学习模型论，何其难？

四十年过去了，在国内学习纯粹数学（例如：模型论）仍然困难重重，甚至无人问津。

什么是数学模型理论？国内学界不发声，不说话，令人很无奈。

为此，我们推荐一篇科普文章，请见本本文附件。

袁萌陈启清 12月14日

附件：
Fundamentals of Model Theory
William Weiss and Cherie D’Mello
Department of Mathematics University of Toronto
c 2015 W.Weiss and C. D’Mello

1
Introduction
Model Theory is the part of mathematics which shows how to apply logic to the study of structures in pure mathematics. On the one hand it is the ultimate abstraction; on the other, it has immediate applications to every-day mathematics. The fundamental tenet of Model Theory is that mathematical truth, like all truth, is relative. A statement may be true or false, depending on how and where it is interpreted. This isn’t necessarily due to mathematics itself, but is a consequence of the language that we use to express mathematical ideas. What at ﬁrst seems like a deﬁciency in our language, can actually be shaped into a powerful tool for understanding mathematics. This book provides an introduction to Model Theory which can be used as a text for a reading course or a summer project at the senior undergraduate or graduate level. It is also a primer which will give someone a self contained overview of the subject, before diving into one of the more encyclopedic standard graduate texts. Any reader who is familiar with the cardinality of a set and the algebraic closure of a ﬁeld can proceed without worry. Many readers will have some acquaintance with elementary logic, but this is not absolutely required, since all necessary concepts from logic are reviewed in Chapter 0. Chapter 1 gives the motivating examples; it is short and we recommend that you peruse it ﬁrst, before studying the more technical aspects of Chapter 0. Chapters 2 and 3 are selections of some of the most important techniques in Model Theory. The remaining chapters investigate the relationship between Model Theory and the algebra of the real and complex numbers. Thirty exercises develop familiarity with the deﬁnitions and consolidate understanding of the main proof techniques. Throughout the book we present applications which cannot easily be found elsewhere in such detail. Some are chosen for their value in other areas of mathematics: Ramsey’s Theorem, the Tarski-Seidenberg Theorem. Some are chosen for their immediate appeal to every mathematician: existence of inﬁnitesimals for calculus, graph colouring on the plane. And some, like Hilbert’s Seventeenth Problem, are chosen because of how amazing it is that logic can play an important role in the solution of a problem from high school algebra. In each case, the derivation is shorter than any which tries to avoid logic. More importantly, the methods of Model Theory display clearly the structure of the main ideas of the proofs, showing how theorems of logic combine with theorems from other areas of mathematics to produce stunning results. The theorems here are all are more than thirty years old and due in great part to the cofounders of the subject, Abraham Robinson and Alfred Tarski. However, we have not attempted to give a history. When we attach a name to a theorem, it is simply because that is what mathematical logicians popularly call it. The bibliography contains a number of texts that were helpful in the preparation of this manuscript. They could serve as avenues of further study and in addition, they contain many other references and historical notes. The more recent titles were added to show the reader where the subject is moving today. All are worth a look. This book began life as notes for William Weiss’s graduate course at the University of Toronto. The notes were revised and expanded by Cherie D’Mello and
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William Weiss, based upon suggestions from several graduate students. The electronic version of this book may be downloaded and further modiﬁed by anyone for the purpose of learning, provided this paragraph is included in its entirety and so long as no part of this book is sold for proﬁt.
Contents
Chapter 0. Models, Truth and Satisfaction 4 Formulas, Sentences, Theories and Axioms 4 Prenex Normal Form 9
Chapter 1. Notation and Examples 11
Chapter 2. Compactness and Elementary Submodels 14 The Compactness Theorem 14 Isomorphisms, elementary equivalence and complete theories 15 The Elementary Chain Theorem 16 The L¨owenheim-Skolem Theorem 19 The L o´s-Vaught Test 20 Every complex one-to-one polynomial map is onto 22
Chapter 3. Diagrams and Embeddings 24 Diagram Lemmas 25 Every planar graph can be four coloured 25 Ramsey’s Theorem 26 The Leibniz Principle and inﬁnitesimals 27 The Robinson Consistency Theorem 27 The Craig Interpolation Theorem 31
Chapter 4. Model Completeness 32 Robinson’s Theorem on existentially complete theories 32 Lindstr¨om’s Test 35 Hilbert’s Nullstellensatz 37
Chapter 5. The Seventeenth Problem 39 Positive deﬁnite rational functions are the sums of squares 39
Chapter 6. Submodel Completeness 45 Elimination of quantiﬁers 45 The Tarski-Seidenberg Theorem 48
Chapter 7. Model Completions 50 Almost universal theories 52 Saturated models 54 Blum’s Test 55
Bibliography 60
Index 61
3
CHAPTER 0
Models, Truth and Satisfaction
We will use the following symbols: • logical symbols: – the connectives ∧ ,∨ , ¬ , → , ↔ called “and”, “or”, “not”, “implies” and “iﬀ” respectively – the quantiﬁers ∀ , ∃ called “for all” and “there exists” – an inﬁnite collection of variables indexed by the natural numbers N v0 ,v1 , v2 , ... – the two parentheses ), ( – the symbol = which is the usual “equal sign” • constant symbols : often denoted by the letter c with subscripts • function symbols : often denoted by the letter F with subscripts; each function symbol is an m-placed function symbol for some natural number m ≥ 1 • relation symbols : often denoted by the letter R with subscripts; each relational symbol is an n-placed relation symbol for some natural number n ≥ 1. We now deﬁne terms and formulas. Definition 1. A term is deﬁned as follows: (1) a variable is a term (2) a constant symbol is a term (3) if F is an m-placed function symbol and t1,...,tm are terms, then F(t1 ...tm) is a term. (4) a string of symbols is a term if and only if it can be shown to be a term by a ﬁnite number of applications of (1), (2) and (3). Remark. This is a recursive deﬁnition. Definition 2. A formula is deﬁned as follows : (1) if t1 and t2 are terms, then (t1 = t2) is a formula. (2) if R is an n-placed relation symbol and t1,...,tn are terms, then (R(t1 ...tn)) is a formula. (3) if ϕ is a formula, then (¬ϕ) is a formula (4) if ϕ and ψ are formulas then so are (ϕ∧ψ), (ϕ∨ψ), (ϕ → ψ) and (ϕ ↔ ψ) (5) if vi is a variable and ϕ is a formula, then (∃vi)ϕ and (∀vi)ϕ are formulas (6) a string of symbols is a formula if and only if it can be shown to be a formula by a ﬁnite number of applications of (1), (2), (3), (4) and (5). Remark. This is another recursive deﬁnition. ¬ϕ is called the negation of ϕ;ϕ ∧ψ is called the conjunction of ϕ and ψ; and ϕ∨ψ is called the disjunction of ϕand ψ.
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0. MODELS, TRUTH AND SATISFACTION 5
Definition 3. A subformula of a formula ϕ is deﬁned as follows: (1) ϕ is a subformula of ϕ (2) if (¬ψ) is a subformula of ϕ then so is ψ (3) if any one of (θ∧ψ), (θ∨ψ), (θ → ψ) or (θ ↔ ψ) is a subformula of ϕ, then so are both θ and ψ (4) if either (∃vi)ψ or (∀vi)ψ is a subformula of ϕ for some natural number i, then ψ is also a subformula of ϕ (5) A string of symbols is a subformula of ϕ, if and only if it can be shown to be such by a ﬁnite number of applications of (1), (2), (3) and (4).
Definition 4. A variable vi is said to occur bound in a formula ϕ iﬀ for some subformula ψ of ϕ either (∃vi)ψ or (∀vi)ψ is a subformula of ϕ. In this case each occurrence of vi in (∃vi)ψ or (∀vi)ψ is said to be a bound occurrence of vi. Other occurrences of vi which do not occur bound in ϕ are said to be free.
Example 1.
F(v3) is a term, where F is a unary function symbol. ((∃v3)(v0 = v3)∧(∀v0)(v0 = v0)) is a formula. In this formula the variable v3 only occurs bound but the variable v0 occurs both bound and free.
Exercise 1. Using the previous deﬁnitions as a guide, deﬁne the substitution of a term t for a variable vi in a formula ϕ. In particular, demonstrate how to substitute the term for the variable v0 in the formula of the example above. Definition 5. A language L is a set consisting of all the logical symbols with perhaps some constant, function and/or relational symbols included. It is understood that the formulas of L are made up from this set in the manner prescribed above. Note that all the formulas of L are uniquely described by listing only the constant, function and relation symbols of L. We use t(v0,...,vk) to denote a term t all of whose variables occur among v0,...,vk. We use ϕ(v0,...,vk) to denote a formula ϕ all of whose free variables occur among v0,...,vk.
Example 2. These would be formulas of any language : • For any variable vi: (vi = vi) • for any term t(v0,...,vk) and other terms t1 and t2: ((t1 = t2) → (t(v0,...,vi−1,t1,vi+1,...,vk) = t(v0,...,vi−1,t2,vi+1,...,vk))) • for any formula ϕ(v0,...,vk) and terms t1 and t2: ((t1 = t2) → (ϕ(v0,...,vi−1,t1,vi+1,...,vk) ↔ ϕ(v0,...,vi−1,t2,vi+1,...,vk))) Note the simple way we denote the substitution of t1 for vi. Definition 6. A model (or structure) A for a language L is an ordered pair hA,Ii where A is a nonempty set and I is an interpretation function with domain the set of all constant, function and relation symbols of L such that: (1) if c is a constant symbol, then I(c) ∈ A; I(c) is called a constant
0. MODELS, TRUTH AND SATISFACTION 6
(2) if F is an m-placed function symbol, then I(F) is an m-placed function on A (3) if R is an n-placed relation symbol, then I(R) is an n-placed relation on A. A is called the universe of the model A. We generally denote models with Gothic letters and their universes with the corresponding Latin letters in boldface. One set may be involved as a universe with many diﬀerent interpretation functions of the language L. The model is both the universe and the interpretation function. Remark. The importance of Model Theory lies in the observation that mathematical objects can be cast as models for a language. For instance, the real numbers with the usual ordering < < < and the usual arithmetic operations, addition + + + and multiplication • • • along with the special numbers 0 and 1 can be described as a model.Let L contain one two-placed (i.e. binary) relation symbol R0, two two-placed function symbols F1 and F2 and two constant symbols c0 and c1. We build a model by letting the universe A be the set of real numbers. The interpretation function I will map R0 to < < <, i.e. R0 will be interpreted as < < <. Similarly, I(F1) will be + + +, I(F2) will be • • •, I(c0) will be 0 and I(c1) will be 1. So hA,Ii is an example of a model for the language described by {R0,F1,F2,c0,c1}. We now wish to show how to use formulas to express mathematical statements about elements of a model. We ﬁrst need to see how to interpret a term in a model.
Definition 7. The value t[x0,...,xq] of a term t(v0,...,vq) at x0,...,xq in the universe A of the model A is deﬁned as follows: (1) if t is vi then t[x0,••• ,xq] is xi, (2) if t is the constant symbol c, then t[x0,...,xq] is I(c), the interpretation of c in A, (3) if t is F(t1 ...tm) where F is an m-placed function symbol and t1,...,tm are terms, then t[x0,...,xq] is G(t1[x0,...,xq],...,tm[x0,...,xq]) where G is the m-placed function I(F), the interpretation of F in A. Definition 8. Suppose A is a model for a language L. The sequencex 0,...,xq of elements of A satisﬁes the formula ϕ(v0,...,vq) all of whose free and bound variables are among v0,...,vq, in the model A, written A |= ϕ[x0,...,xq] provided we have: (1) if ϕ(v0,...,vq) is the formula (t1 = t2), then A |= (t1 = t2)[x0,...,xq] means that t1[x0,...,xq] equals t2[x0,...,xq], (2) if ϕ(v0,...,vq) is the formula (R(t1 ...tn)) where R is an n-placed relation symbol, then A |= (R(t1 ...tn))[x0,...,xq] means S(t1[x0,...,xq],...,tn[x0,...,xq]) where S is the n-placed relation I(R), the interpretation of R in A,(3) if ϕ is (¬θ), then A |= ϕ[x0,...,xq] means not A |= θ[x0,...,xq], (4) if ϕ is (θ∧ψ), then A |= ϕ[x0,...,xq] means both A |= θ[x0,...,xq] and A |= ψ[x0,...xq],
0. MODELS, TRUTH AND SATISFACTION 7
(5) if ϕ is (θ∨ψ) then A |= ϕ[x0,...,xq] means either A |= θ[x0,...,xq] or A |= ψ[x0,...,xq], (6) if ϕ is (θ → ψ) then A |= ϕ[x0,...,xq] means that A |= θ[x0,...,xq] implies A |= ψ[x0,...,xq], (7) if ϕ is (θ ↔ ψ) then A |= ϕ[x0,...,xq] means that A |= θ[x0,...,xq] iﬀ A |= ψ[x0,...,xq], (8) if ϕ is ∀viθ, then A |= ϕ[x0,...,xq] means for every x ∈ A,A |= θ[x0,...,xi−1,x,xi+1,...,xq], (9) if ϕ is ∃viθ, then A |= ϕ[x0,...,xq] means for some x ∈ A,A |= θ[x0,...,xi−1,x,xi+1,...,xq]. Exercise 2. Each of the formulas of Example 2 is satisﬁed in any model A for any language L by any (long enough) sequence x0,x1,...,xq of A. This is where you test your solution to Exercise 1, especially with respect to the term and formula from Example 1.
We now prove two lemmas which show that the preceding concepts are welldeﬁned. In the ﬁrst one, we see that the value of a term only depends upon the values of the variables which actually occur in the term. In this lemma the equal sign = is used, not as a logical symbol in the formal sense, but in its usual sense to denote equality of mathematical objects — in this case, the values of terms, which are elements of the universe of a model. Lemma 1. Let A be a model for L and let t(v0,...,vp) be a term of L. Letx 0,...,xq and y0,...,yr be sequences from A such that p ≤ q and p ≤ r, and letx i = yi whenever vi actually occurs in t(v0,...,vp). Then t[x0,...,xq] = t[y0,...,yr] .
Proof. We use induction on the complexity of the term t. (1) If t is vi then xi = yi and so we have t[x0,...,xq] = xi = yi = t[y0,...,yr] since p ≤ q and p ≤ r. (2) If t is the constant symbol c, then t[x0,...,xq] = I(c) = t[y0,...,yr] where I(c) is the interpretation of c in A.(3) If t is F(t1 ...tm) where F is an m-placed function symbol, t1,...,tm are terms and I(F) = G, then t[x0,...,xq] = G(t1[x0,...,xq],...,tm[x0,...,xq]) and t[y0,...,yr] = G(t1[y0,...,yr],...,tm[y0,...,yr]). By the induction hypothesis we have that ti[x0,...,xq] = ti[y0,...,yr] for 1 ≤ i ≤ m since t1,...,tm have all their variables among {v0,...,vp}. So we have t[x0,...,xq] = t[y0,...,yr].
0. MODELS, TRUTH AND SATISFACTION 8
In the next lemma the equal sign = is used in both senses — as a formal logical symbol in the formal language L and also to denote the usual equality of mathematical objects. This is common practice where the context allows the reader to distinguish the two usages of the same symbol. The lemma conﬁrms that satisfaction of a formula depends only upon the values of its free variables. Lemma 2. Let A be a model for L and ϕ a formula of L, all of whose free and bound variables occur among v0,...,vp. Let x0,...,xq and y0,...,yr (q,r ≥ p) be two sequences such that xi and yi are equal for all i such that vi occurs free in ϕ. Then A |= ϕ[x0,...,xq] iﬀ A |= ϕ[y0,...,yr] Proof. Let A and L be as above. We prove the lemma by induction on the complexity of ϕ. (1) If ϕ(v0,...,vp) is the formula (t1 = t2), then we use Lemma 1 to get: A |= (t1 = t2)[x0,...,xq] iﬀ t1[x0,...,xq] = t2[x0,...,xq] iﬀ t1[y0,...,yr] = t2[y0,...,yr] iﬀ A |= (t1 = t2)[y0,...,yr]. (2) If ϕ(v0,...,vp) is the formula (R(t1 ...tn)) where R is an n-placed relation symbol with interpretation S, then again by Lemma 1, we get: A |= (R(t1 ...tn))[x0,...,xq] iﬀ S(t1[x0,...,xq],...,tn[x0,...,xq]) iﬀ S(t1[y0,...,yr],...,tn[y0,...,yr]) iﬀ A |= R(t1 ...tn)[y0,...,yr]. (3) If ϕ is (¬θ), the inductive hypothesis gives that the lemma is true for θ. So, A |= ϕ[x0,...,xq] iﬀ not A |= θ[x0,...,xq] iﬀ not A |= θ[y0,...,yr] iﬀ A |= ϕ[y0,...,yr]. (4) If ϕ is (θ∧ψ), then using the inductive hypothesis on θ and ψ we get A |= ϕ[x0,...,xq] iﬀ both A |= θ[x0,...,xq] and A |= ψ[x0,...xq] iﬀ both A |= θ[y0,...,yr] and A |= ψ[y0,...yr] iﬀ A |= ϕ[y0,...,yr]. (5) If ϕ is (θ∨ψ) then A |= ϕ[x0,...,xq] iﬀ either A |= θ[x0,...,xq] or A |= ψ[x0,...,xq] iﬀ either A |= θ[y0,...,yr] or A |= ψ[y0,...,yr] iﬀ A |= ϕ[y0,...,yr]. (6) If ϕ is (θ → ψ) then A |= ϕ[x0,...,xq] iﬀ A |= θ[x0,...,xq] implies A |= ψ[x0,...,xq] iﬀ A |= θ[y0,...,yr] implies A |= ψ[y0,...,yr] iﬀ A |= ϕ[y0,...,yr].
0. MODELS, TRUTH AND SATISFACTION 9
(7) If ϕ is (θ ↔ ψ) then A |= ϕ[x0,...,xq] iﬀ we have A |= θ[x0,...,xq] iﬀ A |= ψ[x0,...,xq] iﬀ we have A |= θ[y0,...,yr] iﬀ A |= ψ[y0,...,yr] iﬀ A |= ϕ[y0,...,yr]. (8) If ϕ is (∀vi)θ, then A |= ϕ[x0,...,xq] iﬀ for every z ∈ A,A |= θ[x0,...,xi−1,z,xi+1,...,xq] iﬀ for every z ∈ A,A |= θ[y0,...,yi−1,z,yi+1,...,yr] iﬀ A |= ϕ[y0,...,yr]. The inductive hypothesis uses the sequences x0,...,xi−1,z,xi+1,...,xq and y0,...,yi−1,z,yi+1,...,yr with the formula θ. (9) If ϕ is (∃vi)θ, then A |= ϕ[x0,...,xq] iﬀ for some z ∈ A,A |= θ[x0,...,xi−1,z,xi+1,...,xq] iﬀ for some z ∈ A,A |= θ[y0,...,yi−1,z,yi+1,...,yr] iﬀ A |= ϕ[y0,...,yr]. The inductive hypothesis uses the sequences x0,...,xi−1,z,xi+1,...,xq and y0,...,yi−1,z,yi+1,...,yr with the formula θ.   Definition 9. A sentence is a formula with no free variables. If ϕ is a sentence, we can write A |= ϕ without any mention of a sequence fromA since by the previous lemma, it doesn’t matter which sequence from A we use. In this case we say: • A satisﬁes ϕ • or A is a model of ϕ • or ϕ holds in A • or ϕ is true in A If ϕ is a sentence of L, we write |= ϕ to mean that A |= ϕ for every model Afor L. Intuitively then, |= ϕ means that ϕ is true under any relevant interpretation (model forL). Alternatively, no relevant example (model forL) is a counterexample to ϕ — so ϕ is true. Lemma 3. Let ϕ(v0,...,vq) be a formula of the language L. There is anotherformula ϕ0(v0,...,vq) of L such that (1) ϕ0 has exactly the same free and bound occurrences of variables as ϕ. (2) ϕ0 can possibly contain ¬, ∧ and ∃ but no other connective or quantiﬁer. (3) |= (∀v0)...(∀vq)(ϕ ↔ ϕ0) Exercise 3. Prove the above lemma by induction on the complexity of ϕ. As a reward, note that this lemma can be used to shorten future proofs by induction on complexity of formulas.
Definition 10. A formula ϕ is said to be in prenex normal form whenever (1) there are no quantiﬁers occurring in ϕ, or (2) ϕ is (∃vi)ψ where ψ is in prenex normal form and vi does not occur bound in ψ, or
0. MODELS, TRUTH AND SATISFACTION 10
(3) ϕ is (∀vi)ψ where ψ is in prenex normal form and vi does not occur bound in ψ.
Remark. If ϕ is in prenex normal form, then no variable occurring in ϕ occurs both free and bound and no bound variable occurring in ϕ is bound by more than one quantiﬁer. In the written order, all of the quantiﬁers precede all of the connectives. Lemma 4. Let ϕ(v0,...,vp) be any formula of a language L. There is a formulaϕ ∗ of L which has the following properties: (1) ϕ∗ is in prenex normal form (2) ϕ and ϕ∗ have the same free occurrences of variables, and (3) |= (∀v0)...(∀vp)(ϕ ↔ ϕ∗) Exercise 4. Prove this lemma by induction on the complexity of ϕ.
There is a notion of rank on prenex formulas — the number of alternations of quantiﬁers. The usual formulas of elementary mathematics have prenex rank 0, i.e. no alternations of quantiﬁers. For example: (∀x)(∀y)(2xy ≤ x2 + y2). However, the −δ deﬁnition of a limit of a function has prenex rank 2 and is much more diﬃcult for students to comprehend at ﬁrst sight: (∀ )(∃δ)(∀x)((0 < ∧0 < |x−a| < δ) →|F(x)−L| < ). A formula of prenex rank 4 would make any mathematician look twice.
CHAPTER 1
Notation and Examples
Although the formal notation for formulas is precise, it can become cumbersome and diﬃcult to read. Conﬁdent that the reader would be able, if necessary, to put formulas into their formal form, we will relax our formal behaviour. In particular, we will write formulas any way we want using appropriate symbols for variables, constant symbols, function and relation symbols. We will omit parentheses or add them for clarity. We will use binary function and relation symbols between the arguments rather than in front as is the usual case for “plus”, “times” and “less than”. Whenever a language L has only ﬁnitely many relation, function and constant symbols we often write, for example: L = {<,R0,+,F1,c0,c1} omitting explicit mention of the logical symbols (including the inﬁnitely many variables) which are always in L. Correspondingly we may denote a model A for L as: A = hA,< < <,S0,+ + +,G1,a0,a1i where the interpretations of the symbols in the language L are given by I(<) = < < <, I(R0) = S0, I(+) = + + + , I(F1) = G1, I(c0) = a0 and I(c1) = a1. Example 3. R = hR,< < <,+ + +,•,0,1i and Q = hQ,< < <,+ + +,•,0,1i, where R is thereals and Q the rationals, are models for the language L = {<,+,•,0,1}. Here < is a binary relation symbol, + and • are binary function symbols, 0 and 1 are constant symbols whereas < < <, + + +, •, 0, 1 are the well known relations, arithmetic functionsand constants. Similarly, C = hC,+ + +,•,0,1i, where C is the complex numbers, is a model forthe language L = {+,•,0,1}. Note the exceptions to the boldface convention forthese popular sets. Example 4. Here L = {<,+,•,0,1}, where < is a binary relation symbol, +and • are binary function symbols and 0 and 1 are constant symbols. The following formulas are sentences. (1) (∀x)¬(x < x) (2) (∀x)(∀y)¬(x < y∧y < x) (3) (∀x)(∀y)(∀z)(x < y∧y < z → x < z) (4) (∀x)(∀y)(x < y∨y < x∨x = y) (5) (∀x)(∀y)(x < y → (∃z)(x < z∧z < y)) (6) (∀x)(∃y)(x < y) (7) (∀x)(∃y)(y < x) (8) (∀x)(∀y)(∀z)(x + (y + z) = (x + y) + z) (9) (∀x)(x + 0 = x) 11
1. NOTATION AND EXAMPLES 12
(10) (∀x)(∃y)(x + y = 0) (11) (∀x)(∀y)(x + y = y + x) (12) (∀x)(∀y)(∀z)(x•(y•z) = (x•y)•z) (13) (∀x)(x•1 = x) (14) (∀x)(x = 0∨(∃y)(y•x = 1)) (15) (∀x)(∀y)(x•y = y•x) (16) (∀x)(∀y)(∀z)(x•(y + z) = (x•y) + (y•z)) (17) 0 6= 1 (18) (∀x)(∀y)(∀z)(x < y → x + z < y + z) (19) (∀x)(∀y)(∀z)(x < y∧0 < z → x•z < y•z) (20) for each n ≥ 1 we have the formula (∀x0)(∀x1)•••(∀xn)(∃y)(xn •yn + xn−1 •yn−1 +•••+ x1 •y + x0 = 0∨xn = 0)
where, as usual, yk abbreviates
k z }| { y•y•••••y The latter formulas express that each polynomial of degree n has a root. The following formulas express the intermediate value property for polynomials of degree n: if the polynomial changes sign from w to z, then it is zero at some y between w and z. (21) for each n ≥ 1 we have (∀x0)...(∀xn)(∀w)(∀z)[(xn •wn + xn−1 •wn−1 +•••+ x1 •w + x0)• (xn •zn + xn−1 •zn−1 +•••+ x1 •z + x0) < 0 → (∃y)(((w < y∧y < z)∨(z < y∧y < w)) ∧(xn •yn + xn−1 •yn−1 +•••+ x1 •y + x0 = 0))] The most fundamental concept is that of a sentence σ being true when interpreted in a model A. We write this as A |= σ, and we extend this concept in the following deﬁnitions.
Definition 11. If Σ is a set of sentences, A is said to be a model of Σ, written A |= Σ, whenever A |= σ for each σ ∈ Σ. Σ is said to be satisﬁable iﬀ there is some A such that A |= Σ. Definition 12. A theory T is a set of sentences. If T is a theory and σ is a sentence, we write T |= σ whenever we have that for all A if A |= T then A |= σ. We say that σ is a consequence of T. A theory is said to be closed whenever it contains all of its consequences. Definition 13. If A is a model for the language L, the theory of A, denotedby Th A, is deﬁned to be the set of all sentences of L which are true in A, {σ of L : A |= σ}. This is one way that a theory can arise. Another way is through axioms. Definition 14. Σ ⊆T is said to be a set of axioms for T whenever Σ |= σ forevery σ in T; in this case we write Σ |= T. Remark. We will generally assume our theories are closed and we will often describe theories by specifying a set of axioms Σ. The theory will then be all consequences σ of Σ.
1. NOTATION AND EXAMPLES 13
Example 5. We will consider the following theories and their axioms. (1) The theory of Linear Orderings (LOR) is a theory in the language {<} which has as axioms sentences 1-4 from Example 4. (2) The theory of Dense Linear Orders (DLO) is a theory in the language{<} which has as axioms all the axioms of LOR, and sentences 5, 6 and 7 of Example 4. (3) The theory of Fields (FLD) is a theory in the language {0,1,+,•} which has as axioms sentences 8-17 from Example 4. (4) The theory of Ordered Fields (ORF) is a theory in the language given by {<,0,1,+,•} which has as axioms all the axioms of FLD, LOR and sentences 18 and 19 from Example 4. (5) The theory of Algebraically Closed Fields (ACF) is a theory in the language {0,1,+,•} which has as axioms all the axioms of FLD and all sentences from 20 of Example 4, i.e. inﬁnitely many sentences, one for each n ≥ 1. (6) The theory of Real Closed Ordered Fields (RCF) is a theory in the language {<,0,1,+,•} which has as axioms all the axioms of ORF, and all sentences from 21 of Example 4, i.e. inﬁnitely many sentences, one for each n ≥ 1. Exercise 5. Show that : (1) Q |= DLO (2) R |= RCF using the Intermediate Value theorem (3) C |= ACF using the Fundamental Theorem of Algebra where Q, R and C are as in Example 3.
Remark. The theory of Real Closed Ordered Fields is sometimes axiomatised diﬀerently. All the axioms of ORF are retained, but the sentences from 21 of Example 4, which amount to an Intermediate Value Property, are replaced by the sentences from 20 for odd n and the sentence (∀x)(0 < x → (∃y)y2 = x) which states that every positive element has a square root. A signiﬁcant amount of algebra would then be used to verify the Intermediate Value Property from these axioms.
CHAPTER 2
Compactness and Elementary Submodels
Theorem 1. The Compactness Theorem (Malcev) A set of sentences is satisﬁable iﬀ every ﬁnite subset is satisﬁable.
Proof. There are several proofs. We only point out here that it is an easy consequence of the following theorem which appears in all elementary logic texts:
Proposition. The Completeness Theorem (G¨odel, Malcev) A set of sentences is consistent if and only if it is satisﬁable.
Although we do not here formally deﬁne “consistent”, it does mean what you think it does. In particular, a set of sentences is consistent if and only if each ﬁnite subset is consistent.
Remark. The Compactness Theorem is the only one for which we do not give a complete proof. For the reader who has not previously seen the Completeness Theorem, there are other proofs of the Compactness Theorem which may be more easily absorbed: set theoretic (using ultraproducts), topological (using compact spaces, hence the name) or Boolean algebraic. However these topics are too far aﬁeld to enter into the proofs here. We will use the Compactness Theorem as a starting point — in fact, all that follows can be seen as its corollaries. Exercise 6. Suppose T is a theory for the language L and σ is a sentence of L such that T |= σ. Prove that there is some ﬁnite T0 ⊆T such that T0 |= σ. Recall that T |= σ iﬀ T ∪{¬σ} is not satisﬁable. Definition 15. If L, and L0 are two languages such that L⊆L0 we say that L0 is an expansion of L and L is a reduction of L0. Of course when we say that L⊆L0 we also mean that the constant, function and relation symbols of L remain (respectively) constant, function and relation symbols of the same type in L0. Definition 16. Given a model A for the language L, we can expand it to amodel A0 ofL0, whereL0 is an expansion ofL, by giving appropriate interpretations to the symbols in L0\L. We say that A0 is an expansion of A to L0 and that A is a reduct of A0 to L. We also use the notation A0|L for the reduct of A0 to L. Theorem 2. If a theory T has arbitrarily large ﬁnite models, then it has an inﬁnite model. Proof. Consider new constant symbols ci for i ∈ N, the usual natural numbers, and expand from L, the language of T, to L0 = L∪{ci : i ∈N}. Let Σ = T ∪{¬ci = cj : i 6= j,i,j ∈N}. 14
2. COMPACTNESS AND ELEMENTARY SUBMODELS 15
We ﬁrst show that every ﬁnite subset of Σ has a model by interpreting the ﬁnitely many relevant constant symbols as diﬀerent elements in an expansion of some ﬁnite model of T. Then we use compactness to get a model A0 of Σ. The model that we require is for the language L, so we take A to be the reductof A0 to L.   Definition 17. Two models A and A0 forLare said to be isomorphic whenever there is a bijection f : A → A0 such that (1) for each n-placed relation symbol R ofLand corresponding interpretations S of A and S0 of A0 we have S(x1,...,xn) iﬀ S0(f(x1),...,f(xn)) for all x1,...,xn in A (2) for each n-placed function symbol F of L and corresponding interpretations G of A and G0 of A0 we have f(G(x1,...,xn)) = G0(f(x1),...,f(xn)) for all x1,...,xn in A (3) for each constant symbol c of L and corresponding constant elements a of A and a0 of A0 we have f(a) = a0. We write A ∼ = A0. This is an equivalence relation. Example 6. Number theory is ThhN,+ + +,• • •,< < <,0 0 0,1 1 1i, the set of all sentences of L = {+,•,<,0,1} which are true in hN,+ + +,• • •,< < <,0 0 0,1 1 1i, the standard model which we all learned in school. Any model not isomorphic to the standard model of number theory is said to be a non-standard model of number theory.
Theorem 3. (T. Skolem) There exist non-standard models of number theory. Proof. Add a new constant symbol c to L. Consider ThhN,+,•,<,0,1i∪{ n z }| { 1 + 1 +•••+ 1 < c : n ∈N} and use the Compactness Theorem. The interpretation of the constant symbol c will not be a natural number.   Definition 18. Two models A and A0 for L are said to be elementarily equiv-alent whenever we have that for each sentence σ of L A |= σ iﬀ A0 |= σ We write A ≡ A0. This is another equivalence relation. Exercise 7. Suppose f : A → A0 is an isomorphism and ϕ is a formula suchthat A |= ϕ[a0,...,ak] for some a0,...,ak from A; prove A0 |= ϕ[f(a0),...,f(ak)]. Use this to show that A ∼ = A0 implies A ≡ A0. Definition 19. A model A0 is called a submodel of A, and we write A0 ⊆ A whenever φ 6= A0 ⊆ A and (1) each n-placed relation S0 of A0 is the restriction to A0 of the corresponding relation S of A, i. e. S0 = S ∩(A0)n (2) each m-placed function G0 of A0 is the restriction to A0 of the corresponding function G of A, i. e. G0 = G (A0)m
2. COMPACTNESS AND ELEMENTARY SUBMODELS 16
(3) each constant of A0 is the corresponding constant of A.
Definition 20. Let A and B be two models for L. We say A is an elementary submodel of B and B is an elementary extension of A and we write A ≺ B whenever (1) A ⊆ B and (2) for all formulas ϕ(v0,...,vk) of L and all a0,...,ak ∈ A A |= ϕ[a0,...,ak] iﬀ B |= ϕ[a0,...,ak]. Exercise 8. Prove that: • if A ⊆ B and B ⊆ C then A ⊆ C, • if A ≺ B and B ≺ C then A ≺ C, • if A ≺ B then A ⊆ B and A ≡ B. Example 7. Let N be the usual natural numbers with < < < as the usual ordering. Let B = hN,< < 2. COMPACTNESS AND ELEMENTARY SUBMODELS 17
Claim. If t is a term of the language L and a0,...,ap are in Ak, then the value of the term t[a0,...,ap] in A is equal to the value in Ak.
Proof of Claim. We prove this by induction on the complexity of the term. (1) If t is the variable vi then both values are just ai. (2) If t is the constant symbol c then the values are equal because c has the same interpretation in A and in Ak. (3) If t is F(t1 ...tm) where F is a function symbol and t1,...,tm are terms such that each value ti[a0,...,ap] is the same in both A and Ak, then the value F(t1 ...tm)[a0,...,ap] in A is G(t1[a0,...,ap],...,tm[a0,...,ap]) where G is the interpretation of F in A and the value of
F(t1 ...tm)[a0,...,ap]
in Ak is
Gk(t1[a0,...,ap],...,tm[a0,...,ap]) where Gk is the interpretation of F in Ak. But Gk is the restriction of G to Ak so these values are equal.
In order to show that each Ak ≺ A it will suﬃce to prove the following statement for each formula ϕ(v0,...,vp) of L. “ For all k ∈N and all a0,...,ap in Ak: A |= ϕ[a0,...,ap] iﬀ Ak |= ϕ[a0,...,ap].” Claim. The statement is true whenever ϕ is t1 = t2 where t1 and t2 are terms. Proof of Claim. Fix k ∈N and a0,...,ap in Ak. A |= (t1 = t2)[a0,...,ap] iﬀ t1[a0,...,ap] = t2[a0,...,ap] in A iﬀ t1[a0,...,ap] = t2[a0,...,ap] in Ak iﬀ Ak |= (t1 = t2)[a0,...,ap].
Claim. The statement is true whenever ϕ is R(t1 ...tn) where R is a relation symbol and t1,...,tn are terms. Proof of Claim. Fix k ∈N and a0,...,ap in Ak. Let S be the interpretationof R in A and Sk be the interpretation in Ak; Sk is the restriction of S to Ak. A |= R(t1 ...tn)[a0,...,ap] iﬀ S(t1[a0,...,ap],...,tn[a0,...,ap]) iﬀ Sk(t1[a0,...,ap],...,tn[a0,...,ap]) iﬀ Ak |= R(t1 ...tn)[a0,...,ap]
Claim. If the statement is true when ϕ is θ, then the statement is true when ϕ is ¬θ.
2. COMPACTNESS AND ELEMENTARY SUBMODELS 18 Proof of Claim. Fix k ∈N and a0,...,ap in Ak. A |= (¬θ)[a0,...,ap] iﬀ not A |= θ[a0,...,ap] iﬀ not Ak |= θ[a0,...,ap] iﬀ Ak |= (¬θ)[a0,...,ap].
Claim. If the statement is true when ϕ is θ1 and when ϕ is θ2 then the statement is true when ϕ is θ1 ∧θ2. Proof of Claim. Fix k ∈N and a0,...,ap in Ak. A |= (θ1 ∧θ2)[a0,...,ap] iﬀ A |= θ1[a0,...,ap] and A |= θ2[a0,...,ap] iﬀ Ak |= θ1[a0,...,ap] and Ak |= θ2[a0,...,ap] iﬀ Ak |= (θ1 ∧θ2)[a0,...,ap].
Claim. If the statement is true when ϕ is θ then the statement is true when ϕ is ∃viθ. Proof of Claim. Fix k ∈N and a0,...,ap in Ak. Note thatA = ∪{Aj : j ∈N}. A |= ∃viθ[a0,...,ap] iﬀ A |= ∃viθ[a0,...,aq] where q is the maximum of i and p (by Lemma 2), iﬀ A |= θ[a0,...,ai−1,a,ai+1,...,aq] for some a ∈ A, iﬀ A |= θ[a0,...,ai−1,a,ai+1,...,aq] for some a ∈ Al for some l ≥ k iﬀ Al |= θ[a0,...,ai−1,a,ai+1,...,aq] since the statement is true for θ, iﬀ Al |= ∃viθ[a0,...,aq] iﬀ Ak |= ∃viθ[a0,...,aq] since Ak ≺ Al iﬀ Ak |= ∃viθ[a0,...,ap] (by Lemma 2).
By induction on the complexity of ϕ, we have proven the statement for all formulas ϕ which do not contain the connectives ∨, → and ↔ or the quantiﬁer ∀. To verify the statement for all ϕ we use Lemma 3. Let ϕ be any formula of L. By Lemma 3 there is a formula ψ which does not use ∨, →, ↔ nor ∀ such that | = (∀v0)...(∀vp)(ϕ ↔ ψ). Now ﬁx k ∈N and a0,...,ap in Ak. We have A |= (ϕ ↔ ψ)[a0,...,ap] and Ak |= (ϕ ↔ ψ)[a0,...,ap]. A |= ϕ[a0,...,ap] iﬀ A |= ψ[a0,...,ap] iﬀ Ak |= ψ[a0,...,ap] iﬀ Ak |= ϕ[a0,...,ap] which completes the proof of the theorem.
2. COMPACTNESS AND ELEMENTARY SUBMODELS 19
Lemma 5. (The Tarski-Vaught Condition) Let A and B be models for L with A ⊆ B. The following are equivalent: (1) A ≺ B (2) for any formula ψ(v0,...,vq) and any i ≤ q and any a0,...,aq from A: if there is some b ∈ B such that B |= ψ[a0,...,ai−1,b,ai+1,...,aq] then we have some a ∈ A such that B |= ψ[a0,...,ai−1,a,ai+1,...,aq]. Proof. Only the implication (2) ⇒ (1) requires a lot of proof. We will prove that for each formula ϕ(v0,...,vp) and all a0,...,ap from A we will have: A |= ϕ[a0,...,ap] iﬀ B |= ϕ[a0,...,ap] by induction on the complexity of ϕ using only the negation symbol ¬, the connective ∧ and the quantiﬁer ∃ (recall Lemma 3). (1) The cases of formulas of the form t1 = t2 and R(t1 ...tn) come immediately from the fact that A ⊆ B. (2) For negation: suppose ϕ is ¬ψ and we have it for ψ, then A |= ϕ[a0,...,ap] iﬀ not A |= ψ[a0,...,ap] iﬀ not B |= ψ[a0,...,ap] iﬀ B |= ϕ[a0,...,ap]. (3) The ∧ case proceeds similarly. (4) For the ∃ case we consider ϕ as ∃viψ. If A |= ∃viψ[a0,...,ap], then the inductive hypothesis for ψ and the fact that A ⊆ B ensure that B |= ∃viψ[a0,...,ap]. It remains to show that if B |= ϕ[a0,...,ap] then A |= ϕ[a0,...,ap]. Assume B |= ∃viψ[a0,...,ap]. By Lemma 2, B |= ∃viψ[a0,...,aq] where q is the maximum of i and p. By the deﬁnition of satisfaction, there is some b ∈ B such that B |= ψ[a0,...,ai−1,b,ai+1,...,aq]. By (2), there is some a ∈ A such that B |= ψ[a0,...,ai−1,a,ai+1,...,aq]. By the inductive hypothesis on ψ, for that same a ∈ A, A |= ψ[a0,...,ai−1,a,ai+1,...,aq]. By the deﬁnition of satisfaction, A |= ∃viψ[a0,...,aq]. Finally, by Lemma 2, A |= φ[a0,...,ap].   Recall that |B| is used to represent the cardinality, or size, of the set B. Note that since any language L contains inﬁnitely many variables, |L| is always inﬁnite, but may be countable or uncountable depending on the number of other symbols. We often denote an arbitrary inﬁnite cardinal by the lower case Greek letter κ.
2. COMPACTNESS AND ELEMENTARY SUBMODELS 20
Theorem 5. (L¨owenheim-Skolem Theorem) Let B be a model for L and let κ be any cardinal such that |L|≤ κ < |B|. Then B has an elementary submodel A of cardinality κ. Furthermore if X ⊆ B and |X|≤ κ, then we can also have X ⊆ A. Proof. Without loss of generosity assume |X| = κ. We recursively deﬁne setsX n for n ∈ N such that X = X0 ⊆ X1 ⊆ ••• ⊆ Xn ⊆ ••• and such that for each formula ϕ(v0,...,vp) of L and each i ≤ p and each a0,...,ap from Xn such that B |= ∃viϕ[a0,...,ap] we have x ∈ Xn+1 such that B |= ϕ[a0,...,ai−1,x,ai+1,...,ap]. Since |L|≤ κ and each formula of L is a ﬁnite string of symbols from L, there are at most κ many formulas of L. So there are at most κ elements of B that need to be added to each Xn and so, without loss of generosity each |Xn| = κ. Let A = ∪{Xn : n ∈N}; then |A| = κ. Since A is closed under functions from B and contains all constants from B, A gives rise to a submodel A ⊆ B. The Tarski-Vaught Condition is used to show that A ≺ B.   An interesting consequence of this theorem is that the ordered ﬁeld of real numbers R has a countable elementary submodel containing π and e. Definition 23. A theory T for a language L is said to be complete whenever for each sentence σ of L either T |= σ or T |= ¬σ. Lemma 6. A theory T for L is complete iﬀ any two models of T are elementarily equivalent. Proof. (⇒) easy. (⇐) easy.   Definition 24. A theory T is said to be categorical in cardinality κ whenever any two models of T of cardinality κ are isomorphic. We also say that T is κcategorical. The most interesting cardinalities in the context of categorical theories are ℵ0, the cardinality of countably inﬁnite sets, and ℵ1, the ﬁrst uncountable cardinal. Exercise 9. Show that DLO is ℵ0-categorical. There are two well-known proofs. One uses a back-and-forth construction of an isomorphism. The other constructs, by recursion, an isomorphism from the set of dyadic rational numbers between 0 and 1: { n 2m : m is a positive integer and n is an integer 0 < n < 2m}, onto a countable dense linear order without endpoints. Now use the following theorem to show that DLO is complete.
Theorem 6. (The L o´s-Vaught Test) Suppose that a theory T has only inﬁnite models for a language L and that T is κ-categorical for some cardinal κ ≥|L|. Then T is complete.
2. COMPACTNESS AND ELEMENTARY SUBMODELS 21 Proof. We will show that any two models of T are elementarily equivalent.Let A of cardinality λ1, and B of cardinality λ2, be two models of T. If λ1 > κ use the L¨owenheim-Skolem Theorem to get A0 such that |A0| = κand A0 ≺ A. If λ1 < κ use the Compactness Theorem on the set of sentences ThA∪{cα 6= cβ : α 6= β} where {cα : α ∈ κ} is a set of new constant symbols of size κ, to obtain a model C for this expanded language such that |C|≥ κ. The reduct C0 to the language L has the property that C0 |= ThA and hence A ≡ C0. Now use the L¨owenheim-Skolem Theorem to get A0 such that |A0| = κ and A0 ≺ C0. Either way, we can get A0 such that |A0| = κ and A0 ≡ A. Similarly, we canget B0 such that |B0| = κ and B0 ≡ B. Since T is κ -categorical, A0 ∼ = B0. Hence A ≡ B.   Recall that the characteristic of a ﬁeld is the prime number p such that p z }| { 1 + 1 +•••+ 1 = 0 provided that such a p exists, and, if no such p exists the ﬁeld has characteristic 0. All of our best-loved ﬁelds: Q, R and C have characteristic 0. On the other hand, ﬁelds of characteristic p include the ﬁnite ﬁeld of size p (the prime Galois ﬁeld).
Theorem 7. The theory of algebraically closed ﬁelds of characteristic 0 is complete. Proof. We use the L o´s-Vaught Test and the following Lemma.
Lemma 7. Any two algebraically closed ﬁelds of characteristic 0 and cardinality ℵ1 are isomorphic. Proof. Let A be such a ﬁeld containing the rationals Q = hQ,+ + +,• • •,0,1i as a prime subﬁeld. In a manner completely analogous to ﬁnding a basis for a vector space, we can ﬁnd a transcendence basis for A, that is, an indexed subset {aα : α ∈ I}⊆ A such that A is the algebraic closure of the subﬁeld A0 generated by {aα : α ∈ I} but no aβ is in the algebraic closure of the subﬁeld generated by the rest: {aα : α ∈ I and α 6= β}. Since the subﬁeld generated by a countable subset would be countable and the algebraic closure of a countable subﬁeld would also be countable, we must have that the transcendence base is uncountable. Since |A| = ℵ1, the least uncountable cardinal, we must have in fact that |I| = ℵ1. Now let B be any other algebraically closed ﬁeld of characteristic 0 and size ℵ1. As above, obtain a transcendence basis {bβ : β ∈ J} with |J| = ℵ1 and its generated subﬁeld B0. Since |I| = |J|, there is a bijection g : I → J which we can use to build an isomorphism from A to B. Since B has characteristic 0, a standard theorem of algebra gives that the rationals are isomorphically embedded into B. Let this embedding be: f : Q ,→ B. We extend f as follows: for each α ∈ I, let f(aα) = bg(α), which maps the transcendence basis of A into the transcendence basis of B.
2. COMPACTNESS AND ELEMENTARY SUBMODELS 22
We now extend f to map A0 onto B0 as follows: Each element of A0 is given by p(aα1,...,aαm) q(aα1,...,aαm) , where p and q are polynomials with rational coeﬃcients and the a’s come, of course, from the transcendence basis. Let f map such an element to ¯ p(bg(α1),...,bg(αm)) ¯ q(bg(α1),...,bg(αm)) where ¯ p and ¯ q are polynomials whose coeﬃcients are the images under f of the rational coeﬃcients of p and q. The ﬁnal extension of f to all of A and B comes from the uniqueness of algebraic closures.
Remark. Lemma 7 is also true when 0 is replaced by any ﬁxed characteristic and ℵ1 by any uncountable cardinal. Theorem 8. Let H be a set of sentences in the language of ﬁeld theory which are true in algebraically closed ﬁelds of arbitrarily high characteristic. Then H holds in some algebraically closed ﬁeld of characteristic 0. Proof. A ﬁeld is a model in the language {+,•,0,1} of the axioms of ﬁeld theory. Let ACF be the set of axioms for the theory of algebraically closed ﬁelds; see Example 5. For each n ≥ 2, let τn denote the sentence ¬( n z }| { 1 + 1 +•••+ 1) = 0 Let Σ = ACF∪H∪{τn : n ≥ 2} Let Σ0 be any ﬁnite subset of Σ and let m be the largest natural number such that τm ∈ Σ0 or let m = 1 by default. Let A be an algebraically closed ﬁeld of characteristic p > m such that A |= H; then in fact A |= Σ0. So by compactness there is B such that B |= Σ. B is the required ﬁeld.
Corollary 1. Let C denote, as usual, the complex numbers. Every one-to-one polynomial map f : Cm →Cm is onto. Proof. A polynomial map is a function of the form f(x1,...,xm) = hp1(x1,...,xm),...,pm(x1,...,xm)i where each pi is a polynomial in the variables x1,...,xm. We call max { degree of pi : i ≤ m} the degree of f. Let L be the language of ﬁeld theory and let θm,n be the sentence of L which expresses that “each polynomial map of m variables of degree < n which is one-toone is also onto”. We wish to show that there are algebraically closed ﬁelds of arbitrarily high characteristic which satisfy H = {θm,n : m,n ∈ N}. We will then apply Theorem 8, Theorem 7, Lemma 6 and Exercise 5 and be ﬁnished.
2. COMPACTNESS AND ELEMENTARY SUBMODELS 23
Let p be any prime and let Fp be the prime Galois ﬁeld of size p. The algebraic closure ˜ Fp is the countable union of a chain of ﬁnite ﬁelds Fp = A0 ⊆ A1 ⊆ A2 ⊆•••⊆ Ak ⊆ Ak+1 ⊆••• obtained by recursively adding roots of polynomials. We ﬁnish the proof by showing that each h ˜ Fp,+ + +,•,0,1i satisﬁes H. Given any polynomial map f : ( ˜ Fp m ) → ( ˜ Fp m ) which is one-to-one, we show that f is also onto. Given any elements b1,...,bm ∈ ˜ Fp, there is some Ak containing b1,...,bm as well as all the coeﬃcients of f. Since f is one-to-one, f Am k : Am k → Am k is a one-to-one polynomial map. Hence, since Am k is ﬁnite, f Am k is onto and so there are a1,...,am ∈ Ak suchthat f(a1,...,am) = hb1,...,bmi. Therefore f is onto. Thus, for each prime number p and each m,n ∈ N, θm,n holds in a ﬁeld of characteristic p, i.e. h˜ Fp,+ + +,• • •,0 0 0,1 1 1i satisﬁes H.   The above corollary is the famous Ax-Grothendieck Theorem. It is a signiﬁcant problem to replace “one-to-one” with “locally one-to-one”.
CHAPTER 3
Diagrams and Embeddings
Let A = hA,Ii be a model for a language L. Expand L to the language LA = L∪{ca : a ∈ A} by adding new constant symbols to L. We can expand A to a model AA = hA,I0i for LA by choosing I0 extending I such that I0(ca) = a for each a ∈ A. More generally, if f : X → A, we can expand L to LX = L∪{cx : x ∈ X} and expand A = hA,Ii to hA,I0i where I0 extends I with each I0(cx) = f(x). We denote the resulting model as hA,f(x)ix∈X or AX = hA,xix∈X if f is the identity function. Definition 25. Let A be a model for L. (1) The elementary diagram of A is Th(AA), the set of all sentences of LA which hold in AA. (2) The diagram of A, denoted by 4A, is the set of all those sentences in Th(AA) without quantiﬁers.
Remark. There is a notion of atomic formula, which is a formula of the form (t1 = t2) or (R(t1 ...tn)) where t1,...,tn are terms. Sometimes 4A is deﬁned to be the set of all atomic formulas and negations of atomic formulas which occur in Th(AA). However this is not substantially diﬀerent from Deﬁnition 25, since the reader can quickly show that for any model B, B |= 4A in one sense iﬀ B |= 4A in the other sense. Exercise 10. Let A and B be models for L with X ⊆ A ⊆ B. Prove: (i) A ⊆ B iﬀ AX ⊆ BX iﬀ BA |= 4A. (ii) A ≺ B iﬀ AX ≺ BX iﬀ BA |= Th(AA). Hint: A |= ϕ[a1,...,ap] iﬀ AA |= ϕ∗ where ϕ∗ is the sentence of LA formed by replacing each free occurrence of vi with cai. Definition 26. A is said to be isomorphically embedded into B whenever (1) there is a model C such that A ∼ = C and C ⊆ B or (2) there is a model D such that A ⊆ D and D ∼ = B. Exercise 11. Prove that, in fact, (1) and (2) are equivalent conditions.
Definition 27. A is said to be elementarily embedded into B whenever (1) there is a model C such that A ∼ = C and C ≺ B or (2) there is a model D such that A ≺ D and D ∼ = B. Exercise 12. Again, prove that, in fact, (1) and (2) are equivalent.
24
3. DIAGRAMS AND EMBEDDINGS 25
The next result is extremely useful; the ﬁrst part is called the Diagram Lemma and the second part is called the Elementary Diagram Lemma. Theorem 9. Let A and B be models for L. (1) A is isomorphically embedded into B if and only if B can be expanded to a model of 4A.(2) A is elementarily embedded into B if and only if B can be expanded to a model of Th(AA). Proof. We sketch the proof of (1). (⇒) If f is the isomorphism as in 1 of Deﬁnition 26 above, then hB,f(a)ia∈A |= 4A. (⇐) If hB,baia∈A |= 4A, then C = {ba : a ∈ A} generates C ⊆ B with C ∼ = A.   Exercise 13. Give a complete proof of (2). Exercise 14. Show that if A is a model for the language L and C is a model for the language LA such that C |= 4A then there is a model B such that A ⊆ B and BA ∼ = C.
Exercise 15. The L¨owenheim-Skolem Theorem is sometimes called the Downward L¨owenheim-Skolem Theorem. It’s partner is the Upward L¨owenheim-Skolem Theorem: if A is an inﬁnite model for L and κ is any cardinal such that |L| ≤ κ and |A| < κ, then A has an elementary extension of cardinality κ. Prove it. We now apply these notions to graph theory and to calculus. The natural language for graph theory has one binary relation symbol which we call E (to suggest the word “edge”). Graph Theory has the following two axioms: • (∀x)(∀y)E(x,y) ↔ E(y,x) • (∀x)¬E(x,x). A graph is, of course, a model of graph theory.
Corollary 2. Every planar graph can be four coloured.
Proof. We will have to use the famous result of Appel and Haken that every ﬁnite planar graph can be four coloured. Model Theory will take us from the ﬁnite to the inﬁnite. We recall that a planar graph is one that can be embedded, or drawn, in the usual Euclidean plane and to be four coloured means that each vertex of the graph can be assigned one of four colours in such a way that no edge has the same colour for both endpoints. Let A be an inﬁnite planar graph. Introduce four new unary relation symbols: R,G,B,Y (for red, green, blue and yellow). We wish to prove that there is some expansion A0 of A such that A0 |= σ where σ is the sentence in the expanded language: (∀x)[R(x)∨G(x)∨B(x)∨Y (x)] ∧(∀x)[R(x) →¬(G(x)∨B(x)∨Y (x))]∧... ∧(∀x)(∀y)¬(R(x)∧R(y)∧E(x,y))∧••• which will ensure that the interpretations of R,G,B and Y will four colour the graph.
3. DIAGRAMS AND EMBEDDINGS 26
Let Σ = 4A ∪{σ}. Any ﬁnite subset of Σ has a model, based upon the appropriate ﬁnite subset of A. By the compactness theorem, we get B |= Σ. Since B |= σ, the interpretations of R,G,B and Y four colour it. By the diagram lemma A is isomorphically embedded in the reduct of B, and this isomorphism delivers the four-colouring of A.
A graph with the property that every pair of vertices is connected with an edge is called complete. At the other extreme, a graph with no edges is called discrete. A very important theorem in ﬁnite combinatorics says that most graphs contain an example of one or the other as a subgraph. A subgraph of a graph is, of course, a submodel of a model of graph theory.
Corollary 3. (Ramsey’s Theorem) For each n ∈N there is an r ∈N such that if G is any graph with r vertices, then either G contains a complete subgraph with n vertices or a discrete subgraph with n vertices.
Proof. We follow F. Ramsey who began by proving an inﬁnite version of the theorem (also called Ramsey’s Theorem).
Claim. Each inﬁnite graph G contains either an inﬁnite complete subgraph or an inﬁnite discrete subgraph.
Proof of Claim. By force of logical necessity, there are two possiblities: (1) there is an inﬁnite X ⊆ G such that for all x ∈ X there is a ﬁnite Fx ⊆ X such that E(x,y) for all y ∈ X \Fx, (2) for all inﬁnite X ⊆ G there is a x ∈ X and an inﬁnite Y ⊆ X such that ¬E(x,y) for all y ∈ Y . If (1) occurs, we recursively pick x1 ∈ X, x2 ∈ X\Fx1, x3 ∈ X\(Fx1∪Fx2), etc, to obtain an inﬁnite complete subgraph. If (2) occurs we pick x0 ∈ G and Y0 ⊆ G with the property and then recursively choose x1 ∈ Y0 and Y1 ⊆ Y0 , x2 ∈ Y1 and Y2 ⊆ Y1 and so on, to obtain an inﬁnite discrete subgraph.
We now use Model Theory to go from the inﬁnite to the ﬁnite. Let σ be the sentence, of the language of graph theory, asserting that there is no complete subgraph of size n. (∀x1 ...∀xn)[¬E(x1,x2)∨¬E(x1,x3)∨•••∨¬E(xn−1,xn)]. Let τ be the sentence asserting that there is no discrete subgraph of size n. (∀x1 ...∀xn)[E(x1,x2)∨E(x1,x3)∨•••∨E(xn−1,xn)]. Let T be the set consisting of σ, τ and the axioms of graph theory. If there is no r as Ramsey’s Theorem states, then T has arbitrarily large ﬁnite models. By Theorem 2, T has an inﬁnite model, contradicting the claim.
Ramsey’s Theorem says that for each n there is some r. The proof does not, however, let us know exactly which r corresponds to any given n. There has been considerable eﬀorts made to ﬁnd a more constructive proof. In particular we would
3. DIAGRAMS AND EMBEDDINGS 27
like to know, for each n, the smallest value of r which would satisfy Ramsey’s Theorem, called the Ramsey Number of n. The Ramsey number of 3 is 6; the Ramsey number of 4 is 18; the Ramsey number of 5 is ...unknown; but it’s somewhere between 40 and 50. Even less is known about the Ramsey numbers for higher values of n. Determining the Ramsey numbers may be the most mysterious problem in all of mathematics.
The following theorem of A. Robinson ﬁnally solved the centuries old problem of inﬁnitesimals in the foundations of calculus. Theorem 10. (The Leibniz Principle) There is an ordered ﬁeld ∗R called the hyperreals, containing the reals R and a number larger than any real number such that any statement about the reals which holds in R also holds in ∗R. Proof. Let R be hR,+ + +,• • •,< < <,0 0 0,1 1 1i. We will make the statement of the theoremprecise by proving that there is some model H, in the same language L as R andwith the universe called ∗R , such that R ≺ H and there is b ∈∗R such that a < bfor each a ∈R. For each real number a, we introduce a new constant symbol ca. In addition, another new constant symbol d is introduced. Let Σ be the set of sentences in the expanded language given by: ThRR∪{ca < d : a is a real} We can obtain a model C |= Σ by the compactness theorem. Let C0 be the reduct of C to L. By the elementary diagram lemma R is elementarily embedded in C0, and so there is a model H for L such that C0 ∼ = H and R ≺ H. Take b to be theinterpretation of d in H.   Remark. The element b ∈ ∗R gives rise to an inﬁnitesimal 1/b ∈ ∗R. Anelement x ∈ ∗R is said to be inﬁnitesimal whenever −1/n < x < 1/n for eachn ∈ N. 0 is inﬁnitesimal. Two elements x,y ∈ R are said to be inﬁnitely close, written x ≈ y whenever x−y is inﬁnitesimal, so that x is inﬁnitesimal iﬀ x ≈ 0. An element x ∈ ∗R is said to be ﬁnite whenever −r < x < r for some positive r ∈ R. Else it is inﬁnite. Each ﬁnite x ∈ ∗R is inﬁnitely close to some real number, called the standardpart of x, written st(x). This idea is extremely useful in understanding calculus. To diﬀerentiate f, for each Mx ∈ ∗R generate My = f(x + Mx)−f(x). Then f0(x) = stMy Mx wheneverthis exists and is the same for each inﬁnitesimal Mx 6= 0. This legitimises the intuition of the founders of the diﬀerential calculus and allows us to use that intuition to move from the (ﬁnitely) small to the inﬁnitely small. Proofs of the usual theorems of calculus are now much easier. More importantly, reﬁnements of these ideas, now called non-standard analysis, form a powerful tool for applying calculus, just as its founders envisaged. The following theorem is considered one of the most fundamental results of mathematical logic. We give a detailed proof. Theorem 11. (Robinson Consistency Theorem) Let L1 and L2 be two languages with L = L1∩L2. Suppose T1 and T2 are satisﬁable
3. DIAGRAMS AND EMBEDDINGS 28
theories in L1 and L2 respectively. Then T1∪T2 is satisﬁable iﬀ there is no sentence σ of L such that T1 |= σ and T2 |= ¬σ. Proof. The direction ⇒ is easy and motivates the whole theorem. We begin the proof in the ⇐ direction. Our goal is to show that T1 ∪T2 is satisﬁable. The following claim is a ﬁrst step. Claim. T1 ∪{ sentences σ of L : T2 |= σ} is satisﬁable. Proof of Claim. Using the compactness theorem and considering conjunctions, it suﬃces to show that if T1 |= σ1 and T2 |= σ2 with σ2 a sentence of L, then {σ1,σ2}is satisﬁable. But this is true, since otherwise we would have σ1 |= ¬σ2 and hence T1 |= ¬σ2 and so ¬σ2 would be a sentence of L contradicting our hypothesis. This proves the claim.
The basic idea of the proof from now on is as follows. In order to construct a model of T1 ∪T2 we construct models A |= T1 and B |= T2 and an isomorphism f : A|L → B|L between the reducts of A and B to the language L, witnessing that A|L∼ = B|L. We then use f to carry over interpretations of symbols in L1 \Lfrom A to B , giving an expansion B∗ of B to the language L1 ∪L2. Then, sinceB ∗|L1 ∼ = A and B∗|L2 = B we get B∗ |= T1 ∪T2. The remainder of the proof will be devoted to constructing such an A, B and f. A and B will be constructed as unions of elementary chains of An’s and Bn’s while f will be the union of fn : An ,→ Bn. We begin with n = 0, the ﬁrst link in the elementary chain. Claim. There are models A0 |= T1 and B0 |= T2 with an elementary embeddingf 0 : A0|L ,→ B0|L. Proof of Claim. Using the previous claim, let A0 |= T1 ∪{ sentences σ of L : T2 |= σ} We ﬁrst wish to show that Th(A0|L)A0∪T2 is satisﬁable. Using the compactness theorem, it suﬃces to prove that if σ ∈ Th(A0|L)A0 then T2 ∪{σ} is satisﬁable. For such a σ let ca0,...,can be all the constant symbols from LA0 \L which appear in σ. Let ϕ be the formula of L obtained by replacing each constant symbol cai by a new variable ui. We have A0|L|= ϕ[a0,...,an] and so A0|L|= ∃u0 ...∃unϕ By the deﬁnition of A0, it cannot happen that T2 |= ¬∃u0 ...∃unϕ and so there is some model D for L2 such that D |= T2 and D |= ∃u0 ...∃unϕ. So there are elements d0,...,dn of D such that D |= ϕ[do,...,dn]. Expand D to a model D∗ for L2 ∪LA0, making sure to interpret each cai as di. Then D∗ |= σ, and so D∗ |= T2 ∪{σ}. Let B∗ 0 |= Th(A0|L)A0∪T2. Let B0 be the reduct of B∗ 0 toL2; clearly B0 |= T2.Since B0|L can be expanded to a model of Th(A0|L)A0, the Elementary Diagram Lemma gives an elementary embedding f0 : A0|L ,→ B0|L and ﬁnishes the proof of the claim.
3. DIAGRAMS AND EMBEDDINGS 29
The other links in the elementary chain are provided by the following result. Claim. For each n ≥ 0 there are models An+1 |= T1 and Bn+1 |= T2 with an elementary embedding fn+1 : An+1|L ,→ Bn+1|L such that An ≺ An+1, Bn ≺ Bn+1, fn+1 extends fn and Bn ⊆ range of fn+1. A0 ≺ A1 ≺ ••• ≺ An ≺ An+1 ≺ ••• ↓f0 ↓f1 ↓fn ↓fn+1 B0 ≺ B1 ≺ ••• ≺ Bn ≺ Bn+1 ≺ ••• The proof of this claim will be discussed shortly. Assuming the claim, let A =Sn∈N An, B =Sn∈N Bn and f =Sn∈N fn. The Elementary Chain Theorem gives that A |= T1 and B |= T2. The proof of the theorem is concluded by simply verifying that f : A|L→ B|L is an isomorphism. The proof of the claim is long and quite technical; it would not be inappropriate to omit it on a ﬁrst reading. The proof, of course, must proceed by induction on n. The case of a general n is no diﬀerent from the case n = 0 which we state and prove in some detail. Claim. There are models A1 |= T1 and B1 |= T2 with an elementary embeddingf 1 : A1|L ,→ B1|L such that A0 ≺ A1, B0 ≺ B1, f1 extends f0 andB 0 ⊆ range of f1. A0 ≺ A1 ↓f0 ↓f1 B0 ≺ B1 Proof of Claim. Let A+ 0 be the expansion of A0 to the language L+ 1 = L1∪ {ca : a ∈ A0}formed by interpreting each ca as a ∈ A0; A+ 0 is just another notation for (A0)A0. The elementary diagram of A+ 0 is ThA+ 0 A+ 0 . Let B∗ 0 be the expansion of B0|L to the language L∗ = L∪{ca : a ∈ A0}∪{cb : b ∈ B0} formed by interpreting each ca as f0(a) ∈ B0 and each cb as b ∈ B0. We wish to prove that ThA+ 0 A+ 0 ∪ThB∗ 0 is satisﬁable. By the compactness theorem it suﬃces to prove that ThA+ 0 A+ 0 ∪{σ} is satisﬁable for each σ in ThB∗ 0. For such a sentence σ, let ca0,...,cam,cb0,...,cbn be all those constant symbols occuring in σ but not in L. Let ϕ(u0,...,um,w0,...,wm) be the formula of L obtained from σ by replacing each constant symbol cai by a new variable ui and each constant symbol cbi by a new variable wi. We have B∗ 0 |= σ so B0|L|= ϕ[f0(a0),...,f0(am),b0,...,bn] So B0|L|= ∃w0 ...∃wnϕ[f0(a0),...,f0(am)] Since f0 is an elementary embedding we have : A0|L|= ∃w0 ...∃wnϕ[a0,...,am]
3. DIAGRAMS AND EMBEDDINGS 30
Let ˆ ϕ(w0,...,wn) be the formula of L+ 1 obtained by replacing occurrences of ui in ϕ(u0,...,um,w0,...,wn) by cai; then A+ 0 |= ∃w0 ...∃wn ˆ ϕ. So, of course, A+ 0 A+ 0 |= ∃w0 ...∃wn ˆ ϕ and this means that there are d0,...,dn in A+ 0 = A0 such that (A+ 0 )A+ 0 |= ˆ ϕ[d0,...,dn]. We can now expandA+ 0 A+ 0 to a model D by interpreting each cbi as di to obtain D |= σ and so ThA+ 0 A+ 0 ∪{σ} is satisﬁable. Let E |= ThA+ 0 A+ 0 ∪ThB∗ 0. By the elementary diagram lemma A+ 0 is elementarily embedded into E|L+ 1 . So there is a model A+ 1 for L+ 1 with A+ 0 ≺ A+ 1 and an isomorphism g : A+ 1 → E|L+ 1 . Using g we expand A+ 1 to a model A0 1 isomorphic to E. Let A∗ 1 denote A0 1|L∗; we have A∗ 1 |= ThB∗ 0. We now wish to prove that Th(A∗ 1)A∗ 1 ∪ThB+ 0 B+ 0 is satisﬁable, where B+ 0 is the common expansion of B0 and B∗ 0 to the language L+ 2 = L2 ∪{ca : a ∈ A0}∪{cb : b ∈ B0}. By the compactness theorem, it suﬃces to show that ThB+ 0 B+ 0 ∪{σ} is satisﬁable for each σ in Th(A∗ 1)A∗ 1 . Let cx0,...,cxn be all those constant symbols which occur in σ but are not inL∗. Let ψ(u0,...,un) be the formula ofL∗ obtained from σ by replacing each cxi with a new variable ui. Since (A∗ 1)A∗ 1 |= σ we have A∗ 1 |= ψ[x0,...,xn], and so A∗ 1 |= ∃u0 ...∃unψ. Also A∗ 1 |= ThB∗ 0 and ThB∗ 0 is a complete theory in the language L∗; hence ∃u0 ...∃unψ is in ThB∗ 0. Thus B∗ 0 |= ∃u0 ...∃unψ and so B+ 0 B+ 0 |= ∃u0 ...∃unψ and therefore there are b0,...,bn in B+ 0 = B0 such that B+ 0 B+ 0 |= ψ[b0,...,bn]. We can now expandB+ 0 B+ 0 to a model F by interpreting each cxi as bi; then F |= σ and ThB+ 0 B+ 0 ∪{σ} is satisﬁable. Let G |= Th(A∗ 1)A∗ 1 ∪ThB+ 0 B+ 0 . By the elementary diagram lemma B+ 0 is elementarily embedded into G|L+ 2 . So there is a model B+ 1 for L+ 2 with B+ 0 ≺ B+ 1 and an isomorphism h : B+ 1 → G|L+ 2 . Using h we expand B+ 1 to a model B0 1 isomorphic to G. Let B∗ 1 denote B0 1|L∗. Again by the elementary diagram lemmaA ∗ 1 is elementarily embedded into B∗ 1. Let this be denoted by f1 : A∗ 1 ,→ B∗ 1.
3. DIAGRAMS AND EMBEDDINGS 31
Let a ∈ A0; we will show that f0(a) = f1(a). By deﬁnition we have
B∗ 0 |= (v0 = ca)[f0(a)] and so B+ 0 |= (v0 = ca)[f0(a)]. Since B+ 0 ≺ B+ 1 , B+ 1 |= (v0 = ca)[f0(a)] and so B∗ 1 |= (v0 = ca)[f0(a)]. Now A+ 0 |= (ca = v1)[a] and A+ 0 ≺ A+ 1 so A+ 1 |= (ca = v1)[a] so A∗ 1 |= (ca = v1)[a]. Since f1 is elementary,B ∗ 1 |= (ca = v1)[f1(a)] so B∗ 1 |= (v0 = v1)[f0(a),f1(a)] and so f0(a) = f1(a).Thus f1 extends f0. Let b ∈ B0; we will prove that b = f1(a) for some a ∈ A1. By deﬁnition we have: B∗ 0 |= (v0 = cb)[b] so B+ 0 |= (v0 = cb)[b]. Since B+ 0 ≺ B+ 1 ,B+ 1 |= (v0 = cb)[b]so B∗ 1 |= (v0 = cb)[b]. On the other hand, since (∃v1)(v1 = cb) is always satisﬁed,we have: A∗ 1 |= (∃v1)(v1 = cb) so there is a ∈ A1 such that A∗ 1 |= (v1 = cb)[a]. Sincef 1 is elementary, B∗ 1 |= (v1 = cb)[f1(a)] so B∗ 1 |= (v0 = v1)[b,f1(a)] so b = f1(a).Thus B0 ⊆ range of f1.We now let A1 be A+ 1 |L1 and let B1 be B+ 1 |L2. We get A0 ≺ A1 and B0 ≺ B1and f1 : A1|L→ B1|L remains an elementary embedding. This completes the proof of the claim and the theorem.   Exercise 16. The Robinson Consistency Theorem was originally stated as: Let T1 and T2 be satisﬁable theories in languages L1 and L2 respectively and letT ⊆T1∩T2 be a complete theory in the language L1 ∩L2. Then T1 ∪T2 is satisﬁable in the language L1 ∪L2. Show that this is essentially equivalent to our version in Theorem 11 by ﬁrst proving that this statement follows from Theorem 11 and then also proving that this statement implies Theorem 11. Of course, for this latter argument you are looking for a proof much shorter than our proof of Theorem 11; however it will help to use the ﬁrst claim of our proof in your own proof.
Theorem 12. (Craig Interpolation Theorem) Let ϕ and ψ be sentences such that ϕ |= ψ. Then there exists a sentence θ, called the interpolant, such that ϕ |= θ and θ |= ψ and every relation, function or constant symbol occuring in θ also occurs in both ϕ and ψ.
Exercise 17. Show that the Craig Interpolation Theorem follows quickly from the Robinson Consistency Theorem. Also, use the Compactness Theorem to show that Theorem 11 follows quickly from Theorem 12.
CHAPTER 4
Model Completeness
The quantiﬁer ∀ is said to be the universal quantiﬁer and the quantiﬁer ∃ tobe the existential quantiﬁer. A formula ϕ is said to be quantiﬁer free whenever no quantiﬁers occur in ϕ. A formula ϕ is said to be universal whenever it is of the form ∀x0 ...∀xkθwhere θ is quantiﬁer free. A formula ϕ is said to be existential whenever it is of the form ∃x0 ...∃xkθwhere θ is quantiﬁer free. A formula ϕ is said to be universal-existensial whenever it is of the form ∀x0 ...∀xk∃y0 ...∃ykθ where θ is quantiﬁer free. We extend these notions to theories T whenever each axiom σ of T has the property.
Remark. Note that each quantiﬁer free formula ϕ is trivially equialent to the existential formula ∃viϕ where vi does not occur in ϕ. Exercise 18. Let A and B be models for L with A ⊆ B. Verify the following three statements: (i) A ≺ B iﬀ BA |= Th(AA) iﬀ AA |= Th(BA). (ii) A ⊆ B iﬀ BA |= σ for each existential σ of Th(AA). (iii) A ⊆ B iﬀ AA |= σ for each universal σ of Th(BA). Definition 28. A model A of a theory T is said to be existentially closed if whenever A ⊆ B and B |= T, we have AA |= σ for each existential sentence σ of Th(BA). Remark. If A is existentially closed and A0 ∼ = A then A0 is also existentially closed. Definition 29. A theory T is said to be model complete whenever T ∪4A is complete in the language LA for each model A of T. Theorem 13. ( A. Robinson ) Let T be a theory in the language L. The following are equivalent: (1) T is model complete, (2) T is existentially complete, i.e. each model of T is existentially closed. (3) for each formula ϕ(v0,...,vp) of L there is some universal formula ψ(v0,...,vp) such that T |= (∀v0 ...∀vp)(ϕ ↔ ψ)(4) for all models A and B of T, A ⊆ B implies A ≺ B. Remark. Equivalently, in part (3) of this theorem the phrase “universal formula” could be replaced by “existential formula”. We chose the version which makes the proof smoother.
32
4. MODEL COMPLETENESS 33
Proof. (1) ⇒ (2):Let A |= T and B |= T with A ⊆ B. Clearly AA |= 4A and by Exercise 10 weB A |= 4A. Now by (1),T∪4A is complete and both AA and BA are models of this theory so they are elementarily equivalent. For any sentence σ of LA (existential or otherwise), if BA |= σ then AA |= σ and (2) follows. (2) ⇒ (3): Lemma 4 shows that it suﬃces to prove it for formulas ϕ in prenex normal form. We do this by induction on the prenex rank of ϕ which is the number of alternations of quantiﬁers in ϕ. The ﬁrst step is prenex rank 0. Where only universal quantiﬁers are present the result is trivial. The existential formula case is non-trivial; it is the following claim: Claim. For each existential formula ϕ(v0,...,vp) of L there is a universalformula ψ(v0,...,vp) such that T |= (∀v0)...(∀vp)(ϕ ↔ ψ) Proof of Claim. Add new constant symbols c0,...,cp to L to form L∗ = L∪{c0,...,cp} and to form a sentence ϕ∗ of L∗ obtained by replacing each free occurrence of vi in ϕ with the corresponding ci; ϕ∗ is an existential sentence. It suﬃces to prove that there is a universal sentence γ of L∗ such that T |= ϕ∗ ↔ γ. Let Γ = {universal sentences γ of L∗ such that T |= ϕ∗ → γ} We hope to prove that there is some γ ∈ Γ such that T |= γ → ϕ∗. Note, however, that any ﬁnite conjunction γ1 ∧γ2 ∧•••∧γn of sentences from Γ is equivalent to a sentence γ in Γ which is simply obtained from γ1 ∧γ2 ∧•••∧γn by moving all the quantiﬁers to the front. Thus it suﬃces to prove that there are ﬁnitely many sentences γ1,γ2,...,γn from Γ such that T |= γ1 ∧γ2 ∧•••∧γn → ϕ∗. If no such ﬁnite set of sentences existed, then each T ∪{γ1,γ2,...,γn}∪{¬ϕ∗} would be satisﬁable. By the compactness theorem, T ∪Γ∪{¬ϕ∗} would be satisﬁable. Therefore it just suﬃces to prove that T ∪Γ |= ϕ∗. In order to prove that T ∪Γ |= ϕ∗, let A be any model of T ∪Γ for the language L∗. Let Σ = T ∪{ϕ∗}∪4A. be a set of sentences for the language L∗ A; we wish to show that Σ is satisﬁable. By the compactness theorem it suﬃces to consider T ∪{ϕ∗,τ} where τ is a conjunction of ﬁnitely many sentences of, and hence in fact a single sentence of, 4A. Let θ be the formula obtained from τ by exchanging each constant symbol from L∗ A \L∗ occurring in τ for a new variable ua. So A |= ∃ua0 ...∃uamθ(ua0,...,uam).
4. MODEL COMPLETENESS 34
But then A is not a model of the universal sentence ∀ua0 ...∀uam¬θ(ua0,...,uam). Recalling that A |= Γ, we are forced to conclude that this universal sentence is not in Γ and so not a consequence of T ∪{ϕ∗}. Therefore T ∪{ϕ∗}∪{∃ua0 ...∃uamθ(ua0,...,uam)} must be satisﬁable, and any model of this can be expanded to a model ofT∪{ϕ∗,τ} and so Σ is satisﬁable. Let C |= Σ. By the Exercise 14, there is a model B for L∗ such that A ⊆ Band BA ∼ = C ; in particular: BA |= Σ.Since A ⊆ B the interpretation of each of c0,...,cp in B is the same as theinterpretation in A; let’s denote these by a0,...,ap. Let σ denote this sentence: (∃v0 ...∃vp)(ϕ∧v0 = ca0 ∧v1 = ca1 ∧•••∧vp = cap) which is equivalent to an existential sentence of LA. Since (B|L)A |= σ we can apply (2) to A|L and B|L to get that (A|L)A |= σ. By our choice of a0,...,ap we get that A |= ϕ∗. This means T ∪Γ |= ϕ∗ and ﬁnishes the proof of the claim.
We will now do the general cases for the proof of the induction on prenex rank. There are two cases, corresponding to the two methods available for increasing the number of alternations of quantiﬁers: (a) the addition of universal quantiﬁers (b) the addition of existential quantiﬁers. For the case (a), suppose ϕ(v0,...,vp) is ∀w0 ...∀wmχ(v0,...,vp,w0,...,wm)and χ has prenex rank lower than ϕ so that we have by the inductive hypothesis that there is a quantiﬁer free formula θ(v0,...,vp,w0,...,wm,x0,...,xn) with new variables x0,...,xn such that T |= (∀v0 ...∀vp∀w0 ...∀wm)(χ ↔∀x0 ...∀xnθ) Therefore, case (a) is concluded by noticing that this gives us T |= (∀v0 ...∀vp)(∀w0 ...∀wmχ ↔∀w0 ...∀wm∀x0 ...∀xnθ). Exercise 19. Check this step using the deﬁnition of satisfaction. For case (b), suppose ϕ(v0,...,vp) is ∃w0 ...∃wnχ(v0,...,vp,w0,...,wm) andχ has prenex rank less than ϕ. Here we will use the inductive hypothesis on ¬χ which of course also has prenex rank less than ϕ. We obtain a quantiﬁer free formula θ(v0,...,vp,w0,...,wm,x0,...,xn) with new variables x0,...,xn such that T |= (∀v0 ...∀vp∀w0 ...∀wm)(¬χ ↔∀x0 ...∀xnθ) So T |= (∀v0 ...∀vp)(∀w0 ...∀wm¬χ ↔∀w0 ...∀wm∀x0 ...∀xnθ) And T |= (∀v0 ...∀vp)(∃w0 ...∃wmχ ↔∃w0 ...∃wm∃x0 ...∃xn¬θ) Now ∃w0 ...∃wm∃x0 ...∃xn¬θ is an existential formula, so by the claim there is a universal formula ψ such that T |= (∀v0 ...∀vp)(∃w0 ...∃wm∃x0 ...∃xn¬θ ↔ ψ). Hence T |= (∀v0 ...∀vp)(∃w0 ...∃wnχ ↔ ψ) which is the ﬁnal result that we needed.
4. MODEL COMPLETENESS 35
(3) ⇒ (4)Let A |= T and B |= T with A ⊆ B. Let ϕ be a formula of L and let a0,...,ap bein A such that B |= ϕ[a0,...,ap]. Obtain a universal formula ψ such that T |= (∀v0 ...∀vp)(ϕ ↔ ψ). Hence B |= ψ[a0,...,ap]. Since A ⊆ B by an argument similar to Exercise 18 we have A |= ψ[a0,...,ap]. So A |= ϕ[a0,...,ap]. Therefore A ≺ B. (4) ⇒ (1)Let A |= T. We will show that T ∪4A is complete by showing that T ∪4A |= Th(AA). Let C |= T ∪4A. By Exercise 14 there is a model B such that A ⊆ B and BA ∼ = C. By (4) we have that A ≺ B. By Exercise 10 we get that BA |= Th(AA). Thus C |= Th(AA).   Example 9. We will see later that the theory ACF is model complete. But ACF is not complete because the characteristic of the algebraically closed ﬁeld can vary among models of ACF and the assertion that “I have characteristic p” can easily be expressed as a sentence of the language of ACF. Exercise 20. Suppose that T is a model complete theory in L and that either (1) any two models of T are isomorphically embedded into a third or (2) there is a model of T which is isomorphically embedded in any other. Then prove that T is complete. Example 10. Let N be the natural numbers and < the usual ordering. Let B = hN, Theorem 14. (Lindstr¨om’s Test) Let T be a theory in a countable language L such that (1) all models of T are inﬁnite, (2) the union of any chain of models of T is a model of T, and (3) T is κ-categorical for some inﬁnite cardinal κ. Then T is model complete. Proof. W.L.O.G. we assume that T is satisﬁable. We use conditions (1) and (2) to prove the following: Claim. T has existentially closed models of each inﬁnite size κ. Proof of Claim. By the L¨owenheim-Skolem Theorems we get A0 |= T with |A0| = κ. We recursively construct a chain of models of T of size κ A0 ⊆ A1 ⊆ ... ⊆ An ⊆ An+1 ⊆••• with the property that if B |= T and An+1 ⊆ B and σ is an existential sentence of Th(BAn), then (An+1)An |= σ.
4. MODEL COMPLETENESS 36
Suppose An is already constructed; we will construct An+1. Let Σn be a maximally large set of existential sentences of LAn such that for each ﬁnite Σ0 ⊆ Σn there is a model C for LAn such that C |= Σ0∪T ∪4An By compactness T ∪Σn ∪4An has a model D and without loss of generosityA n ⊆ D. By the Downward L¨owenheim-Skolem Theorem we get E such thatA n ⊆ E, |E| = κ and E ≺ D. Let An+1 = E|L; we will show that An+1 has the required properties. SinceE ≡ D, E |= T ∪4An and so An ⊆ An+1 (See Exercise 18). Let B |= T with An+1 ⊆ B and σ be an existential sentence of Th(BAn); we will show that (An+1)An |= σ. Since Σn consists of existential sentences and D ≡ E ≡ (An+1)An ⊆ BAn we have (see Exercise 18) that BAn |= Σn. The maximal property of Σn then forces σ to be in Σn because if σ / ∈ Σn then there must be someﬁnite Σ 0 ⊆ Σn for which there is no C such that C |= Σ0∪{σ}∪T ∪4An; but BAnis such a C! Now since σ ∈ Σn and E ≡ D |= Σn we must have E = (An+1)An |= σ.Now let A be the union of the chain. By hypothesis A |= T. It is easy to checkthat |A| = κ. To check that A is existentially closed, let B |= T with A ⊆ B andlet σ be an existential sentence of Th(BA). Since σ can involve only ﬁnitely many constant symbols, σ is a sentence of LAn for some n ∈ N. Thus An+1 ⊆ A ⊆ B gives that (An+1)An |= σ. Since σ is existential (see Exercise 18 again) we get that AA |= σ. This completes the proof of the claim. We now claim that T is model complete using Theorem 13 by showing that every model A of T is existentially closed. There are three cases to consider: (1) |A| = κ (2) |A| > κ (3) |A| < κ where T is κ-categorical. Case (1). Let A∗ be an existentially closed model of T of size κ. Then there is an isomorphism f : A → A∗. Hence A is existentially closed. Case (2). Let σ be an existential sentence ofLA and B |= T such that A ⊆ Band BA |= σ. Let X = {a ∈ A : ca occurs in σ}. By the Downward L¨owenheimSkolem Theorem we can ﬁnd A0 such that A0 ≺ A, X ⊆ A0 and |A0| = κ. Now by Case (1) A0 is existentially closed and we have A0 ⊆ B and σ in LA0 so A0A0 |= σ. But since σ ∈ Th(A0A0) and A0 ≺ A we have AA |= σ. Case (3). Let σ and B be as in case (2). By the Upward L¨owenheim-Skolem Theorem we can ﬁnd A0 such that A ≺ A0 and |A0| = κ. By case (1) A0 is existentially closed. Claim. There is a model B0 such that A0 ⊆ B0 and BA ≡ B0A. Assuming this claim, we have B0 |= T and B0A |= σ and by the fact that A0 is existentially closed we have A0A0 |= σ. Since A ≺ A0 we have AA |= σ. The following lemma implies the claim and completes the proof of the theorem.   Lemma 8. Let A, B and A0 be models for L such that A ⊆ B and A ≺ A0. Then there is a model B0 for L such that A0 ⊆ B0 and BA ≡ B0A.
4. MODEL COMPLETENESS 37
Proof. Let A, B, A0 and L be as above. Let τ be a sentence from 4A0. Let {dj : 0 ≤ j ≤ m} be the constant symbolsfrom LA0 \LA appearing in τ. Obtain a quantiﬁer free formula ϕ(u0,...,um)of LA by exchanging each dj in τ with a new variable ui. Since A0A0 |= τ wehave A0A |= ∃u0 ...∃umϕ. Since A ≺ A0, Exercise 10 gives us AA ≺ A0A and soA A |= ∃u0 ...∃umϕ. Also by Exercise 10 we have AA ⊆ BA, so BA |= ∃u0 ...∃umϕ. Hence forsome b0,...,bm in B, BA |= ϕ[b0,...,bm]. Expand BA to be a model B∗ A for the language LA ∪{dj : 0 ≤ j ≤ m} by interpreting each dj as bj. Then B∗ A |= τ andso Th( BA)∪{τ} is satisﬁable. This shows that Th(BA)∪Σ is satisﬁable for each ﬁnite subset Σ ⊆4A0. By the Compactness Theorem there is a model C |= Th(BA)∪4A0. Using the Diagram Lemma for the language LA we obtain a model B0 for L such that A0A ⊆ B0A and B0A ∼ = C|LA. Hence B0A |= Th(BA) and so B0A ≡ BA.   Exercise 21. Suppose A ≺ A0 are models for L. Prove that for each sentenceσ of LA, if 4A0 |= σ then 4A |= σ. Exercise 22. Prove that if T has a universal-existential set of axioms, then the union of a chain of models of T is also a model of T. Remark. The converse of this last exercise is also true; it is usually called the Chang - L o´s - Suszko Theorem. Theorem 15. The following theories are model complete: (1) dense linear orders without endpoints. (DLO) (2) algebraically closed ﬁelds. (ACF) Proof. (DLO): This theory has a universal existential set of axioms so that it is closed under unions of chains. It is ℵ0-categorical (by Exercise 9) so Lindstr¨om’s test applies. (ACF): We ﬁrst prove that for any ﬁxed characteristic p, the theory of algebraically closed ﬁelds of characteristic p is model complete. The proof is similar to that for DLO, with ℵ1-categoricity (Lemma 7 ). Let A ⊆ B be algebraically closed ﬁelds. They must have the same character-istic p. Therefore A ≺ B.   Corollary 4. Any true statement about the rationals involving only the usual ordering is also true about the reals. Proof. Let A = hQ,<1 <1 <1i and B = hR,<2 <2 <2i where <1 <1 <1 and <2 <2 <2 are the usual orderings. The precise version of this corollary is: A ≺ B. This follows from Theorem 13 and Theorem 15 and the easy facts that A |= DLO, B |= DLO and A ⊆ B. The reader will appreciate the power of these theorems by trying to prove A ≺ B directly, without using them.   The model completeness of ACF can be used to prove Hilbert’s Nullstellensatz. The result below is the heart of the matter. Corollary 5. Let Σ be a ﬁnite system of polynomial equations and inequations in several variables with coeﬃcients in the ﬁeld A. If Σ has a solution in some ﬁeld extending A then Σ has a solution in the algebraic closure of A.
4. MODEL COMPLETENESS 38
Proof. Let σ be the existential sentence of the language LA which asserts the fact that there is a solution of Σ. Suppose Σ has a solution in a ﬁeld B with A ⊆ B. Then BA |= σ. So B0A |= σ where B0 is the algebraic closure of B. Let A0 be the algebraic closure of A. Since A ⊆ B, we have A0 ⊆ B0. By Theorem 15, ACF is model complete, so A0 ≺ B0. Hence A0A ≡ B0A andA 0 A |= σ.   The usual form of the (weak) Nullstellensatz can now be obtained from the algebraic fact that the ideal I of the polynomial ring A[x1,...,xn] generated by Σ is proper exactly when I has a solution in the ﬁeld A[x1,...,xn]/I0 for some maximal ideal I0 containing I. Remark. We cannot apply Lindstr¨om’s Test to the theory of real closed ordered ﬁelds (RCF) because RCF is not categorical in any inﬁnite cardinal. This is because, as demonstrated in Theorem 10, RCF neither implies nor denies the existence of inﬁnitesimals. Nevertheless, as we shall later prove, RCF is indeed model complete.
Exercise 23. Use Exercise 20 and the fact that RCF is model complete to show that RCF is complete. Step 0: the integers, step 1: the rationals, step 2: the real algebraic numbers, step 3: ...
CHAPTER 5
The Seventeenth Problem
We will give a complete proof later that RCF, the theory of real closed ordered ﬁelds, is model complete. However, by assuming this result now, we can give a solution to the seventeenth problem of the list of twenty-three problems of David Hilbert’s famous address to the 1900 International Congress of Mathematicians in Paris.
Corollary 6. (E. Artin) Let q(x1,...,xn) be a rational function with real coeﬃcients, which is positive definite. i.e. q(a1,...,an) ≥ 0 for all a1,...,an ∈R Then there are ﬁnitely many rational functions with real coeﬃcients f1(x1,...,xn), ..., fm(x1,...,xn) such that
q(x1,...,xn) =
m X j=1
(fj(x1,...,xn))2
We give a proof of this theorem after a sequence of lemmas. The ﬁrst lemma just uses calculus to prove the special case of the theorem in which q is a polynomial in only one variable. This result probably motivated the original question.
Lemma 9. A positive deﬁnite real polynomial is the sum of squares of real polynomials.
Proof. We prove this by induction on the degree of the polynomial. Let p(x) ∈ R[x] with degree deg(p) ≥ 2 and p(x) ≥ 0 for all real x. Let p(a) = min{p(x) : x ∈R}, so p(x) = (x−a)q(x) + p(a) and p0(a) = 0 for some polynomial q. But p0(a) = [(x−a)q0(x) + q(x)]

x=a = q(a)so q(a) = 0 and q(x) = r(x)(x−a) for some polynomial r(x). So p(x) = p(a) + (x−a)2r(x). For all real x we have (x−a)2r(x) = p(x)−p(a) ≥ 0. Since r is continuous, r(x) ≥ 0 for all real x, and deg(r) = deg(p) − 2. So, by induction r(x) =Pn i=1 (ri(x))2 where each ri(x) ∈R[x]. So p(x) = p(a) + n X i=1 (x−a)2 (ri(x))2 . 39
5. THE SEVENTEENTH PROBLEM 40 i.e. p(x) =hpp(a)i2 + n X i=1 [(x−a)ri(x)]2 .
The following lemma shows why we deal with sums of rational functions rather than sums of polynomials. Lemma 10. x4y2 + x2y4 − x2y2 + 1 is positive deﬁnite, but not the sum of squares of polynomials.
Proof. Let the polynomial be p(x,y). A little calculus shows that the minimum value of p is 26 27 and conﬁrms that p is positive deﬁnite. Suppose
p(x,y) =
l X i=1
(qi(x,y))2
where qi(x,y) are polynomials, each of which is the sum of terms of the form axmyn. First consider powers of x and the largest exponent m which can occur in any of the qi. Since no term of p contains x6 or higher powers of x, we see that we must have m ≤ 2. Considering powers of y similarly gives that each n ≤ 2. So each qi(x,y) is of the form: aix2y2 + bix2y + cixy2 + dix2 + eiy2 + fixy + gix + hiy + ki for some coeﬃcients ai,bi,ci,di,ei,fi,gi,hi and ki. Comparing coeﬃcients of x4y4 in p and the sum of the q2 i gives
0 =
l X i=1
a2 i
so each ai = 0. Comparing the coeﬃcients of x4 and y4 gives that each di = 0 = ei. Now comparing the coeﬃcients of x2 and y2 gives that each gi = 0 = hi. Now comparing the coeﬃcients of x2y2 gives
−1 =
l X i=1
f2 i
which is impossible.

The next lemma is easy but useful.
Lemma 11. The reciprocal of a sum of squares is a sum of squares.
Proof. For example
1 A2 + B2
=
A2 + B2 (A2 + B2)2
= A A2 + B2 2 + B A2 + B2 2

The following lemma is an algebraic result of E. Artin and O. Schreier, who invented the theory of real closed ﬁelds.
5. THE SEVENTEENTH PROBLEM 41
Lemma 12. Let A = hA,+ + +,• • •,< < 5. THE SEVENTEENTH PROBLEM 42
Lemma 13. Every ordered ﬁeld can be embedded as a submodel of a real closed ordered ﬁeld.
Proof. It suﬃces to prove that for every ordered ﬁeld A there is an ordered ﬁeld B such that A ⊆ B and for each natural number n ≥ 1, B |= σn where σn is the sentence in the language of ﬁeld theory which formally states: If p is a polynomial of degree at most n and w < y such that p(w) < 0 < p(y) then there is an x such that w < x < y and p(x) = 0. Consider the statement called IH(n): For any ordered ﬁeld E there is an ordered ﬁeld F such that E ⊆ F and F |= σn. IH(1) is true since any ordered ﬁeld E |= σ1. We will prove below that for eachn , IH(n) implies IH(n + 1). Given our model A |= ORF, we will then be able to construct a chain of models: A ⊆ B1 ⊆ B2 ⊆ ... ⊆ Bn ⊆ Bn+1 ⊆••• such that each Bn |= ORF∪{σn}. Let B be the union of the chain. Since the theory ORF is preserved under unions of chains (see Exercise 22), B |= ORF. Furthermore, the nature of the sentences σn allows us to conclude that for each n, B |= σn and so B |= RCF. All that remains is to prove that for each n, IH(n) implies IH(n + 1). We ﬁrst make a claim: Claim. If E |= ORF∪{σn} and p is a polynomial of degree at most n+1 with coeﬃcients from E and a < d are in E such that p(a) < 0 < p(d) then there is a model F such that E ⊆ F, F |= ORF and there is b ∈ F such that a < b < d and p(b) = 0.
Let us ﬁrst see how this claim helps us to prove that IH(n) implies IH(n+1). Let E |= ORF; we will use the claim to build a model F such that E ⊆ F and F |= σn+1. We ﬁrst construct a chain of models of ORF E = E0 ⊆ E1 ⊆ ... ⊆ Em ⊆ Em+1 ⊆••• such that for each m and each polynomial p of degree at most n+1 with coeﬃcients from Em and each pair of a, d of elements of Em such that p(a) < 0 < p(d) there is a b ∈ Em+1 such that a < b < d and p(b) = 0. Suppose Em has been constructed; we construct Em+1 as follows: let Σm be the set of all existential sentences of LEm of the form (∃x)(ca < x∧x < cd ∧p(x) = 0) where p is a polynomial of degree at most n + 1 and such that ca, cd and the coeﬃcients of the polynomial p are constant symbols from LEm and (Em)Em |= p(ca) < 0∧0 < p(cd) We claim that ORF∪4Em ∪Σm is satisﬁable.
5. THE SEVENTEENTH PROBLEM 43
Using the Compactness Theorem, it suﬃces to ﬁnd, for each ﬁnite subset {τ1,...,τk} of Σm, a model C such that Em ⊆ C and C |= ORF∪{τ1,...,τk}. By IH(n), obtain a model F1 such that Em ⊆ F1 and F1 |= ORF∪{σn}. By the claim, obtain a model F2 such that F1 ⊆ F2 and F2 |= ORF∪{τ1}. Again by IH(n), obtain F3 such that F2 ⊆ F3 and F3 |= ORF∪{σn}. Again by the claim, obtain F4 such that F3 ⊆ F4 and F4 |= ORF∪{τ2}. Continue in this manner, getting models of ORF Em ⊆ F1 ⊆ ... ⊆ F2k with each F2j |= τj. Since each τj is existential, we get that F2k is a model of each τj (see Exercise 18). Let D |= ORF∪4Em ∪Σm and then use the Diagram Lemma to get Em+1 such that Em ⊆ Em+1, Em+1 |= ORF and Em+1 |= Σm, thus satisfying the required property concerning polynomials from Em. Let F be the union of the chain. Since ORF is a universal-existential theory, F |= ORF (see Exercise 22) and F |= σn+1 by construction. So IH(n+1) is proved. We now ﬁnish the entire proof by proving the claim. Proof of Claim. Suppose that p(x) = q(x) • s(x) with the degree of q atmost n. Since E |= σn we are guaranteed c ∈ E with a < c < d and q(c) = 0. Hencep (c) = 0 and we can let F = E. So we can assume that p is irreducible over E. Introduce a new element b to E where the place of b in the ordering is given by: b < x iﬀ p(y) > 0 for all y with x ≤ y ≤ d. Note that b < d since p(d) > 0. The fact that p is irreducible over E means that we can extend hE,+ + +,• • •,0 0 0,1 1 1iby quotients of polynomials in b of degree ≤ n in the usual way to form a ﬁeld hF,+ + +,• • •,0 0 0,1 1 1i in which p(b) = 0. We leave the details to the reader, but point out that the construction cannot force q(b) = 0 for any polynomial q(x) with coeﬃcients from E of degree ≤ n. This is because we could take such a q(x) of lowest degree and divide p(x) by it to get p(x) = q(x)•s(x) + r(x) where degree of r is less than the degree of q. This means that r(x) = 0 constantly and so p could have been factored over E. Now we must expand hF,+ + +,• • •,0 0 0,1 1 1i to an ordered ﬁeld F while preserving theorder of E. We are aided in this by the fact that if q is a polynomial of degree at most n with coeﬃcients from E then there are a1 and a2 in E such that a1 < b < a2 and q doesn’t change sign between a1 and a2; this comes from the fact that E |= σn.

Proof of the Corollary. Using Lemma 11 we see that it suﬃces to prove the corollary for a polynomial p(x1,...,xn) such that p(a1,...,an) ≥ 0 for all a1,...,an ∈R.
5. THE SEVENTEENTH PROBLEM 44 Let B = hR(x1,...,xn),+ + +,•,0,1i be the ﬁeld of “rational functions”. Notethat B contains the reduct of R to {+,•,0,1} as a subﬁeld, where R is deﬁned as in Example 3 as the usual real numbers. By Lemma 12, if p is not the sum of squares in B, then we can ﬁnd an ordering CHAPTER 6
Submodel Completeness
Definition 30. A theory T is said to admit elimination of quantiﬁers in L whenever for each formula ϕ(v0,...,vp) of L there is a quantiﬁer free formula ψ(v0,...,vp) such that: T |= (∀v0 ...∀vp)(ϕ(v0,...,vp) ↔ ψ(v0,...,vp)) Remark. There is a ﬁne point with regard to the above deﬁnition. If ϕ is actually a sentence of L there are no free variables v0,...,vp. So T |= ϕ ↔ ψ for some quantiﬁer free formula with no free variables. But if L has no constant symbols, there are no quantiﬁer free formulas with no free variables. For this reason we assume that L has at least one constant symbol, or we restrict to those formulas ϕ with at least one free variable. This will become relevant in the proof of Theorem 16 for (2) ⇒ (3). Definition 31. A theory T is said to be submodel complete whenever T ∪4A is complete in LA for each submodel A of a model of T. Exercise 25. Use Theorem 13 and the following theorem to ﬁnd four proofs that every submodel complete theory is model complete. Theorem 16. Let T be a theory of a language L. The following are equivalent: (1) T is submodel complete (2) If B and C are models of T and A is a submodel of both B and C, then every existential sentence which holds in BA also holds in CA. (3) T admits elimination of quantiﬁers (4) whenever A ⊆ B, A ⊆ C, B |= T and C |= T there is a model D such that A ⊆ D and both BA and CA are elementarily embedded in DA. Proof. (1) ⇒ (2)Let B |= T and C |= T with A ⊆ B and A ⊆ C. Then BA |= T ∪4A andC A |= T ∪4A. So (1) and Lemma 6 give BA ≡ CA. Thus (2) is in fact proved for all sentences, not just existential ones. (2) ⇒ (3) We will proceed as we did in the proof of Theorem 13. Lemma 4 shows that it suﬃces to prove (3) for formulas in prenex normal form. We do this by induction on the prenex rank of ϕ using the following claim. Claim. For each existential formula ϕ(v0,...,vp) of L there is a quantiﬁer freeformula ψ(v0,...,vp) such that T |= (∀v0 ...∀vp)(ϕ ↔ ψ) Proof of Claim. Add new constant symbols c0,...,cp to L to form L∗ = L∪{c0,...,cp} 45
6. SUBMODEL COMPLETENESS 46
and to form the existential sentence ϕ∗ of L∗ obtained by replacing each free occurrence of vi in ϕ with the corresponding ci. It suﬃces to prove that there is a quantiﬁer free sentence γ of L∗ such that T |= ϕ∗ ↔ γ. Let Γ = { quantiﬁer free sentences γ of L∗ : T |= ϕ∗ → γ}. It suﬃces to ﬁnd some γ in Γ such that T |= γ → ϕ∗. Since a ﬁnite conjunction of sentences of Γ is also in Γ, it suﬃces to ﬁnd γ1,...,γn in Γ such that T |= γ1 ∧•••∧γn → ϕ∗. If no such ﬁnite subset {γ1,...,γn} of Γ does exist, then each T ∪{γ1,...,γn}∪{¬ϕ∗} would be satisﬁable. So by compactness it suﬃces to prove that T ∪Γ |= ϕ∗. Let C |= T ∪Γ with intent to prove that C |= ϕ∗. Let A be the smallest submodel of C in the sense of the language L∗. That is, every element of A is the interpretation of a constant symbol from L∗ or built from these using the functions of C. Let ∆ = {δ : δ is quantiﬁer free sentence of L∗ and A |= δ}. We wish to show that T ∪{ϕ∗}∪ ∆ is satisﬁable. By compactness, it suﬃces to consider only T ∪{ϕ∗,τ} where τ is a single sentence in ∆. If this set is not satisﬁable then T |= ϕ∗ →¬τ so that by deﬁnition of Γ we have ¬τ ∈ Γ and hence C |= ¬τ. But this is impossible since A ⊆ C means that C |= ∆. Let B0 |= T ∪{ϕ∗}∪∆. The interpretations of the constant symbols in L∗ generate a submodel of A0 ⊆ B0 isomorphic to A. So by Exercise 11, there is a model B for L∗ such that B ∼ = B0 and A ⊆ B.Since A ⊆ B and A ⊆ C the interpretation of each of c0,...,cp in A is the same as the interpretation in B or in C; let’s denote these by a0,...,ap. Let σ denote the sentence (∃v0 ...∃vp)(ϕ∧v0 = ca0 ∧v1 = ca1 ∧•••∧vp = cap) which is equivalent to an existential sentence of LA. In order to invoke (2) we use the restrictions of A, B and C to the language L. We have B|L|= T, C|L|= T, A|L⊆ B|L and A|L⊆ C|L. Since B0 |= ϕ∗ we have that (B|L)A |= σ. So by (2), (C|L)A |= σ and ﬁnally this gives C |= ϕ∗ which completes the proof of the claim.
We now do the general cases for the proof of the induction on prenex rank. There are two cases, corresponding to the two methods available for increasing the number of alternations of quantiﬁers: (a) the addition of universal quantiﬁers (b) the addition of existential quantiﬁers. For case (a), suppose ϕ(v0,...,vp) is ∀w0 ...∀wmχ(v0,...,vp,w0,...,wm) andχ has prenex rank lower than ϕ. Then ¬χ also has prenex rank lower than ϕ
6. SUBMODEL COMPLETENESS 47
and we can use the inductive hypothesis on ¬χ to obtain a quantiﬁer free formula θ1(v0,...,vp,w0,...,wm) such that T |= (∀v0 ...∀vp)(∀w0 ...∀wm)(¬χ ↔ θ1) So T |= (∀v0 ...∀vp)(∃w0 ...∃wm¬χ ↔∃w0 ...∃wmθ1) By the claim there is a quantiﬁer free formula θ2(v0,...,vp) such that T |= (∀v0 ...∀vp)(∃w0 ...∃wmθ1 ↔ θ2) So T |= (∀v0 ...∀vp)(∃w0 ...∃wm¬χ ↔ θ2) So T |= (∀v0 ...∀vp)(∀w0 ...∀wmχ ↔¬θ2) and so ¬θ2 is the quantiﬁer free formula equivalent to ϕ. For case (b), suppose ϕ(v0,...,vp) is ∃w0 ...∃wmχ(v0,...,vp,w0,...,wm) andχ has prenex rank lower than ϕ. We use the inductive hypothesis on χ to obtain a quantiﬁer free formula θ1(v0,...,vp,,w0,...,wm) such that T |= (∀v0 ...∀vp)(∀w0 ...∀wm)(χ ↔ θ1) So T |= (∀v0 ...∀vp)(∃w0 ...∃wmχ ↔∃w0 ...∃wmθ1) By the claim there is a quantiﬁer free formula θ2(v0,...,vp) such that T |= (∀v0 ...∀vp)(∃w0 ...∃wmθ1 ↔ θ2) So T |= (∀v0 ...∀vp)(∃w0 ...∃wmχ ↔ θ2) and so θ2 is the quantiﬁer free formula equivalent to ϕ. This completes the proof. (3) ⇒ (4) Let A ⊆ B, A ⊆ C, B |= T and C |= T. Using the Diagram Lemmas it will suﬃce to show that Th(BB)∪ Th(CC) is satisﬁable. Without loss of generosity, we can ensure that LB ∩LC = LA. By the Robinson Consistency Theorem, it suﬃces to show that there is no sentence σ of LA such that both: Th(BB) |= σ and Th(CC) |= ¬σ Suppose σ is such a sentence and let {ca0,...,cap} be the set of constant symbols from LA \L appearing in σ. Let ϕ(u0,...,up) be obtained from σ by exchanging each cai for a new variable ui. Let ψ(u0,...,up) be the quantiﬁer free formula from (3): T |= (∀u0,...,∀up)(ϕ ↔ ψ) Let ψ∗ be the result of substituting cai for each ui in ψ. ψ∗ is also quantiﬁer free. Since BB |= σ, B |= ϕ[a0,...,ap]. Since B |= T, B |= ψ[a0,...,ap] and soB A |= ψ∗. Since ψ∗ is quantiﬁer free and AA ⊆ BA we have AA |= ψ∗; sinceA A ⊆ CA we then get that CA |= ψ∗. Hence C |= ψ[a0,...,ap] and then sinceC | = T we then get that C |= ϕ[a0,...,ap]. But then this means that CA |= σ andso CC |= σ so σ is in Th(CC) and we are done. (4) ⇒ (1)Let B |= T and A ⊆ B; we show that T ∪4A is complete. Since BA |= T ∪4A, we see that it suﬃces by Lemma 6 to show that BA ≡ C0 for each C0 |= T ∪4A. For each such C0, by Exercise 14, there is a model C for L such that A ⊆ C andC A ∼ = C0. Then C |= T so by (4) there is a D with A ⊆ D such that both BA andC A are elementarily embedded into DA.
6. SUBMODEL COMPLETENESS 48 In particular BA ≡ DA ≡ CA so we are done.   Example 11. (Chang and Keisler) Let T be the theory in the language L = {U,V,W,R,S} where U, V and W are unary relation symbols and R and S are binary relation symbols having axioms which state that there are inﬁnitely many things, that U ∪V ∪W is everything, that U, V and W are pairwise disjoint, that R is a one-to-one function from U onto V and that S is a one-to-one function from U ∪V onto W. Exercise 26. Show that T above is complete and model complete but not submodel complete. Hints: For completeness, use the L o´s-Vaught test and for model completeness use Lindstr¨om’s test. For submodel completeness use (2) of the theorem with B |= T and A ⊆ B where a ∈ A = {b ∈ B : B |= W(v0)[b]} along with the sentence (∃v0)(U(v0)∧S(v0,ca)). We will prove in the next chapter that each of the following theories admits elimination of quantiﬁers: (1) dense linear orders with no end points (DLO) (2) algebraically closed ﬁelds (ACF) (3) real closed ordered ﬁelds (RCF) C. H. Langford proved elimination of quantiﬁers for DLO in 1924. The cases of ACF and RCF were more diﬃcult and were done by A. Tarski. Thus, by Exercise 25, we will have model completeness of RCF which was promised at the beginning of Chapter 5. Exercise 27. LetT be a theory in the languageLwhich is submodel complete.Expand L to L+ by only adding new constant symbols. Show that T admits elimination of quantiﬁers inL+. Use this and the fact that DLO admits elimination of quantiﬁers in its own language to show that in the languageL = {<,c1,c2}where c1 and c2 are constant symbols, DLO is submodel complete but not complete.
Exercise 28. Suppose A is the reduct of a real closed ordered ﬁeld to the language of ﬁeld theory. Show that A[√−1] is algebraically closed. You may use the Fundamental Theorem of Algebra. Hint: Show that any polynomial with coeﬃcients from A has a quadratic factor.
Corollary 7. (The Tarski-Seidenberg Theorem) The projection of a semi-algebraic set in Rn to Rm for m < n is also semialgebraic. The semi-algebraic sets of Rn are deﬁned to be all those subsets of Rn which can be obtained by repeatedly taking ﬁnite unions and intersections of sets of these two forms {hx1,...,xni∈Rn : p(x1,...,xn) = 0} {hx1,...,xni∈Rn : q(x1,...,xn) < 0} where p and q are polynomials with real coeﬃcients.
Proof. We ﬁrst need a simple result which we state as an exercise. Let R = hR,+ + +,• • •,< < <,0,1i be the usual model of the reals. Let T be RCFconsidered as a theory in the language LR.
6. SUBMODEL COMPLETENESS 49 Exercise 29. A set X ⊆ Rn is semi-algebraic iﬀ there is a quantiﬁer freeformula ϕ(v1,...,vn) of LR such that X = {hx1,...,xni : RR |= ϕ[x1,...,xn]}. Now, in order to prove the corollary, let X ⊆ Rn be semi-algebraic and let ϕ be its associated quantiﬁer free formula. The projection Y of X into Rm is {hx1,...,xmi : for some xm+1,...,xn hx1,...,xm,xm+1,...,xni∈ X} So Y = {hx1,...,xmi : RR |= ∃vm+1 ...∃vnϕ[x1,...,xm]} From the assumption that RCF is submodel complete and Exercise 27 we have that T admits elimination of quantiﬁers. So there is a quantiﬁer free formula θ of LR such that T |= (∀v1 ...∀vm)(∃vm+1 ...∃vnϕ ↔ θ) Hence for all x1,...,xm RR |= ∃vm+1 ...∃vnϕ[x1,...,xm] iﬀ RR |= θ[x1,...,xm] So Y = {hx1,...,xmi : RR |= θ[x1,...,xm]} and by the exercise, Y is semi-algebraic.   As an application of quantiﬁer elimination of ACF we have the following result of A. Tarski.
Corollary 8. The truth value of any algebraic statement about the complex numbers can be determined algebraically in a ﬁnite number of steps. Proof. Let C be the complex numbers in the language of ﬁeld theory L; letσ be a sentence of LC. Then let A be the ﬁnite subset {a0,...,ap}⊆C consisting of those elements of C (other than 0 or 1) which are mentioned in σ. Let ϕ be the formula of L formed by exchanging each cai for a new variable wi. Then ACF |= ∀w0 ...∀wp(ϕ ↔ ψ) for some quantiﬁer free ψ. Hence C |= σ iﬀ C |= ϕ[a0,...,ap] iﬀ C |= ψ[a0,...,ap] but checking this last statement amounts to evaluating ﬁnitely many polynomials in a0,...,ap.   Tarski’s original proof actually gave an explicit method for ﬁnding the quantiﬁer free formulas and this led, via the argument above, to an eﬀective decision procedure for determining the truth of elementary algebraic statements about the complex numbers.
CHAPTER 7
Model Completions
Closely related to the notions of model completeness and submodel completeness is the idea of a model completion. Definition 32. Let T ⊆T∗ be two theories in a language L. T∗ is said to bea model completion of T whenever T∗∪4A is satisﬁable and complete in LA for each model A of T. Lemma 14. Let T be a theory in a language L. (1) If T∗ is a model completion of T, then for each A |= T there is a B |= T∗ such that A ⊆ B.(2) If T∗ is a model completion of T, then T∗ is model complete.(3) If T is model complete, then it is a model completion of itself.(4) If T∗ 1 and T∗ 2 are both model completions of T, then T∗ 1 |= T∗ 2 and T∗ 2 |= T∗ 1 . Proof. (1) Easy. (2) Easier. (3) Easiest. (4) This needs a proof. Let A |= T∗ 2 . It will suﬃce to prove that A |= T∗ 1 . Let A0 = A. since A0 |= T and T∗ 1 is a model completion of T we obtain, from(1), a model A1 |= T∗ 1 such that A0 ⊆ A1. Similarly, since A1 |= T and T∗ 2 is a model completion of T we obtain A2 |= T∗ 2 such that A1 ⊆ A2.Continuing in this manner we obtain a chain: A0 ⊆ A1 ⊆ A2 ⊆ ... ⊆ An ⊆ An+1 ⊆••• Let B be the union of the chain,∪{An : n ∈N}. For each n ∈N we have A2n |= T∗ 2 . By part (2) of this lemma and by part (4) of Theorem 13 we get that for each n, A2n ≺ A2n+2. By the Elementary Chain Theorem A0 ≺ B. Similarly A1 ≺ B. So A0 ≡ A1 and hence A |= T∗ 1 .
Remark. Part (4) of the above lemma shows that model completions are essentially unique. That is, if model completions T∗ 1 and T∗ 2 of T are closed theoriesin the sense of Deﬁnition 12 then T∗ 1 = T∗ 2 . Since there is no loss in assuming that model completions are closed theories, we speak of the model completion of a theory T. Theorem 17. Suppose T ⊆T∗ are theories for a language L. T∗ is the model completion of T iﬀ the following two conditions are satisﬁed. (1) For each A |= T there is a B |= T∗ with A ⊆ B. (2) For each A |= T, B |= T∗ and C |= T∗ such that A ⊆ B and A ⊆ C we have a model D such that BA is isomorphically embedded into DA and C ≺ D. 50
7. MODEL COMPLETIONS 51
Proof. First, assume that T∗ is the model completion of T. Condition (1) is part (1) of the previous lemma. Now, let A, B and C be as in Condition (2). By Exercise 10, BA |= T∗∪4Aand CA |= T∗∪4A. Without loss of generosity, we may assume thatLB∩LC = LA. By assumption T∗ ∪4A is a complete theory in LA. Therefore BA ≡ CA. By the Robinson Consistency Theorem, the set of sentences ThBB ∪ ThCC is satisﬁable. Let E |= ThBB ∪ ThCC The Elementary Diagram Lemma now gives us a model D such that CA ≺ DA and DA ∼ = E|LA. By the Diagram lemma BA is isomorphically embedded into E|LA and hence also into DA.
Now assume that conditions (1) and (2) hold. We ﬁrst show that T∗ is model complete using Theorem 13; we show that T∗ is existentially complete. Let A |= T∗; we show that A is existentially closed. Let B |= T∗ such that A ⊆ B and let σ be an existential sentence of LA with BA |= σ; our aim is to prove that AA |= σ. We invoke condition (2) with C = A to get a model D such that A ≺ D andB A is isomorphically embedded into DA. Referring to Exercise 11 we get a model E for LA with BA ⊆ E and DA ∼ = E. Since σ is existential, By Exercise 18 we have that E |= σ; and by Exercise 7, DA |= σ. Now A ≺ D implies that AA ≡ DA so AA |= σ and T∗ is model complete. We now show that T∗ is the model completion of T. Let A |= T; condition (1) gives that T∗ ∪4A is satisﬁable. We show that T∗ ∪4A is complete in LA by showing that for each B |= T∗ and C |= T∗ with A ⊆ B and A ⊆ C we have BA ≡ CA. Letting B and C be as above, we invoke condition (2) to obtain a model D such that BA is isomorphically embedded into DA and C ≺ D. C ≺ D gives that D |= T∗. The isomorphic embedding gives us a model E such that B ⊆ E and DA ∼ = EA. So E |= T∗ . Using the model completeness of T∗ and Theorem 13 we can infer that B ≺ E. We have: BA ≡ EA ≡ DA ≡ CA and we are done.
Let’s compare the deﬁnitions of model completion and submodel complete. Let T∗ be the model completion of T. Then T∗ will be submodel complete provided that every submodel of a model of T∗ is a model of T. Since T ⊆T∗, it would be enough to show that every submodel of a model of T is again a model of T. And this is indeed the case whenever T is a universal theory, that is, whenever T has a set of axioms consisting of universal sentences. Unfortunately, this is not always the case. Our aim is to show that DLO, ACF and RCF are submodel complete by showing that these theories are the model completions of LOR, FLD and ORF respectively. See Example 5 to recall the axioms for these theories. Well, LOR is a universal theory but FLD and ORF are not. The culprits are the axioms asserting the existence of inverses:
∀x∃y(x + y = 0) and ∀x((x 6= 0) →∃y(y•x = 1))
7. MODEL COMPLETIONS 52
.
In fact, a submodel A of a ﬁeld B is only a commutative semi-ring, not necessarily a subﬁeld. Nevertheless, A generates a subﬁeld of B in a unique way. This motivates the following deﬁnition. Definition 33. A theory T is said to be almost universal whenever A ⊆ B,B | = T and A ⊆ C, C |= T imply there are models D and E such that D |= T,A ⊆ D ⊆ B and E |= T, A ⊆ E ⊆ C and DA ∼ = EA. Example 12. LOR is almost universal since any universal theory T is almost universal — just let D = E = A and note A |= T. Example 13. FLD is almost universal — just let D and E be the subﬁelds of B and C, respectively, generated by A. The isomorphism DA ∼ = EA is the natural one obtained from the identity map on A.
Example 14. ORF is almost universal — again just let D and E be the ordered subﬁelds of B and C, respectively, generated by A. The extension of the identity map on A to the isomorphism DA ∼ = EA is aided by the fact that the order placement of the inverse of an element a is completely determined by the order placement of a. Theorem 18. Let T and T∗ be theories of the language L such that T is almost universal and T∗ is the model completion of T. Then T∗ is submodel complete. Proof. We show that condition (2) of Theorem 16 is satisﬁed. Let B and C be models of T∗ with A a submodel of both B and C; we will show that BA ≡ CA. Now T ⊆ T∗ so B |= T and C |= T. Since T is almost universal there aremodels D and E of T such that A ⊆ D ⊆ B, A ⊆ E ⊆ C and DA ∼ = EA. So BD |= T∗∪4D and CE |= T∗∪4E. Now BD is a model for the language LD whereas CE is a model for LE. We wish to obtain a model C0 for LD which “looks exactly like” CE. We just let C0 be C and in fact let C0|LA = CE|LA. The interpretation of a constant symbol cd ∈LD \LA is the interpretation of ce ∈LE \LA in CE where the isomorphism DA ∼ = EA takes d to e. Now D |= T and since T∗ is the model completion of T, T∗∪4D is complete. The isomorphism DA ∼ = EA ensures that C0 |= T∗ ∪4D. So BD ≡ C0. HenceB D|LA ≡ C0|LA; that is, BA ≡ CA.
The way to show that DLO, ACF and RCF admit elimination of quantiﬁers is now clear: ﬁrst use Theorem 16 and Theorem 18. They reduce our task to showing that DLO, ACF and RCF are the model completions of LOR, FLD and ORF respectively. To do this we use Theorem 17; we will show that each pair of theories satisﬁes both conditions (1) and (2) of Theorem 17. We begin with condition (1): if A |= T then there is a B |= T∗ such that A ⊆ B. The caseT = LOR andT∗ = DLO is easy; every linear order can be enlarged to a dense linear order without endpoints by judiciously placing copies of the rationals into the linear order. The case T = FLD and T∗ = ACF is just the well known fact that every ﬁeld has an algebraic closure. The case T = ORF and T∗ = RCF is just Lemma 13.
7. MODEL COMPLETIONS 53
So all that remains of the quest to prove elimination of quantiﬁers for DLO, ACF and RCF is to verify condition (2) of Theorem 17 in each of these cases. At this point the reader may already be able to verify this condition for one or more of the pairs T = LOR and T∗ = DLO, T = FLD and T∗ = ACF, or T = ORF and T∗ = RCF. However the remainder of this chapter is devoted to a uniform method. Definition 34. Let L be a language and Σ(v0) a set of formulas of L in the free variable v0. A model A for L is said to realise Σ(v0) whenever there is some a ∈ A such that A |= ϕ[a] for each ϕ(v0) in Σ(v0). Definition 35. The set of formulas Σ(v0) in the free variable v0, is said to be a type of the model A whenever (1) every ﬁnite subset of Σ(v0) is realised by A (2) Σ(v0) is maximal with respect to (1).
Remark. Every set of formulas Σ(v0) having property (1) of the deﬁnition of type can be enlarged to also have property (2). Lemma 15. Suppose A is a model for a language L. Let X ⊆ A and let Σ(v0) be a type of AX in the language LX. Then there is a B such that A ≺ B and BX realises Σ(v0). Proof. Let T = ThAA ∪Σ(c) where c is a new constant symbol and Σ(c) = {ϕ(c) : ϕ ∈ Σ(v0)} and of course ϕ(c) is ϕ(v0) with c replacing v0. By the deﬁnition of type, for each ﬁnite T0 ⊆ T, there is an expansion A0 ofA such that A0 |= T0. The Compactness Theorem and the Elementary Diagram Lemma will complete the proof.   Lemma 16. Suppose A is a model for a language L. There is a model B for L such that A ≺ B and BA realises each type of AA in the language LA. Proof. Let {Σα(v0) : α ∈ I} enumerate all types of AA in the language LA. For each α ∈ I introduce a new constant symbol cα and let Σα(cα) = {ϕ(cα) : ϕ ∈ Σα(v0)}. Let Σ = ∪{Σα(cα) : α ∈ I}. Let Σ0 ⊆ Σ be any ﬁnite subset. Claim. Σ0∪ThAA is satisﬁable for the language LA ∪{cα : α ∈ I}. Proof of Claim. Let Σα1(v0),...,Σαn(v0) be ﬁnitely many types such that Σ0 ⊆ Σα1(c0)∪Σα2(c1)∪•••∪Σαn(cαn). By Lemma 15 there is a model A1 such that A ≺ A1 and (A1)A realises Σα1(v0). Using Lemma 15 repeatedly, we can obtain A ≺ A1 ≺ A2 ≺•••≺ An such that each (Aj)A realises Σαj(v0). Now A ≺ An so (An)A |= ThAA. It is easy to check that since each Aj ≺ An,A n realises each Σαj(v0) and furthermore so does (An)A. So we can expand (An)A to the language LA ∪{cα1,...,cαn} to satisfy Σ0∪ThAA.
7. MODEL COMPLETIONS 54
By the claim and the Compactness Theorem, there is a model C |= Σ∪ThAA. By the Elementary Diagram Lemma, A is elementarily embedded into C|L, the restriction of C to the language L. Therefore there is a model B for L such that A ≺ B and BA ∼ = C|LA. It is now straightforward to check that BA realises eachtype Σ α(v0).
Definition 36. A model A for L is said to be κ-saturated whenever we have that for each X ⊆ A with |X| < κ, AX realises each type of AX. Recall that for any set X we denote by |X| the cardinality of X. The notationκ + is used for the cardinal number just larger than the cardinal κ. So a model A will be κ+-saturated whenever we have that for each X ⊆ A with |X| ≤ κ, AX realises each type of AX. In particular, if B is any set, A will be |B|+-saturated whenever we have that for each X ⊆ A with |X|≤|B|, AX realises each type of AX. Remark. A model A is said to be saturated whenever it is |A|-saturated,where |A| is the size of the universe of A. For example, hQ,< < 7. MODEL COMPLETIONS 55
recursively as follows. At stage β, suppose we have already constructed Cα for each α ∈ I with α < β. The union of the chain up to β E = ∪{Cα : α ∈ I and α < β} falls under the scope of an upgraded Elementary Chain Theorem (which is proved exactly as Theorem 4) and so Cα ≺ E for each α ∈ I with α < β. We now use Lemma 16 to get Cβ such that E ≺ Cβ and (Cβ)E realises each type of EE. As before, let D = ∪{Cα : α ∈ I} be the union of the entire chain and by the upgraded Elementary Chain Theorem C ≺ D. Also as before, DX realises each type of DX for each X ⊆ D such that X ⊆ Cα for some α ∈ I. We can now complete the proof of the lemma by choosing our well ordered set hI, |B|.
Definition 37. We say that B is a simple extension of A whenever (1) A ⊆ B and (2) there is some b ∈ B such that no smaller submodel of B contains A∪{b}.
Theorem 19. (Blum’s Test) Suppose T ⊆T∗ are theories of a language L. Suppose further that: (1) T is an almost universal theory, (2) for each A |= T there is a B |= T∗ with A ⊆ B, and (3) for each A |= T and each simple extension B of A which is a submodel of a model of T, and for each C |= T∗ with A ⊆ C such that C is |B|+saturated, there is an isomorphic embedding f : B → C such that f A is the identity on A. Then: (4) T∗ is the model completion of T, (5) T∗ is submodel complete, and (6) T∗ admits elimination of quantiﬁers. Proof. With Theorems 16, 17 and 18, statements (4), (5) and (6) all follow from (1), (2) and condition (2) of Theorem 17. We will therefore only need to prove that for each A |= T, B |= T∗ and C |= T∗ such that A ⊆ B and A ⊆ C we have a model D such that BA is isomorphically embedded into DA and C ≺ D, Let A, B and C be as above. Using Lemma 17 we obtain a |B|+-saturatedmodel D such that C ≺ D. We wish to prove that BA is isomorphically embeddedinto DA. Since A ⊆ D, the following collection E of functions is nonempty: {e : for some A ⊆ E ⊆ B e: EA ,→ DA is an isomorphic embedding} and so has a maximal member f in the sense that no other e ∈E extends f. From f : FA ,→ DA and Exercise 11 we get G with F ⊆ G and an isomorphism g: G → D extending f.
7. MODEL COMPLETIONS 56
Claim. F |= T Proof of Claim. We have both F ⊆ B and F ⊆ G. By condition (1), there are models H and J of T with F ⊆ H ⊆ B and F ⊆ J ⊆ G such that HF ∼ = JF. This gives an isomorphic embedding h: H ,→ G such that h F is the identity on F. The composition g◦h: H ,→ D is an isomorphic embedding with the property that for all x ∈ F: (g◦h)(x) = g(x) = f(x). By the maximality of f, f = g◦h. Hence F = H and F |= T, ﬁnishing the proof of the claim.
Claim. F = B Proof of Claim. If not, pick b ∈ B\F and form the simple extension F0 ofF by b. Since G ∼ = D, G is also |B|+− saturated so that we can apply condition (3)to F, F0 and G. We obtain an isomorphic embedding f0: F0 ,→ G such that f0 F is the identity on F. But now g◦f0 contradicts the maximality of f and completes the proof of the claim.
Therefore f isomorphically embeds BA into DA.
The following lemma completes the proofs that each of the theories DLO, ACF and RCF admits elimination of quantiﬁers. Lemma 18. Each of the following three pairs of theories T and T∗ satisfy condition (3) of Blum’s Test. (1) T = LOR, theory of linear orderings. T∗ = DLO, theory of dense linear orderings without endpoints. (2) T = FLD, theory of ﬁelds. T∗ = ACF, theory of algebraically closed ﬁelds. (3) T = ORF, theory of ordered ﬁelds. T∗ = RCF, theory of real closed ordered ﬁelds. Proof of (1). Let A and B be linear orders, with B = A∪{b} and A ⊆ B. Let C be a |B|+-saturated dense linear order without endpoints with A ⊆ C. We wish to ﬁnd an isomorphic embedding f : B → C which is the identity on A. Consider a type of CA containing the following formulas: ca < v0 for each a ∈ A such that a < b v0 < ca for each a ∈ A such that b < a Since C is a dense linear order without endpoints each ﬁnite subset of the type can be realised in CA. Saturation now gives some t ∈ C realising this type. We set f(b) = t and we are ﬁnished.
Proof of (2). Let A be a ﬁeld and B a simple extension of A witnessed by b such that B is a submodel of a ﬁeld (a commutative semi-ring). Let C be a |B|+-saturated algebraically closed ﬁeld such that A ⊆ C. We wish to ﬁnd an isomorphic embedding f : B → C which is the identity on A. There are two cases: (I) b is algebraic over A, (II) b is transcendental over A.
7. MODEL COMPLETIONS 57
Case(I). Let p be a polynomial with coeﬃcients from A such that p(b) = 0 but b is not the root of any such polynomial of lower degree. Since C is algebraically closed there is a t ∈ C such that p(t) = 0. We extend the identity map f on A to make f(b) = t. We extend f to the rest of B by letting f(r(b)) = r(t) for any polynomial r with coeﬃcients from A. It is straightforward to show that f is still a well-deﬁned isomorphic embedding.
Case (II). Let us consider a type of CA containing the following set of formu
las:
{¬(p(v0) = 0) : p is a polynomial with coeﬃcients in {ca : a ∈ A}} Since C is algebraically closed, it is inﬁnite and hence each ﬁnite subset is realised in CA. Saturation will now give some t ∈ C such that t realises the type. We set f(b) = t. Since t is transcendental over A, the extension of f to all of B comes easily from the fact that every element of B\A is the value at b of some polynomial function with coeﬃcients from A.
Proof of (3). Let A be an ordered ﬁeld and B be a simple extension of A witnessed by b such that B is a submodel of an ordered ﬁeld (an ordered commutative semi-ring). Let C be a |B|+-saturated real closed ﬁeld such that A ⊆ C. We wish to ﬁnd an isomorphic embedding f : B → C which is the identity on A. There are two cases: (I) b is algebraic over A. (II) b is transcendental over A.
Case (I). Since b is algebraic over A we have a polynomial p with coeﬃcients in A such that p(b) = 0. All other elements of the universe of the simple extension B are of the form q(b) where q is a polynomial with coeﬃcients in A. Before beginning the main part of the proof we need some algebraic facts.
Claim. Let D be a real closed ordered ﬁeld and q(x) be a polynomial over D of degree n. Then for any e ∈ D we have:
q(x) =
n X m=0
q(m)(e) m!
(x−e)m
where q(m) stands for the polynomial which is the m-th derivative of q.
Proof of Claim. This is Taylor’s Theorem from Calculus; unfortunately we cannot use Calculus to prove it because we are in D, not necessarily the reals R. However the reader can check that the Binomial Theorem gives the identity for the special cases of q(x) = xn and that these special cases readily give the full result.
Claim. Let D be a real closed ordered ﬁeld and q(x) a polynomial over D with e ∈ D and q(e) = 0. If there is an a < e such that q(x) > 0 for all a < x < e then q0(e) ≤ 0. If there is an a > e such that q(x) > 0 for all e < x < a then q0(e) ≥ 0. Here q0 is the ﬁrst derivative of q.
7. MODEL COMPLETIONS 58
Proof of Claim. From the previous claim we get q(x)−q(e) x−e = q0(e) + (x−e) n X m=2 q(m)(e) m!
(x−e)m−2! for any x 6= e in D. By choosing x close enough to e we can ensure that the entire right hand side has the same sign as q0(e). A proof by contradiction now follows readily.
Claim. Let D be a real closed ordered ﬁeld and q(x) be a polynomial over D with e ∈ D and q(e) = 0. If w and z are in D such that w < e < z and q(w)•q(z) > 0 then there is a d in D such that w < d < z and q0(d) = 0.
Proof of Claim. Without loss of generosity q(w) > 0 and q(z) > 0. Since q has only ﬁnitely many roots, we can pick d1 to be the least x such that w < x ≤ e and q(x) = 0. Since q(x) 6= 0 for all w < x < d1, the Intermediate Value Property of Real Closed Ordered Fields shows that q cannot change sign here and so q(x) > 0 for all w < x < d1. By the previous claim, q0(d1) ≤ 0. A similar argument with z shows that there is a d2 such that e ≤ d2 < z and q0(d2) ≥ 0. If d1 = e = d2 take d = e. If d1 < d2 the Intermediate Value Property gives a d with the required properties.
Claim. Let D be a real closed ordered ﬁeld with an ordered ﬁeld E ⊆ D. Letf : E → C be an isomorphic embedding into a real closed ordered ﬁeld. Let q be a polynomial with coeﬃcients in E such that {x ∈ D : q0(x) = 0}⊆ E. Let d ∈ D\E be such that q(d) = 0 but d is not a root of a polynomial with coeﬃcients from E which has lower degree. Then f can be extended over the subﬁeld of D generated by E∪{d}. Proof of Claim. Since the ﬁnitely many roots of q0 from D actually lie in E, we can get e1 and e2 in E such that e1 < d < e2 and q0(x) 6= 0 for all x in D such that e1 < x < e2. Furthermore for all x in E we have q(x) 6= 0. We can now apply the previous claim to get that q(w)•q(z) < 0 for all w and z in E such that e1 < w < d < z < e2. We now move to the real closed ordered ﬁeld C and the isomorphic embedding f. For each w and z in E such that e1 < w < d < z < e2 we have f(w) < f(z) and q(f(w)) • q(f(z)) < 0. By the Intermediate Value property of C we get, for each such w and z, a y ∈ C such that f(w) < y < f(z) and q(y) = 0. Since q has only ﬁnitely many roots there is some t ∈ C such that q(t) = 0, f(w) < t for all e1 < w < d and t < f(z) for all d < z < e2. We now extend f by letting f(d) = t and f(r(d)) = r(t) for any polynomial r with coeﬃcients from E. It is straightforward to check that the extension is a well-deﬁned isomorphic embedding of the simple extension of E by d into C. We use the fact that ORF is almost universal to extend the isomorphic embedding to all of the subﬁeld of D generated by E∪{d}, since we can rephrase the deﬁnition of almost universal as follows: Whenever C |= T, D |= T, E0 ⊆ D and f : E0 ,→ C is an isomorphic embedding there is a model E00 |= T such that E0 ⊆ E00 ⊆ D and f extends over E00.
7. MODEL COMPLETIONS 59
It is now time for the main part of the proof of this case. Using Lemma 13, let D be a real closed ordered ﬁeld with B ⊆ D. We have a polynomial p with coeﬃcients from A such that p(b) = 0. By induction on the degree of p, we can show that there is a sequence of elements d0,...,dm = b of elements of D, a sequence of subﬁelds of D: A = E0 ⊆ E1 ⊆ ... ⊆ Em+1 with each dj ∈ Ej+1 \Ej and corresponding isomorphic embeddings fj : Ej → C coming from the previous claim and having the property that f0 is the identity and fj+1 extends fj. In this way we extend the identity map f0 : A0 ,→ C until we reach fm+1 : Em+1 ,→ C. We then note that since b ∈ Em+1 we have B ⊆ Em+1 and we are ﬁnished.
Case (II). Let us consider a type of CA containing the following formulas: ca < v0 for all a ∈ A with a < b v0 < ca for all a ∈ A with b < a ¬(p(v0) = 0) for all polynomials p with coeﬃcients in {ca : a ∈ A} Since each interval of C is inﬁnite, each ﬁnite subset of this type is realised by CA. Saturation now gives t ∈ C which realises this type. We put f(b) = t. We can now extend f on the rest of B\A, since each such element is the valueat b of a polynomial function with coeﬃcients from A.
Bibliography
1. J. Barwise (ed.), Handbook of Mathematical Logic, North Holland, 1997. 2. J. L. Bell and A. B. Slomson, Models and Ultraproducts, North Holland, 1969. 3. Felix E. Browder (ed.), Mathematical developments arising from Hilbert’s problems, Proceedings of Symposia in Pure Mathematics, vol. XXVIII - Parts 1 and II, AMS, 1976. 4. S. Buechler, Essential Stability Theory, Perspectives in Mathematical Logic, Springer, 1996. 5. C. C. Chang and H. J. Keisler, Model Theory, second ed., North Holland, 1977. 6. G. Cherlin, Model Theoretic Algebra, Selected Topics, LNM521, Springer-Verlag, 1976. 7. N. Jacobson, Basic Algebra I, W. H. Freeman, 1985. 8. H. J. Keisler, Foundations of Inﬁnitesimal Calculus, Prindle, Weber and Schmidt, 1976. 9. M. D. Morley (ed.), Studies in Model Theory, MAA Studies in Mathematics, MAA, 1973. 10. A. Robinson, Introduction to Model Theory and the Metamathematics of Algebra, second ed., North-Holland, 1965. 11. , Non Standard Analysis, North Holland, 1966. 12. G. E. Sacks, Saturated Model Theory, W. A. Benjamin Inc., 1972. 13. D. H. Saracino and V. B. Weispfenning (eds.), Model theory and algebra: A Memorial Tribute to Abraham Robinson, LNM498, Springer Verlag, 1975. 14. S. Shelah, Classiﬁcation Theory and the Numbers of Nonisomorphic models, second ed., North Holland, 1990. 15. J. R. Shoenﬁeld, Mathematical Logic, Addison-Wesley, 1967. 16. A. Tarski, A Descision Method for Elementary Algebra and Geometry, second ed., The Rand Corporation, 1957.
60
Index
A ⊆ B, 15 A = ∪{An : n ∈N}, 16 ThA, 12 ThAA, 24 | =, 12 ≺, 16 A0|L0, 14 ⊆, 14 t[x0,...,xq], 6 hC,+ + +,•,0,1i, 11 hQ,< < <,+ + +,•,0,1i, 11 hR,< < <,+ + +,•,0,1i, 11 hN,+ + +,• • •,< < <,0 0 0,1 1 1i, 15 A. Robinson, 32 ACF, 13 submodel complete, 51 algebraically closed ﬁelds axioms,theory of, 13 almost universal, 52, 58 axioms, 12
Blum’s Test, 55 bound variable, 5
categorical κ-categorical theory, 20 chain of models, 16 elementary, 16 Compactness Theorem, 14 complete theory, 20 Completeness Theorem, 14 complex, 11 Craig Interpolation Theorem, 31
dense linear orders without endpoints axioms,theory of, 13 diagram lemmas, 25 DLO, 13 submodel complete, 51 Downward L¨owenheim-Skolem Theorem, 25
elementarily embedded model, 24 elementarily equivalent models, 15
Elementary Chain Theorem, 16 elementary diagram, 24 elementary extension, 16 elementary submodel, 16 elimination of quantiﬁers, 45 existentially closed, 32 expansion language, 14 model, 14
ﬁelds axioms,theory of, 13 FLD, 13 almost universal, 52 formula, 4 free variable, 5
isomorphic models, 15 isomorphically embedded model, 24
language, 5 Leibniz Principle, 27 Lindstr¨om’s Test, 35 linear orders axioms,theory of, 13 LOR, 13 almost universal, 52 L o´s-Vaught Test, 20 Lowenheim-Skolem Theorem, 20
model, 5 satisﬁes, 6 model complete theory, 32 model completion, 50 submodel complete, 52
Number Theory, 15 number theory non-standard models, 15
ordered ﬁeld, 42 ordered ﬁelds axioms,theory of, 13 ORF, 13 almost universal, 52
61
INDEX 62
prenex normal form, 9
rational numbers, 11 RCF, 13 submodel complete, 51 real closed ordered ﬁeld, 42 real closed ordered ﬁelds axioms, theory, 13 axioms,theory of, 13 Intermediate Value Property, 13 real numbers, 11 realize, 53 reduction language, 14 model, A0|L0, 14 Robinson Consistency Theorem, 27
satisfaction A |= Σ, 12 saturated κ-saturated model, 54 sentence, 9 simple extension, 55 subformula, 5 submodel, 15 submodel complete, 45 submodel complete theory, 45
T. Skolem, 15 Tarski’s Elementary Chain Theorem, 16 Tarski-Vaught Condition, 19 term, 4 theory, 12 almost universal, 52 model completion, 50 theory of A, 12 type, 53
Upward L¨owenheim-Skolem Theorem,

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