ τn θ ]foreach m,1 ∈m

Ts+1 = Ts θ(c ,...,c ) ∪{ 1 n } and, for all m 1, ≥ s s+1 ϕm if 1 m n, ϕm = s+1 ≤ ≤ ( τm if m>n.

s+1 s Otherwise do nothing, i.e. Ts+1 = Ts and ϕm = ϕm for all m 1. It is clear that the construction is recursive and hence produces≥ a recursively axiomatizable theory T . We shall now prove a sequence of claims leading to the conclusion that∞ this theory is complete and gives rise to an atomic Henkin model. s s Claim 1: T = τm ϕm. | → 0 0 s+1 s This is clear by induction on s since ϕm = τm and ϕm =eitherϕm or s+1 τm . s s ϕm Claim 2: τm p . ∈ 0 We prove this by induction on s.Fors = 0 it is trivial since τm = 0 0 ϕm ϕm p . If no action is taken at stage s + 1, the induction step is trivial. Consider∈ a stage s + 1 at which action is taken. Let n and θ be as in the s+1 s construction. For m

s s+1 s s+1 s+1 s s+1 ϕm ϕm T = τm τm .Alsoϕm = ϕm so by induction τm p = p .For s s | → s+1 s s ϕn ∈ ϕn m = n,wehaveτn = τn θ and τn p by induction, and θ p by ∧s s+1 ∈ ∈ s+1 ϕn ϕn s+1 s hypothesis. Hence τn p = p since ϕn = ϕn. Finally, for m>n, s+1 s+1 s+1 ∈ϕm we have τm = ϕm p . ∈ s0 s s Claim 3: n s s0 sϕn = ϕn. (We then deﬁne ϕn = lims ϕn.) ∀ ∃ ∀ s ≥ ϕn Claim 4: n s (τn is a generator of p ). We prove∀ claims∃ 3 and 4 simultaneously by induction on n. Assume that 3 nd 4 hold for all m

s s T = τ vm+1 vn [ τ θ ] . | m →∃ ··· n ∧ s Thus, unless T = τn θ already, we have Ts+1 = Ts θ(c1,...,cn) .In |s+1 → s+1 ϕn ∪{ } any case T = τn θ so τn is a generator of p . This completes the proofofclaims3and4.| →

From claim 4 it follows that T = s ω Ts generates a complete decidable ∞ ∈ extension of T0 whose asociated Henkin model is atomic. This completes the proofofthetheorem. S

4. Remark. In chapter 11 we shall apply the above theorem to show that the differential closure of a computable differential field of characteristic 0 is computable. This was Harrington’s original application.

5. Exercise. Prove the following theorem of Morley3:

3Israel J. Math. vol. 25 pp. 233-240 108 CHAPTER 10. PRIME MODELS (COUNTABLE CASE)

Theorem (Morley). A complete decidable theory T has a decidable saturated model if and only if there is a uniform recursive enumeration of

n ω Sn(T ). ∈ S For more information on decidable models, see e.g. Millar, Annals of Math. Logic vol. 13, pp. 45-72.

6. Exercise. Use theorem 10.4.3 to prove that the algebraic closure of a computable ﬁeld is computable, and the real closure of a computable ordered ﬁeld is computable. Chapter 11

Diﬀerentially closed ﬁelds of characteristic 0

11.1 Simple extensions

1. The theory of diﬀerential ﬁelds, DF, has the following set of axioms:

(a) ﬁeld axioms

(b) x y (x + y)0 = x0 + y0 ∀ ∀

(c) x y (x y)0 = x0 y + x y0 ∀ ∀ · · ·

Here 0 is a 1-place operation symbol. A differential field is a model of these axioms. A differential domain is a model of the axioms for domains plus (b), (c). A differential ring is defined similarly. The characteristic of a diffrential field is the characteristic of the underlying field.

2. Examples. (1) Let be any ﬁeld. The polynomial ring [x]canbemade A A into a diﬀerential ring by interpreting 0 as the formal derivative, i.e. if f = n n 1 anx + + a1x + a0, ai ,thenf 0 = nanx − + + a1. ··· ∈|A| ··· 109 110 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0

We can extend the derivative to the ﬁeld of rational functions (x)= ﬁeld of quotients of [x] in the usual way: A A

f 0 f 0 g f g0 = · − · . g g2 Then (x) becomes a differential field. A (2) Any field can be made into a differential field in a trivial way by A putting a0 =0foralla . ∈|A| Note that this is the only way to make the rational field =(Q, +, , , 0, 1) into a differential field. Thus the differential field is embeddedQ in− every· differential field of characteristic 0. Q (3) Let D be a connected open set in the complex z-plane (or more generally any Riemann surface). Let be the field of meromorphic functions on D, i.e. the field of quotients of theM ring of holomorphic functions on D. Thiscanberegardedasadifferentialfieldintheobviousway: f 0 = derivative of f = df / d z . (4) Let be the set of all functions which are meromorphic in some open set containingN 0 in the z-plane. (Identify two such functions if they are equal on some open set containing 0.) This is again a differential field in an obvious way.

3. Remark. Ritt and Seidenberg1 have shown that every differential field which is finitely generated over the reational field is isomorphic to a differential subfield of . Thus everything we shall do in this chapter is meaningful for analysis. N

4. We now study simple extensions of diﬀerential ﬁelds of characteristic 0.

Lemma. Let be a differential domain and let be its field of quotients, = c/d : c, dD ,d =0 . Then there is one andF only one way to make F into{ a differential∈|D| field6 extension} of . F D 1Proc. AMS 9 (1958) 159-164; 23 (1969) 689-691. 11.1. SIMPLE EXTENSIONS 111

Proof. We must have

c 0 c 0 c c0 = d = d + d0 d · d · d · so 

c c0 d c d0 0 = · − · . d d2 It is easy to check that this makes a diﬀerential ﬁeld, etc. F 5.

Lemma. Let be a differential field of characteristic 0. Let = (b0,...,bn 1,bn) be a finitely generatedA field extension of such that B A − A (i) for each i

Before proving this lemma, we give an example. Let b0 be transcendental 2 over and let b1 satisfy b b0 =0.Then (b0,b1)canbemadeintoa A 1 − A differential field extension of in which b0 is a solution of the first order 2 A differential equation (y0) y = 0 and is not the solution of any algebraic equation over . − A 7.

Proofoflemma5. Consider the ring [~y]= [y0,...,yn 1,yn]. A typical A A − element g of [y0,...,yn 1,yn] looks like A − j g(~y)= aj~y j X j j0 jn 1 j − n where aj = aj0j1...jn and ~y = y0 yn 1 yn .Letuswrite ∈|A| ··· − j g(~y)= aj0 ~y j X e 112 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0

and ∂g/∂yi = the formal partial derivative of g(y0,...,yn 1,yn) with respect − to yi.Giveng(b0,...,bn 1,bn) [b0,...,bn 1,bn] we are forced to deﬁne − ∈A − n ∂g g(b0,...,bn 1,bn)0 = g(b0,...,bn 1,bn)+ (b0,...,bn 1,bn) bi+1 − − ∂y − · i=0 i X where however bn+1 remainse to be determined. Let f(b0,b1,...,bn 1,yn) (b0,...,bn 1)[yn] be irreducible such that − ∈A − = (b0,...,bn 1)[yn]/f (by theorem 6.1.12). Multiplying f by a suitable B ∼ A − element of [b0,...,bn 1],wemaysafelyassumethatf(y0,...,yn 1,yn) A − − ∈ [y0,...,yn 1,yn]. As in the previous paragraph, we want A − n ∂f 0=f(b0,b1,...,bn 1,bn)0 = f(b0,...,bn 1,bn)+ (b0,...,bn 1,bn) bi+1 . − − ∂y − · i=0 i X e Since ∂f/∂yn is of lower degree than f in yn,wemusthave ∂f (b0,...,bn 1,bn) =0. ∂yn − 6 Hence we are forced to deﬁne

n 1 ∂f f(b0,...,bn 1,bn)+ i=0− (b0,...,bn 1,bn) bi+1 − ∂yi − · bn+1 = ∂f . (b ,...,bn ,bn) ∂yn P0 1 e − −

We must check that 0 is well deﬁned. We have f(b0,...,bn 1,bn)=0.Also − f(b0,...,bn 1,bn)0 = 0 by choice of bn+1. Suppose g1,g2 [y0,...,yn 1,yn] − ∈A − are such that g1(b0,...,bn 1,bn)=g2(b0,...,bn 1,bn). Then − −

g1(b0,...,bn 1,yn) g2(b0,...,bn 1,yn)modf(b0,...,bn 1,yn) − ≡ − − in the polynomial ring (b0,...,bn 1)[yn]. Hence g1 g2 = f h where A − − · h (y0,...,yn 1)[yn]. Letting h = p/q,wehave ∈A −

q (g1 g2)=f p · − · where p [y0,...,yn 1,yn] q [y0,...,yn 1], q 0. Substituting bi for ∈A − ∈A − 6≡ yi and diﬀerentiating both sides we have ~ ~ ~ ~ ~ ~ ~ q(b) (g1(b)0 g2(b)0)=f(b)0 p(b)+f(b p(b)0 · − · · =0 11.1. SIMPLE EXTENSIONS 113

~ and q(b) = 0, hence g1(b0,...,bn 1,bn)0 = g2(b0,...,bn 1,bn)0. This shows 6 − − that 0 is well defined on the ring [b0,...,bn 1,bn]. It is easy to check the sum and product rules, so we now knowA that there− is one and only one way to make [b0,...,bn 1,bn] into a differential ring extension of . By lemma 11.1.4 A − A we get the same conclusion for the field of quotients = (b0,...,bn 1,bn). This completes the proof. B A −

8. Remark. Let be a differential field. We have considered four different kinds of “simpleA extensions” of ,namely: A [b] = ring generated by b , A |A| ∪ { } (b) = field generated by b , A |A| ∪ { } b = differential ring generated by b , A{ } |A| ∪ { } b = differential field generated by b . Ah i |A| ∪ { } Our goal is to classify the simple differential field extensions b . Ah i 9. Definition. Let be a differential field. A differential polynomial f(y)= (n) A f(y,y0,...,y ) is an element of the differential ring

(n) y = [y,y0,...,y ,...] A{ } A

(n) 00···0 where of course y = yn times and y is a differential indeterminate. (Alter- natively, we could define a differential polynomial to be an equivalence class of terms t(y) in the language|{z} of the diagram of , as in 6.1.4.) Note that y is a differential domain andA y , its field of quotients, is a differentialA{ field} by lemma 11.1.4. Ah i

10. Definition. If f(y) y is a nonzero differential polynomial over ,we define the order of f∈A{to be} the largest n such that y(n) occurs in f,andtheA 114 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0 degree of f to be the largest k such that (y(n))k occurs in f,wheren is the order of f. (3) 4 5 (3) For example, the differential polynomial (y ) +(y0) y y has order 3 and degree 4. · −

11. We now survey all simple differential field extensions b of a differential field of characteristic 0. Ah i CaseA 1: f(b) = 0 for some nonzero f y . Choose such an f of lowest possible order n, and then of lowest possible∈A{ degree} k for that order. Note that ∂f/∂y(n) has lower order or else the same order and lower degree. Hence

∂f (b) =0. ∂y(n) 6 Hence, as in 11.1.7, we have

n 1 ∂f (i+1) f(b)+ i=0− (i) (b) b b(n+1) = ∂y · ∂f (b) e P− ∂y(n) (n+1) (n 1) (n) (j) so in particular b (b,...,b − ,b ). From this it follows that b (n 1) (n) ∈A (n 1) (n) ∈ (b,...,b − ,b ) for all j, and hence the field (b,...,b − ,b )isclosed A (n 1) (n) A under 0. Hence b = (b,...,b − ,b ) and its structure is as in lemma 11.1.5. In this caseAh i we sayA that b is differentially algebraic over . A Case 2: negation of case 1. In this case it is easy to see that b = y , A{ } ∼ A{ } hence the quotient fields b ∼= y by lemma 11.1.4. In this case we say that b is differentially transcendentalAh i Ah iover . This completes our survey of the simpleA differential field extensions of . A 12. From the above survey of simple extensions we immediately obtain the following result.

Theorem. Let be a differential field of characteristic 0. Up to isomorphism over ,thereareexactlyA distinct simple differential field extensions b . A kAk Ah i 11.2. DIFFERENTIALLY CLOSED FIELDS 115 11.2 Differentially closed fields

Definition (Blum). Let be a differential field of characteristic 0. We say that is differentiallyB closed if is algebraically closed and, for all f,g yB with 0 order(g) < order(Bf), there exists b with f(b)=0, g(b)∈B{=0. } ≤ ∈|B| 6

Lemma. Let be a differential field of characteristic 0. Then there exists a differential fieldA such that (i) is differentially closed, (ii) is differentially algebraicB⊇A over , i.e. everyBb is differentially algebraicB over . A ∈|B| A

Proof. By transﬁnite induction, it suﬃces to show:

Sublemma. Given f y of order > 0, there exists a simple differential field extension b such∈A{ that} f(b)=0andg(b) =0forall g y with 0 order(g) < Ahorder(i f). 6 ∈A{ } ≤ (In this event b is called a generic solution of the differential equation f(y)= 0over .) A

To prove the sublemma, let n be the order of f and let = (b0,...,bn 1) A∗ A − be a ﬁeld extension of in which each bi, i

4. Remark. In the statement of the lemma is not unique. See 11.3.4 below. B

5. Theorem (Robinson, Blum). The theory of differentially closed fields of characteristic 0 (DCF(0)) is the model completion of the theory of differential fields of characteristic 0 (DF(0)).

Proof. By the previous lemma plus theorem 8.3.2 it suﬃces to show the following:

Let where is a DF(0), a DCF(0), κ+-saturated whereA⊆Bκ = . ThenA for any simpleB differentialB field extension c of , wekAk can find b such that b = c over . Ah i A ∈|B| Ah i ∼ Ah i A To prove this, we consider two cases. Case 1: c is differential algebraic over , i.e. f(c) = 0 for some nonzero f y .Takef to be of least order, andA of least degree for that order. Let∈A{p bethe1-typeover} generated by the differential equation f(y)=0 and the inequations g(y)A=0,g y ,0 order(g) < order(f). Since is differentially closed, every6 finite∈A{ subset} of≤p is realized in . Hence by B B saturation p is realized in ,saybyb . We claim that b ∼= c over . This is clear from 11.1.5B and 11.1.11.∈|B| Ah i Ah i A Case 2: c is differentially transcendental over . Consider the 1-type p over generated by g(y) =0forallg y ,0A order(g). Again, since is differentiallyA closed, each6 finite subset∈A{ of }p is realized≤ in . Hence p is realizedB in ,saybyb .Thenb is differentially transcendentalB over , B ∈|B| A so b = c . This completes the proof. Ah i ∼ Ah i

6. Corollary. DCF(0) is complete and decidable.

Proof. By the previous theorem DCF(0) (diagram of Q) is complete, but ∪ any DCF(0) has Q as a uniquely embedded diﬀerential subﬁeld, so DCF(0) is complete. Decidability follows by theorem 4.2.3. 11.3. DIFFERENTIAL CLOSURE (COUNTABLE CASE) 117

7. Corollary (Seidenberg). Let S be a system of finitely many algebraic differential equations in n unknown functions y1,...,yn and their derivatives, with rational coefficients. Then we can recursively decide whether S has a solution in some differential field of characteristic 0 (equivalently by 11.1.3 in some field of meromorphic functions).

8. Corollary. DCF(0) admits elimination of quantiﬁers.

Proof. From theorem 11.2.5 and lemma 11.1.4 it follows that DCF(0) is the model completion of the theory of diﬀerential domains of characteristic 0. The latter theory is universal so quantiﬁer elimination follows by theorem 8.2.3.

9. Corollary (Seidenberg). There exists an elimination theory for ﬁnite systems of algebraic diﬀerential equations and inequations (cf. 8.2.6).

11.3 Diﬀerential closure (countable case)

1. Definition. Let be a DF(0). A differential closure of is a differential A A field such that (i) is differentially closed, and (ii) if 1 is any B⊇A B B differentially closed field containing ,then is embeddable into 1 over . A B B A In view of theorem 11.2.5 this is equivalent to saying that is a prime model of the complete theory generated by DCF(0) (diagramB of ). ∪ A

2. Theorem (Blum). Let be a countable DF(0). Then has a diﬀerential closure, denoted . FurthermoreA is unique up to isomorphismA over . A A A 118 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0

(We shall see in chapters 12 and 13 that the hypothesis of countability can be dropped.)

Proof. Let T be the complete countable theory generated by DCF(0) (diagram of ). By quantifier elimination, a complete n-type over T ∪is essentially theA same thing as an isomorphism type of an n-fold simple differential field extension b1,...,bn over . Hence by theorem 11.1.12 it Ah i A follows that Sn(T ) is countable for all n. Hence by Vaught’s theorem 10.2.12 it follows that T has a prime model = . Furthermore, the uniqueness theorem for prime models 10.2.8 impliesB thatA is unique up to isomorphism over . A A 3. Remark. By lemma 11.2.2 it follows that the differential closure is differentially algebraic over . However, this fact alone does not sufficeA to characterize among all differentiallyA closed fields . For example,A let b be a simple extension in whichB⊇Ab is a generic solution Ah i of y0 = 0. In other words, b is transcendental over but b0 =0.Put = b .Then is differentially closed, and differentiallyA algebraic over B , butAh wei claim thatB is not isomorphic to over . To see this, let p be Athecomplete1-typeoverB T = DCF(0) (diagramA ofA ) realized by b . Clearly p is nonprincipal. Hence by theorem∪ 10.2.7 itA follows that p ∈|B|is not realized in . A 4. In order to prove our next theorem, we need the following algebraic result which we state without proof.

Lemma. Let be a differential field of characteristic 0. Let S be an infinite A set of algebraic differential equations in finitely many unknowns y1,...,yn with coefficients in . Then there exists a finite subset S0 of S such that |A| every solution of S0 (in some differential field extension of ) is a solution of S. A

This is an immediate corollary of the Ritt basis theorem, i.e. theorem 11.4.4 below. For a proof, see Kaplansky, Introduction to Diﬀerential Algebra. 11.3. DIFFERENTIAL CLOSURE (COUNTABLE CASE) 119

5. Theorem (Harrington). Let be a computable differential field of characteristic 0. Then , the differentialA closure of , is computable. A A Proof. is by definition the prime model of the complete theory T whose axiomsA are DCF(0) (diagram of ). Since is computable, T is recursively axiomatizable, and∪ hence decidable.A But theoremA 10.4.3 gives a necessary and sufficient condition for a complete decidable theory to have a decidable prime model. Thus, in order to show that is computable, it suffices to show that T verifies condition (b) of that theorem.A

Given ϕ Fn(T ) consistent with T , by quantifier elimination we may ∈ safely assume that ϕ(y1,...,yn) consists of finitely many algebraic diferential equations and inequations in n unknowns y1,...,yn with coefficients in . |A| Let ψk(y1,...,yn):k ω be a recursive enumeration of all algebraic { ∈ } ϕ differential equations in n unknowns y1,...,yn with coefficients in .Letp |A| be the n-type over T generated by ϕk : k ω where ϕ0 = ϕ, ϕk+1 = ϕk ψk { ∈ } ∧ ϕ if this is consistent with T , ϕk+1 = ϕk otherwise. The passage from ϕ to p is clearly recursive, and by quantifier elimination, pϕ is complete. It remains to show that pϕ is principal. Let S be the set of all algebraic ϕ differential equations in p . By lemma 11.3.4 there exists a finite set S0 S ⊆ such that S0 is equivalent to S over T .Letψk1 ,...,ψkm be the elements of ϕ S0. Then clearly ϕ ψk1 ψkm is a generator of p . This completes the proof. ∧ ∧···∧

Corollary. The diﬀerential closure of the rational ﬁeld Q is computable.

Remark. An explicit presentation of the diﬀerential closure of Q is lacking. It is unknown whether the following decision problem has a positive solution:

Given a finite system of differential equations and inequations in n unknowns y1,...,yn with rational coefficients, to decide whether the principal n-type generated by the system is complete. 120 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0

In the absence of such a decision procedure, the only known construction of the diﬀerential closure of Q is by means of Harrington’s priority construction in the proof of theorem 10.4.3.

11.4 Ritt’s Nullstellensatz

Let be a differential field of characteristic 0. Let = y1,...,yn be the A R A{ } ring of differential polynomials in n indeterminates y1,...,yn over .Let be the differential closure of . A A A 2.

Theorem (Ritt’s Nullstellensatz). Let f,g1,...,gm be such that every ∈R common zero of g1,...,gm in is a zero of f. Then there exist positive integers k and l such that A

m l k (j) f = pijgi i=1 j=1 X X (j) where pij and g is the jth derivative of gi. ∈R i Proof. Let I be the set of all f for which the conclusion holds. Suppose f/I.Thenwehave ∈R ∈ (i) g,h I g + h I ∈ ⇒ ∈ (ii) g I,h g h I ∈ ∈R⇒ · ∈ (iii) 0 I,1 / I ∈ ∈

(iv) g I g0 I ∈ ⇒ ∈ i.e. I is a diﬀerential ideal. Here (i), (ii), (iii) are proved as in the proof of Hilbert’s Nullstellensatz 6.4.2. For (iv), suppose g I,saygk 0 ∈ ≡ mod g1,...,gm. Diﬀerentiate and multiply by 1/k to get

k 1 g − g0 0modg1,...,gm . ≡ 11.4. RITT’S NULLSTELLENSATZ 121

This is the case i = 1 of the statement

k i 2i 1 g − (g0) − 0modg1,...,gm ≡ which we prove by induction on i,1 i k. Diﬀerentiating and multiplying ≤ ≤ by g0 we get

k i 1 2i+1 k i 2i 1 (k i)g − − (g0) +(2i 1)g − (g0) − g00 0 . − − ≡ 0 ≡ Multiplying by 1/(k i)weget | {z } − k i 1 2i+1 g − − (g0) 0modg1,...,gm ≡ so the induction step is proved. Finally for i = k we get

2k 1 (g0) − 0modg1,...,gm ≡ so (iv) is proved. It is also clear that I is a radical differential ideal, i.e. I satisfies (v) gk I g I, ∈ ⇒ ∈ and f/I. By Zorn’s lemma let J I be a radical differential ideal such that f/∈ J and maximal with this property.⊇ We∈ claim that J is prime, i.e. g/J, h / J imply g h/J. Suppose not. ∈ ∈ · ∈ 2 Note that g h J implies (g0 h + g h0) g0 h J whence (g0 h) J, · ∈ (i)· (j) · · · ∈ · ∈ whence g0 h J. Similarly g h J for all i, j 0. If g/J then the radical differential· ∈ ideal generated· by J∈ and g must contain≥ f,say∈

k (i) f = s + pig i X where s J, pi . Similarly if h/J then ∈ ∈R ∈ l (j) f = t + qjh j X where t J, qj . Hence ∈ ∈R k+l l k (i) (j) f = s f + t f + piqjg h · · i j X X J ∈ | {z } 122 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0 contradicting the fact that J is radical and f/J.Thisprovestheclaim. ∈ Now let 1 = /J = the quotient ring of mod J. Clearly 1 is R R R R a differential domain. Let be the field of quotients of 1.Then is a B R B differential field extension of . Furthermore, there exist b1,...,bn A ∈|B| (corresponding to y1,...,yn ) such that for all g , g(b1,...,bn)=0 ∈|R| ∈R if and only if g J. In particular f(b1,...,bn) = 0 while g1(b1,...,bn)= ∈ 6 = gm(b1,...,bn) = 0. Hence by model completeness of DCF(0) we can ··· find a1,...,an such that f(a1,...,an) =0andg1(a1,...,an)= = ∈|A| 6 ··· gm(a1,...,an) = 0. This completes the proof of the theorem.

3. Remark. As in the case of Hilbert’s Nullstellensatz, we can deduce a version of Ritt’s Nullstellensatz with eﬀective bounds. We leave to the reader the formulation and proof of this result. (Compare 6.4.3.)

4. An interesting theorem related to Ritt’s Nullstellensatz is the following.

Theorem (Ritt’s basis theorem). Let be as above and let I be a radical differential ideal in .ThenI has a Rfinite basis, i.e. there exists a finite R set g1,...,gr I such that I is the radical differential ideal generated by ∈ g1,...,gr, i.e. the set of all f such that ∈R r l k (j) f = pijgi i=0 j=0 X X for some k, l ω and pij . ∈ ∈R Proof. See Kaplansky, Introduction to Differential Algebra.

5. Remark. The Ritt Basis Theorem tends to clarify several aspects of the proof of the Ritt Nullstellensatz. In the first place, the Ritt Basis Theorem shows that the use of Zorn’s lemma could have been eliminated. In the second place, since the radical differential ideal J has a finite basis g1,...,gr, 11.4. RITT’S NULLSTELLENSATZ 123

it follows by quantifier elimination that the complete n-type of b1,...,bn over is generated by the differential equations g1(y1,...,yn)= = |A| ··· gr(y1,...,yn)=0andthesingleinequationf(y1,...,yn) = 0. In particular this n-type is principal2 and hence already realized in . In6 other words, the A differential field = b1,...,bn is already embeddable into .(Wecould have used Hilbert’sB BasisAh Theoremi to make similar remarks aboutA the proof of the Hilbert Nullstellensatz.)

2The algebraists express this fact by saying that J is constrained. See Wood, Israel J. Math. vol. 25, p. 331. 124 CHAPTER 11. DIFF. CLOSED FIELDS OF CHAR. 0 Chapter 12

Totally transcendental theories

12.1 Stability

1. 12.2 Rank of an element type 12.3 Indiscernibles 12.4 Existence of saturated models

125 126 CHAPTER 12. TOTALLY TRANSCENDENTAL THEORIES Chapter 13

Prime models (uncountable case)

13.1 Strongly atomic models

1. 13.2 Normal sets 13.3 Uniqueness and characterization of prime models

127