Mo del Theory

J van Oosten

Department of Mathematics

Utrecht University

Spring

Contents

Intro duction and Notations

Homomorphisms Emb eddings and Diagrams

Elementary Emb eddings and Elementary Diagrams

Directed Systems of Lstructures

Theorems of Robinson Craig and Beth

Preservation Theorems

Filters and Ultrapro ducts

Quantier Elimination and Mo del Completeness

Quantier Elimination Real Closed Fields

Mo del Completeness and Mo del Companions

Mo del Completions Amalgamation Quantier Elimination

Countable Mo dels

End extensions of mo dels of PA

Atomic Theories and Atomic Mo dels

Saturated Mo dels

Stone Duality

Literature

Intro duction and Notations

In the course Foundations of Mathematics we have made our acquaintance with

the notions of a language L in predicate logic and a structure A for the lan

guage L A language is a collection of symb ols divided into three groups we

have constants function symbols and relation or predicate symbols Given a

language L we have dened the class of Lterms

A

A structure A for the language L is a pair A A where A is a

A

nonempty set and is an interpretation function dened on the symb ols in

A

L such that for every constant c of L c is an element of A for every nplace

A n

predicate symb ol R R is a subset of the set A of ntuples of elements of A

A n

and for every nary function symb ol F of L F is a function from A to A

The set A is called the domain universe or underlying set of the structure A

A A

In practice we shall often omit brackets writing F instead of F and ditto

for R

In these notes structures will b e denoted by Gothic symb ols A B their

domains will b e denoted by the corresp onding Latin characters A B

Given a structure A for the language L we consider the language L for

A

every element a of A we add a constant a to the language and we assume

that the new constants a are dierent from constants that are already in the

A

language L The structure A b ecomes a L structure by putting a a for

A

each a A

We have also dened the relation A j for L sentences rst for closed

A

A

L terms t we dene their meaning t in A Then we dened by recursion on

A

the L sentence

A

A A

A j t s if and only if t s

A A A

A j Rt t if and only if t t R

n n

A j if and only if A j and A j

A j if and only if A j or A j or b oth

A j if and only if A j

A j xx if and only if for some a A A j a

A j xx if and only if for all a A A j a

Furthermore we have learnt the notions of L or L theory this is a collection

A

of L or L sentences A is a model of the theory T if A j for every T

A

The relation A j is pronounced as A satises is true in A or holds

in A

We also recall one of the most imp ortant theorems we proved in the course

Foundations of Mathematics

Theorem Compactness Theorem A theory T has a model if and only

if every nite subset of T has a model

In this course we shall give an indep endent pro of of Theorem that is

indep endent of the Completeness Theorem

Often we shall say that a theory is consistent in this course this is synony

mous with has a mo del

We shall also use the notation T j where T is an Ltheory and and

Lsentence this means that is true in every mo del of T

A theory T is complete if for every sentence in the language of T either

T j or T j holds

Homomorphisms Emb eddings and Diagrams

The purp ose of Mo del Theory is the study of theories by means of their classes

of mo dels A very useful to ol of Mo del Theory is the p ossibility of varying the

language Supp ose we have two languages L L If B is an L structure then

restricting its interpretation function to L gives an Lstructure A We say that

A is the Lreduct of B and that B is an L expansion of A In our denition

of A j we have already seen the expansion of A to L

A

We start by giving some structure on the class of Lstructures Let A

A B

A and B B b e Lstructures A function f A B is called a

homomorphism of Lstructures if it commutes with the interpretation functions

That is

A B

For every constant c of L f c c

for every nary function symb ol F of L and every ntuple a a of

n

A B

elements of A we have f F a a F f a f a

n n

for every nplace predicate symb ol R of L and ntuple a a of el

n

A

ements of A we have if a a R then f a f a

n n

B

R

Examples If L is the language of groups and A and B are groups a homo

morphism of Lstructures is nothing but a homomorphism of groups If L is the

language fg of partial orders and A B are partial orders a homomorphism is

nothing but an orderpreserving map Similar for rings graphs etcetera

We note immediately that if f A B and g B C are homomorphisms

of Lstructures then so is the comp osition g f A C Moreover the identity

function A A is always a homomorphism of Lstructures A homomorphism

f A B is an isomorphism if there is a homomorphism g B A inverse

to f so the comp ositions g f and f g are the identity functions on A and B

resp ectively

Exercise Prove if f A B is an isomorphism of Lstructures then for

any Lformula x x and any ntuple a a from A we have

n n

A j a a B j f a f a

n n

A consequence of this exercise is that isomorphic Lstructures satisfy the same

Lsentences check this Two Lstructures that satisfy the same Lsentences

are called elementarily equivalent Notation A B The notation for isomor

phic structures is Summarizing

A B A B

The converse implication do es not hold

Exercise Prove an Ltheory T is complete if and only if every pair of mo dels

of T is elementarily equivalent

One sees that the notion of completeness can b e characterized by a prop erty of

the class of mo dels of the theory

A homomorphism f A B is called injective if the function f is f

is an embedding if f is injective and moreover for every relation symb ol R of

B

L and every ntuple a a from A if f a f a R then

n n

A

a a R Clearly if L contains no relation symb ols every injec

n

tive homomorphism is an emb edding but in the general case the notions are

dierent

A substructure or submodel of an Lstructure B is an Lstructure A such

that A is a subset of B and the interpretation function of A is the restriction of

the one of B to A Hence an emb edding f A B denes an isomorphism

b etween A and a submo del of B

If B is an Lstructure and X B there is a least subset A of B which

B

contains X and the elements c c a constant of L and is closed under the

interpretations in B of the function symb ols of L if A A is the domain of

a submo del A of B the submo del generated by the set X which we denote by

h X i

Exercise Show that the domain of hX i is the set

B

ft x x j n IN tv v an Lterm x x X g

n n n

A structure B is nitely generated if B hX i for some nite subset X of B

Recall that a formula is called atomic if it contains no connectives or quan

tiers in other words if it is of from t t or Rt t

n

Exercise Let f A B an emb edding Then for every atomic Lformula

x x and every ntuple a a from A

n n

A j a a B j f a f a

n n

Prove also that this equivalence in fact holds for every quantierfree formula

Exercise Let L fg b e the language of partial orders Find out when

a monotone map b etween partial orders is an injective Lhomomorphism and

when it is an emb edding Give an example of an injective homomorphism which

is not an emb edding

Let A b e an Lstructure with its natural expansion to L We dene a few

A

L theories in connection to A

A

The positive diagram of A is the collection of all atomic L sentences

A

A

that are true in A

The diagram of A is the collection of all atomic L sentences and all

A A

negations of atomic L sentences that are true in A

A

The notion of positive diagram generalizes the idea of multiplication table of a

group

Exercise Show that giving an Lhomomorphism A B is equivalent to

giving an expansion of B to L which is a mo del of the p ositive diagram of

A

A Show also that giving an emb edding A B is equivalent to giving an

expansion of B to L which is a mo del of the diagram of A

A

We now give an example in order to apply our denitions to a little mathemat

ical theorem

Theorem Order Extension Principle Let A be a partial ly or

dered set Then there is a linear order on A which extends equiva

lently there is an injective monotone function from A into a linearly ordered

set B

Pro of The equivalence stated in the theorem is clear b ecause if B is a

linear order and f A B an injective orderpreserving map then dening

on A by x y i f x f y gives a linear order on A which extends

We prove the theorem in two steps

a First we do it for A nite This is easy induction on the numb er of elements

of A If jAj we are done Now if A fa a g cho ose a A such

n i

that a is maximal wrt the order on A By induction hyp othesis Anfa g has

i i

a linear order which extends the one of A relativized to Anfa g put a back in

i i

as greatest element

b The general case Let L fg We use the idea of diagrams and the

Compactness Theorem Let T b e the L theory where A is the partial order

A

A seen as Lstructure consisting of the axioms

The p ositive diagram of A

The axioms of a linear order

The axioms a b for a b A

In every nite subset D of T there is only a nite numb er of constants from the

set A say a a Let fa a g b e the subpartial order of A on these

n n

elements This has by a an extension to a linear order so D has a mo del By

the Compactness Theorem T has a mo del B since B is a mo del of

and there is an injective monotone map from A to B

Exercise As an example of theorem dene an injective orderpreserving

map from P IN to R where P IN is the p owerset of IN ordered by the subset

relation

Elementary Emb eddings and Elementary Di

agrams

An emb edding f A B of Lstructures is called elementary if for ev

ery Lformula x x with free variables x x and every ntuple

n n

a a of elements of A we have

n

A j a a B j f a f a

n n

This means that the elements of A have the same prop erties with resp ect to A as

to B For example if we consider Q and R as elds or rings the emb edding is

not elementary since the element of Q is a square in R but not in Q However

if we consider Q and R just as ordered structures the emb edding is elementary

as we shall see later

The elementary diagram E A of A is the collection of all L sentences which

A

are true in A

Exercise Giving an elementary emb edding f A B is equivalent to giving

an L expansion of B which is a mo del of E A

A

A is called an elementary submodel of B notation A B if A is a substructure

of B and the inclusion is an elementary emb edding Therefore an elementary

emb edding A B is an isomorphism b etween A and an elementary submo del

of B We also say that B is an elementary extension of A

Note that A B implies A B The converse implication however is by

no means true not even if A is a submo del of B Example let for L fg

A INnfg and B IN Then A B but the inclusion is not an isomorphism

and not an elementary emb edding check

Exercise A B if and only if A is a submo del of B and for every L

A

sentence of the form y y which is true in B there is a b A such that b

is true in B

Hint use induction on formulas

The notion of elementary submo del gives rise to the following imp ortant the

orems called the LowenheimSkolemTarski Theorems roughly together they

say that to innite structures there are elementarily equivalent structures of

almost arbitrary innite cardinality That is predicate logic has nothing to say

ab out innite cardinalities

Theorem Upward LowenheimSkolemTarski Theorem Every in

nite Lstructure has arbitrarily large elementary extensions

Pro of Let A b e innite ie A is innite By arbitrarily large we mean

for every set X there is an elementary extension B of A such that there is

an injective function from X into B The pro of is a simple application of the

Compactness Theorem

Given X we cho ose for every x X a new constant c not in L Let

x A

L L fc j x X g Consider the L theory

A x

E A fc c j x y X x y g

x y

Since A is innite every nite subset of has an interpretation in A simply

interpret the constants c by dierent elements of A and hence has a mo del By

x

the Compactness Theorem has a mo del B Since B is a mo del of E A there

is an elementary emb edding A B so we can identify A with an elementary

B

submo del of B And the assignment x c is an injective function from X

x

into B

The Downward LowenheimSkolemTarski Theorem is a little bit more in

volved First we recall that jjLjj is dened as the maximum of the cardinal

numb ers and jLj is the cardinality of the set IN Recall also that jjLjj is

the cardinality of the set of all Lformulas

Theorem Downward LowenheimSkolemTarski Theorem Let B

be an Lstructure such that jjLjj jB j and suppose X B is a set with jjLjj

jX j Then there is an elementary submodel A of B such that X A and

jX j jAj

Pro of Given X let L L X elements of X as new constants as usual

X

we take this union to b e disjoint Note that jjL jj jX j For every L

X X

sentence x x which is true in B we cho ose an element b B such that

B j c Let X b e the union of X and all the bs so chosen Again since we

b

add at most jjL jj many new elements jX j jX j Now rep eat this with X

X

etcetera obtaining X and this innitely often in the place of X and L

X

1

obtaining a chain

X X X X

S

Let A X By induction one proves easily that jX j jX j for all n so

n n

nN

jAj jX j since X is innite

Now if x x is an L formula which is true in B then for some n

A

formula so by construction there is a b X A such that is an L

n X

n

B j b In particular this holds for formulas xx F a a so A is

n

closed under the interpretations in B of the function symb ols of L hence A is

the domain of a submo del A of B But by the same token A is an elementary

submo del exercise

An immediate application of the LowenheimSkolemTarski theorems is the so

called LosV aught test A theory is called categorical for some cardinal numb er

if for every pair A B of mo dels of T with jAj jB j we have A B

Theorem LosV aught Test Suppose T is a theory which has only in

nite models and T is categorical for some cardinal number jjLjj L is the

language of T Then T is complete

Pro of Supp ose A and B are mo dels of T Then b oth A and B are innite by

hyp othesis so if is the maximum of f jAj jB jg then by b oth A and B

have elementary extensions A B resp ectively with jA j jB j By

A and B have elementary submo dels A and B resp ectively of cardinality

Since T is categorical we have

B B B A A A

Hence A B for any two mo dels A B of T hence T is complete exercise

Examples of categorical theories

a Let L fg and T the theory of dense linear orders without endp oints

By the familiar backandforth construction T is categorical T is not

categorical b ecause and are nonisomorphic mo dels

of T

b Torsionfree divisible ab elian groups

Let L b e the language of groups say L f g and T the Ltheory

with axioms

Axioms of an ab elian group

torsionfree fxx x x j n g

z

n times

divisible fxy y y x j n g

z

n times

T is not categorical but T is categorical for uncountable cardinalities

Sketch of pro of a divisible torsionfree ab elian group is a Qvector space

If such a space has uncountable cardinality then since Q is countable

it must have a basis of cardinality but any two vector spaces over the

same eld with bases of the same cardinality are isomorphic as vector

spaces hence as ab elian groups

If the cardinality of the vector space is however then its basis may b e

nite or countably innite

c Algebraically closed elds of a given xed characteristic

Again this is categorical in every uncountable cardinality but not

Q categorical consider the elds Q and QX Their algebraic closures

and QX are b oth countable but not isomorphic Why

d Let L consist of one place function symb ol F Let T b e the theory with

axioms

xy F x F y x y

xy F y x

n

fxF x x j n g

A mo del of T is nothing but a set X with a Zaction on it which action is

free So X naturally decomp oses into a numb er of orbits which are all in

corresp ondence with Z Therefore if jX j the numb er of orbits is

equal to jX j So T is categorical in every uncountable cardinality However

if X is countable the numb er of orbits may b e either nite or countably

innite

In this list we have seen examples of theories that are categorical but not cate

gorical in higher cardinalities or the converse categorical in every uncountable

cardinal but not in of course there are also theories which are categorical

in every cardinality the empty theory That this is not a coincidence is the

content of the famous Morley Categoricity Theorem if a theory in a countable

language is categorical in an uncountable cardinality it is categorical in every

uncountable cardinality This is a deep result of Mo del Theory and the starting

p oint of a whole branch of Mo del Theory Stability Theory

Directed Systems of Lstructures

Let K b e a partially ordered set or poset for short K is called directed if

K is nonempty and every pair of elements of K has an upp er b ound in K that

is K satises xy z x z y z

A directed system of Lstructures consists of a family A of Lstructures

k k K

indexed by K together with homomorphisms f A A for k l These

k l k l

homomorphisms should satisfy

f is the identity homomorphism on A

k k k

if k l m then f f f

k m lm k l

Given such a directed system we dene its colimit as follows

First take the disjoint union of all the sets A

k

G

A fk a j k K a A g

k k

k K

Then dene an equivalence relation on this by putting k a l b if there is

m k l in K such that f a f b see for yourself how the directedness

k m lm

F

A of K is used to show that this relation is transitive Let A

k

k K

b e the set of equivalence classes we denote the equivalence class of k a by

k a

We dene an Lstructure A with underlying set A as follows For an nary

function symb ol F of L we put

A A

k

F k a k a k F f a f a

n n k k k k n

1 n

where by directedness of K k is chosen so that k k k k all hold Of

n

course we must show that this denition makes sense that it do es not dep end

on the choice of representatives or the choice of k This is done in the following

exercise

Exercise Prove if k a k a k a k a and k

n n

n n

k k then

n

A A

0

k

k

0

0 0

0

f a f a k F f a k F a f

k k k k n k k

k k

1 n n

n

1

If R is an nplace relation symb ol of L and k a k a are n elements

n n

A

of A we let k a k a R if and only if there is k k k in

n n n

K such that

A

k

f a f a R

k k k k n

1 n

Again check that this do es not dep end on representatives

Lemma The function f A A dened by f a k a is a homo

k k k

morphism of Lstructures If k l the diagram

f

kl

A A l k / A AA ~~

AA ~~

f f l k A ~

A ~~

A

of Lstructures and homomorphisms is commutative Moreover if g A

k k

B is another K indexed system of homomorphisms satisfying g f g

k K l k l k

whenever k l there is a unique homomorphism f A B such that f f g

k k

for al l k K

Exercise Prove lemma

Exercise In the notation of lemma and ab ove if all the maps f

k l

A A are injective then all the maps f A A are injective The same

k l k k

holds with injective replaced by an emb edding

Lemma Elementary System Lemma If al l f A A are ele

k l k l

mentary embeddings so are al l the maps f

k

Pro of So we supp ose all f A A are elementary emb eddings Now by

k l k l

induction we prove for an Lformula x x

n

For all k K and all a a A

n k

A j a a A j f a f a

k n k k n

Note that the universal quantier for all k K o ccurs inside the induction

hyp othesis

I give only the quantier step in fact the step for atomic is included in

exercise and the steps for the prop ositional connectives are easy So let

x

If A j x x a a so for some a A A j a a a

k n k k n

then by induction hyp othesis A j f a f a f a whence A j

k k k n

x x f a f a

k k n

Conversely supp ose A j x x f a f a Then for some k a

k k n

A we have A j k a f a f a Take by directedness of K a

k k n

00 00 0 0 00 00

f and k a f f f f k K with k k k Since f

k k k k k k k k

0 00

a we have k f

k k

00 00 00 00 0 00 0 0

a f a f f a f f A j f

n k k k k k k k k k

By induction hyp othesis we have

00 00 0 00 00

a a f a f j f A

n k k k k k k k

00 00 00 00

a Now we use the fact that f a f j x x f whence A

n k k k k k k k

is an elementary emb edding to deduce that A j x x a a and we

k n

are done

Theorems of Robinson Craig and Beth

We shall see an application of the Elementary System Lemma b elow in the

pro of of Theorem First a lemma whose pro of is another application of the

Compactness Theorem

Lemma Let L L A an Lstructure and B an L structure Suppose

moreover that A is elementarily equivalent to the Lreduct of B Then there is

an L structure C and a diagram of embeddings

A ?? ? f ?? ??

? C

B /

g

where g is an elementary embedding of L structures and f is an elementary

embedding of Lstructures

where we take the constants L L Pro of We consider the language L

A

B AB

consider the theory T E A E B E A from A and B disjoint In L

AB

theories Then theory so b oth are L is an L theory and E B is an L

A

AB B

any mo del C of T gives us the required diagram by exercise

So for contradiction supp ose T has no mo del By Compactness some

nite subset of T has no mo del so taking conjunctions we may assume that

for some a a E A and b b E B has no

n m

structure which satises b b can b e ex mo del But then no L

m

B

panded with interpretations for a a such that the expansion satises

n

this means that every L structure which satises b b will satisfy

m

B

x x x x This is a contradiction since this is an Lsentence

n n

A j x x x x and A is elementarily equivalent to the Lreduct

n n

of B

Exercise Prove that for two Lstructures A and B the following are equiv

alent

i A B

ii A and B have a common elementary extension

Theorem Robinsons Consistency Theorem Let L and L be two

languages L L L Suppose that T is a complete Ltheory and let T be

an L theory T an L theory which both extend T If both T and T have a

model so has T T

Pro of Let A b e a mo del of T B a mo del of T Then A and B are also

mo dels of T and hence since T is complete their Lreducts are elementarily

equivalent By lemma there is a diagram A

CC f

C 0 CC CC

C!

B B

/

h

0

with h an elementary extension of L structures and f elementary as a map

b etween Lstructures The Lreducts of A and B are still elementarily equiv

alent so applying the same lemma in the other direction gives a diagram

k

0

A A / C

CC O C g C 0

g C 0 C

C!

B B

/

h

0

with g an elementary map of Lstructures and k elementary of L structures

We can pro ceed in this way obtaining a directed system

k k

0 1

A A A

/ / / C C

CC O CC O

g C g C 1

CC 0 CC

f f 1 0 C C

C! C!

B B B

/ / /

h h

0 1

in which the k s are elementary maps of L structures the f s and g s of L

structures and the hs of L structures Since the colimit C of this system is

equally the colimit of the As and the colimit of the Bs it has an L L

structure into which b oth A and B emb ed elementarily by lemma So C

is the required mo del of T T

The following theorem states a prop erty that is commonly known as the

amalgamation prop erty for elementary emb eddings

Theorem Every diagram B ~? f ~~ ~~ ~~ A @ @@ @@ g @@

@

C

consisting of elementary embeddings between Lstructures can be completed to

a commutative diagram B ~? @@ f ~~ @@ ~~ @@

~~ @ D A @ > @@ ~~ @@ ~~ g @@ ~~

@ ~~

C

of elementary embeddings between Lstructures

Pro of Formulate the elementary diagram E B in the language

L L fb j b f Ag

A

simply replacing the constants f a by a and similarly write E C in

L L fc j c g Ag

A

Then L L L In L we have the complete theory E A and E B and

A A

E C are extensions of it in L L resp ectively By E B E C has a

mo del D Check that this gives a diagram of the required form

Exercise Give also a direct pro of of theorem not using theorem but

along the lines of the pro of of lemma

A more serious application of Robinsons Consistency Theorem is Craigs Inter

p olation Theorem

Theorem Craig Interp olation Theorem Suppose and are

Lsentences such that j that is every Lstructure in which is true

satises also Then there is an Lsentence with the properties

i j and j

ii Every nonlogical symbol function symbol constant or relation symbol

which occurs in also occurs both in and

Pro of Such a as in the theorem is called a Craig interpolant for and

We assume for a contradiction that no such Craig interp olant exists

Then has a mo del for otherwise xx x is a Craig interp olant Simi

larly has a mo del for otherwise xx x were a Craig interp olant

Let L the set of nonlogical symb ols in and L of those in let L

L L Let b e the set of L sentences such that either j or j

Supp ose fg has no mo del Then by compactness there are sentences

and such that j and j and f g has no mo del Then

since j f g has no mo del whence

j and j

contradicting our assumption that no Craig interp olant exists In a similar way

f g is consistent

Now apply Zorns Lemma to nd a set of L sentences which is maximal

with resp ect to the prop erties that and b oth fg and f g are

consistent I claim that is a complete L theory

For supp ose is not complete then there is a sentence such that

and f g is consistent Then by maximality of either f g or

f g is inconsistent In the rst case there is a sentence such

that j Then is by denition of in since also

we see that f g is inconsistent contradicting the choice of In the other

case one obtains a similar contradiction So is complete

We are now in the p osition to apply Robinsons Consistency Theorem

to the complete L theory and its consistent extensions fg and

f g The conclusion is that f g is consistent but this contradicts the

assumption j

Our last theorem in this section is due to Beth We consider a language L and

a new nplace relation symb ol P not in L Let T b e an L fP gtheory T is

said to dene P implicitly if for any Lstructure A there is at most one way to

expand A to an L fP gstructure which is a mo del of T This can b e said in a

dierent way let P b e another new nplace relation symb ol and consider the

theory T P where all P s are replaced by P s Then T denes P implicitly if

and only if

T T P j x x P x x P x x

n n n

On the other hand we say that T denes P explicitly if there is an Lformula

x x such that

n

T j x x P x x x x

n n n

Clearly if T denes P explicitly then T denes P implicitly the converse is

known as Beths Denability Theorem

Theorem Beth Denability Theorem If T denes P implicitly then

T denes P explicitly

Pro of Add new constants c c to the language By the remark ab ove

n

we have

T T P j P c c P c c

n n

By Compactness we can nd nite T and T P such that

j P c c P c c

n n

Taking conjunctions we can in fact nd an L fP gformula such that T j

and

P j P c c P c c

n n

Taking the P s to one side and the P s to another we get

P P c c j P P c c

n n

By the Craig interp olation theorem there is an Lformula such that P

P c c j c c and c c j P P c c Re

n n n n

placing P by P again in this second entailment and using that T j P

we nd that c c is in T equivalent to P c c so since the

n n

c c are arbitrary new constants

n

T j x x x x P x x

n n n

and we are done

Exercise In this section we have proved the Craig Interp olation Theorem

from Robinsons Consistency Theorem Now assume the Craig Interp olation

Theorem and use it to give another pro of of the Robinson Consistency Theorem

Preservation Theorems

Let T b e an Ltheory We consider the class of mo dels of T as sub class of the

class of Lstructures if the class of mo dels of T is closed under certain op erations

of Lstructures we also say that T is preserved under those op erations In this

section we shall see that preservation under certain op erations can b e related to

the syntactical structure of axioms for T For example theL osTarski Theorem

b elow Theorem says that a theory is preserved under submo dels if and

only if it is equivalent to a theory consisting solely of sentences of the form

x x with quantierfree

n

We shall now dene some notions precisely A set of axioms for T is a theory

T which has exactly the same mo dels as T hence is logically equivalent to T

An Lformula of the form x x with quantierfree is called a

n

formula a formula x x with quantierfree is called a formula

n

Generally a formula is if it is of form x x with a formula

n m n

and dually it is if it is of form x x with a formula One

n m n

also says meaning is a formula etc

n n

Lemma Let T be an Ltheory and a set of Lsentences which is closed

under disjunction Suppose that the fol lowing holds whenever A is a model of

T and B an Lstructure which satises al l sentences in which are true in A

then B is a model of T Then T has a set of axioms which is a subset of

Pro of Let b e the set f j T j g We show that every mo del of is a

mo del of T so that is a set of axioms for T

Supp ose B is a mo del of Let f j B j g I claim that

T is consistent For supp ose not then for some we would

n

have T j ie T j Since is closed under

n n

disjunctions We obtain a contradiction since on the one

n

hand B is a mo del of on the other hand B j

n

Let A b e a mo del of T Then for we have if A j then

hence B j so B j By the assumption in the lemma and the fact that

A is a mo del of T we get that B is a mo del of T Since we started with an

arbitrary mo del B of we see that is a set of axioms for T

A theory T is said to b e preserved under substructures if any substructure of a

mo del of T is again a mo del of T For example any substructure of a group is

a group but not every subring of a eld is a eld hence the theory of groups is

preserved under substructures but the theory of elds isnt

Theorem LosT arski A theory is preserved under substructures if and

only if it has a set of axioms consisting of sentences

Pro of Every sentence which is true in A is true in every substructure of

A check so one direction is obvious

For the other we shall apply Lemma observing that the set of L

sentences which are equivalent to sentences is closed under disjunction

So let A b e a mo del of T and supp ose every sentence which is true in

A also is true in B Now consider the theory T in L If this theory is

B B

inconsistent then there is a quantierfree sentence b b in such

n B

that T fg has no mo del This means that A which is a mo del of T cannot b e

expanded to an L fb b gstructure which mo dels from which it follows

n

that A j x x x x But this is a sentence so by assump

n n

tion we must have B j x x x x which is a contradiction

n n

since b b is an element of

n B

We conclude that T has a mo del C and B emb eds into C Because T

B

is preserved under submo dels B is a mo del of T

Lemma now tells us that T has a set of axioms consisting of formulas

Exercise Strengthen the argument in the pro of of Theorem to prove

if A and B are Lstructures such that every sentence which is true in A is

true in B then B is a substructure of an elementary extension of A

Exercise Use the previous exercise to prove the following result Let T b e

the set of sentences which are consequences of T Then every mo del of T

is a substructure of a mo del of T Conversely every substructure of a mo del of

T is a mo del of T

Exercise Relativise Theorem to supp ose T is a subtheory of T and

for every pair A B of mo dels of T with B a submo del of A we have that if A

is a mo del of T then so is B Then there is a set of sentences such that

T is a set of axioms for T

Hint consider the set of sentences for which there is a sentence such

that T j

Our next theorem characterizes theories which have a set of axioms We say

that a theory T is preserved under directed unions if for any directed system

A of Lstructures as in section such that all homomorphisms f for

k k K k l

k l are emb eddings if all A are mo dels of T then so is the colimit of the

k

system

Theorem ChangLosSuszk o A theory T is preserved under directed

unions if and only if T has a set of axioms consisting of sentences

Pro of First we do the easy direction let A b e the colimit Then every

sentence which is true in every A is also true in A for let x x y y

k n m

b e such a sentence with quantierfree and take a a A Then

n

by construction of the colimit since K is directed there is k K such that

a f a a f a for suitable a a A Then in A we can

k n k k k

n n

nd b b such that A j a a b b Because f A A

m k m k k

n

is an emb edding we have

A j a a f b f b

n k k m

Hence A j x x y y

n m

For the other direction we use again Lemma observing that the disjunction

of two sentences is equivalent to a sentence So let A b e a mo del of T

T b e preserved under directed unions and B satises every sentence which

is true in A We construct a directed system of emb eddings in fact a chain

B B A B A B

with the prop erties

i A A

n

ii the comp osed emb edding B B is elementary

n n

which is true in B is also true iii every sentence in the language L

n B

n

structure in the in A regarding A via the emb edding as an L

n n B

n

obvious way

The construction will pro ceed inductively and we shall assume each of the

conditions iiii as induction hyp othesis as we go along We start by putting

B B this was easy

If B has b een constructed in order to make A we consider the theory

n n

where ThA is the set of all Lsentences which are true in A and ThA

B

n

which are true in B note is the set of sentences in the language L

n B B

n n

that every quantierfree sentence is trivially a sentence so

B B

n n

Supp ose this theory is inconsistent then for some sentence

y y b b y y

n m n

we have that ThA f g has no mo del As b efore we see then with

B

n

that

A j x x y y

m n

Since this is a sentence it must by assumption on A and B b e true in B

which by induction hyp othesis is an elementary submo del of B but this is in

n

contradiction with the fact that is true in B check

n

is consistent and we let A b e a mo del of it Therefore ThA

n B

n

Clearly A A and condition iii also holds for B and A

n n n

in the If A has b een constructed we consider the theory E B

n n A

n+1

since B is a submo del of A we can take E B to b e a language L

n n n A

n+1

theory in this language

Supp ose this theory is inconsistent Then for some formulas E B and

n

we have j Now is of form b b a a with

k l A

n+1

a a A nB and b b B We see that

l n n k n

B j x x b b x x

n l k l

By induction hyp othesis iii then which is a sentence in the language L

B

n

A j x x b b x x

n l k l

which is in a now familiar way a contradiction

Then the emb edding B We let B b e a mo del of E B

n n n A

n+1

B is elementary which is the only induction hyp othesis we have to check at

n

this stage This nishes the construction of the sequence

B A B A

In order to nish the argument let C b e the colimit of this chain Then C is

equally the colimit of the chain

A A

which is a chain of mo dels of T so since T is preserved by directed unions C is

a mo del of T On the other hand C is also colimit of the chain

B B B

which is a chain of elementary emb eddings by the Elementary System Lemma

B C is an elementary emb edding It follows that B is a mo del of T

which was to b e proved A nal application of Lemma now yields the result

Exercise Use Theorem to prove the following equivalence T has a set

of axioms consisting of sentences if and only if the following prop erty holds

whenever A is an Lstructure and for every nite subset A A there is a

substructure B of A such that A B and B is a mo del of T then A is a mo del

of T

Exercise Supp ose that T satises the following prop erty whenever A is a

mo del of T and B C are substructures of A which are also mo dels of T and

B C then B C the substructure with domain B C is also a mo del

of T

Show that T has a set of axioms consisting of sentences

I mention one more preservation theorem without pro of Call a formula positive

if it do es not contain the symb ols or

A theory T is said to b e preserved under homomorphic images if whenever

A is a mo del of T and f A B a surjective homomorphism then B is a

mo del of T

Theorem A theory is preserved under homomorphic images if and only if

it has a set of positive axioms

Exercise Prove the easy part of theorem that is every p ositive sentence

is preserved under homomorphic images

Filters and Ultrapro ducts

Let I b e a set A lter over I is a subset U of P I the p owerset of I satisfying

the prop erties

i I U

ii if U U and U V I then V U

iii U V U implies U V U for any U V I

iv U

Examples For every nonempty subset J of I there is the principal lter

U fA I j J Ag

J

If I is innite there is the Frechet lter or conite lter

U fA I j I nA is niteg

F

Exercise Let A b e a set of nonempty subsets of I such that for every nite

sub collection fA A g of A the intersection

n

A A

n

is nonempty Show that the collection

fU I j A A AA A U g

n n

is a lter it is said to b e the lter generated by A

Exercise Filters and Congruence Relations A congruence relation on

P I is an equivalence relation on P I such that satises the two prop

erties

A B C A C B

A B I nA I nB

Show every lter U over I determines a congruence relation on P I by

A B i I nA B A I nB U

Conversely show that for every congruence relation on P I such that I

fA j A I g is a lter over I

Let U b e a lter over the set I and supp ose that A is a family of sets

i iI

We dene the reduced product mo dulo U written A as follows The set

U i

A will b e

U i

Y

A

i

iI

Q

where A is the pro duct of all the sets A that is the set of all I indexed

i i

iI

sequences x such that x A for all i and is the equivalence relation

i iI i i

given by

x y i fi I j x y g U

i iI i iI i i

This is an equivalence relation since I U we have x x and obviously

i i

the relation is symmetric If x y and y z then if U fi

i i i i

I j x y g and V fi I j y z g then clearly U V fi I j x z g so

i i i i i i

x z

i i

Exercise Show that if U is the principal lter U for J I then A is

J U i

Q

A in bijective corresp ondence with

i

iJ

An ultralter is a maximal lter ie a lter that cannot b e extended to a larger

lter

Exercise Show that for an ultralter U over I

a U U or I nU U for every U I

b if U V U then U U or V U

Exercise Show that in the corresp ondence of Exercise an ultralter

corresp onds to a congruence relation with exactly two classes

Exercise Show

a If an ultralter is a principal lter U J is a singleton set

J

b if an ultralter is not principal it contains every conite set

Lemma Every lter is contained in an ultralter

Exercise Prove Lemma using Zorns Lemma

Exercise Rene Lemma to the following statement supp ose U is a

lter and A a collection of subsets of I such that no element of U is a subset

of a nite union of elements of A Then U is contained in an ultralter that is

disjoint from A

Q

If U is an ultralter the reduced pro duct A is called an ultraproduct

i

U

Exercise Let I IN and A f g for all i IN Show that for every

i

Q

ultralter U over IN the ultrapro duct A has cardinality

i

U

Exercise Let again I IN and A f ig Let U b e an ultralter

i

over IN Show

Q

A is nite a If U is principal

i

U

Q Q

A is uncountable A is nonprincipal b if

i i

U U

We shall now extend the reduced pro duct construction to Lstructures

n

First if we are given for each i an nary function f A A we have

i i i

Q Q

n

a function f A A dened by

i i

iI iI

n n

f a a f a a

i i i i

i i i i

n n

Now if U is a lter over I and a b a b in the equivalence

i i i i

i i i i

Q

relation dening the reduced pro duct A then for each k n we have

i

U

k k

fi j a b g U

i i

so since U is closed under nite intersections

k k

fi j k k n a b g U

i i

n n

This clearly implies fi j f a a f b b g U so

i i

i i i i

n n

f a a f b b

i i i i

i i i i

Q Q

n

Hence f determines a function A A which we shall also denote

i i

U U

by f By the ab ove we may put

n n

f a a f a a

i i i i

i i i i

Secondly if in every A an element c is chosen this determines an obvious

i i

Q

element c of A

i i i

U

n

Thirdly if for every i we have an nary relation R A we dene an

i i

Q

nary relation R on A by

i

U

n

R fa a j

i i

i i

n

fi j a a R g U g

i

i i

Exercise Show that R is welldened

Q

A In this way we see that if every A is the domain of an Lstructure A

i i i

U

Q

A is the domain of an Lstructure

i

U

Q

Exercise If t is an Lterm with variables x x and A A and

n i

U

Q

n

a a are elements of A then

i i i

i i

U

n A n A

i

a a t a a t

i i i

i i i i

From this exercise and the denition of the interpretation of relation symb ols

of L in the reduced pro duct we see

Q

Prop osition Let A A as above Then every atomic Lformula

i

U

has the fol lowing property if has n free variables x x then for every

n

Q

n

A ntuple a a of elements of

i i i

i i

U

n

A j a a

i i

i i

n

fi j A j a a g U

i

i i

Lemma The col lection of Lformulas which have the property in Propo

sition is closed under conjunction and existential quantication

Pro of Let us drop the constants from the notation Since always U V U

if and only if b oth U U and V U we have for

A j

fi j A j g U and fi j A j g U

i i

fi j A j g U

i

so is in

For existential quantication supp ose x again may have more

free variables which we suppress If A j x so A j a then by

i i

assumption fi j A j a g U so certainly fi j A j xg U since this is a

i i i

bigger set

Conversely if fi j A j xg U U pick for each i U an a A

i i i

such that A j a For i U let a A arbitrary Then a satises

i i i i i i

fi j A j a g U so by assumption A j a hence A j x

i i i i

Theorem Los Fundamental Theorem for Ultrapro ducts

a If U is an ultralter every Lformula has the property of proposition

b hence for Lsentences

A j fi I j A j g U

i

Pro of a From Prop osition and Lemma we know that the collection

of formulas satisfying the mentioned prop erty contains all atomic formulas

and is closed under and Because every formula is logically equivalent to

a formula that contains only f g it suces to show that when U is an

ultralter the collection is closed under

But we know for an ultralter U that U U if and only if I nU U so

implies

A j

fi j A j g U

i

fi j A j g U

i

so

b This is an immediate consequence of a

Exercise a Supp ose A A for all i I and U is an ultralter over I

i

Q

A is then called an ultrapower of A Show that the The ultrapro duct

U

Q

map A A given by a a is an elementary emb edding

i

U

b Supp ose A B are two I indexed collections of Lstructures and let

i i i i

f A B b e a homomorphism for all i Given an ultralter U over I

i i i

Q Q

dene a homomorphism f A B and show if every f is an

i i i

U U

elementary emb edding so is f

We turn now to some applications of the Fundamental Theorem for Ultra

pro ducts and the Compactness Theorem

Let K b e a class of Lstructures We say that the class K is denable if there

is an Lsentence such that K is the class of mo dels of K is elementary if

there is an Ltheory T such that K is the class of mo dels of T The following

theorem characterizes denable and elementary classes

Theorem a A class K of Lstructures is denable if and only if both

K and its complement are elementary

b K is elementary if and only if K is closed under elementary equivalence

and ultraproducts

Pro of a If K is the class of mo dels of then the complement of K is the class

of mo dels of so b oth classes are elementary Conversely if K is the class of

mo dels of an Ltheory T and its complement is the class of mo dels of another

theory T then clearly T T is inconsistent so assuming T and T to b e closed

under conjunctions by Compactness there are sentences T T such

that has no mo del Clearly then every mo del of is a mo del of T and

conversely since T and every mo del of is a mo del of T and vice versa

So K is denable

b One direction follows directly from the Fundamental Theorem if K is the

class of mo dels of T A is a family of elements of K and U is an ultralter

i iI

Q

over I then the ultrapro duct A is a mo del of T hence in K And obviously

i

U

K is closed under elementary equivalence

Conversely supp ose K is closed under ultrapro ducts and elementary equiv

alence Let T the set of Lsentences which are true in all elements of K We

show that K is the class of mo dels of T So let B b e a mo del of T Let

f g b e a nite subset of ThB the set of Lsentences true in B

n

Then there is an element of K in which every element of is true for otherwise

would b e an element of T But this cannot b e since B is a

n

mo del of T Cho ose for every such a mo del A of from K So I is the set

of all nite subsets of ThB For every ThB let U b e f j A j g

Then the collection fU j ThBg is closed under nite intersections hence

generates a lter U which can b e extended to an ultralter F let C b e the

Q

ultrapro duct A Just as in the ultrapro duct pro of of the Compactness

F

theorem one sees that C is a mo del of ThB hence is elementarily equivalent

to B Moreover C is an element of K since K is closed under ultrapro ducts so

B is an element of K since K is closed under elementary equivalence Since we

started with an arbitrary mo del B of T we see that K contains every mo del of

T

In a way quite similar to the pro of of Theorem b we can use the Funda

mental Theorem for Ultrapro ducts to prove the Compactness Theorem

Supp ose is a set of Lsentences such that every nite subset of has a

mo del A We let I f j niteg and for each let U f

I j g Since f g U U for all the collection fU j g

is contained in an ulttralter F over I

Q

Exercise Prove yourself that the ultrapro duct A is a mo del of

F

The following exercise gives some examples of the use of Theorem

Exercise Examples a Let I IN A f ig and F a nonprin

i

Q

cipal ultralter over I Use the ultrapro duct A to show

i

F

i The class of cyclic ab elian groups is not elementary

ii the class of wellfounded linear orders is not elementary a linear

order is wel lfounded if there is no innite chain a a a

iii the class of connected graphs is not elementary a graph is connected

if for every pair of p oints a b there is a nite path a a a

a b

n

b Let I IN b e the set of primes A f i g F a nonprincipal

i

Q

ultralter over I Show that A is a eld of characteristic zero and

i

F

conclude that the class of elds of characteristic is not elementary

c An ab elian group G is divisible if for every n

x G j xy y y

z

n times

Construct yourself an ultrapro duct example to show that the class of non

divisible ab elian groups is not elementary hence the theory of divisible

groups not nitely axiomatizable

A strengthening of Exercise is given by the following theorem which we state

without pro of

Theorem KeislerShelah Let A and B be Lstructures Then A B if

and only if there exist a set I and an ultralter F over I such that the ultrapowers

Q Q

A and B are isomorphic

F F

Exercise Use Theorem to obtain the following renement of Theo

rem A class K of Lstructures is elementary if and only if K satises the

prop erties

a K is closed under isomorphism and under ultrapro ducts

b Whenever some ultrap ower of A is in K then A is in K

I nish this chapter with a theorem whose pro of is an application of the idea

of ultralters It is an imp ortant theorem in the eld of innite combinatorics

and nds applications in advanced Mo del Theory though p ossibly not in this

course

Theorem Ramsey Let I be an innite set Write P I for the col lec

n

tion of subsets of I with exactly n elements If P I A B then there is an

n

innite subset J of I such that either P J A or P J B

n n

Pro of If I is uncountable we may take any countable subset of I so without

loss of generality we may assume I IN The theorem is trivial for n

check so assume n For a nite subset of IN we write k for

m m k

Let F b e a nonprincipal ultralter over IN We dene for r n sets

r r n n r

A B P IN as follows let A A and B B If for r n A

r

r

and B have b een dened we put

r r

A f P IN j fm j fmg A g F g

r

r r

B f P IN j fm j fmg B g F g

r

r r r r r

Then if P IN A B also P IN A B if P IN A

r r r

r

then fm j fmg A g F Since F is a nonprincipal ultralter and

r

fm j mg is nite so not in F we must have fm j fmg B g F

In particular we have ffng j n INg A B and therefore either fn j fng

A g F or fn j fng B g F assume fn j fng A g F the other case is

dealt with in a symmetric way

We dene J as follows Let j b e the least n such that fng A Inductively

supp ose j j j have b een dened such that for all r n and

k

r

all P fj j g A

r k

Then for all r n and all P fj j g we have

r k

r

U fm j fmg A g F

r

Now there are only nitely many pairs r with r n and

P fj j g so the set

r k

U fn j fng A g

r

r n

P fj j g

r 0 k

is an element of F Let j b e the least element in this set which is j

k k

Convince yourself that again for each r n and P fj j g

r k

r

A

The set J fj j g thus constructed has the prop erty that P J A

n

Quantier Eliminati on and Mo del Complete

ness

We say that an Ltheory T admits elimination of quantiers or T has quantier

elimination if for every Lformula x x there is a quantierfree L

n

formula with at most the variables x x free such that

n

T j x x x x

n

In particular every Lsentence is in T equivalent to a quantierfree sentence

Therefore if T has quantier elimination and T j or T j for every

quantierfree Lsentence then T is complete

Exercise a Show T has quantier elimination if and only if every for

mula of form xA where A is quantierfree is equivalent to a quantier

free formula having at most the same free variables

b show that we can simplify further to of form xA where A is a con

junction of atomic formulas and negations of atomic formulas

Hint for a use induction on the number of quantiers for b by disjunc

tive normal form every formula of form in a is equivalent to a disjunction of

formulas of the form in b

An immediate consequence of the denition is the following prop osition

Prop osition Suppose T has quantier elimination Then for any two mod

els A B of T if A is a substructure of B then A is an elementary substructure

of B

Pro of Let x x an Lformula and a a A Let x b e

n n

quantierfree such that T j xx x Then A j a a i

n

A j a a since A is a mo del of T i B j a a since A is

n n

a substructure of B and is quantierfree i B j a a since B is

n

a mo del of T

The prop erty stated in Prop osition is weaker than quantier elimination

and is called model completeness of the theory T a theory T is mo del complete

if for any two mo dels A B of T A B implies A B

Examples The theory of elds or even elds of characteristic zero is not

mo del complete Q is a subeld of R but not elementary as we have seen The

theory of torsionfree divisible ab elian groups is mo del complete

Exercise

a The following two statements are equivalent for an Ltheory T

i For every Lformula x there is a formula x such that T j

xx x

ii for every Lformula x there is a formula x such that T j

xx x

b Show that T is mo del complete if and only if T satises the equivalent

conditions of a

Hint use the renement of theLosT arski Theorem in Exercise

Exercise Robinsons Test Show that T is mo del complete if and only

if for every emb edding A B of mo dels of T every sentence of the language

L which holds in B also holds in A

A

Exercise Use the ChangL osSuszko Theorem Theorem and the Ele

mentary System Lemma Lemma to show that every mo del complete theory

has a set of axioms

Exercise Show that T is mo del complete if and only if for every mo del A

of T T is a complete L theory

A A

We shall return to mo del completeness later for the moment we fo cus on

quantier elimination for a while We give a general lemma which characterizes

quantier elimination in a mo deltheoretic way and we shall prove for two

theories that they admit quantier elimination the theories of dense linear

orders without endpoints and real closed elds Before starting however we

have to clear up a triviality ab out the logic

Linguistic detail From now on we assume that in the predicate logic we are

using there is an atomic sentence which is never true in a mo del Of course

we have then that j xx x so is redundant in a sense but without

it there may b e no quantierfree sentences at all if L has no constants or ary

relation symb ols

Lemma The fol lowing three conditions are equivalent for an Ltheory T

i For any model B of T and any nitely generated substructure A of B

T is a complete L theory

A A

ii T has quantier elimination

f g

iii For every pair A B A C of embeddings of Lstructures where B

and C are models of T and A is nitely generated there is a commutative

diagram

f A

/ B

g k

  D

C /

h

where h k are elementary embeddings

Pro of We prove iiiiiiiiii

iiii this is easy A mo del B of T is nothing but an emb edding of A

A

into a mo del of T By iii every two such mo dels have a common elementary ex

tension such that the diagram commutes this means that they are elementarily

equivalent L structures Hence T is complete by Exercise

A A

iii This is somewhat similar to the pro of of the L osTarski Theorem

Let x x b e an Lformula Pick new constants c c and let

n n

b e the set of all quantierfree L fc c gsentences such that T j

n

c c

n

Supp ose B is a mo del of T Let A b e the substructure of B generated

B B

by c c so A is the set

n

B B B

ft c c j tx x an Ltermg

n

n

Supp ose B j c c We regard c c as an L sentence Since

n n A

T is complete we have T j c c It follows that for some

A A n

sentence a a c c

m n A

T j a a c c c c

m n n

Here the constants a are the constants from A dierent from the c It follows

i i

that T j y y y y c c c c

m m n n

B

Now pick for each a an Lterm t such that a t c Then

i i i

i

T j t c t c c c

m

so t c t c c is an element of and therefore true in B but this is

m

B B

a contradiction since a t c We conclude that B j c since B was an

i

i

arbitrary mo del of T we have T j c and hence for some c

T j c c so T j c c so T j x x x as required

iii Let a a b e an L sentence where A is a substructure of a

n A

mo del B of T Since T admits quantier elimination

T j a a a a

n n

for some quantierfree a a If a a then T j

n n A A

a a and if a a then T j a a So

n n A A n

T is complete

A

iiii Since b oth B and C are mo dels of the complete theory T they

A

are L elementarily equivalent and have therefore a common L elementary

A A

extension by Exercise The extension is then also Lelementary and the fact

that it is a common L elementary extension entails that the diagram in iii

A

commutes

We shall now use this lemma to prove that the theory of dense linear orders

without endp oints has quantier elimination Lets recall the axioms of this

theory the language L is fg and the axioms are

irreexivity x x

transitivity x y y z x z

linearity x y x y y x

density x y w x w w y

no endp oints w z w x x z

We shall call this theory dlo

Lemma Let A be a nite linear order a a Suppose A is em

m

bedded in both B and C where B and C are models of dlo Consider B C as

L structures Then for any L formula x x we have if two ntuples

A A n

b b B c c C satisfy the conditions

n n

i m j n a b a c b a c a

i j i j j i j i

i j n b b c c

i j i j

then B j b b if and only if C j c c

n n

Pro of We use induction on The case for atomic is left to you as are the

induction steps for and Now supp ose the lemma is true for x x x

n

the tuples b b B c c C satisfy the condition in the lemma and

n n

B j xx b b Then for some b B B j b b b It sues

n n

now to nd c C such that the n tuples b b b B c c c C

n n

satisfy the conditions in the lemma but it is easy to see that thanks to the

axioms of dlo this can always b e achieved

Corollary dlo is model complete

Pro of If B C are mo dels of dlo apply Lemma with A empty

Theorem dlo has quantier elimination

Pro of We prove prop erty iii of Lemma for dlo

So supp ose A is a substructure of B and of C where B C are mo dels of dlo

We regard B and C as L structures Supp ose dlo is inconsistent as

A B C

L theory Then for some quantierfree b b and c c

ABC n m

such that B j b b and C j c c we have

n m

dlo j b c

hence dlo j y b y Now in and together there are nitely

many constant from A say a a By the axioms of dlo pick in B elements

k

b b such that the mtuples b b B c c C satisfy the

m

m m

conditions in Lemma with resp ect to the linear order on fa a g Then

k

Clearly a contradiction is b by that lemma we must have B j b

m

obtained

Therefore dlo is consistent as L theory and has a mo del

B C ABC

D This means we have a commutative diagram B

A / k

 h  D

C /

with h k emb eddings But we have already seen that dlo is mo del com

plete so the emb eddings h k are elementary This proves prop erty iii

of Lemma for dlo which therefore has quantier elimination

Quantier Elimination Real Closed Fields

In this section we review a famaous theorem by Tarski that the theory of real

closed elds has quantier elimination The pro of is not extremely dicult but

rather long if one do esnt want to assume deep results from algebra and the

techniques used in it have little to do with the rest of these notes Therefore

I have decided to put it in a separate section which may b e skipp ed without

disturbing ones reading of these notes

There are a numb er of equivalent ways to formulate the theory of real closed

elds We shall use the language L f g which we x for this entire

section The theory rcf is given by the following axioms

The axioms for a eld

the axioms for a linear order

x x y y x

x x x x x x

n n

n

for all n

n n

y y xx y x y x y

n n n

for all o dd n

A real closed eld is a mo del of rcf Examples are R and various subelds of

R such as the algebraic closure of Q in R the algebraic closure of Qe in R

Another example is the eld of recursive reals that is the real numb ers r

which are the limit of a Cauchy sequence in Q that is a recursive function from

IN into Q

Exercise a Show that every real closed eld is an ordered eld this

means that the following sentences are true in it

xy z x y x z y z

xy z x y z xz y z

b Show that in a real closed eld the ordering is dense and has no end

p oints

c Every real closed eld has characteristic zero

Denable functions Supp ose x x y is an Lformula for which

n

rcf j xy x x y

n

Then denes an nary function on any real closed eld We may intro duce a

function symb ol f or f and consider the theory rcf in the language L ff g

f

which is rcf together with the axiom

x x x x f x x

n n n

The theory rcf is conservative over rcf This means every Lsentence which

f

is a consequence of rcf is also a consequence of rcf This follows from the easy

f

observation that every real closed eld has a unique expansion to an L ff g

structure which is a mo del of rcf

f

We shall call f a denable function when is understo o d We can have

more than one denable function if f f are denable functions we shall

n

which extends rcf by all the dening axioms for have the theory rcf

f f

1 n

f f The pro cess can b e iterated if x x y is now an L

n m

ff f gformula such that

n

j xy x y rcf

f f

1 n

However where f is dened by in rcf we can have rcf

f f f f f

1 n 1 n

it is an easy exercise that f can already b e dened in rcf We express this as

follows denable functions are closed under composition

Let f b e an nary denable function We shall call f eliminable if for every

quantierfree Lformula B y u u there exists another quantierfree

m

Lformula C x x u u such that

n m

rcf j xu B f x u C x u

f

We note that also the eliminable functions are closed under comp osition if

f f f are eliminable g x is dened as f f x f x and B y u

n n

is quantierfree then in rcf

g f f f

1 n

B g x u C f x f x u

n

C f x f x x u

n

C x u

n

for suitable quantierfree formulas C C C

n

Polynomial relations Every term tx x of L denotes a p olynomial in

n

indeterminates x x and integer co ecients

n

A formula tx where t is an Lterm is called a polynomial relation Ev

ery quantierfree formula is in rcf equivalent to a prop ositional combination

of p olynomial relations eg t s is equivalent to t s s t

Therefore for testing whether a denable function is eliminable it suces

to lo ok at p olynomial relations

We shall make use of the following theorem familiar from elementary analysis

the pro of is omitted

Theorem Rolles Theorem for Real Closed Fields Let K be a real

closed eld and P K X be a polynomial with coecients in K Let P denote

its derivative then

K j xy x y P x P y z x z y P z

The following lemma will take up most of this section

Lemma Let tx x x be an Lterm regarded as a polynomial whose

m

degree in x is n Then

a There are eliminable functions x x x x such

m n m

that the fol lowing statements are consequences of rcf

1 m

x x x x

m n

for al l x x the function x tx x is either constant or

m

it is strictly monotonic increasing or decreasing on each interval

x x x x x x

n n n

b There are eliminable functions k x x x such that the fol low

n

ing statements are consequences of rcf

k

1 n

k x k x n

x x

n

For each j f ng k x j implies that x x are

j

exactly the zeros of tx x

k x implies that tx x has no zeros

k x n implies that tx x is the constant zero polynomial

Pro of I have given the statements in lemma in informal language and leave

it to the reader to see that these statements can b e expressed by Lformulas

Of course for a natural numb er j j is the Lterm

z

j times

We shall prove the lemma by induction on n We write tx x as

n

u xx u xx u x

n n

For n tx x u x and there is nothing to prove for a for b we let

if u x

k x

if u x

Then k x is eliminable for a p olynomial relation py v we have

pk x v u x p v

u x p v

For n tx x u xx u x Induction hyp othesis a means now that

the statement that either u x or tx x is monotonic on is a

consequence of rcf This is easy to see and left to you For induction hyp othesis

b we let

if u x

if u x u x

k x

if u x u x

Again k x is eliminable We let

if u x

x

u xu x otherwise

And also x is eliminable and has the right prop erties

Now supp ose n and we have proved the lemma for all n n Let t x x

b e the derivative of tx x with resp ect to x Since the degree of t in x is n by

induction hyp othesis b we have eliminable functions k x x x

n

satisfying b for t x x

To prove a for tx x we take the s for the s Now either k x n

tx x is constant or tx x is monotonic on each interval of form as in a

this follows from Rolles Theorem for rcf

To prove b we dene formulas C x C x as follows

n

C x t x x t x xt x x

C x t x x

i

i

t x xt x x

i i

for i n

C x t x x

n

n

t x xt x x t x x

n n n

Now C x means that either t x x and t x x t x x

or t x x and t x x t x x by the induction hyp othesis

and the axioms of rcf this means either tx x is constant or tx x has a

zero in the interval x In a similar way for i n C x

i

expresses that tx x has a zero in the halfop en interval x x and

i i

C x expresses that tx x has a zero in x Note that all C are

n i

n

quantierfree formulas in the s

Again using Rolles theorem one sees that if tx x is nonconstant each

of the intervals contains at most one zero of tx x Therefore we can dene a

formula K x j expressing that tx x has exactly j zeros in a quantierfree

way by

K x j C x C x

k k

k A k A

Af ng

jAjj

Let Lx b e the quantierfree formula u x u x then we dene

n

k x as n if Lx and j for the unique j such that K x j otherwise Every

p olynomial relation pk x v is now equivalent to a disjunction

n

k x j pj v

j

so k is eliminable

We have to dene the functions for tx x I do this in words

n

x is dened as if k x or k x n and as the least zero of tx x

otherwise

Supp ose x has b een dened then x is dened as x if

j j j

k x j or k x n otherwise it is the least zero which is greater than

x

j

We are left to prove that the functions are eliminable Every p olynomial

j

relation p x v is equivalent to the disjunction

j

n

k x i p x v

j

i

Now for i or i n k x i p x v is equivalent to k x

j

i pj v

0

is eliminable Then if i j Supp ose we have shown that for j j

j

k x i p x v is equivalent to k x i p x j i v

j i

So we are left with the case that x is a real zero of tx x For simplicity

j

we assume j the other cases involve bigger formulas but are essentially

similar So t x x t is not constant zero and we have to consider

p x v

By division with remainder in p olynomial rings there are p olynomials f x x v

and g x x v such that

px v f x x v tx x g x x v

the degree of g in x is less than n

Since is a real zero of t we have that p x v g x x v Let the

degree of g in x b e r n then we may apply induction hyp othesis b to g and

assume there are eliminable functions

l x v x v x v

r

for g as in b ie giving numb er of zeros and list of p ossible zeros

We also have the eliminable functions x x satisfying a for

n

tx x Let us from now on suppress the extra variables and just write

pu g u

r n

The statement p is as we have seen equivalent to g This

is equivalent to a disjunction distinguishing cases according to the relative

p osition of among the s for example if then g is

So equivalent to g

g g

g

g

g

r r

etc are By eliminability of the s all parts g g

equivalent to quantierfree formulas not involving the s So we are left with

the formulas We do the case This time we distinguish cases

according to the relative p osition of among the s we have

Take for example the case Recall that is the rst zero of t

Now this zero o ccurs if and only if

either t t t t

or t t t t

By eliminability of the s and s these formulas are equivalent to quantier

free formulas not involving these symb ols

All other cases are dealt with in a similar way This completes the pro of of

lemma

Theorem Tarski rcf has quantier elimination

Pro of By exercise it suces to consider Lformulas of form xAx x

where A is a conjunction of atomic formulas and negations of atomic formulas

By the axioms for a linear order we can eliminate the negations t s is

W

equivalent to t s s t So A is equivalent to a disjunction A where

i

i

every A is a conjunction of formulas of form px x or px x hence it

i

suces to consider the formulas xA so we may assume that A is of this form

i

p

For each p we have from lemma eliminable functions k x and x

p

p

x such that px x is equivalent to the disjunction

n

p

n

p

p

x k x i x

p

i

i

and px x is equivalent to the disjunction

p p

x x x p x

W

n

p p p p

p

x x x p x x x

i i i i i

p p

x x x p x

n n

p p

W

p p

n i k x x x p x x

p p

i i

i

So again we can reduce to the case of a formula xA where A is a conjunction

p

s and some other of statements of form x is in a certain interval b ounded by

i

side conditions which dont dep end on x

Then xA is equivalent to the statement that the side conditions hold which

by eliminability of the s is equivalent to a quantierfree Lformula and that

the intersection of the intervals is nonempty this is equivalent to a quantier

free Lformula

Mo del Completeness and Mo del Companions

We now return to mo del complete theories Robinsons Test Exercise sug

gests to lo ok at socalled existential ly closed mo dels of a theory T Let A B

b e an emb edding of Lstructures A is called existential ly closed in B if for

every sentence in the language L if B j then A j A is existentially

A

closed for T if whenever A B and B is a mo del of T A is existentially closed

in B

Exercise Show that T is mo del complete if and only if every mo del of T is

existentially closed for T

Now let T b e an arbitrary Ltheory not necessarily mo del complete Supp ose

that the class of Lstructures which are mo dels of T and existentially closed for

T is elementary so equal to the class of mo dels of a theory T Then T T

and T is mo del complete

There are several examples of mathematical theories T for which the class of

existentially closed mo dels of T is indeed the class of mo dels of such an extension

T and moreover every mo del of T can b e completed that is emb edded in a

mo del of T

Every torsionfree ab elian group can b e emb edded in a divisible torsion

free ab elian group

every integral domain can b e emb edded in an algebraically closed eld

every distributive lattice can b e emb edded in an atomless Bo olean algebra

every ordered eld can b e emb edded in a real closed eld

In Mo del Theory this situation is describ ed with the notion of model companion

Let T and T b e Ltheories T is called a model companion of T if the following

conditions hold

i Every mo del of T can b e emb edded in a mo del of T and vice versa

ii T is mo del complete

Exercise Show that condition i ab ove is equivalent to T T where T

is the set of all sentences which are consequences of T

Hint use Exercise

In this section we shall prove for two pairs of theories T and T that T is a mo del

companion of T distributive latticesatomless Bo olean algebras and integral

domainsalgebraically closed elds

Denition A lattice is a partial order which has a least and greatest ele

ment and in which for each pair of elements x y the inmum or meet x u y and

the supremum or join x t y exist these elements are dened by the conditions

z x t y z x z y z

z z x u y z x z y

A lattice is called distributive if moreover the distributive law holds

xy z x u y t z x u y t x u z

Exercise a Show that the conditions which x u y and x t y are required

to satisfy indeed determine these elements uniquely

b show that the distributive law implies its dual

xy z x t y u z x t y u x t z

The most immediate examples of distributive lattices are collections of subsets

of a given set X which contain and X and are closed under union and inter

section with inclusion of subsets as the partial order We shall so on see that

every distributive lattice is isomorphic to one of this form

A lattice which is not distributive is the following partial order

======

=

c a > b >> ¡¡ >> ¡¡ >> ¡¡

> ¡¡

We now dene formally the theory of distributive lattices

Denition The theory of distributive lattices dl is formulated in the

language f u tg and has the following axioms

xy x u y y u x xy x t y y t x

xy z x u y u z x u y u z xy z x t y t z x t y t z

x u x x x t x x

x u x x t x

xx u x x xx t x x

xy x u y t x x xy y t x u y y

together with the distributive law

xy z x u y t z x u y t x u z

A distributive lattice is a mo del of dl

Exercise Show that denitions and agree In particular given a

mo del of dl if we dene

x y i x u y x

we get a partial order such that x u y is the meet of x and y and x t y the join

is the least element and the greatest

Let A u t b e a distributive lattice a A A complement of a in A is

an element b satisfying a u b a t b If a has a complement it is

unique This follows by distributivity if b oth b and b are complements of a

then b b u b u a t b b u a t b u b t b u b b u b so b b

similarly b b Note that this also implies that complements are preserved by

any homomorphism of distributive lattices

A distributive lattice in which every element has a complement is called a

Boolean algebra Note that a Bo olean algebra is a mo del of dl together with

the axiom

xy x u y x t y

We call this the theory of Boolean algebras Since dl has a set of axioms

the theory of Bo olean algebras has a set of axioms

Examples of Bo olean algebras are the p owerset of any set where comple

ment is real complement if T is a theory we can consider the set of equiva

lence classes of sentences in the language of T with i T j for

equivalence classes we have u t and

the complement of is This Bo olean algebra is called the Lindenbaum

algebra of the theory T Note that T is complete if and only if its Lindenbaum

algebra has exactly two elements An imp ortant case is where we just consider

equivalence classes of formulas in prop ositional logic with prop ositional vari

ables p p The resulting Lindenbaum algebra is the free Boolean algebra

on countably many generators

Let A b e a distributive lattice and a A a is called an atom in A if a

and for every b a we have b or b a A Bo olean algebra which contains

no atoms is called atomless Note that an atomless Bo olean algebra is a Bo olean

algebra satisfying the axiom

xy x y y x y y u x

So also the theory of atomless Bo olean algebras has a set of axioms The

free Bo olean algebra on countably many generators is an example of an atomless

Bo olean algebra

Theorem Every distributive lattice can be embedded in a Boolean algebra

and every Boolean algebra can be embedded in an atomless Boolean algebra

Therefore every existential ly closed distributive lattice is an atomless Boolean

algebra

Pro of Let us rst show the last statement assuming the rst if A is a

distributive lattice and a A the statement a has a complement can b e

expressed by a sentence of L So since every distributive lattice can b e

A

emb edded in a Bo olean algebra if A is existentially closed it is a Bo olean algebra

Similarly the statement a is not an atom can b e expressed by a sentence

of L so if A is existentially closed it must b e atomless

A

For the rst statement we use the notion of lter and prime lter In a

distributive lattice A a lter is a subset U of A with the prop erties U

a U a b b U a b U a u b U and U Note that if a the

set a fb A j a bg is a lter

A prime lter is a lter U which moreover has the prop erty that whenever

a t b U a U or b U

First we prove if in a distributive lattice a b there is a prime lter U

which contains a but not b This is done with the help of Zorns Lemma since

a b a so a is a lter containing a but not b So the partially ordered

set of all lters which contain a but not b ordered by inclusion is nonempty

One easily sees that it is closed under unions of chains By Zorns Lemma it

has a maximal element U Supp ose that c t d U but c U d U Then by

maximality of U there must b e u u U such that c u u b and d u u b

Then u u u u U and we have

u u c t d u u c t u u d b

so b U contradiction

Now let X b e the set of all prime lters of the distributive lattice A Dene

f A P X by

f a fU X j a U g

By what we just proved it follows that f is and that f a t b f a f b

Clearly f X f and f a u b f a f b so f is an emb edding

of distributive lattices and P X is a Bo olean algebra

Next we show that every Bo olean algebra can b e emb edded in an atomless

Bo olean algebra Since every Bo olean algebra is a distributive lattice it can

b e emb edded in a Bo olean algebra of the form P X so it suces to see that

every P X can b e emb eeded in an atomless Bo olean algebra For this we

observe that for every function f Y X of sets the inverse image function

f P X P Y is a homomorphism of distributive lattices Moreover if f

is a surjective function then f is

We dene a chain of Bo olean algebras and emb eddings

1 1

0 1

P X P X

by letting X X X X f g and X X the pro jection We

i i i i i

may assume X so all are surjective and the therefore injective The

i

i

atoms in P X are the singleton subsets of X Now for y X fy g

i i i

i

fy y g so sends every atom to a nonatom

i

Let B b e the colimit of the chain ab ove Since the theory of Bo olean algebras

has a set of axioms it is preserved under unions of chains so B is a Bo olean

algebra in which all P X emb ed Let b B Then by the construction of

i

the colimit b comes from an element of some P X say Y X If Y is not

i i

an atom in P X then b is certainly not an atom in B but if Y is an atom

i

Y is a nonatom and gives the same element b in the colimit So b cannot

i

b e an atom and B is an atomless Bo olean algebra in which all P X emb ed

i

This completes the pro of

Next we wish to prove that every atomless Bo olean algebra is existentially

closed for the theory of distributive lattices Equivalently that the theory of

atomless Bo olean algebras is mo del complete This will follow from Lindstrms

Test Lemma b elow once we have proved that the theory of atomless

Bo olean algebras is categorical So we do that rst

Lemma a Let A be an atomless Boolean algebra and a A a

Then for each n IN there are elements b b in A such that b

n i

b u b for i j and b t t b a

i j n

b Let A be an atomless Boolean algebra and C D be an embedding of nite

Boolean algebras Then every embedding C A extends to an embedding

D A

Pro of a is proved by induction on n For n take b a For n since

a is not an atom there is b a in A then if c is the complement of

n

b a u c Moreover b t a u c a So apply the induction hyp othesis

n n

to nd b b for a u c

n

For b we note that every nite Bo olean algebra is isomorphic to P X for

a nite set X Now every emb edding f P X P Y for nite X Y is

determined by its values on the singleton subsets of X since it must preserve

unions and every subset of X is a nite union of singletons If X fx x g

n

we see that f fx g f fx g for i j since f preserves and intersection

i j

S

n

f fx g Y since f preserves f fx g since f is an emb edding and

i i

i

and unions So f divides Y into n parts Now supp ose g P X A is

an emb edding For each y Y there is a unique i with y f fx g supp ose

i

f fx g fy y g By a cho ose b b in A for g fx g and send

i m m i

fy g to b This extends to an emb edding h P Y A such that g hf

j j

Corollary The theory of atomless Boolean algebras is categorical

Pro of Let A fa a a g and B fb b b g b e two countable

atomless Bo olean algebras We dene a chain of isomorphisms f A B

i i i

such that A A A is a chain of nite Bo olean algebras with union

A and B B B is a chain of nite Bo olean algebras with union B

and f A B extends f for each i The construction is very similar

i i i i

to Cantors backandforth construction for dense linear orders

Let A b e the subBo olean algebra of A containing just the elements and

B likewise the subBo olean algebra of B with two elements and f A B

the obvious isomorphism

Supp ose f A B is constructed Let C b e the subBo olean algebra

i i i i

of A generated by A fa g Then C is nite and A C Since f gives

i i i i i i

an emb edding of A into the atomless Bo olean algebra B by lemma f

i i

extends to an emb edding g C B Let D b e the image of g and B the

i i i i i

subBo olean algebra of B generated by D fb g Since g is an emb edding its

i i i

inverse is an emb edding of D in A which again by lemma extends to an

i

emb edding h B A Let A b e the image of h Then the inverse of h

i i i i i

is an isomorphism f A B which extends f Since a A the

i i i i i i

union of the A s is A and similarly the union of the B s is B and the union

i i

of the f s is an isomorphism A B

i

The following two general lemmas nd frequent application in the study of mo del

completeness

Lemma Let T be an Ltheory with a set of axioms Then for any

model A of T there is an embedding of A into a model B of T which is exis

tential ly closed for T Moreover if jAj jjLjj we may assume that jB j jAj

Pro of We construct a chain of mo dels of T

A A A A

as follows if A has b een dened let b e a maximal set of sentences in

k k

is consistent such a set exists by such that T the language L

k A A

k k

then A A Zorns Lemma Let A b e a mo del of T

k k k k A

k

Let B b e the colimit of this chain Since T has a set of axioms B is a

mo del of T Supp ose is a sentence of L which holds in some extension of

B

B which is a mo del of T By construction of the colimit is already an L

A

k

is consistent whence by maximality sentence for some k then T

k A

k

of and holds in A so holds in B So B is existentially closed

k k k

for T

To prove the nal statement if jAj jjLjj we may take all A such that

k

jA j jAj Then jB j jAj

k

Lemma Lindstrms Test Suppose T is an Ltheory with a set of

axioms which only has innite models and is categorical for some cardinal

number jjLjj Then T is model complete

Pro of If A is a mo del of T of cardinality then by the previous lemma A

may b e extended to an existentially closed mo del B of cardinality Since T

is categorical A B so A is existentially closed for T So every mo del of T

of cardinality is existentially closed for T

Now let A B b e an arbitrary emb edding of mo dels of T We have to show

that A is existentially closed in B

First supp ose jAj Then for any sentence of L which holds in B

A

there is by downward LowenheimSkolemTarski an elementary submo del

A of A of cardinality which contains all constants o ccurring in then A is

existentially closed so holds in it so holds in A to o So A is existentially

closed for T

If jAj we view B as a mo del for the language L fX g where X is a new

B

ary relation symb ol putting X A

Consider the language L L fX g fc j g where is a set of

cardinality and the c are new constants let b e the L theory consisting of

the elementary diagram of B as L fX gstructure together with the set of

B

axioms

0

j g fX c j g fc c

Since A is innite b ecause T has only innite mo dels every nite subset of

has an interpretation in B hence is consistent and has a mo del C Then

C

C is a mo del of T and an elementary extension of B and X is the domain

of a submo del A of C which is an elementary extension of A This is seen as

X

follows for an L sentence let b e obtained by replacing each quantier

A

X

x by xX x and each x by xX x Then A j i B j

X

i C j i A j

Note that A has cardinality at least so by what we have already proved

A is existentially closed in C But now it is easy to deduce that A is existentially

closed in B

Corollary The theory of atomless Boolean algebras is model complete

hence it is a model companion of the theory of distributive lattices

Pro of Every atomless Bo olean algebra is innite this follows at once from

lemma and the theory of atomless Bo olean algebras is categorical as

we have seen so this follows from Lindstrms Test

As a nal example of a mo del companion we show that the theory of alge

braically closed elds is a mo del companion of the theory of integral domains

commutative rings with having no zero divisors By elementary algebra

every integral domain is emb edded in a eld and every eld is emb edded in an

algebraically closed eld

Since for an integral domain A the statement that a A is either or has

n

a multiplicative inverse and the statement that the p olynomial a X

a X a has a ro ot where a a A are sentences of L which

n n n A

hold in some extension of A every existentially closed integral domain is an

algebraically closed eld

Conversely let A b e an algebraically closed eld and x x where

n

is a conjunction of atomic L formulas and negations of such by Exercise

A

it suces to consider sentences of this form First we use the axioms of a eld

to reduce formulas t s to y y t s so is equivalent to

x x where is a conjunction of statements P x P a p olynomial

m

with co ecients in A

In order to show that A is existentially closed we need to show for such

that if holds in some extension of A it holds in A This follows from the

following elementary theorem from commutative algebra

Theorem Hilb ert Nullstellensatz Let K be an algebraical ly closed

eld and f f be polynomials in X X with coecients in K If

n m

there is no mtuple a a K such that

m

f a a f a a

m n m

then there are polynomials g g in X X and coecients in K such

n m

that

f g f g in K X X

n n m

Let P X X P X X b e the system of p olynomials in the

m n m

formula and supp ose that this system has a common zero b b in an

m

extension L of A Then the b b induce a ring homomorphism

m

AX X L by f f b b

m m

On the other hand if the system has no common zero in A by the Nullstellen

P

n

satz there are Q Q with P Q in AX X Combining

n i i m

i

this gives in L a contradiction

Conclusion

Corollary The theory of algebraical ly closed elds is a model companion

of the theory of integral domains

Mo del Completions Amalgamation Quantier Elim

ination

In this section we collect some miscellaneous facts and denitions ab out mo del

completeness and mo del companions

Let us rst observe that if a theory T has a mo del companion it is unique

up to equivalence of theories For supp ose T and T are mo del companions of

T Then T T are mo del complete and every mo del of T can b e emb edded in

a mo del of T and vice versa So given any mo del A of T we can form a chain

A A A A

where A A A are mo dels of T and A A are mo dels of T If B

is the colimit then since b oth theories are mo del complete we have

A A A B

A A A B

So A is also a mo del of T By symmetry T and T have the same mo dels

Denition Let T b e an Ltheory

a A mo del A of T is said to b e algebraical ly prime if A can b e emb edded in

every mo del of T

b T is said to have the joint embedding property if for every two mo dels A

and B of T there is a mo del C of T and emb eddings A C B C

c T is said to have the amalgamation property if every diagram B A /



C

of emb eddings of mo dels of T can b e completed to a commutative diagram B A /

  D

C /

of emb eddings b etween mo dels of T

Exercise Let T b e mo del complete

i Prime Mo del Test If T has an algebraically prime mo del T is com

plete

ii T is complete if and only if T has the joint emb edding prop erty

The notion of model completion is a renement of that of mo del companion If

T is a mo del companion of T every mo del of T can b e emb edded in a mo del of

T If T is a mo del completion such an extension is unique up to elementary

equivalence in parameters of the mo del one starts with The formal denition

is

Denition T is called a model completion of T if T is a mo del companion

of T and for every mo del A of T T is complete

A

Exercise Let T b e a mo del companion of T Show that the following two

statements are equivalent

i T is a mo del completion of T

ii T has the amalgamation prop erty

Exercise Let T b e a mo del complete theory Show that the following are

equivalent

i T is a mo del completion of T

ii T has the amalgamation prop erty

iii T has quantier elimination

Exercise a Prove that the theory of integral domains has the amalga

mation prop erty

b Deduce from a that the theory of algebraically closed elds has quantier

elimination

Countable Mo dels

In this section we shall b e concerned with countable languages and countable

structures So L is countable in this chapter and if A is a countable L

structure of course L is countable to o

A

If is a set of Lformulas with at most the variables x x free we write

n

x x or x An Lstructure A realizes x if there is an ntuple

n

a a A such that A j a a for every x x x If

n n n

A do es not realize A is said to omit

Let T b e an Ltheory T is said to realize local ly if there is an Lformula

x x such that T fx x x g is consistent and T j x x

n n

x for every x

Theorem Omitting Typ es Theorem Let T be a consistent Ltheory

If T and for each m IN let be a set of formulas in variables x x

m k

m

does not realize any local ly then T has a countable model which omits each

m

m

Pro of First we add a countable set C fc c g of new constants to the

language L let L L C Fix an enumeration of all L sentences

in such a way that every L sentence o ccurs innitely often in this enumeration

We shall build a chain of L theories

T T T T

such that the following hold

Every T is a consistent extension of T by nitely many L sentences

n

if T f g is consistent then T

n n n n

if x and T then c T for some c C

n n n n

for each m n and each k tuple c of elements of C which o ccur in T

m n

such that c T there is a formula x x

m n k

m

Now supp ose we have constructed T T with these prop erties let

S

T T Then T is consistent and has a mo del A but by construction

n

n

A

A has a submo del B with underlying set fc j c C g and B is actually an

elementary submo del so is a countable mo del of T And B omits each by

m

construction

To construct our chain we start by putting T T T was assumed consis

tent which is all there is to check at this stage

Supp ose T has b een constructed We build T in stages

n n

Stage We check If T f g is consistent we put in T If not we

n n n n

do nothing in this stage

Stage If x was put into T at stage let c b e the rst constant

n n

in the enumeration of C which do esnt o ccur in T which contains only nitely

n

many constants from C by induction hyp othesis and put c into T

n

Stage Let C b e the nite set of constants from C which so far o ccur in

T

n

For each m n let L fc c g b e a list of all k tuples of

m m

U m

elements of C We work through each m and each j j U m as follows

we start with m j At substage m j we have added a nite numb er

of L formulas to T let C b e the conjunction of these Write C as

mj mj

d d c c

u k

m

is the tuple c and d d are the other constants from where c c

j u k

m

C Since T do es not lo cally realize there is a formula x x

m m k

m

such that

y y x x y y x x T j x x

u k u k k

m m m

to T and pro ceed to substage m j if j Now add c c

n k

m

U m otherwise to m if m n otherwise stage is completed and the

construction of T to o

n

The Omitting Typ es Theorem is often applied in order to construct count

able mo dels which have to b e small in some sense there are no elements or

tuples realizing any of countably many sets of formulas

An example of this is the construction of endextensions another one is in

the construction of atomic models

End extensions of mo dels of PA

The theory of Peano Arithmetic PA is formulated in the language L f g

and its axioms are

xx xy x y x y

xx x xx

xy x y x y xy xy xy x

x x y y x y x y y x

The last axiom is meant to b e an axiom for every Lformula y x These

axioms are called induction axioms

Any mo del A of PA has the prop erties that and are commutative and

asso ciative that is distributive over that the formula z x z y

denes a linear order x y for which is the least element and such that

every element x has a successor that is a least element greater than x

Let A B b e an emb edding of mo dels of PA B is called an end extension

of A or A an initial segment of B if for a A and b B if b a then b A

We say that the emb edding is prop er if A B

We shall use the Omitting Typ es Theorem to show that every countable

mo del of PA has a countable prop er elementary end extension

Exercise Let A b e a mo del of PA and u x b e an L formula such that

A

A j z y xu x y u x u z

Show using the induction axioms of PA that

A j z y xu x y u x u z

Prop osition Every countable model of PA has a countable proper elemen

tary end extension

Pro of Let A b e a countable mo del of PA let L L fcg where c is a new

A

constant Consider the L theory

E A fc a j a Ag

Clearly T is consistent by the Compactness Theorem and every mo del of T is

a prop er elementary extension of A

Now consider for each a A the set of formulas

x fx ag fx b j b Ag

a

Convince yourself that a prop er elementary end extension of A is nothing but a

mo del of T which omits each x

a

Since there are only countably many sets x b ecause A is assumed count

a

able the Omitting Typ es Theorem gives us such a countable mo del provided

we can show that no x is lo cally realized by T

a

So supp ose x is an L formula such that

T j x x x a

T j x x x b for all b A

Write x x c where is an L formula From we deduce using the

A

Compactness Theorem that there is some a A such that

A j xuu a x u x a

From we deduce in the same way that for each b A there is b A such

that A j xuu b x u x b in other words

A j z w xuu w x u x z

Applying Exercise we nd that

A j z w xuu w x u x z

However combining this with we get a contradiction unless A j xu x u

but this means that T fx xg is inconsistent

Therefore no x is lo cally realized by T and we are done

a

Atomic Theories and Atomic Mo dels

If A is an Lstructure and a a A the set of Lformulas

n

fx x j A j a a g

n n a a

1 n

is maximal wrt the prop erty that it is realized in some Lstructure

the type of the tuple a a We call

n a a

1 n

In general if T is an Ltheory and x x a set of Lformulas in

n

variables x x which is maximal wrt the prop erty that it is realized in

n

some mo del of T we call a type of T

Exercise If a typ e x of T is lo cally realized by T there is a formula x

such that

x fx j T j x x xg

In this case we say that the typ e is principal for T and that is generated

by x

Exercise Let L x b e the set of equivalence classes of Lformulas in vari

T

ables x where x x if and only if T j x x x

Show that L x is a Bo olean algebra Show that x is realized in some

T

mo del of T if and only if the set f j g is contained in a lter on L x

T

Show that x is a typ e of T if and only if f j U g for some ultralter

U on L x

T

We say that T is an atomic theory if every Lformula x such that T

fxxg is consistent is an element of a principal typ e in x of T

An Lstructure A is called atomic if for every ntuple a a the typ e

n

is principal for ThA

a a

1 n

Exercise Call a Bo olean algebra B atomic if for every b B if b there

is some atom in B which is b

Show that in a Bo olean algebra b is an atom if and only if the lter b is an

ultralter

Show that a theory T is atomic if and only if for each tuple x of variables

the Bo olean algebra L x is atomic

T

Exercise Let A b e an Lstructure such that for each a A there is an

Lformula x such that

A j x x x a

Show that A is atomic

Theorem Let T be a complete Ltheory

a If for every n IN T has only countably many types in x x T has

n

a countable atomic model

b If A is a countable atomic model of T A is elementarily embedded in every

model of T

c If A and B are countable atomic models of T they are isomorphic

Pro of For a we use the Omitting Typ es Theorem Since there are only

countably many typ es altogether there are certainly only countably many non

principal typ es So by the Omitting Typ es Theorem let A b e a countable

mo del of T which omits each nonprincipal typ e we cannot omit a principal

is principal for typ e why Then for each a a A the typ e

n a a

1 n

T ThA so A is atomic

For b supp ose A is a countable atomic mo del of T and let B b e any mo del

is principal for ThA T there is a of T List A as a a Since

a

0

fx j T j x x x g Since T is formula x such that

a

0

complete T j x x so B j b for some b B

Now supp ose we have dened b b B such that b b realizes

n n

is generated by some formula x x in B The typ e

n n a a a a

0 n+1 0 n

so B j b b Then the formula x is an element of

n n n n a a

0 n

for some b Then b b realizes

n n a a

0 n+1

The map a b is then an elementary emb edding A B

n n

c is proved by p erforming the construction of b in two directions this is similar

to the backandforth metho d in the pro of of Corollary and is left to you

Exercise Work out the pro of of c ab ove

Exercise Theorem b has a converse if A is a countable structure such

that A is elementarily emb edded in every structure that is a mo del of ThA

then A is atomic

must be local ly realized by ThA Hint for a a A show that

n a a

1 n

using the Omitting Types Theorem So it must be principal

Exercise Show for a complete theory T that T has a countable atomic

mo del if and only if T is an atomic theory

Hint in one direction if A is an atomic model of T and xx consistent

with T show that x is realized by a tuple a in A so x which is a

a

principal type

For the other direction let for each n be the set of formulas x x

n n

for such that generates a principal type for T Use the Omitting Types

Theorem to show that T has a countable model which omits each and hence

n

is atomic

Exercise Let L b e the language of partial orders together with a countably

innite set of new constants c c Let T b e the Ltheory

dlo fc c j i INg

i i

Describ e up to isomorphism the countable mo dels of T Do es T have an

atomic mo del

Exercise For those of you who are familiar with Godels Incompleteness

Theorems

a Show that Peano Arithmetic has an atomic mo del the standard model

N Show that the mo del N is not elementarily emb edded in every other

mo del of PA Why is this not in conict with Theorem

b The theory rcf of real closed elds has an atomic mo del What is it

Show that rcf has uncountably many typ es in one variable So the

converse to c is false

I nish this chapter with a theorem giving a characterization of categorical

theories in terms of the numb er of typ es Its pro of uses atomic mo dels

Theorem Let T be a complete theory which has innite models Then T

is categorical if and only if for every n IN T has only nitely many types

in x x

n

Pro of First supp ose T is categorical let A b e a countably innite mo del of T

Every mo del of T is innite check hence has a countable elementary submo del

by downwards LowenheimSkolemTarski therefore since T is categorical A

is elementarily emb edded in every mo del of T So A is atomic by exercise

Let x b e a typ e of T Then x is realized in some countable mo del of

T so it is realized in A which is atomic it follows that x for some

a

a A so x is principal So every typ e of T is principal

Consider now the set

x f x j x generates a typ e of T g

Then x cannot b e extended to a typ e b ecause every typ e is principal By

the compactness theorem there are nitely many formulas x x

k

generators of typ es such that

T j x x x

k

We see that the typ es generated by the x are the only typ es of T which

i

therefore has only nitely many typ es

Conversely if T has only nitely many typ es in x for each tuple x then each

Bo olean algebra L x must b e nite for if T j x x x there

T

is a typ e which contains x but not x It follows that every typ e of T

is principal so that every mo del of T is atomic Then T is categorical by

Theorem c

Exercise Show that a Bo olean algebra B has only nitely many ultralters

if and only if B itself is nite Show moreover that every innite Bo olean

algebra has a nonprincipal ultralter

Exercise Describ e the typ es of dlo in variables x x

n

! Saturated Mo dels

The notion of a saturated mo del is dual to the notion of atomic mo del saturated

mo dels are large they realize as many typ es as p ossible In this chapter where

we fo cus on countable mo dels we restrict ourselves to the notion of saturated

mo del

Denition Let A b e an Lstructure We say that A is saturated if

for every nitely generated substructure B of A and every set of L formulas

B

x x in free variables x x the following holds if is realized in

n n

an L elementary extension of A then is realized in A

B

Exercise Show that every nite structure is saturated

Lets recall that if B is generated by a a we may replace L by L

m B

fa a g

1 m

Exercise Supp ose A and B are Lstructures and a a A b b

n n

A

B We can then see A and B as L fc c gstructures by putting c a

n i i

B

and c b of course the constants c are new

i i i

if the L fc c gstructures thus dened B We write A

n b b a a

1 n 1 n

are elementarily equivalent

Prove A is saturated if and obly if for each Lstructure B the following

hold

If A B then for every b B there is an a A such that A B

a b

B for tuples a a A and b b B such that A

b b n n a a

1 n 1 n

B and every b B there is a A such that A

b b n n a a

1 n+1 1 n+1

Theorem Let T be a complete Ltheory

a If A is an saturated model of T every countable model of T is elemen

tarily embedded in A

b T has a countable saturated model if and only if for each n IN T has

only countably many types in x x

n

c if T has a countable saturated model T has an atomic model

Pro of a Supp ose A is an saturated mo del of T and B a countable mo del

of T Enumerate B as fb b g Since T is complete A B Therefore

applying Exercise we can nd a sequence a in A such that for each

n n

The assignment b a now denes an elementary B n A

n n b b a a

1 n 1 n

emb edding of B into A

b If T has a countable saturated mo del A then there can b e at most

countably many typ es since every typ e is realized in A

Conversely supp ose T has only countably many typ es in x x for each

n

n Let A b e a countable mo del of T Then we can enumerate all pairs a

i i

of elements of A and a where a is a nite p ossibly empty tuple a

n i n

i 1

such that the set y y x is a typ e in variables x

k n i n

i i 1

j xy g y y a a f a

i k n i i n

i i 1

is realized in some L elementary extension of A We construct a chain of

a

i

mo dels

A A A A

as follows supp ose A is constructed Let B b e a countable L elementary

n n

a

n

extension of A such that a is realized in B Then certainly A B

n n n n

so A B let A b e a countable common elementary extension of them

n n n

One sees by induction that for each n A realizes a for all i n

n i i

Let A b e the colimit of the chain A A Then A is countable

A A and moreover the following prop erty holds

For every nitely generated substructure B of A and every set x x

n

of L formulas which is realized in some L elementary extension of A

B B

is realized in A

Now we iterate this pro cedure innitely often

A A A

n n n

where each A is countable and each A has the prop erty wrt A

n

Let C b e the colimit of the chain A Now it is easy to see that C is

countable and saturated

c follows at once from b and Theorem a

Exercise Show that if A and B are two countable saturated mo dels of a

complete theory T then A B

Stone Duality

In this chapter we shall establish a natural corresp ondence b etween Bo olean

algebras and a certain typ e of top ological spaces the socalled Stone spaces

The corresp ondence when applied to the Lindenbaum algebra L of a theory

T

T or to the algebras L x gives another p ersp ective on the Compactness

T

Theorem

Let us rst recall and collect the facts ab out lters on a Bo olean algebra

that we have already seen at dierent places or that are easy to derive

A lter on a Bo olean algebra B is an upwards closed nonempty prop er

subset of B which is closed under nite meets inma

an ultralter is a maximal lter

in a Bo olean algebra the notions of prime lter and ultralter coincide

if a b in a Bo olean algebra B there is an ultralter on B which contains

a but not b

a lter U is an ultralter if and only if for each b B exactly one element

c c

of fb b g b elongs to U here b denotes the complement of b

if f A B is a homomorphism of Bo olean algebras and U is an ultralter

on B then f U is an ultralter on A

if A is a subset of a Bo olean algebra B there is an ultralter on B which

contains A if and only if for each nite subset fa a g of A a u

n

u a

n

Let B b e a Bo olean algebra We denote the set of ultralters on B by S B

We give S B a top ology

Let for b B U b e fF S B j b F g

b

I claim that the collection fU j b B g forms a basis for a top ology on S B

b

S

Indeed it is clear that S B U and you can check for yourself that

b

bB

0 0

U U U

bub b b

We now examine the top ological space S B with the top ology generated

by the sets U

b

Prop osition The space S B is a compact Hausdor space which has a

basis of clopen closed and open sets

c

So S B has a basis Pro of Every set U is clop en since its complement is U

b b

of clop en sets

c

If F G in S B there is b B with b F and b G so F U G U

b b

which are disjoint op en neighb orho o ds of F and G resp ectively So S B is

Hausdor

j i I g covers S B then every ultralter Finally supp ose the collection fU

b

i

c

on B contains some b It follows that the set fb j i I g is not a subset of any

i i

u u b ultralter But then we must have for some i i I that b

i n i

n 1

g covers S B So S B is compact U in B but this implies that fU

b b

i i

n

1

At this generality Prop osition says all there is to say ab out S B This

follows from

Prop osition Let X be a nonempty compact Hausdor space which has

a basis of clopen sets Then there is a Boolean algebra B X unique up to

isomorphism such that X is homeomorphic to S B X Consequently if B is

a Boolean algebra then B S B is isomorphic to B

Pro of For any top ological space X the set B X of clop en subsets of X

contains and X and is closed under nite unions and intersections as well as

under taking complements Therefore it is a Bo olean algebra

Now consider an ultralter F on B X In particular this is a collection of

closed subsets of X which is closed under nite intersections so if X is compact

T

the intersection F is nonempty and if X is Hausdor the intersection is a

singleton fxg by maximality of the lter So we have a function h S B X

X which is easily seen to b e a bijection It is also continuous for if hF x

then F fU X j U clop en x U g so for a clop en basiselement A we have

h A fF S B X j A F g U

A

which is a basic op en in S B X Since b oth X and S B X are compact

Hausdor we are done and h is an homeomorphism

Now lets lo ok at B S B Clearly there is a homomorphism of Bo olean

algebras B B S B given by b U This is an injective map Now

b

take any clop en subset W of S B Then b oth W and its complement are

unions of clop en basis elements Since S B is compact we must have W

and we see that the for some b b B so W U U U

n b ttb b b

1 n n 1

map B B S B is also surjective It is easy to see that its inverse is also a

homomorphism of Bo olean algebras

Denition A Stone space is a compact Hausdor space with a basis of

clop en sets

Exercise Examples i Show that Cantor space is a Stone space Show

that under the corresp ondence of prop osition it corresp onds to the

Bo olean algebra P IN

j n IN n g viewed as a subspace of R is ii Show that the set fg f

n

a Stone space To which Bo olean algebra do es it corresp ond

We shall now extend the corresp ondence of prop osition also to maps

Supp ose f X Y is a continuous function b etween top ological spaces

Then f P Y P X maps clop en sets to clop en sets and commutes with

nite unions and intersections as well as complements So it is a homomorphism

of Bo olean algebras B Y B X

Now if X and Y are Stone spaces every homomorphism B Y B X

is f for a unique continuous f X Y For we have seen that p oints

of a Stone space are in corresp ondensce with ultralters on its asso ciated

Bo olean algebra Since sends ultralters on B X to ultralters on B Y

it determines a map f X Y such that f and it is left to you to see

that f is continuous

Summing up we have established that for Stone spaces X and Y there is

a corresp ondence b etween continuous maps X Y and Bo olean homo

morphisms B Y B X If we take Y X the the identity map on X

f g

corresp onds to the identity map on B X and given X Y Z the com

p osition of f and g corresp onds to the comp osition of the maps corresp onding

to g and f In the language of Category Theory we say that the categories of

Stone spaces and Bo olean algebras are dual ly equivalent or dual to each other

the word dual refers to the fact that the direction of the arrows is reversed

For the record

Theorem Stone Duality Theorem The category of Stone spaces is

dual to the category of Boolean algebras

Exercise The Bo olean algebra f f g has very sp ecial features

a for every Bo olean algebra B ultralters on B are in corresp ondence

with Bo olean homomorphisms B f

b for every Bo olean algebra B there is exactly one Bo olean homomorphism

f B

Applying Stone Duality interpret these facts for the corresp onding Stone space

Exercise Let T b e a theory and consider the Bo olean algebra L Describ e

T

the p oints of the Stone space corresp onding to L in terms of T

T

Exercise Show without using Zorns Lemma or any of its equivalents that

the Compactness Theorem is equivalent to the statement that each of the spaces

S L is compact

T

Exercise Let X b e an arbitrary top ological space Show that there is a

Stone space T X and a continuous map X T X such that the following

holds for every continuous function f from X to a Stone space Y there is a

unique continuous function f T X Y such that f f

Literature

There are two standard reference works for general Mo del Theory

CC Chang and HJ Keisler Mo del Theory Amsterdam North Holland

rd edition

W Ho dges Mo del Theory Cambridge Cambridge University Press

Both b o oks are exp ensive and not very suitable for a rst course The rst is

moreover despite the additions in the edition rather oldfashioned

A nice compact intro duction is

K Do ets Basic Mo del theory Stanford CSLI Publications

However its emphasis is quite dierent from that of these notes Has nothing

on quantier elimination or mo del completeness

General Mathematical Logic b o oks such as Sho enelds Mathematical Logic a

classic or Bell Machovers A Course in Mathematical logic usually have a

chapter on Mo del Theory where basic notions are given

Material on Real Closed Fields can b e found in

N Jacobson Lectures in Abstract Algebra vol I I I Princeton Van Nos

trand

The pro of of Quantier Elimination for Real Closed Fields in section is based

on the pap er

PJ Cohen Decision Pro cedures for Real and pAdic Fields in Commu

nications on Pure and Applied Mathematics XXI I pp

For a detailed pro of of Hilb erts Nullstellensatz lo ok at

H Matsumura Commutative Ring Theory Cambridge University Press

There are several directions in which to study Mo del Theory further The

hottest elds are Stability Theory and oMinimality Theory A few intro ductory

texts

A Pillay Geometric Stability Theory New York Clarendon Press

D Marker M Messmer and A Pillay Mo del Theory of Fields Berlin

Springer

L van den Dries Tame Top ology and ominimal structures Cambridge

University Press

However without some further intro duction in Mo del Theory it is not advisable

to start on these b o oks right away

A nice and accessible sp ecialization of Mo del Theory is the topic of mo dels of

Peano Arithmetic Two texts

R Kaye Mo dels of Peano Arithmetic Oxford Clarendon

J van Oosten Intro duction to Peano Arithmetic Godel Incompleteness

and Nonstandard Mo dels Communications of the Mathematical Institute

Utrecht University

Index

E A connected graph

L x Craig Interp olation Theorem

T

S B

denable

T j

class of structures

T

dene

A

U i

explicitly

n

implicitly

n

diagram

elementary

A B

p ositive

A j

directed

A B

p oset

Q

system of Lstructures

A

i

iI

divisible ab elian group

L

A

domain of structure

ThA

dually equivalent categories

jjLjj

amalgamation

elementarily equivalent structures

for elementary emb eddings

elementary

amalgamation prop erty

class of structures

atom

elementary extension

in distributive lattice

Elementary System Lemma

atomic

elimination of quantiers

Bo olean algebra

emb edding

structure

elementary

theory

end extension

atomic formula

existentially closed structure

atomless Bo olean algebra

expansion

axioms

set of

lter

conite

Beth Denability Theorem

Frechet

Bo olean algebra

generated by

in distributive lattice

categorical

over a set

principal

ChangL osSuszko Theorem

Fundamental Theorem

colimit of directed system

for ultrapro ducts

Compactness Theorem

complement

Hilb ert Nullstellensatz

in distributive lattice

homomorphism of Lstructures

congruence relation

on P I

induction axioms

Robinsons Consistency Theorem isomorphism of Lstructures

Robinsons Test

joint emb edding prop erty

Rolles Theorem

KeislerShelah Theorem

saturated mo del

formula

LowenheimSkolemTarski Theorem

Stone Space

downward

structure

upward

nitely generated

lattice

submo del

distributive

elementary

Lindenbaum algebra

generated by

Lindstrms Test

substructure

L osTarski Theorem

L osVaught test

Tarskis Theorem

on quantier elimination for rcf

mo del

algebraically prime

theory

mo del companion

complete

mo del complete theory

consistent

mo del completion

true in a mo del

mo del of a theory

typ e

Morleys Theorem

generated by

of a theory

omit a set of formulas

of a tuple of elements

Omitting Typ es Theorem

principal

Order Extension Principle

ultralter

Peano Arithmetic PA

ultrap ower

formula

ultrapro duct

p oset

universe of a structure

p ositive formula

preserved

wellfounded linear order

under directed unions

under homomorphic images

under substructures

prime lter

in distributive lattice

Prime Mo del Test

Ramseys Theorem

real closed eld

realize

a set of formulas

lo cally

reduced pro duct

reduct