Subdirectly irreducible algebras in varietiesof universal algebras

Dorien Zwaneveld

August 15, 2014

Bachelorthesis

Supervisors: dr. N. Bezhanishvili, J. Ilin Msc

Korteweg-de Vries Institute for Mathematics

Institute for Logic, Language and Computation

Faculty of Science

University of Amsterdam

Abstract

In this thesis we give a characterization of subdirectly irreducible algebras within thevarieties of Boolean algebras and of Heyting algebras. In order to do this we will startwith an introduction into universal algebra and in particular in lattice theory. Wewill define congruences and we will define varieties via the operations of homomorphicimages, subalgebras and products. Then we will prove Birkhoff’s Theorem which statesthat a class of algebras is an equational class if and only if it is a variety. In the lastchapter we will define subdirectly irreducible algebras. We will give characterizations ofsubdirectly irreducible universal algebras via congruences. For subdirectly irreducibleBoolean and Heyting algebras we will work out a more convenient characterization viafilters. As a result we obtain a direct characterization of subdirectly irreducible Booleanand Heyting algebras.

Title: Subdirectly irreducible algebras in varieties of universal algebrasAuthor: Dorien Zwaneveld, tcam.zwaneveld@gmail.com, 6114571Supervisors: dr. N. Bezhanishvili, J. Ilin MscSecond Grader: Prof. dr. Y. VenemaDeadline: August 15, 2014

Korteweg-de Vries Institute for MathematicsUniversity of AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math

Institute for Logic, Language and ComputationUniversity of AmsterdamScience Park 107, 1098 XG Amsterdamhttp://www.illc.uva.nl

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Acknowledgements

During my research for this thesis I was an intern at a high school, on my way tobecoming a teacher, like most of the people who contributed to the theory discussed inthis thesis. This gave me a lot of stress and therefore I would like to thank my boyfriend,Jelle Spitz, for supporting me, loving me and keeping faith even when I went crazy anddid not have much faith myself. I would like to thank my teachers and friends at theILO for supporting me and Chris Zaal for cutting me some slack during these stressfultimes. I would also like to thank my friends, Hanneke van der Beek and Timo de Vriesfor getting up early so many times during the summer and supporting me in writingthis thesis. Furthermore I would like to thank my parents and sisters for their love andsupport.Most of all I give many thanks to my supervisors Nick Bezhanishvili and Julia Ilin,whose doors were always open and who pointed me in the right direction when I gotstuck. Without these people I would have never made it this far.

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Contents

1 Introduction 5

2 An introduction to universal algebra 62.1 Universal algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Lattice theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Boolean and Heyting algebras . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Varieties of universal algebras 173.1 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.2.1 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.2.2 Subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2.3 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2.4 Tarski’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 Term algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.4 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.5 The K-free algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4 Subdirectly irreducible algebras 364.1 Building blocks of varieties . . . . . . . . . . . . . . . . . . . . . . . . . . 364.2 Determining subdirectly irreducible algebras . . . . . . . . . . . . . . . . 40

5 Conclusions 53

6 Popular summary 54

Bibliography 56

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1 Introduction

During a typical bachelors in mathematics one learns about groups and rings and it getscalled ‘algebra’. But what is an algebra? A question raised by many mathematiciansduring the 19th century, but it took until 1933 that the first definition of a universalalgebra was given. During the studies of algebraic structures (or algebras) one finds thatcertain theorems do not only hold for groups, but they also hold for rings and for otherstructures as well. (Think of the homomorphism theorem for example.) In universal al-gebra one does not only look at groups, or rings, or . . . , one looks at the whole picture.In this approach one is taking a step back and is looking at things from a distance. Itunderlines general patterns that various algebraic structures have in common. This issomething logicians do more often than mathematicians. Since this thesis is written atthe ILLC, taking a step back and looking at the whole picture is what I will be doingin the first part of my thesis. I will be looking at the concept of universal algebra andin particular lattice theory. Universal algebra is a relatively new part of mathematicsfounded by Garrett Birkhoff (1911-1996) during the 1930s when he was teaching at theHarvard University. Garrett Birkhoff did not have a doctoral degree or even a mastersdegree, but still he is one of the main influences in the mathematical branch called uni-versal algebra. He was the one who gave the definition of a universal algebra and healso contributed greatly to lattice theory which is why we will come by his name fairlyoften in this thesis.We will also take a look at some algebraic structures that are less familiar to mathemati-cians, but more familiar to logicians: Boolean algebras and Heyting algebras. Booleanalgebras are named after George Boole (1815-1864), who was an English mathematicianas well as logician and philosopher. Boolean algebras are models for classical proposi-tional logic. Because of this, Boolean algebras have applications in computer sciencethat turn out to be very useful. At the end of the thesis we briefly discuss the connec-tion between Boolean algebras and classical propositional logic and we will see that the2-element Boolean algebra plays a special role for classical logic.We will also discuss Heyting algebras and show that every Boolean algebra is a Heyt-ing algebra. These algebras were named after a Dutch mathematician and logician:Arend Heyting (1898-1980). Heyting’s calculus formalizes the ideas of the Dutch math-ematician and philosopher Luitzen Egbertus Jan Brouwer (1881-1966) on constructivemathematics. They both studied and taught mathematics at the University of Amster-dam. In short, Heyting algebras are to intuitionistic propositional logic what Booleanalgebras are to classical propositional logic.Boolean and Heyting algebras form varieties (as do groups and rings). In this thesiswe will take a closer look at these algebraic structures with the main goal as to give acharacterization of the building blocks for the varieties of Heyting and Boolean algebras.

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2 An introduction to universalalgebra

The goal of this first chapter is to define and develop an understanding of varieties. Todo this, we will start with a universal algebra. What is a universal algebra? Do wealready know some examples? We will answer these question and then we will moveon to lattice theory. This is an area of logic and universal algebra one is unlikely toencounter in a bachelors in mathematics. It is also very important for the contents ofthis thesis, because this is one of the major building blocks for the rest of the theory.After discussing the basics of lattice theory, we will turn to some examples of algebrasarising from logic: Boolean algebras and Heyting algebras.

2.1 Universal algebras

The reader may have encountered examples of algebras in various contexts. But in mostof these contexts we often do not know what an algebra actually is. So let us start withgiving a definition of an algebra. But first let us define the type of an algebra.

Definition 2.1. A type of algebras is a set F of function symbols. Each of these functionsymbols has a nonnegative integer n assigned to it. This integer is called the arity ofthe symbol and we say that the function symbol is n-ary.

Definition 2.2. A (universal) algebra A of type F is an ordered pair 〈A,F 〉 where Ais a nonempty set and F is a family of finitary operations on A indexed by the type Fsuch that every n-ary function symbol corresponds to an n-ary operation on A.We call the set A the underlying set of A.

Note that the set A is closed under the finitary operations in F . So the operations inan algebra are the interpretation of the function symbols of the type.Now let us look at some examples of universal algebras. The reader may have seen someof these before.

Example 2.3.

• GroupsA group is a set G together with the binary operation ·, the unary operation −1

and the nullary operation 1. Notation: 〈G, ·,−1 , 1〉.

• RingsA ring 〈R,+, ·,−, 0〉 is a set R with two binary operations + and ·, one unaryoperation − and one nullary operation 0.

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• Sigma-algebrasAlso studying the mathematical field of measure theory one encounters algebras.Given a set Σ of elements one can define different sigma-algebras. For let S be thesome collection of subsets of Σ, then 〈S,∩,∪,′ 〉 is one of the sigma-algebras on Σ.

• LatticesThis is an algebra we have not yet seen before. A lattice has two binary operations〈L,∧,∨〉. Lattices are important for the contents of this thesis and therefore wewill look at them more closely.

• Boolean algebrasA Boolean algebra is a lattice together with some extra properties and operations〈B,∧,∨,′ , 0, 1〉. We will find out more about these algebras further in this thesis.

• Heyting algebrasWe will also discuss some properties and examples of Heyting algebras. A Heytingalgebra is also a lattice with the same type as Boolean algebras, but they satisfydifferent axioms 〈H,∧,∨,→, 0, 1〉.

In these algebras there are usually finitely many operations, we put the operation withthe biggest arity first and the operation with the smallest arity last.

Definition 2.4. Let G and G′ be two groups. A map f : G→ G′ is called a homomor-phism if for all x, y ∈ G

(i) f(x · y) = f(x) · f(y);

(ii) f(x−1) = f(x)−1;

(iii) f(1) = 1′.

Here 1′ is the nullary operation in G′.

For two rings R and R′, a mapping f : R → R′ is called a homomorphism if for allx, y ∈ R we have

(i) f(x+ y) = f(x) + f(y);

(ii) f(x · y) = f(x) · f(y);

(iii) f(−x) = −f(x);

(iv) f(0) = 0′.

As with the groups, 0′ is the nullary operation in R′.

Now we will generalize these to any universal algebra.

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Definition 2.5. When we have two algebras A and B of the same type, a mappingα : A→ B is called a homomorphism if

α(fA(a1, . . . , an)) = fB(α(a1), . . . , α(an)).

Here a1, . . . , an are elements in A and f is an n-ary operation. fA is the interpretationof f on A.

Now that we have the concept of a homomorphism it is convenient to extend this ideaand specify a couple of different kinds of homomorphisms.

Definition 2.6. An injective (one-to-one) homomorphism is called an embedding or amonomorphism. When a homomorphism is surjective (onto) it is an epimorphism. Ahomomorphism that is both injective and surjective is called an isomorphism.

With these homomorphisms we prove some results later on.

2.2 Lattice theory

In the previous section we have seen that a lattice is an algebra. In this section wewill see some different definitions of a lattice and we will prove that these definitions areindeed equivalent. We will also look at some special lattices and some of their properties.We will start by giving a definition of a lattice as given in the first paragraph.

Definition 2.7. A lattice is an algebra 〈L,∧,∨〉 with two binary operations which satisfythe identities:

(commutative laws) x ∧ y ≈ y ∧ xx ∨ y ≈ y ∨ x

(associative laws) x ∧ (y ∧ z) ≈ (x ∧ y) ∧ zx ∨ (y ∨ z) ≈ (x ∨ y) ∨ z

(idempotent laws) x ∧ x ≈ x

x ∨ x ≈ x

(absorption laws) x ≈ x ∧ (x ∨ y)

x ≈ x ∨ (x ∧ y)

The binary operation ∧ is also called the ‘meet’-operation and ∨ is called the ‘join’.Since a lattice is also an algebra it is implicit that our set L is non-empty.

The reader may have seen these operation symbols before, namely in the propositionallogic. There ∧ is called and or conjunction and ∨ is called or or disjunction. When welook at it this way, the properties from propositional logic form the laws. We model thisby taking the propositions as elements of a set and note that this actually forms a lattice.

Recall that a partially ordered set is a set with a partial order on it. A partial or-der is an ordering which is reflexive, antisymmetric and transitive. We can use partialsorders to get another definition of a lattice.

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Definition 2.8. A lattice is a partially ordered set L for which the following holds:

∀a, b ∈ L : ∃ sup{a, b}, inf{a, b} ∈ L.

Here sup{a, b} is the least upper bound of a and b. Thus sup{a, b} = p ⇔ a ≤ p, b ≤ pand for all c such that a ≤ c and b ≤ c we have p ≤ c. Here ≤ is the partial order onthe set.The infimum is the greatest lower bound i.e. inf{a, b} = p ⇔ a ≥ p, b ≥ p and for all csuch that a ≥ c and b ≥ c we have p ≥ c.

So a partially ordered set L is a lattice if the supremum and infimum of every 2-elementset exist in L. We have now seen two different definitions of a lattice. But are they infact equivalent? We will show this in the following theorem.

Theorem 2.9. Definition 2.7 and Definition 2.8 are equivalent.

Proof. Suppose that L is a lattice by Definition 2.7. We have to create a partial orderon L. Define:

a ≤ b⇐⇒ a = a ∧ b.

Now all we have to do is show that the ≤-relation defined above is a partial order andthat by this definition the supremum and infimum of every 2-element set exist.By the idempotent law we find that a = a∧a and therefore we have established reflexivity.Now suppose we have a ≤ b and b ≤ a. We find a = a ∧ b = b ∧ a = b. Thus by thecommutative law and the above we also have antisymmetry.For the transitivity, suppose a ≤ b and b ≤ c. The above tells us that a = a ∧ b andb = b∧c thus a = a∧(b∧c). Now we can use the associative law to find that a = (a∧b)∧c.But a = a ∧ b so this tells us that a = a ∧ c. Therefore we have a ≤ c and we haveproved the transitivity.Now suppose that a and b are two elements from L. We want to find an element p suchthat p = sup{a, b}. Since by absorption a∧ (a∨ b) = a and b∧ (a∨ b) = b holds, we finda ≤ a ∨ b and b ≤ a ∨ b. Thus a ∨ b is an upper bound for both a and b. Now supposethere exists some element c such that a ≤ c and b ≤ c. Then a ∧ c = a and b ∧ c = b soa ∨ c = (a ∧ c) ∨ c = c and b ∨ c = (b ∧ c) ∨ c = c by absorption again. Now

(a ∨ b) ∨ c = (a ∨ b) ∨ (c ∨ c) (idempotent)

= (a ∨ c) ∨ (b ∨ c) (associative)

= c ∨ c= c

Thus we find (a ∨ b) ∧ c = (a ∨ b) ∧ ((a ∨ b) ∨ c) = a ∨ b by using absorption. Thereforea ∨ b ≤ c and sup{a, b} = a ∨ b. Now we use the fact that a lattice is an algebra andtherefore L is closed under ∨ thus sup{a, b} exists in L.We also want to find an element q such that q = inf{a, b}. Since sup{a, b} = a ∨ b itseems to be a rather safe assumption that a ∧ b = inf{a, b}. So let us see why this istrue. (a ∧ b) ∧ a = a ∧ b by associativity. In the same way we have (a ∧ b) ∧ b = a ∧ b.

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So we find a ∧ b ≤ a and a ∧ b ≤ b. Now assume there exists some element d such thatd ≤ a and d ≤ b. Then a ∧ d = d = b ∧ d. Now

(a ∧ b) ∧ d = (a ∧ b) ∧ (d ∧ d) (idempotent)

= (a ∧ d) ∧ (b ∧ d) (associative)

= d ∧ d= d

Thus we find d ≤ a ∧ b and therefore inf{a, b} = a ∧ b as we claimed.So our first definition (Definition 2.7) implies our second definition (Definition 2.8). Nowlet us assume that we have a partially ordered set L such that for every 2-element set{a, b} ⊆ L their infimum and supremum exist within L.For a, b ∈ L define sup{a, b} = a ∨ b and inf{a, b} = a ∧ b.

a ∧ b = inf{a, b} = inf{b, a} = b ∧ a (commutative laws)

a ∨ b = sup{a, b} = sup{b, a} = b ∨ aa ∧ (b ∧ c) = inf{a, inf{b, c}} = inf{a, b, c} (associative laws)

= inf{inf{a, b}, c} = (a ∧ b) ∧ ca ∨ (b ∨ c) = sup{a, sup{b, c}} = sup{a, b, c}

= sup{sup{a, b}, c} = (a ∨ b) ∨ ca ∧ a = inf{a, a} = a (idempotent laws)

a ∨ a = sup{a, a} = a

a ∧ (a ∨ b) = inf{a, sup{a, b}} = a (absorption laws)

a ∨ (a ∧ b) = sup{a, inf{a, b}} = a

So the partially ordered set satisfies the identities from Definition 2.7. It becomes analgebra by noticing that L is also closed under ∧ and ∨ since the supremum and infimumare in L for every 2-element set. Thus the definitions are indeed equivalent.

In Figure 2.1 one will find some examples of lattices while in Figure 2.2 one will find apicture of a Hasse diagram that is not a lattice. This is because sup{c, d} and inf{a, b}do not exist. By Definition 2.8 we deduce that Figure 2.2 is not a lattice.

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b = a ∧ b

a = a ∨ b

c = a ∧ c = b ∧ c

b = a ∧ b = b ∨ c

a = a ∨ c = a ∨ c

a ∧ b

a b

a ∨ b

a ∧ c

a b c

a ∨ b

(a ∧ b) ∧ c

c a ∧ b

a b

a ∨ b

c ∧ d

c d a ∧ b

a b c ∨ d

a ∨ b

a ∧ b

a b

a ∨ b

d a ∧ b

c

b

a

b ∨ c

Figure 2.1: some lattices

c d

a b

Figure 2.2: not a lattice

Definition 2.10. A lattice is called distributive if it satisfies one of the distributive laws:

(distributive laws) x ∧ (y ∨ z) ≈ (x ∧ y) ∨ (x ∧ z)

x ∨ (y ∧ z) ≈ (x ∨ y) ∧ (x ∨ z)

Definition 2.11. A lattice is called modular if it satisfies the modular law:

(modular law) x ≤ y → x ∨ (y ∧ z) ≈ y ∧ (x ∨ z)

In Figure 2.3 one of the lattices is modular but not distributive and the other lattice isneither modular nor distributive. These two lattices are called M5 and N5, respectively.We will use these to characterize non-modular and non-distributive lattices.

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0

x y z

1

(a) M5

0

x

z

y

1

(b) N5

Figure 2.3

To see that M5 is not distributive let us take x, y and z as in Figure 2.3a. Now

x ∧ (y ∨ z) = x ∧ 1 = x

and(x ∧ y) ∨ (x ∧ z) = 0 ∨ 0 = 0

therefore x ∧ (y ∨ z) 6= (x ∧ y) ∨ (x ∧ z) and we deduce that M5 is not distributive. ButM5 is modular, there are ten cases to distinguish, and checking them takes a little timebut is routine.As stated before, N5 is not distributive, and not modular. To see this, let us take x, yand z as in Figure 2.3b. We find x∨ (y ∧ z) = x∨ 0 = x 6= y = y ∧ 1 = (x∨ y)∧ (x∨ z)therefore N5 cannot be distributive. Also, x ≤ y so for N5 to be modular we should havex ∨ (y ∧ z) = y ∧ (x ∨ z). But we have

x ∨ (y ∧ z) = x ∨ 0 = x 6= y = y ∧ 1 = y ∧ (x ∨ z).

Thus N5 is neither a modular nor distributive lattice.This gives rise to a question: does the implication ‘distributive ⇒ modular’ hold forevery lattice?

Theorem 2.12. Any distributive lattice is modular.

Proof. Suppose L is a distributive lattice and x, y, z ∈ L. Then x ≤ y means x ∧ y = xbut we have also seen that it means x ∨ y = y. Now the distributive law statesx ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) = y ∧ (x ∨ z). And thus we have satisfied the modularlaw.

The next theorem proved by R. Dedekind gives an easy way to identify modular lat-tices using N5 as in Figure 2.3. He also uses the fact that an embedding of a latticeL′ into another lattice L can be seen as if L contains a copy of L′ where the latticeoperations of L hold.

Theorem 2.13 (Dedekind). L is a nonmodular lattice if and only if N5 can be embeddedinto L.

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Proof. [1, Theorem 3.5]

G. Birkhoff used this theorem to prove another theorem which gives a characterizationof distributive lattices. Since we will be looking at these lattices, we will state thetheorem here.

Theorem 2.14 (Birkhoff). L is a nondistributive lattice if and only if M5 or N5 can beembedded into L.

Proof. [1, Theorem 3.6]

2.3 Boolean and Heyting algebras

Now that we introduced distributive lattices, let us look at some special distributivelattices: Boolean and Heyting algebras. We will start this section with the definitionsand then we will discuss some properties of these algebras. At the end of the thesis wewill look at classes of Boolean and Heyting algebras.

Definition 2.15. A Boolean algebra is an algebra 〈B,∧,∨,′ , 0, 1〉 with two binary, oneunary, and two nullary operations which satisfy:

• 〈B,∧,∨〉 is a distributive lattice

• x ∧ 0 ≈ 0; x ∨ 1 ≈ 1

• x ∧ x′ ≈ 0; x ∨ x′ ≈ 1.

In short, a Boolean algebra is a distributive lattice with a top element 1 and a bottomelement 0 and where every element has a unique complement. In Figure 2.4 someexamples are shown.

0

1

0

a a′

1

0

c′ b′ a′

a b c

1

Figure 2.4: Some Boolean algebras

Now that we have seen some Boolean algebras, let us define Heyting algebras.

Definition 2.16. An algebra 〈H,∧,∨,→, 0, 1〉 with three binary and two nullary oper-ations is a Heyting algebra if it satisfies:

(i) 〈H,∧,∨〉 is a distributive lattice

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(ii) x ∧ 0 ≈ 0; x ∨ 1 ≈ 1

(iii) x→ x ≈ 1

(iv) (x→ y) ∧ y ≈ y; x ∧ (x→ y) ≈ x ∧ y

(v) x→ (y ∧ z) ≈ (x→ y) ∧ (x→ z); (x ∨ y)→ z ≈ (x→ z) ∧ (y → z).

The binary operation → sends two elements x and y to the element x→ y. In Figure2.5 we determine this element for some cases.

0 = 1→ 0

1

0

x = 1→ x

1

0

x = y → x y = x→ 0

1

0

y = x→ 0 z

w x = y → z

1

0

1 1

0

Figure 2.5: Some Heyting algebras

When determining a→ b when given two elements a and b the following theorem canbe very useful. To prove it we first need some auxiliary lemmas.

Lemma 2.17. In any Heyting algebra the following holds:

(i) x→ (y ∧ x) ≈ x→ y

(ii) a ≤ b implies x→ a ≤ x→ b.

Proof. (i) We have

x→ (y ∧ x) ≈ (x→ y) ∧ (x→ x) (by (v) of 2.16)

≈ (x→ y) ∧ 1 (by (iii) of 2.16)

≈ x→ y (Since for all a we have a ≤ 1 thus a ∧ 1 = a)

(ii) Suppose a ≤ b then we have that x→ a ≈ x→ (a∧ b) since a ≈ a∧ b follows fromthe fact that a ≤ b. Now by (v) of Definition 2.16 it follows that a ≤ b implies(x→ a) ∧ (x→ b).

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With these properties we can prove the following theorem.

Theorem 2.18. If 〈H,∧,∨ →, 0, 1〉 is a Heyting algebra and a, b ∈ H then a→ b is thelargest element c of H such that a ∧ c ≤ b.

Proof. Let us start with the easy part. One of the properties of a Heyting algebra isx∧ (x→ y) ≈ x∧ y. So the first part follows directly from the properties of the Heytingalgebra.Now we use the previous lemma to prove the other direction. So let us assume that(a ∧ c) ≤ b. Then by Lemma 2.17 (ii) a → (a ∧ c) ≤ (a → b). But by Lemma 2.17 (i)this means that (a→ c) ≤ (a→ b). Now, by the same property of the Heyting algebrawe used before, we find c ≤ (a→ b).Thus a→ b is indeed the largest element c in H such that a ∧ c ≤ b.

Lemma 2.19. If 〈H,∧,∨ →, 0, 1〉 is a Heyting algebra and a, b ∈ H then

a ≤ b if and only if a→ b = 1.

Proof. If a ≤ b then a = a ∧ b. Since (a ∧ b) ∧ 1 = a ∧ b we have that (a ∧ b) ∧ 1 ≤ b.Since 1 is the biggest element in H we find by Theorem 2.18 that a→ b = 1.Another way to prove this is the following:

a→ b = (a→ b) ∧ 1

= (a→ b) ∧ (a→ a)

= a→ (b ∧ a)

= a→ a

= 1

Now suppose that a→ b = 1 then again by Theorem 2.18 we find that a = a∧1 ≤ b.

In a Heyting algebra the infinite distributive law also holds as we see in the followinglemma. We will need this result later in the thesis when we determine subdirectlyirreducible Heyting algebras.

Lemma 2.20. For any Heyting algebra A, a set S ⊆ A and x ∈ A, the following holds:

x ∧∨

S =∨{x ∧ s | s ∈ S}.

Proof. The proof of the lemma can be found in [2, Proposition 2.2.7].

As the reader may have noticed, examples of Boolean algebras are also examples ofHeyting algebras. This is a fact that holds for every Boolean algebra.

Lemma 2.21. Let 〈B,∧,∨,′ , 0, 1〉 be a Boolean algebra. Then 〈B,∧,∨,→, 0, 1〉 is aHeyting algebra with a→ b = a′ ∨ b for a, b ∈ B.

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Proof. We have to check the five axioms (or identities as we will call them later) thathold for any Heyting algebra.Suppose x, y, z ∈ B then we can say the following:

(i) It follows from the definition of a Boolean algebra that 〈B,∧,∨〉 is a distributivelattice.

(ii) x ∧ 0 ≈ 0 and x ∨ 1 ≈ 1 are also axioms of Boolean algebras, so they hold.

(iii) x→ x ≈= x′ ∨ x = 1. This follows from the definition of ′ in a Boolean algebra.

(iv) Since 〈B,∧,∨〉 is a distributive lattice, we can use the absorption law to find(x→ y) ∧ y = (x′ ∨ y) ∧ y = y.Now for the other part, we use the distributivity of the lattice and the definitionof the complement ′ and the nullary operation 0 to see thatx ∧ (x→ y) = x ∧ (x′ ∨ y) = (x ∧ x′) ∨ (x ∧ y) = 0 ∨ (x ∧ y) = x ∧ y.

(v) Again from the distributivity we havex→ (y ∧ z) = x′ ∨ (y ∧ z) = (x′ ∨ y) ∧ (x′ ∨ z) = (x→ y) ∧ (x→ z).The last part of the last property uses the next claim.

Claim: In a Boolean algebra B : (x ∨ y)′ ≈ x′ ∧ y′.To see this, just checking the properties of the complement ′ in a Boolean algebrais enough. So suppose x, y ∈ B then we have

(x ∨ y) ∨ (x′ ∧ y′) = ((x ∨ y) ∨ x′) ∧ ((x ∨ y) ∨ y′)= ((x ∨ x′) ∨ y) ∧ (x ∨ (y ∨ y′))= 1.

and we obtain

(x ∨ y) ∧ (x′ ∧ y′) = (x ∧ (x′ ∧ y′)) ∨ (y ∧ (x′ ∧ y′))= ((x ∧ x′) ∧ y′) ∨ (x′ ∧ (y ∧ y′))= 0.

So now we can prove the last property of the Heyting algebras.This prove uses the claim as stated and it uses the distributivity of the latticeagain:(x∨ y)→ z = (x∨ y)′ ∨ z = (x′ ∧ y′)∨ z = (x′ ∨ z)∧ (y′ ∨ z) = (x→ z)∧ (y → z).

16

3 Varieties of universal algebras

We have introduced Boolean and Heyting algebras and we have discussed some of theirproperties. So now let us take a step back again and go back to universal algebra. Inthis chapter we will define the notion of varieties. The goal of this chapter is to obtaintwo characterizations of varieties, proved by Alfred Tarski and by Garrett Birkhoff,respectively In order to do this we will define congruences in the first section and wewill discuss the K-free algebra with some of its properties in the last.

3.1 Congruences

In this section we will discuss the notion of congruences. We will define them and we willdiscuss some of their properties. We will see that all congruences on a universal algebraform a lattice where ⊆ is the order. We will also give some examples of congruencelattices on Heyting algebras. We will end this section by defining the quotient algebra.

Definition 3.1. An equivalence relation on a set V is a subset R of V × V such thatfor all x, y, z ∈ V :

(reflexivity) (x, x) ∈ R;

(symmetry) If (x, y) ∈ R then (y, x) ∈ R;

(transitivity) If (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R.

Definition 3.2. Let A be an algebra. An equivalence relation θ on A is a congruence iffor every n-ary fundamental operation of A and a1, . . . , an, b1, . . . , bn ∈ A the followingholds:

〈ai, bi〉 ∈ θ =⇒ 〈fA(a1, . . . , an), fA(b1, . . . , bn)〉 ∈ θ.

Let us denote the set of all congruences on A by Con(A).

So a congruence on an algebra A is an equivalence relation on A with the extraproperty that a congruence is compatible with the operations of A. Now let us definethe meet and join of two congruences in order to deduce that the set of all congruencesCon(A) forms a lattice.

Definition 3.3. For two congruences θ and φ on an algebra A let us define

• θ ∧ φ := θ ∩ φ

• θ ∨ φ := θ ∪ (θ ◦ φ) ∪ (θ ◦ φ ◦ θ) ∪ (θ ◦ φ ◦ θ ◦ φ) ∪ . . . .

17

Here ◦ is the relational product defined by

〈a, b〉 ∈ θ ◦ φ⇐⇒ ∃c ∈ A : 〈a, c〉 ∈ θ, 〈c, b〉 ∈ φ.

Then Con(A) = 〈Con(A),∧,∨〉 is the congruence lattice of A.Every congruence lattice has a largest and smallest element. These are the trivial con-gruences on the algebra. On an algebra A, the smallest congruence is ∆ = {〈a, a〉 ∈ A2}and the largest one is ∇ = A× A.

In Figure 3.1 we show some of the Heyting algebras we have seen in the previoussection together with their congruence lattices.

0

1

∆

∇

0

x

1

∆

θ1

∇

0

x y

1

∆

θ2 θ3

∇

z

x y

1

0 ∆

θ5 θ6

θ4

∇

18

0

z y

x

1

θ9

θ7 θ8

∇

∆

0

z y

x w

1

∆

θ12 θ13

θ10 θ11

∇

Figure 3.1: Some Heyting algebras on the left with their congruences lattices on theright.

In Figure 3.1 θ1 to θ13 are defined as in Table 3.1.

θ1 ∆ ∪ {〈1, x〉, 〈x, 1〉}

θ2 ∆ ∪ {〈1, x〉, 〈x, 1〉, 〈y, 0〉, 〈0, y〉}θ3 ∆ ∪ {〈1, y〉, 〈y, 1〉, 〈x, 0〉, 〈0, x〉}

θ4 ∆ ∪ {〈a, b〉 | a, b ≥ z}θ5 ∆ ∪ {〈1, x〉, 〈x, 1〉, 〈y, z〉, 〈z, y〉}θ6 ∆ ∪ {〈1, y〉, 〈y, 1〉, 〈x, z〉, 〈z, x〉}

θ7 ∆ ∪ {〈a, b〉 | a, b ≥ y} ∪ {〈z, 0〉, 〈0, z〉}θ8 ∆ ∪ {〈a, b〉 | a, b ≥ z} ∪ {〈y, 0〉, 〈0, y〉}θ9 ∆ ∪ {〈1, x〉, 〈x, 1〉}

θ10 ∆ ∪ {〈a, b〉 | a, b ≥ y} ∪ {〈c, d〉 | c, d ≤ x}θ11 ∆ ∪ {〈a, b〉 | a, b ≥ z} ∪ {〈y, 0〉, 〈0, y〉}θ12 ∆ ∪ {〈1, w〉, 〈w, 1〉, 〈x, z〉, 〈z, x〉}θ13 ∆ ∪ {〈1, x〉, 〈x, 1〉, 〈w, z〉, 〈z, w〉}

Table 3.1

We know that an equivalence relation on a group or ring gives rise to a quotient.Since a congruence is an equivalence relation this also gives rise to a quotient. We willgeneralize this concept and define the quotient algebra via a congruence:

19

Definition 3.4. The quotient algebra of A by θ (notation: A/θ) is the algebra withA/θ, the set of all congruence classes of A by θ, as its underlying set. For any elementa ∈ A, we denote the congruence class of a via θ by a/θ. The operations of the quotientalgebra satisfy:

fA/θ(a1/θ, . . . , an/θ) = fA(a1, . . . , an).

Here a1, . . . , an ∈ A and f is an n-ary operation of A.

Let us end this section with a useful property of the congruence, the proof of thislemma is straightforward see, e.g. the book A Course in Universal Algebra by StanleyBurris and H.P. Sankappanavar [1, Theorem 5.9].

Lemma 3.5. Let A be an algebra and suppose θ, φ ∈ Con(A). Then the following areequivalent:

(i) θ ◦ φ = φ ◦ θ

(ii) θ ∧ φ = φ ◦ θ

(iii) θ ◦ φ ⊆ φ ◦ θ.

3.2 Varieties

In this section we will take another look at homomorphisms, discuss some of their prop-erties and define homomorphic images. Then we will define the notion of a subalgebraand after that we will define products of algebras. We will need these three notions inorder to give a definition of a variety. We will end this section with Tarski’s Theoremwhich states that whenever we have a class of algebras of the same type, we can first takeall products, then all subalgebras and then all homomorphic images to get the varietygenerated by this class.

3.2.1 Homomorphisms

When we think of homomorphisms, the first theorem that comes to mind is of coursethe homomorphism theorem. So in this subsection we will generalize this theorem forall universal algebras. Let us start with a recap of this theorem in group theory.

Theorem 3.6 (Homomorphism theorem (for groups)). Let G and G′ be groups, andlet f : G → G′ be a surjective homomorphism. Then there exists an isomorphismh : G/ker(f)→ G′ defined by f = h ◦ g. Here g is the natural map from G to G/ker(f).

In order to generalize this for all universal algebras we have to generalize the notionof the kernel and of the natural map. Let us start with the latter together with one ofits properties.In the previous section we have defined congruences. When given a congruence on auniversal algebra A we can define the natural map as follows.

20

Definition 3.7. Let A be an algebra and let θ be a congruence on A. The natural mapνθ : A→ A/θ is defined by νθ = a/θ. It sends an element a to its congruence class by θ.

Lemma 3.8. The natural map from an algebra to the quotient of the algebra by acongruence is a surjective homomorphism.

Proof. Recall that a mapping α : A→ B is called a homomorphism if α(fA(a1, . . . , an)) =fB(α(a1), . . . , α(an)). Given a congruence θ on A it is easy to see that the natural mapνθ as defined above is onto. So all we have to show is that νθ is a homomorphism. Tosee this, suppose f is an n-ary operation in the type of A and B and a1, . . . , an ∈ A.Then we find

νθ(fA(a1, . . . , an)) = fA(a1, . . . , an)/θ

= fA/θ(a1/θ, . . . , an/θ)

= fA/θ(νθ(a1), . . . , νθ(an)).

Thus νθ is a surjective homomorphism.

Now that we have defined the natural map, let us define the kernel of a homomorphism.

Definition 3.9. Let α : A → B be a homomorphism. Then the kernel of α is definedby

ker(α) = {〈a, b〉 ∈ A2 | α(a) = α(b)}

Lemma 3.10. Let α : A→ B be a homomorphism. Then ker(α) is a congruence on A.

Proof. Suppose A and B are algebras of the same type and α : A → B is a homomor-phism. To see that ker(α) is a congruence we have to check the following properties:reflexivity, symmetry, transitivity and compatibility with the operations of A. In orderto check this, suppose a, b, c ∈ A.

(reflexivity) Trivially we have α(a) = α(a) thus 〈a, a〉 ∈ ker(α).

(symmertry) If 〈a, b〉 ∈ ker(α) then α(a) = α(b) and of course α(b) = α(a) so we find 〈b, a〉 ∈ ker(α).

(transitivity) Suppose 〈a, b〉 ∈ ker(α) and 〈b, c〉 ∈ ker(α). Then α(a) = α(b) and α(b) = α(c)thus α(a) = α(c) and therefore 〈a, c〉 ∈ ker(α).

(compatibility) Now suppose f is an n-ary operation in the type of A and B. Also let us supposethat we have 〈ai, bi〉 ∈ ker(α) where 1 ≤ i ≤ n. Then we have α(ai) = α(bi) for alli and therefore we have

α(fA(a1, . . . , an)) = fB(α(a1), . . . , α(an))

= fB(α(b1), . . . , α(bn))

= α(fA(b1, . . . , bn)).

Therefore 〈fA(a1, . . . , an), fA(b1, . . . , bn)〉 ∈ ker(α).

21

Now we have obtained that ker(α) is indeed a congruence on A.

Now that we have generalized the notions of natural map and kernel we can state thehomomorphism theorem for universal algebras.

Theorem 3.11 (Homomorphism Theorem). Suppose α : A → B is an onto homomor-phism. Then there is an isomorphism β : A/ker(α)→ B defined by α = β ◦ ν. Here ν isthe natural map from A to A/ker(α).

Proof. A proof of this theorem can be found in [1, Theorem 6.12].

So when we take the natural homomorphism from an algebra A to A/ker(α) for anysurjective homomorphism α from A to another algebra B, then we obtain an isomorphismbetween A/ker(α) and B. This is exactly what the homomorphism theorem tells us.Now let us define a homomorphic image of an algebra.

Definition 3.12. Let α : A→ B be a homomorphism. Then α(A) is the homomorphicimage of A by α.

Note that when α is in addition onto, then α(A) = B and we have that B is ahomomorphic image of A. We denote the set of all homomorphic images of an algebraA by H(A). Analogous we denote the set of all isomorphic images of an algebra A byI(A).

3.2.2 Subalgebras

The second ingredient for creating a variety is the notion of a subalgebra. In this sectionwe give a definition of a subalgebra and we give an example of how this notion can beused.

Definition 3.13. Let A and B be two algebras of type F . Then B is a subalgebra ofA if B ⊆ A and for every n-ary f ∈ F and b1, . . . , bn ∈ B we have fB(b1, . . . , bn) =fA(b1, . . . , bn) i.e. every operation of B is the restriction of the corresponding operationof A.

We can relate the definition of homomorphic image and subalgebra as follows.

Lemma 3.14. Suppose α : A→ B is a one-to-one homomorphism. Then the homomor-phic image of A is a subalgebra of B.

Proof. By definition of α it follows that α(A) ⊆ B. For the second property, supposef is an n-ary operation of the type of A and B and suppose a1, . . . , an ∈ A. Thenα(a1), . . . , α(an) ∈ α(A) and thus α(a1), . . . , α(an) ∈ B. Now we find

fα(A)(α(a1), . . . , α(an)) = fB(α(a1), . . . , α(an))

and hence α(A) is a subalgebra of B.

Let us denote the set of all subalgebras of an algebra A by S(A).

22

3.2.3 Products

The last notion we need in order to start defining varieties is products. We will start thissubsection by defining a (direct) product of two algebras together with the projectionmap. Then we will generalize this to the product of more (possibly infinitely many)algebras. At the end of this subsection we will show one of the properties of a productwe are going to need later in this thesis.

Definition 3.15. Let A1 and A2 be two algebras of the same type. Define the (direct)product A1 × A2 to be the algebra whose set is A1 × A2, and such that for every n-aryoperation and ai ∈ A1, a

′i ∈ A2 we have

fA1×A2(〈a1, a′1〉, . . . , 〈an, a′n〉) = 〈fA1(a1, . . . , an), fA2(a′1, . . . , a

′n)〉.

The product defined above gives rise to a map that projects an element to its ith

coordinate. We call this map the projection of the algebra.

Definition 3.16. For i ∈ {1, 2} the mapping

πi :A1 × A2 → Ai

〈a1, a2〉 7→ ai

is called the projection map on the ith coordinate of the product.

Lemma 3.17. The projection map is an onto homomorphism.

Proof. The fact that πi is onto follows directly from the definition. To see that πi is ahomomorphism, let a1, . . . , an ∈ A1, a′1, . . . , a

′n ∈ A2 and suppose f is an n-ary operation

then for i = 1 we find

π1(fA1×A2(〈a1, a′1〉, . . . , 〈an, a′n〉)) = π1(〈fA1(a1, . . . , an), fA2(a′1, . . . , a

′n)〉)

= fA1(a1, . . . , an)

= fA1(π1(〈a1, a′1〉), . . . , π1(〈an, a′n〉)).

We can do the same for every i = 2 and then obtain that the projection map is indeeda homomorphism.

Now let us generalize the notion of product to a product of any family of algebras.

Definition 3.18. Let {Ai}i∈I be an indexed family of algebras of the same type. The(direct) product A =

∏i∈I Ai is an algebra with set

∏i∈I Ai and with the operations

defined coordinate-wise:

fA(a1, . . . , an)(i) = fAi(a1(i), . . . , an(i)).

Here i ∈ I, f is n-ary and a1, . . . , an ∈∏

i∈I Ai.And we also have projection maps as defined before:

πj :∏i∈I

Ai → Aj

23

Given an indexed family of algebras of the same type K = {Ai}i∈I , let us define theset of all products of algebras in K by P (K). We will need the next property when wegive a characterization of subdirectly irreducible algebras in the last chapter.

Lemma 3.19. For an indexed family of maps αi : A → Ai, i ∈ I, the following areequivalent:

(i) The map α : A→∏

i∈I Ai defined coordinate-wise by α(a)(i) = αi(a) is injective.

(ii)⋂i∈I ker(αi) = ∆.

Proof. Suppose a1, a2 ∈ A such that a1 6= a2, then

α is injective ⇔ α(a1) 6= α(a2)

⇔ ∃i ∈ I : α(a1)(i) 6= α(a2)(i)

⇔ ∃i ∈ I : αi(a1) 6= αi(a2)

⇔ ∃i ∈ I : 〈a1, a2〉 6∈ ker(ai)

⇔⋂i∈I

ker(αi) = ∆.

Thus α is injective if and only if⋂i∈I ker(αi) = ∆ which is what we had to show.

3.2.4 Tarski’s Theorem

Now that we have established the definitions of homomorphic images, subalgebras andproducts, we can define the notion of varieties. In this subsection we will discuss differentclasses of algebras and some of their properties that will lead up to Tarski’s Theorem.

Definition 3.20. A nonempty class K of algebras of the same type is called a varietyif it is closed under homomorphic images, subalgebras and direct products.

In the previous subsections we already have encountered the following notions in thespecial case where the class consists of only one algebra. We will now generalize thisnotion to classes consisting of more algebras.

Definition 3.21. Let K be a class of algebras of the same type. ThenA ∈ H(K) if and only if A is a homomorphic image of some member of K.A ∈ I(K) if and only if A is isomorphic to some member of K.A ∈ S(K) if and only if A is a subalgebra of some member of K.A ∈ P (K) if and only if A is a direct product of a nonempty family of algebras in K.

From one class we can get bigger classes by applying H, I, S, or P to another classand in this way we can expand a class by first applying one operation and then anotheretc. In the following lemma we will find that some of the operations create bigger classesthan others.

Lemma 3.22. For any class K of algebras of the same type we have

24

• SH(K) ⊆ HS(K)

• PH(K) ⊆ HP (K)

• PS(K) ⊆ SP (K)

• For the operations H, S and IP the idempotent law holds.

Proof. Suppose K is a class of algebras of the same type.

• SH(K) ⊆ HS(K)Let A an algebra in SH(K). Then there exists an algebra B ∈ H(K) such that A isa subalgebra of B. Since B ∈ H(K) there also exists an algebra C ∈ K such that Bis a homomorphic image of C i.e. there exists an onto homomorphism α : C→ B.We have to prove that A ∈ HS(K) i.e. there exists an algebra D ∈ S(K) and anonto homomorphism β : D→ A and there exists an algebra E ∈ K such that D isa subalgebra of E.Claim: α−1(A) is a subalgebra of C.It is easy to see that α−1(A) = {c ∈ C | α(c) ∈ A} ⊆ C. So for the secondcondition, let f be an n-ary operation and c1, . . . , cn ∈ α−1(A) then

α(fα−1(A)(c1, . . . , cn)) = fA(α(c1), . . . , α(cn)) (α is a homomorpism)

= fB(α(c1), . . . , α(cn)) (A is a subalgebra of B)

= α(fC(c1, . . . , cn)). (α is a homomorpism)

So we have proved the claim. Now since α(α−1(A)) = A and α is a homomorpism,we have established that A ∈ HS(K).

• PH(K) ⊆ HP (K)Suppose A is an algebra in PH(K). Then there exist Ai ∈ H(K) such thatA =

∏Ai. Thus there are Bi ∈ K such that there are αi : Bi → Ai that are onto

homomorphisms. Now let us define

α : B→ Aα(b)(i) 7→ αi(b(i)).

Where B =∏

Bi. Now if α is an onto homomorphism, then A ∈ HP (K). Sincefor all i, αi is a surjective homomorphism, we find that α is also surjective and wehave for b1, . . . , bn ∈ B:

α(fB(b1, . . . , bn))(i) = αi(fBi(b1(i), . . . , bn(i)))

= fAi(αi(b1(i)), . . . , αi(bn(i)))

= fAi(α(b1)(i), . . . , α(bn)(i))

= fA(α(b1), . . . , α(bn))(i).

Thus α is an onto homomorphism.

25

• PS(K) ⊆ SP (K)Given an algebra A ∈ PS(K) then there exist Ai ∈ S(K) such that A =

∏Ai.

This yields that there exist Bi ∈ K such that for every i, Ai is a subalgebra of Bi.Now let us define B to be

∏Bi then A ∈ SP (K) if A is a subalgebra of B. Since

for every i we have Ai is a subalgebra of Bi thus for every i we have Ai ⊆ Bi whichimplies A =

∏Ai ⊆

∏Bi = B and for f an n-ary operation we have

fA(a1, . . . , an)(i) = fAi(a1(i), . . . , an(i))

= fBi(a1(i), . . . , an(i))

= fB(a1, . . . , an)(i).

Here a1, . . . , an ∈ A. Therefore we have that A ∈ SP (K).

• H, S and IP are idempotentIn order to see this, let A be an algebra.First suppose A ∈ HH(K) then there exists an algebra B ∈ H(K) and an ontohomomorphism α : B→ A i.e. there exists an algebra C ∈ K and a homomorphismβ : C→ B that is onto. Now let us show that α◦β is also an onto homomorphism,because then A ∈ H(K) and we find that for the operation H the idempotent lawholds.Since α and β are both onto we have that α ◦ β is onto as well. Now supposec1, . . . , cn ∈ C then from the fact that both α and β are homomorphisms it followsthat

(α ◦ β)(fC(c1, . . . , cn)) = α(β(fC(c1, . . . , cn)))

= α(fB(β(c1), . . . , β(cn)))

= fA(α(β(c1)), . . . , α(β(cn)))

= fA((α ◦ β)(c1), . . . , (α ◦ β)(cn).

Thus α ◦ β is also a surjective homomorphism and A ∈ H(K).Now suppose A ∈ SS(K) then there exists an algebra B ∈ S(K) such that A is asubalgebra of B. Thus there is an algebra C ∈ K such that B is a subalgebra of Cwhile A is a subalgebra of B. Now if we can show that A is also a subalgebra of Cthen we have established that SS is idempotent as well. Since A is a subalgebraof B and B is a subalgebra of C we have A ⊆ B and B ⊆ C thus by transitivity of⊆ we also have A ⊆ C. Now suppose a1, . . . , an ∈ A then for f an n-ary operationwe have fA(a1, . . . , an) = fB(a1, . . . , an) = fC(a1, . . . , an) hence A is a subalgebraof C.To see that IP is also idempotent, suppose that A ∈ IPIP (K). Then there existsan algebra B ∈ PIP (K) that is isomorphic to A. Thus there exist Bi ∈ IP (K)such that B =

∏Bi. Therefore there exist Ci ∈ P (K) such that for every i,

Ci∼= Bi. Therefore there exist Cij ∈ K such that for every i we have Ci =

∏Cij.

So we haveA ∼= B =

∏Bi ∼=

∏Ci =

∏∏Cij.

26

Hence A ∼=∏∏

Cij. Since the product of a product is again a product we nowhave A ∈ IP (K).

All algebras of the same type form a variety and since the intersection of any classof varieties is also a variety we know that there exists a smallest variety. Let us denoteV (K) for the smallest variety that contains K. We will end this section with a theoremfirst proved by Alfred Tarski in 1946 [3, pp 163-165].

Theorem 3.23 (Tarski). Let K be a class of algebras of the same type then we have

V (K) = HSP (K).

Proof. In order to prove this theorem we have to show two things: V (K) ⊆ HSP (K)and V (K) ⊇ HSP (K) where K is a class of algebras of the same type. Since V (K) is avariety we have that V is closed under H, S and P and thus HSP (K) ⊆ V (K) followsdirectly from the definition.To see that HSP (K) is a variety (and thus V (K) ⊆ HSP (K)) we have to show thatHSP (K) is closed under homomorphic images, subalgebras and direct products. ByLemma 3.22 we know that the operation H is idempotent thus HHSP (K) = HSP (K)and we find that HSP (K) is closed under homomorphic images.From Lemma 3.22 we also know that the operation S is idempotent and SH(K) ⊆ HS(K).Therefore we have SHSP (K) ⊆ HSSP (K) = HSP (K). Thus HSP (K) is also closedunder subalgebras.In order to see that HSP (K) is closed under products as well we use Lemma 3.22 againand find

PHSP (K) ⊆ HPSP (K) (PH(K) ⊆ HP (K))

⊆ HSPP (K) (PS(K) ⊆ SP (K))

⊆ HSIPIP (K) ( The identity map is an isomorphism)

= HSIP (K) (IP is idempotent)

⊆ HSHP (K) (any isomorphism is a homomorphism)

⊆ HHSP (K) (SH(K) ⊆ HS(K))

= HSP (K).

Hence HSP (K) is closed under homomorphic images, subalgebras and products andthus HSP (K) is a variety. Now since HSP (K) ⊆ V (K) and V (K) is the smallestvariety that contains K we have established the equality V (K) = HSP (K).

3.3 Term algebras

Now that we have defined varieties, we will use the rest of this chapter to characterizethese classes as Garrett Birkhoff did in his paper in 1935 [4, pp 433-454]. He stated that

27

a class of algebras K is a variety if and only if K is an equational class. This theoremis known as one of Birkhoff’s most famous theorems. In this paper we call it Birkhoff’sTheorem.In order to fully understand what the theorem states we have to give a definition of anequational class. For this we first have to define terms in order to be able to define termalgebras. We will use this to construct identities and then we can give a definition of anequational class.For the full proof of Birkhoff’s Theorem we also need to understand the notion of thefree algebra and some of its properties. For now, let us start by defining terms.

Definition 3.24. Given a type of algebras F . Let X be a set of variables. The setT (X) of terms of F over X is the smallest set such that

(i) X ∪ {f | f is a nullary operation in F} ⊆ T (X)

(ii) T (X) is closed under all the operations of F .

For p in T (X) we say that p is n-ary if there are n or less variables occurring in p andwe write p(x1, . . . , xn).

The notion of terms is syntactic, just like the operations in a type: only the interpre-tation within an algebra has a meaning. So let us look at the meaning of a term in analgebra.

Definition 3.25. Given an algebra A and a term p(x1, . . . , xn) ∈ T (X) of the sametype, let us define the map pA : An → A inductively:

(i) If p(x1, . . . , xn) = xi for 1 ≤ i ≤ n, then

pA(a1, . . . , an) = ai

and pA is the ith projection map.

(ii) If p = f(p1(x1, . . . , xn), . . . , pk(x1, . . . , xn)) for f a k-ary operation, then

pA = fA(pA1 (x1, . . . , xn), . . . , pAk (x1, . . . , xn)).

pA is the interpretation of the term p in the algebra A.

Now let us state some properties of terms. The proof can be found in [1, Theorem10.3].

Lemma 3.26. For two algebras A and B of the same type and p ∈ T (X) n-ary, thefollowing holds:

(i) Let θ be a congruence on A and suppose 〈ai, bi〉 ∈ θ for 1 ≤ i ≤ n. Then

〈pA(a1, . . . , an), pA(b1, . . . , bn)〉 ∈ θ.

28

(ii) If α : A→ B is a homomorphism, then

α(pA(a1, . . . , an)) = pB(α(a1), . . . , α(an)).

The set of terms also gives rise to an algebra: the term algebra.

Definition 3.27. Given a set X of variables and a type F of algebras, if T (X) 6= ∅ thenT(X) is the term algebra of F over X. The underlying set of this algebra is T (X) andthe operations satisfy:

fT(X) : T (X)n → T (X)

(p1, . . . , pn) 7→ f(p1, . . . , pn).

Here f ∈ F is n-ary and p1, . . . , pn ∈ T (X).

Note that for T(X) to exist, either X or the set of nullary operations within the typehas to be non-empty.Remark that T(X) is generated by the set X of variables i.e. T(X) is the smallestalgebra such that the underlying set of T(X) contains X. The term algebra is one ofthe algebras for which the universal mapping property holds. Let us first define thisproperty and finish this section with a prove of this statement.

Definition 3.28. Let K be a class of algebras of the same type F and let U(X) be analgebra generated by a set X. The algebra U(X) also has type F . If for every A ∈ Kand for every map

α : X → A

there is a homomorphism

β : U(X)→ Ax 7→ α(x) for all x ∈ X.

Then we say that β extends α and that U(X) has the universal mapping property for Kover X. Moreover, X is called the set of free generators of U(X), and U(X) is said tobe freely generated by X.

Note that for the universal mapping property to hold the map α need not be a homo-morphism, α can be any map.

Theorem 3.29. For any type of algebras F and any set of variables X, the term algebraT(X) has the universal mapping property for the class of all algebras of type F over X.

Proof. Suppose A is an algebra of type F and α : X → A is a map. Now let us defineanother map β : T(X)→ A as follows:

β(x) = α(x) for all x ∈ X;

β(p(x1, . . . , xn)) = p(β(x1), . . . , β(xn));

β(fT(X)(p1(x1, . . . , xn), . . . , pk(x1, . . . , xn))) = fA(β(p1(x1, . . . , xn)), . . . , β(pk(x1, . . . , xn))).

By definition β is a homomorphism.

29

3.4 Identities

In some literature, identities are called ‘equations’ or ‘axioms’, in this thesis however, wewill stick to the term ‘identities’. In this section we will define the notion of identities anddiscuss some properties. We will end this section with a proof of one side of Birkhoff’sTheorem.

Definition 3.30. Given a type of algebras F and a set X of variables. Let p, q ∈ T (X).An identity of F over X is an expression of the form

p ≈ q.

Let us define Id(X) to be the set of all identities of the type over X.

Now we will define when an identity holds in an algebra.

Definition 3.31. An identity p ≈ q holds in an algebra A if the interpretations of pand q are the same in A. Thus

A |= p(x1, . . . , xn) ≈ q(x1, . . . , xn)⇔ pA(a1, . . . , an) = qA(a1, . . . , an)

for (a1, . . . , an) ∈ A.When an identity holds in an algebra we write

A |= p ≈ q

and we say that p ≈ q is true in A. A class K of algebras satisfies p ≈ q if each memberof K satisfies p ≈ q, notation: K |= p ≈ q.For a set of identities Σ we say that K satisfies Σ if K |= p ≈ q for every identityp ≈ q ∈ Σ and we write K |= Σ.Given a class of algebras K and a set X of variables, define

IdK(X) = {p ≈ q ∈ Id(X) | K |= p ≈ q}.

We can characterize this as follows:

Lemma 3.32. Given a class of algebras K of type F , a set of variables X and anidentity of F over X, p ≈ q, then

K |= p ≈ q

if and only if for every A ∈ K and for every homomorphism α : T(X)→ A we have

α(p) = α(q).

Proof. Suppose A ∈ K, p and q are n-ary terms and α : T(X)→ A is a homomorphism.Then by definition of |= we have K |= p ≈ q if and only if for every B ∈ K we haveB |= p ≈ q. Since A ∈ K this implies that A |= p ≈ q. Now this tells us that for all

30

a1, . . . , an ∈ A we have pA(a1, . . . , an) = qA(a1, . . . , an). Since α is a homomorphism, by(ii) of Lemma 3.26 we have

α(pT(X)(x1, . . . , xn)) = pA(α(x1), . . . , α(xn))

= qA(α(x1), . . . , α(xn))

= α(qT(X)(x1, . . . , xn)).

Hence α(p) = α(q).Now for the other direction again suppose A ∈ K and p and q are n-ary terms. Further-more assume that a1, . . . , an ∈ A. Now by Theorem 3.29 there exists a homomorphismα : T(X)→ A such that α(xi) = ai for 1 ≤ i ≤ n. Then we have

pA(a1, . . . , an) = pA(α(x1), . . . , α(xn))

= α(pT(X)(x1, . . . , xn))

= α(p)

= α(q)

= α(qT(X)(x1, . . . , xn))

= qA(α(x1), . . . , α(xn)

= qA(a1, . . . , an).

Hence K |= p ≈ q.

Now that we have defined identities (or equations as some might say), we can defineequational classes:

Definition 3.33. Given a type of algebras F and a set of identities Σ of F . Define

M(Σ) = {A | A |= Σ}.

A class K of algebras is an equational class if there is a set of identities Σ such thatK = M(Σ). In this case we say that K is defined or axiomatized by Σ.

This is all we need to understand Birkhoff’s Theorem which states that for every classof algebras K we have that K is an equational class if and only if K is a variety. Wewill prove one direction right now. In order to be able to prove the other direction wewill need some more theory. We will prove it at the end of this chapter.

Theorem 3.34 (Birkhoff part 1). Every equational class is a variety.

Proof. Let K be an equational class. Thus there exists a set of identities Σ such thatM(Σ). A variety is a class of algebras which is closed under the operations H, S andP . So we have to show that M(Σ) is closed under these operations.

31

H Suppose A ∈ HM(Σ) i.e. there exists a B ∈ M(Σ) and a homomorphismα : B→ A which is onto.What we want is to show that A satisfies Σ. So let p ≈ q ∈ Σ and a1, . . . , an ∈ A.Since α is onto, we have that there exist b1, . . . , bn ∈ B such that αbi = ai for1 ≤ i ≤ n. B ∈ M(Σ) thus pB(b1, . . . , bn) = qB(b1, . . . , bn) holds. So by Lemma3.32 we have

pA(a1, . . . , an) = pA(αb1, . . . , αbn)

= αpB(b1, . . . , bn)

= αqB(b1, . . . , bn)

= qA(αb1, . . . , bn)

= qA(a1, . . . , an).

Thus A |= Σ and therefore A ∈M(Σ).

S Suppose A ∈ SM(Σ) i.e. there exists a B ∈ M(Σ) such that A is a subalgebraof B. So there exists an embedding ι : A → B defined by the restriction of theidentity map to A.The identity map is a homomorphism thus again by Lemma 3.32 we have

pA(a1, . . . , an) = ι(pA(a1, . . . , an))

= pB(a1, . . . , an)

= qB(a1, . . . , an)

= ιqA((a1, . . . , an))

= qA(a1, . . . , an).

Now A |= Σ and thus A ∈M(Σ).

P Suppose A ∈ PM(Σ) i.e. there exist Ai ∈ M(Σ) such that A =∏

i∈I Ai forsome set I. Since for every i ∈ I the projection map πi : A → Ai is a surjectivehomomorphism we have

pA(a1, . . . , an)(i) = πi(pA(a1, . . . , an))

= pAi(πi(a1), . . . , πi(an))

= pAi(a1(i), . . . , an(i))

= qAi(a1(i), . . . , an(i))

= qAi(πi(a1), . . . , πi(an))

= πi(qA(a1, . . . , an))

= qA(a1, . . . , an)(i).

This holds for every i ∈ I so A |= p ≈ q and A ∈M(Σ).

Thus M(Σ) is closed under homomorphic images, subalgebras and product and now wehave found that every equational class is indeed a variety.

32

3.5 The K-free algebra

In order to prove the second part of Birkhoff’s Theorem, we need K-free algebras. Justlike the term algebras, these algebras have the universal mapping property. We will alsodiscuss some other properties we will need in order to prove the theorem. We will endthis section by proving the second part of Birkhoff’s Theorem. For now let us start withthe K-free algebras.

Definition 3.35. Let K be a class of algebras of the same type. Given a set X ofvariables define the congruence θK(X) on T(X) to be

θK(X) =⋂{φ ∈ Con(T(X)) | T(X)/φ ∈ IS(K)}.

Now define the K-free algebra over X/θK(X), write X̄, by

FK(X̄) = T(X)/θK(X).

The K-free algebra is the algebra satisfying precisely the same identities as the classK. The K-free algebra is of the same type as the class K and has T (X)/θK(X) as itsunderlying set. It satisfies the universal mapping property for K over X̄ which was firstproved by Garrett Birkhoff. One proof of this theorem can be found in [1, Theorem10.10].

Theorem 3.36 (Birkhoff). Suppose T(X) exists. Then FK(X̄) has the universal map-ping property for K over X̄.

The next corollary follows from this fact. It states that for any algebra A in a class Kconsisting of algebras of the same type, there is a surjective homomorphism α : FK(X̄)→A. Thus if K is a variety, then K = V (FK(X̄)) = H(FK(X̄)).

Corollary 3.37. For a class K of algebras of the same type, A ∈ K, and a large enoughset of variables X, we have A ∈ H(FK(X̄)).

Proof. Suppose |X̄| ≥ |A| and let α : X̄ → A be a surjective map. Since FK(X̄)has the universal mapping property for K over X̄, there exists a homomorphism β :FK(X̄) → A that extends α. Because of this, β is also surjective and we deduce thatA ∈ H(FK(X̄)).

The following property of the K-free algebra, also first proved by Garrett Birkhoff,states that for every variety V , the V -free algebra FV (X̄)), is an element of V . One canfind the proof in [1, Theorem 10.12]

Theorem 3.38 (Birkhoff). If T(X) exists, then for any nonempty set K of algebras ofthe same type, FK(X̄) ∈ ISP (K).

We stated before that the K-free algebra satisfies exactly the same identities as Kdoes. We will prove this among other facts in the following theorem.

33

Theorem 3.39. Given a class K of algebras of type F and terms p, q ∈ T (X) also oftype F , then we have

K |= p ≈ q

⇔FK(X̄) |= p ≈ q

⇔p/θK(X) = q/θK(X) in FK(X̄)

⇔〈p, q〉 ∈ θK(X).

Proof. Suppose p = p(x1, . . . , xn), q = q(x1, . . . , xn) ∈ T (X). Then K |= p ≈ q ifand only if for every A ∈ K we have A |= p ≈ q. Since by Theorem 3.38 we havethat the K-free algebra is an element from the variety generated by K. From thisit follows that FK(X̄) |= p ≈ q. (One can find this proof in [1, Lemma 11.3].) Forr1, . . . , rn ∈ T (X) we find that FK(X̄) |= p ≈ q if and only if pFK(X̄)(r1/θK , . . . , rn/θK) =qFK(X̄)(r1/θK , . . . , rn/θK). From (i) of Lemma 3.26 we have that terms are compatiblewith congruences and therefore we have pFK(X̄)(r1, . . . , rn)/θK = qFK(X̄)(r1, . . . , rn)/θKand hence p/θK = q/θK in FK(X̄). Thus in FK(X̄) we have that the congruence classby θK of p and of q are the same. Therefore 〈p, q〉 ∈ θK . The last part of the proof(〈p, q〉 ∈ θK ⇒ K |= p ≈ q follows from Lemma 3.32 and the second isomorphismtheorem which lies out of the scope of this thesis. This theorem with its proof can befound in [1, Theorem 6.15].

Now that we have established this we are able to prove the second part of Birkhoff’sTheorem.

Theorem 3.40 (Birkhoff part 2). Every variety is an equational class.

Proof. Let K be a variety. We want to show that there exists a set of identities Σ suchthat K = M(Σ).To see this we have to show two things: K ⊆ M(Σ) and K ⊇ M(Σ) for some set Σ ofidentities. Let us start with the first of these two. For any set of variables X we knowthat K |= IdK(X), so let us take an infinite set of variables for X. Since K |= IdK(X)by definition, we know that K ⊆M(IdK(X)).Now if K ⊇M(IdK(X)) also holds, then we have found our Σ and we are done.So suppose A ∈ M(IdK(X)). By Theorem 3.34 we know that M(IdK(X)) is a variety.By Corollary 3.37 we find that A ∈ H(FK(X̄)) since we have chosen our X to be infinitelylarge.If we can prove that IdK(X) = IdM(IdK(X))(X) then by Theorem 3.39 we find that

〈p, q〉 ∈ θK ⇐⇒ FK(X̄) |= p ≈ q

⇐⇒ K |= p ≈ q ⇐⇒M(IdK(X)) |= p ≈ q

⇐⇒ FM(IdK(X)) |= p ≈ q

⇐⇒ 〈p, q〉 ∈ θM(IdK(X)).

Thus θK = θM(IdK(X)) and now it follows that FK = FM(IdK(X)).Since A ∈ H(FM(IdK(X))) we now have that A ∈ H(FK). By assumption K is a variety,

34

so we can use Theorem 3.38 to find that A ∈ K, which means that we are done.So all that is left to do is to prove that IdK(X) = IdM(IdK(X))(X).First let us show that IdK(X) ⊆ IdM(IdK(X))(X).For this, let p ≈ q ∈ IdK(X), then we find that K |= p ≈ q so for all A ∈ K we have thatA |= p ≈ q. But then p ≈ q ∈ {p ≈ q | A |= IdK(X) ⇒ A |= p ≈ q} since A |= IdK(X)and A |= p ≈ q. We also have

{p ≈ q | A |= IdK(X)⇒ A |= p ≈ q} = {p ≈ q | {A|A |= IdK(X)} |= p ≈ q}= {p ≈ q |M(IdK(X) |= p ≈ q}= IdM(IdK(X))

So we find that whenever p ≈ q ∈ IdK(X), we find that p ≈ q ∈ M(IdK(X)) andtherefore the first part is proved.Now for the second part, IdK(X) ⊇ IdM(IdK(X))(X), we know thatM(IdK(X)) |= (IdM(IdK(X))(X)which means that for all A ∈ M(IdK(X)) we have A |= IdM(IdK(X))(X). In the firstpart of the proof we have shown that K ⊆ M(IdK(X)) so it follows that for all A ∈ Kwe also have A |= IdM(IdK(X))(X). We now have established that K |= IdM(IdK(X))(X)and since K |= IdK(X), we now know that IdM(IdK(X))(X) ⊆ IdK(X).Therefore IdM(IdK(X))(X) = IdK(X) and K = M(IdK(X)) thus K is an equationalclass.

Together with the first part of Birkhoff’s Theorem, this leads to the full theorem.

Corollary 3.41. For a class K of algebras of the same type we have:

K is a variety⇐⇒ K is an equational class

Proof. This result follows directly from Theorems 3.34 and 3.40.

35

4 Subdirectly irreducible algebras

In this chapter we will look at the building blocks of varieties: the subdirectly irreduciblealgebras. We will give a characterization of subdirectly irreducible universal algebras interms of congruences. After that we will give a concrete characterization of subdirectlyirreducible Boolean and Heyting algebras.In the first section we will define them and find out why they are useful. In the nextsection we will give a couple of characterizations of subdirectly irreducible algebras invarieties of Boolean and Heyting algebras. We will look at characterizations via thecongruence lattice (as defined in Section 3.1) and then we will define the notion offilters and use these obtain a characterization of the subdirectly irreducible Boolean andHeyting algebras. At the end of this section the reader should be able to determinesubdirectly irreducible Boolean and Heyting algebras by just looking at their Hassediagrams.

4.1 Building blocks of varieties

In the first chapter we have introduced direct products. Before we can start definingsubdirectly irreducible algebras we need to define the subdirect product. From thiswe can define a subdirect embedding, which is what we need for the definition of asubdirectly irreducible algebra. The definition is as follows:

Definition 4.1. An algebra A is a subdirect product of some family of algebras (Ai)i ∈ Iif the following holds:

(i) A is a subalgebra of∏

i∈I Ai and

(ii) for every i ∈ I : πi(A) = Ai.

We call an embedding subdirect α : A →∏

i∈I Ai if α(A) is a subdirect product of(Ai)i∈I .

A2

A1

A A2

A1

A A2

A1

A

Figure 4.1: The second property of the subdirect product only holds in the last case.

36

To get a better understanding of the second condition of the subdirect product, let usturn to Figure 4.1. There one can find a schematic drawing of a product of two algebrasA1 and A2 with an algebra A in it in grey. On the left the projection of A is not equalto any of the Ai for i = 1, 2. In the middle drawing the property also does not holdbecause the projection has to be equal to Ai for both i = 1 and i = 2. On the right theprojection is equal to both algebras of the product and therefore the second propertyholds.Now we know how the second property works, let us look at an example of a subdirectproduct.

Example 4.2. Consider the Heyting algebra A:

z

x y

1

0

Figure 4.2: A

and let A1 and A2 the following Heyting (Boolean) algebras:

0

1

(a) A1

0

x y

1

(b) A2

Figure 4.3

Then the product A1 × A2 is the following Heyting (Boolean) algebra:

0

b z c

x a y

1

Figure 4.4: A1 × A2

37

Now we find that A ⊆ A1 × A2 and for 1, x, y, z, 0 ∈ A1 × A2 we have

1 ∧ 1 = 1 1 ∨ 1 = 1 1→ 1 = 1 z → 1 = 11 ∧ x = x 1 ∨ x = 1 1→ x = x z → x = 11 ∧ y = y 1 ∨ y = 1 1→ y = y z → y = 11 ∧ z = z 1 ∨ z = 1 1→ z = z z → z = 11 ∧ 0 = 0 1 ∨ 0 = 1 1→ 0 = 0 z → 0 = 0x ∧ x = x x ∨ x = x x→ 1 = 1 0→ 1 = 1x ∧ y = z x ∨ y = 1 x→ x = 1 0→ x = 1x ∧ z = z x ∨ z = x x→ y = y 0→ y = 1x ∧ 0 = 0 x ∨ 0 = x x→ z = z 0→ z = 1y ∧ y = y y ∨ y = y x→ 0 = 0 0→ 0 = 1y ∧ z = z y ∨ z = y y → 1 = 1y ∧ 0 = 0 y ∨ 0 = y y → x = xz ∧ z = z z ∨ z = z y → y = 1z ∧ 0 = 0 z ∨ 0 = z y → z = z0 ∧ 0 = 0 0 ∨ 0 = 0 y → 0 = 0

Thus A is closed under ∧, ∨ and→ as a subset of A1×A2 and therefore A is a subalgebraof A1 × A2. So now we only have to check the second condition.It is easy to see that π1(A) = A1 and π2(A) = A2. And therefore A is a subdirectembedding of A1 and A2.

We have now reached the point that we can define the main subject of this thesis:subdirectly irreducible algebras.

Definition 4.3. An algebra A is subdirectly irreducible if there exists an i ∈ I such that

πi ◦ α : A→ Ai

is an isomorphism.Here α : A→

∏i∈I Ai is any subdirect embedding.

In his book [5, Proposition 4.4, page 576] Pierre Antoine Grillet shows that only thecyclic groups of order pn where p is a prime number and n an integer or the Pruffergroup are subdirectly irreducible Abelian groups. The theory we need to see this liesout of the scope of this thesis. So instead we will look at the subdirectly irreducibleBoolean and Heyting algebras. In the next section we will prove some characterizations.

The proof of the following result can be found in an article written by Garrett Birkhoffin 1944 [6, pp. 764-768].

Theorem 4.4 (Birkhoff). Let A be an algebra. Then A is isomorphic to a subdirectproduct of subdirectly irreducible algebras which are homomorphic images of A.

The theorem states the following:Let KA

si denote the set of subdirectly irreducible algebras that are homomorphic imagesof A thus

KAsi = {Asi ∈ H(A) | Asi is subdirectly irreducible}.

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Then there exist Ai ∈ KAsi such that A ∼= B where B is a subalgebra of

∏Ai and for all

i, πi(B) = Ai.We will now concentrate at a very useful corollary of Birkhoff’s Theorem.

Corollary 4.5. Every variety is determined by its subdirectly irreducible members.

Proof. Given a variety V let us define

Vsi = {A ∈ V | A is subdirectly irreducible}

the set of the subdirectly irreducible members of V . We need to prove that V =HSP (Vsi).In order to prove this, we only have to show that V ⊆ HSP (Vsi) since V ⊇ HSP (Vsi)follows from the fact that Vsi ⊆ V and that V is a variety and thus closed under homo-morphic images, subalgebras and direct products.Now suppose V is a variety and that A ∈ V . Then by the previous theorem we havethat there exist Ai ∈ KA

si such that A ∼= B where B is a subalgebra of∏

Ai and for alli, πi(B) = Ai.Claim: KA

si ⊆ Vsi.To see this, let us suppose that C ∈ KA

si. Then the subdirectly irreducible C is a homo-morphic image of A i.e. C ∈ H(A). Since V is closed under homomorphic images andA ∈ V , we deduce that C ∈ Vsi.So for all Ai we have that Ai ∈ Vsi. By definition, B is a subalgebra of

∏Ai and thus

B ∈ SP (Vsi). Now since A is isomorphic to B, we have obtained that A ∈ HSP (Vsi).

Birkhoff’s Theorem (Theorem 3.41) states that every variety is an equational class.So any set of algebras of the same type that is defined by a set of identities is a varietyand every variety has a set of identities that define it.

Corollary 4.6. Let V be a variety an suppose p ≈ q is an identity of the same type asthe elements in the variety, then

V |= p ≈ q if and only if Vsi |= p ≈ q.

Proof. ⇒ follows from the fact that Vsi ⊆ V , so we only need to show ⇐.But that is a direct result of the fact that identities are preserved under homomorphicimages, subalgebras and products, this finishes the proof.

So when we want to know whether an identity holds on a variety, it is enough to knowwhether the identity holds on all subdirectly irreducible members of this variety. Thisis why we would like to know which algebras are subdirectly irreducible and which onesare not. A characterization is given in the next section.

39

4.2 Determining subdirectly irreducible algebras

The goal of this section is to find subdirectly irreducible algebras in different varieties.So we would like to find out when an algebra is subdirectly irreducible. In order to dothis we can use the following characterization.

Theorem 4.7. An algebra A is subdirectly irreducible if an only if A is trivial or thereis a minimum congruence in Con(A)−∆.

Proof. (⇒) We will prove this by contraposition. So suppose that A is a non-trivialalgebra such that there is no minimum congruence in Con(A)−∆. So if we takethe intersection of all congruences except for ∆ we get

⋂(Con(A)−∆) = ∆. Now

for θi ∈ Con(A)−∆, let us define:

βi :A→ A/θia 7→ a/θi.

Then for all i, βi is a natural map and by Lemma 3.8 we find that every βi is asurjective homomorphism and ker(βi) = θi. Thus⋂

(Con(A)−∆) =⋂

(θi)

=⋂

(ker(βi))

= ∆.

Now we can use Lemma 3.19 to see that the map

β :A→∏

θi∈Con(A−∆)

A/θi

β(a)(i) = βi(a)

is injective. Therefore β as defined above is an embedding and β(A) is a subalgebraof

∏θi∈Con(A−∆) A/θi i.e. β(A) ≤

∏θi∈Con(A−∆) A/θi. Since for each i we have

πi(β(A) = A/θi by definition of β, we also have that β(A) is a subdirect productof

∏θi∈Con(A−∆) A/θi and we find that βi is surjective for every i. But since βi is

not one-to-one, A is not subdirectly irreducible.

(⇐) Suppose A is an algebra, then there are two cases to consider for the proof:

(i) A is trivial and

(ii) A is non-trivial but there exists a minimum congruence in Con(A)−∆.

So let us start with the case that A is trivial.

(i) A is trivial.This means that A consists of only one element, thus A = {a}. Now supposeα : A→

∏i∈I Ai is a subdirect embedding for some set I. Then by definition

40

πi(α(A)) = Ai and since A = {a} we have that πi(α(a)) = Ai. This meansthat Ai also consists of only one element and thus Ai is a trivial algebra. Nowit is easy to wee that (πi ◦ α) : A → Ai is an isomorphism for every i andfrom that it follows that A is subdirectly irreducible.

(ii) A is non-trivial but there exists a minimum congruence in Con(A)−∆.Then

⋂(Con(A)−∆) 6= ∆ so let us define θ to be

θ =⋂

(Con(A)−∆).

Now for α : A →∏

i∈I Ai a subdirect embedding, we know the following: αis injective, α is a homomorphism and ∀i : πi(α(A)) = Ai.Thus for all 〈a, b〉 ∈ θ with a 6= b we have α(a) 6= α(b) hence there exists ani ∈ I such that α(a)(i) 6= α(b)(i) and thus for this i we have (πi ◦ α)(a) 6=(πi ◦α)(b). Therefore 〈a, b〉 6∈ ker(πi ◦α) and since 〈a, b〉 ∈ θ we now find thatθ 6⊆ ker(πi ◦α). But θ =

⋂(Con(A)−∆) thus ker(πi ◦α) = ∆ must hold and

πi ◦ α is injective. Now by definition of the subdirect embedding α we findthat for some i ∈ I, πi ◦ α is an isomorphism and we have shown that A issubdirectly irreducible.

Figure 4.5

What the theorem states is that the congruence lattice of a subdirectly irreduciblealgebra looks like the one in Figure 4.5. So all we need to do in order to determinewhether an algebra A is subdirectly irreducible is to calculate the congruence lattice ofA.Let us prove a property of congruences which makes this an easier task to perform.

Lemma 4.8. Let θ be a congruence on an algebra A. For a, b, c ∈ A with a ≤ c ≤ b and〈a, b〉 ∈ θ we have 〈a, c〉 ∈ θ.

Proof. By assumption 〈a, b〉 ∈ θ holds and by reflexivity we also have 〈c, c〉 ∈ θ. Nowby the compatibility of the congruence we have 〈a ∧ c, b ∧ c〉 ∈ θ. But since a ≤ c ≤ bwe now find 〈a, c〉 ∈ θ.

We would like to determine the subdirectly irreducibility of some Boolean and Heytingalgebras. We have seen that we could do this via the congruence lattice. But calculatingthese lattices can be an extensive procedure. This is why we introduce the followingdefinition.

41

Definition 4.9. A filter on a Boolean or Heyting algebra A is a nonempty subset ofF ⊆ A such that the following holds for a, b ∈ A:

(i) a, b ∈ F ⇒ a ∧ b ∈ F

(ii) a ∈ F and a ≤ b⇒ b ∈ F .

Lemma 4.10. The set of filters on a Boolean or Heyting algebra A forms a lattice where

F1 ∧ F2 = F1 ∩ F2

F1 ∨ F2 = {a ∈ A | a ≥ a1 ∧ a2 for some a1 ∈ F1 and a2 ∈ F2}.

The lattice of all filters of A will be denoted by Fil(A).

Proof. To see this we have to check (i)-(iv) from Definition 2.7. Let F1, F2 and F3 befilters on a Boolean or Heyting algebra. Then for ∧ the first three properties are trivial.So let us focus on the properties for ∨ and then check the last property also for ∧.The first property follows from the fact that in a lattice (thus in a Boolean or Heytingalgebra) the commutative law holds so we have:

F1 ∨ F2 = {a ∈ A | ∃a1 ∈ F1, a2 ∈ F2 such that a ≥ a1 ∧ a2}= {a ∈ A | ∃a1 ∈ F1, a2 ∈ F2 such that a ≥ a2 ∧ a1}= F2 ∨ F1.

In the same way, the associativity of the filters follows from the associativity in thelattice.For the third property, we have the following:

F1 ∨ F1 = {a ∈ A | ∃a1 ∈ F1, a′1 ∈ F1 such that a ≥ a1 ∧ a′1}

= {a ∈ A | ∃a′′1 ∈ F1 such that a ≥ a′′1}= F1.

This follows from the fact that a filter is upwards closed.The absorption laws are the hardest to prove, so we will prove these in parts. The firstlaw F1 ∧ (F1 ∨ F2) ⊆ F1 follows straight from Definition 4.10 (i). Now let us proveF1 ∧ (F1 ∨ F2) ⊇ F1 by contraposition. So assume that a 6∈ F1 ∧ (F1 ∨ F2) then eithera 6∈ F1 and we are done or ∀a1 ∈ F1, a2 ∈ F2 we have a < a1 ∧ a2. Assume thata ∈ F1 then we find a < a2 ∧ a which is a contradiction. So a 6∈ F1 and we haveF1 ∧ (F1 ∨ F2) = F1.The other absorption law we will prove the same way: first we will prove F1∨(F1∧F2) ⊆F1 and then we will show F1 ∨ (F1 ∧ F2) ⊇ F1.

a ∈ F1 ∨ (F1 ∧ F2) =⇒ ∃a1 ∈ F1, a′ ∈ F1 ∩ F2 such that a ≥ a1 ∧ a′

=⇒ ∃a1 ∈ F1, a′ ∈ F1 such that a ≥ a1 ∧ a′

=⇒ ∃a′1 ∈ F1 such that a ≥ a′1=⇒ a ∈ F1.

42

So F1 ∨ (F1 ∧ F2) ⊆ F1 follows by Definition 4.9 (ii).Now we will prove the other side by contraposition again. So suppose a 6∈ F1∨ (F1∧F2)then for every a1 ∈ F1 and a′ ∈ F1 ∩ F2 we have a < a1 ∧ a′. But since a1 ∧ a′ ≤ a1 wenow find a < a1 for all a1 ∈ F1, which means that a is a proper lower bound of F1 thusa 6∈ F1.Since we have now proven all the properties of a lattice for the filters, we have establishedthe fact that the filters on Boolean or Heyting algebras form a lattice too.

Because the filters form a lattice we can use a Hasse diagram to see what the latticeof filters looks like. Let us do this for some Boolean and Heyting algebras:

0

1

{1}

{1, 0}

0

x

1

{1}

{1, x}

{1, x, 0}

0

x y

1

{1}

{1, x} {1, y}

{1, x, y, 0}

z

x y

1

0 {1}

{1, x} {1, y}

{1, x, y, z}

{1, x, y, z, 0}

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0

z y

x

1

{1, x}

{1, x, y} {1, x, z}

{1, x, y, z, 0}

{1}

0

z y

x w

1

{1}

{1, w} {1, x}

{1, w, x, y} {1, x, z}

{1, w, x, y, z, 0}

Figure 4.6: Some Boolean and Heyting algebras on the left with their filter lattices onthe right.

Note that the linear Heyting algebras look exactly like their filter lattices. It is easyto see that this in fact holds for all linear Heyting algebras.As the reader may have noticed, the filter lattices look similar to the congruence latticesof these algebras as in Section 3.1. Since the filters are much easier to calculate than thecongruences we would like to use this fact in determining whether a Heyting or Booleanalgebra is subdirectly irreducible or not. In the proof of the theorem we will need thefollowing lemma:

Lemma 4.11. Let A be a Heyting algebra. For w, x, y, z ∈ A the following holds:

(i) (x→ y) ∧ (y → z) ≤ (x→ z);

(ii) (w → x) ∧ (y → z) ≤ (w ∧ y)→ (x ∧ z);

(iii) (w → x) ∧ (y → z) ≤ (w ∨ y)→ (x ∨ z);

(iv) (w → x) ∧ (y → z) ≤ (x→ y)→ (w → z).

Proof. Suppose w, x, y, z ∈ A and S ⊆ A. By Theorem 2.18 we have that x → y isthe largest element z such that x ∧ z ≤ y and by Lemma 2.20 we can say x ∧

∨S =∨

{x ∧ s | s ∈ S}. We will use this to prove all of the properties above:

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(i)

(x→ y) ∧ (y → z) =∨{a | a ∧ x ≤ y} ∧

∨{b | b ∧ y ≤ z}

=∨{a ∧ b | a ∧ x ≤ y, b ∧ y ≤ z}

≤∨{c | c ∧ x ≤ z}

= x→ z.

(ii)

(w → x) ∧ (y → z) =∨{a ∧ b | a ∧ w ≤ x, b ∧ y ≤ z}

≤∨{c | (w ∧ y) ∧ c ≤ x ∧ z}

= (w ∧ y)→ (x ∧ z).

(iii)

(w → x) ∧ (y → z) =∨{a ∧ b | a ∧ w ≤ x, b ∧ y ≤ z}

≤∨{a ∧ b | a ∧ w ≤ x ∨ z, b ∧ y ≤ x ∨ z}

=∨{a | a ∧ w ≤ x ∨ z} ∧

∨{b | b ∧ y ≤ x ∨ z}

= (w → (x ∨ z)) ∧ (y → (x ∨ z))

= (w ∨ y)→ (x ∨ z).

(iv)

(w → x) ∧ (y → z) =∨{a ∧ b | a ∧ w ≤ x, b ∧ y ≤ z}

≤∨{c | {d ∧ c | d ∧ x ≤ y} ⊆ {e | e ∧ w ≤ z}}

≤∨{c |

∨{d ∧ c | d ∧ x ≤ y} ≤

∨{e | e ∧ w ≤ z}}

=∨{c |

∨{d | d ∧ x ≤ y} ∧ c ≤

∨{e | e ∧ w ≤ z}}

=∨{c | (x→ y) ∧ c ≤ (w → z)}

= (x→ y)→ (w → z).

Here the first ≤ follows from the fact that given d ∧ x ≤ y, and b ∧ y ≤ z we have

a ∧ w ≤ x⇒ (a ∧ w) ∧ d ≤ x ∧ d⇒ (a ∧ w) ∧ d ≤ y

⇒ ((a ∧ w) ∧ d) ∧ b ≤ y ∧ b⇒ (d ∧ (a ∧ b)) ∧ w = ((a ∧ w) ∧ d) ∧ b ≤ z.

45

And we know that

a ∧ b ∈ {c | {d ∧ c | d ∧ x ≤ y} ⊆ {e | e ∧ w ≤ z}}⇔ {d ∧ (a ∧ b) | d ∧ x ≤ y} ⊆ {e | e ∧ w ≤ z}⇔ (d ∧ (a ∧ b)) ∧ w ≤ z.

Thus (w → x) ∧ (y → z) ≤ (x→ y)→ (w → z) holds in any Heyting algebra.

Theorem 4.12. Let A be a Boolean or Heyting algebra. Then

Con(A) ∼= Fil(A).

Proof. To see this we will create two well-defined maps α : Con(A) → Fil(A) andβ : Fil(A)→ Con(A) and show that these maps are each others inverses i.e. β ◦ α =IdCon(A) and α ◦ β = IdFil(A). After that we will show that both α and β are homomor-phisms which enforces Con(A) ∼= Fil(A).So let us start with defining

α :Con(A)→ Fil(A) β :Fil(A)→ Con(A)

θ 7→ 1/θ F 7→ θF .

Here we define θF by〈a, b〉 ∈ θF ⇐⇒ a↔ b ∈ F.

Now we first have to show that α and β are both well-defined i.e. given a congruence θ onA, then 1/θ defines a filter on A and given a filter F on A, then θF defines a congruenceon A. Note that α(∆) = {1} and α(∇) = A, also β({1}) = ∆ and β(A) = ∇.

• α : Con(A)→ Fil(A) is well-defined.Suppose θ is a congruence on A, we want to show that the congruence class of1, 1/θ is a filter. We will prove this by checking the definition. A filter F is anonempty subset of A such that

i) F is closed under the operation ∧ and

ii) F is upwards closed.

It is trivial that 1/θ is a nonempty subset of A so let us check the other conditions.

i) If a, b ∈ 1/θ then we find that 〈a, 1〉 ∈ θ and 〈b, 1〉 ∈ θ. Since θ is a congruence,we know that θ is compatible with ∧ and thus 〈a∧ b, 1∧1〉 ∈ θ. By definitionof the nullary operation 1 we now have 〈a ∧ b, 1〉 ∈ θ which means thata ∧ b ∈ 1/θ thus θ is closed under the ∧-operation.

ii) If a ∈ 1/θ then it follows directly from Lemma 4.8 that for any b ≥ a, b ∈ 1/θalso.

Thus 1/θ the congruence class of 1 in θ on A is a filter and α is therefore a well-defined map.

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• β : Fil(A)→ Con(A) is well-defined.Suppose F is a filter on A, then we want θF to be a congruence on A. We willshow this by checking the properties of a congruence:

(reflexivity) This follows by definition since a↔ a = 1 and 1 ∈ F thus 〈a, a〉 ∈ θ.(symmetry) Suppose 〈a, b〉 ∈ θF thus a↔ b ∈ F and therefore b↔ a ∈ F so we find that

〈b, a〉 ∈ θF which is what we wanted.

(transitivity) Given 〈a, b〉 ∈ θF and 〈b, c〉 ∈ θF . So we know that a↔ b ∈ F and b↔ c ∈ F .But that means that (a → b) ∧ (b → a) ∈ F and (b → c) ∧ (c → b) ∈ F andthus ((a → b) ∧ (b → a)) ∧ ((b → c) ∧ (c → b)) ∈ F . By commutativity wenow have ((a → b) ∧ (b → c)) ∧ ((c → b) ∧ (b → a)) ∈ F . The result nowfollows from Lemma 4.11 (i).

And we find that θF is an equivalence relation on A. We wanted θF to be acongruence so we still have to check the compatibility of θF with ∧, with ∨ andwith →.

(compatibility ∧) Suppose 〈a, b〉 ∈ θF and 〈c, d〉 ∈ θF . Then a→ b ∈ F , b→ a ∈ F , c→ d ∈ Fand d → c ∈ F . Since F is a filter, it is closed under ∧ and therefore(a→ b) ∧ (c→ d) ∈ F and (b→ a) ∧ (d→ c) ∈ F . Now by Lemma 4.11 (ii)we find that (a ∧ c) → (b ∧ d) ∈ F and (b ∧ d) → (a ∧ c) ∈ F which meansthat 〈a ∧ c, b ∧ d〉 ∈ θF . And therefore θF is compatible with ∧.

(compatibility ∨) Now suppose 〈a, b〉 ∈ θF and 〈c, d〉 ∈ θF . Then we know that (a→ b) ∧ (c→ d)∈ F and (b → a) ∧ (d → c) ∈ F and by Lemma 4.11 (iii) we now find that(a ∨ c)→ (b ∨ d) ∈ F and (b ∨ d)→ (a ∨ c) ∈ F . We now have showed that〈a ∨ c, b ∨ d〉 ∈ θF and therefore θF is compatible with ∨.

(compatibility →) When given 〈a, b〉 ∈ θF and 〈c, d〉 ∈ θF , we know (b→ a) ∧ (c→ d) ∈ F and(a→ b) ∧ (d→ c) ∈ F . Now we can use Lemma 4.11 (iv) and find that also(a → c) → (b → d) ∈ F and (b → d) → (a → c) ∈ F which means that〈a→ c, b→ d〉 ∈ θF and thus that θF is compatible with →.

From this it follows that θF as defined above is a congruence and thus β : Fil(A)→ Con(A)is well-defined.Now that we know that both α and β are well-defined, we will show that they are eachothers inverses. So let us begin with proving β ◦ α = IdCon(A):

• β ◦ α = IdCon(A).For θ a congruence on A, we want (β◦α)(θ) = θ1/θ to be the congruence we startedwith: θ. We will prove this in two parts: first we will show θ ⊆ θ1/θ and then wewill show θ1/θ ⊆ θ.For a, b ∈ A, such that 〈a, b〉 ∈ θ we also have 〈a, a〉 ∈ θ and 〈b, b〉 ∈ θ by re-flexivity of the congruence. Since the congruence is compatible with → we nowfind 〈a → b, b → b〉 ∈ θ and 〈a → a, b → a〉 ∈ θ. Recall that x → x ≈ 1 holdsin a Heyting algebra, so we now have found 〈a → b, 1〉 ∈ θ and 〈1, b → a〉 ∈ θ.Any congruence on A is symmetric and closed under ∧ therefore we now know

47

〈(a → b) ∧ (b → a), 1〉 ∈ θ thus (a → b) ∧ (b → a) ∈ 1/θ. We have just seen that1/θ is a filter on A and thus 〈a, b→〉 ∈ θ1/θ if and only if (a→ b)∧ (b→ a) ∈ 1/θ.So now we have shown that θ ⊆ θ1/θ.Now for the other direction, suppose 〈a, b〉 ∈ θ1/θ. Then by definition we havea→ b ∈ 1/θ and b→ a ∈ 1/θ thus we have 〈a → b, 1〉 ∈ θ and 〈b → a, 1〉 ∈ θ.Again by reflexivity and compatibility of θ with ∧ we now deduce 〈(a→ b) ∧ a, 1 ∧ a〉∈ θ and 〈(b → a) ∧ b, 1 ∧ b〉 ∈ θ. By definition of the Heyting algebra we have(x→ y) ∧ x ≈ x ∧ y and 1∧x ≈ x. So now we use this and find that 〈a∧ b, a〉 ∈ θand 〈a ∧ b, b〉 ∈ θ. Now by symmetry and transitivity of θ we have shown that〈a, b〉 ∈ θ.With this we have proven that θ = θ1/θ and thus that β ◦ α = IdCon(A).

• α ◦ β = IdFil(A).We have to show that F = 1/θF for any filter F on A. So suppose F is a filter onA, and a ∈ A, then we have

a ∈ 1/θF ⇔ 〈a, 1〉 ∈ θF⇔ a↔ 1 ∈ F⇔ (a→ 1) ∧ (1→ a) ∈ F⇔ 1 ∧ a ∈ F⇔ a ∈ F

and we find that F = 1/θF . Now we have shown α ◦ β = IdFil(A).

Now we know that the maps α and β are each others inverses, all we need to do nowis prove that they are homomorphisms to get the result we want. So let us start byshowing that α is a homomorphism of lattices.

• α is a lattice homomorphism.For α to be a homomorphism we have to show for θ, φ ∈ Con(A): α(θ ∧ φ) =α(θ) ∧ α(φ) and α(θ ∨ φ) = α(θ) ∨ α(φ).It is not hard to see that

α(θ ∧ φ) = α(θ ∩ φ)

= 1/(θ ∩ φ)

= 1/θ ∩ 1/φ

= α(θ) ∧ α(φ).

The second equality follows from the fact that a ∈ 1/(θ ∩ φ) if and only if 〈a, 1〉 ∈(θ ∩ φ). But this holds if and only if 〈a, 1〉 ∈ θ and 〈a, 1〉 ∈ φ and thereforea ∈ 1/θ ∩ 1/φ.Thus we only need to show now is α(θ ∨ φ) = α(θ) ∨ α(φ).We know that α(θ ∨ φ) = 1/((θ) ∪ (θ ◦ φ) ∪ (θ ◦ φ ◦ θ) ∪ . . . ) and α(θ) ∨ α(φ) ={x ∈ A | x ≥ a∧b, a ∈ 1/θ, b ∈ 1/φ}. Therefore for x ∈ A, x ∈ α(θ∨φ) if and only

48

if x ∈ 1/((θ)∪(θ◦φ)∪(θ◦φ◦θ)∪. . . ) if and only if 〈x, 1〉 ∈ ((θ)∪(θ◦φ)∪(θ◦φ◦θ)∪. . . ).So for x ∈ A, x ∈ α(θ ∨ φ) if and only if either 〈x, 1〉 ∈ θ or 〈x, 1〉 ∈ θ ◦ φ or〈x, 1〉 ∈ θ ◦ φ ◦ θ or . . . .In the first case x ∈ 1/θ and since 1 ∈ 1/φ we have x ∧ 1 = x ∈ 1/θ ∨ 1/φ.The second case, 〈x, 1〉 ∈ θ ◦ φ means that there exists some c ∈ A such that〈x, c〉 ∈ θ and 〈c, 1〉 ∈ φ. Now x ↔ c ∈ 1/θ and c ↔ 1 ∈ 1/φ and 1 ∈ 1/φand we know that (x ↔ c) ∧ ((c ↔ 1) ∧ 1) ∈ (1/θ ∨ 1/φ). We also know that(x ↔ c) ∧ ((c ↔ 1) ∧ 1) = (x ↔ c) ∧ (c ∧ 1) = (x ↔ c) ∧ c = x ∧ c. And sincex ≥ x ∧ c, we now have that x ∈ 1/θ ∨ 1/φ.We can universalize this fact and say the following:

x ∈ 1/((θ) ∪ (θ ◦ φ) ∪ (θ ◦ φ ◦ θ) ∪ . . . )⇔ ∃c1, . . . , cn = 1 such that

(x↔ c1) ∧∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 1, i even } ∈ 1/θ and∧

{ci ↔ ci+1 | 1 ≤ i ≤ n− 1, i odd } ∈ 1/φ.

From this we know that

((x↔ c1)∧∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 1, i even })∧∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 1, i odd }

= (x↔ c1) ∧∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 1}

= (x↔ c1) ∧ (∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 1} ∧ 1)

= ((x↔ c1) ∧∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 2}) ∧ ((cn−1 ↔ 1) ∧ 1)

= ((x↔ c1) ∧∧{ci ↔ ci+1 | 1 ≤ i ≤ n− 2}) ∧ (cn−1 ∧ 1)

= . . .

= x ∧∧{ci | 1 ≤ i ≤ n}

≤ x

And we have found that x ∈ α(θ ∨ φ) = 1/((θ) ∪ (θ ◦ φ) ∪ (θ ◦ φ ◦ θ) ∪ . . . ) if andonly if x ∈ {x ∈ A | x ≥ a ∧ b, a ∈ 1/θ, b ∈ 1/φ} = α(θ) ∨ α(φ).Therefore α is a homomorphism of lattices.

• β is a lattice homomorphism.Suppose F1 and F2 are filters on A then we have

β(F1 ∧ F2) = β(F1 ∩ F2)

= θF1 ∩ θF2

= θF1 ∧ θF2

= β(F1) ∧ β(F2).

49

Since α is the inverse of β and we have already seen that α is a homomorphism,we also have

β(F1 ∨ F2) = β(α(θ1) ∨ α(θ2))

= β(α(θF1 ∨ θF2)

= θF1 ∨ θF2

= β(F1) ∨ β(F2).

Therefore β is also a homomorphism.

Since the lattice of filters of a Heyting algebra is isomorphic to the congruence latticeof the same Heyting algebra, the Hasse diagram of the filter lattice must be the same asin Figure 4.5.

Corollary 4.13. A Heyting algebra A is subdirectly irreducible if and only if A is trivialor there is a minimum filter in Fil(A)− {1}.

Proof. This follows directly from Theorem 4.7 and Theorem 4.12

Theorem 4.14. A Heyting algebra A is subdirectly irreducible if and only if A is trivialor there is a maximum element in A− 1.

Proof. Using contraposition we will prove that a Heyting algebra A is subdirectly irre-ducible implies that either A is trivial or there is a maximum element in A− 1.So suppose A is a nontrivial Heyting algebra with no maximum element in A− 1. Thenthere are two cases to consider:

(i) There are two elements a, b ∈ A− 1 such that a 6≤ b, b 6≤ a and for all x ∈ A− 1we have x > a implies x = 1 and x > b implies x = 1.

(ii) For all a ∈ A− 1, there is a b ∈ A− 1 such that b > a.

So let us start with the first case.

(i) There are two elements a, b ∈ A− 1 such that a 6≤ b, b 6≤ a and for all x ∈ A− 1we have x > a implies x = 1 and x > b implies x = 1. Then both {1, a} and {1, b}are filters on A and there are no filters smaller than these in Fil(A) − {1}. Nowby Corollary 4.13 we find that A cannot be subdirectly irreducible.

(ii) For all a ∈ A−1, there is a b ∈ A−1 such that b > a i.e. there is a chain a1, a2, . . .such that an < an+1 and

∨n∈N an = 1. But then there is a chain of filters Fn

defined byFn = {x ∈ A | x ≥ an}.

We find that Fn ⊃ Fn+1 for all n and that⋂Fn = {1}. Now again by Corollary

4.13 we have that A is not subdirectly irreducible.

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For the other direction, when A is trivial we are done (we can use Corollary 4.13 orTheorem 4.7 to see this).So suppose A is a non-trivial Heyting algebra such that there exists a maximum in A−1.Let us call this maximum a. Then Fa = {a, 1} is a filter on A. Since for any x ∈ A− 1we have x ≤ a thus for any filter F on A we have

x ∈ F ⇒ a ∈ F⇒ Fa ≤ F.

Therefore there exists a minimum filter in Fil(A)− {1}, namely Fa. By Corollary 4.13we now find that A is subdirectly irreducible.

Now we can identify the subdirectly irreducible Boolean and Heyting algebras in onelook:

0

x y

1

z

x y

1

0 0

z y

x w

1

Figure 4.7: Some examples of Boolean and Heyting algebras that are not subdirectlyirreducible

0

1

0

x

1

0

z y

x

1

Figure 4.8: Some subdirectly irreducible Boolean and Heyting algebras.

As we have seen the 2-element Boolean algebra is subdirectly irreducible. The follow-ing corollary states that it is the only subdirectly irreducible Boolean algebra.

Corollary 4.15. The only non-trivial subdirectly irreducible Boolean algebra is

51

Proof. Let B be a non-trivial subdirectly irreducible Boolean algebra.From Theorem 4.14 we know that a Heyting or Boolean algebra A is subdirectly irre-ducible if and only if there is a maximum element in A− 1.Suppose B is subdirectly irreducible. Let us denote the maximum element in B− 1 as xand suppose that x 6= 0. In a Boolean algebra every element has a unique complementso there exists an element x′such that x ∧ x′ = 0 and x ∨ x′ = 1. Now there are twopossible cases to consider for this x′: x′ = 1 and x′ ≤ x. In the first case we find thatx∧x′ = x and since x 6= 0 this leads to a contradiction. Therefore we must have x′ ≤ x.But then x ∧ x′ = x and since x ∈ B − 1 this also leads to a contradiction. Now wehave found that B is subdirectly irreducible if and only if x = 0 i.e. the only non-trivialsubdirectly irreducible Boolean algebra is the 2-element Boolean algebra.

Corollary 4.6 now states that whenever we want to know if an identity holds in thevariety of Boolean algebras, we only have to check whether it holds in the 2-elementBoolean algebra. This result has the following corollary for classical propositional logic.

Corollary 4.16. Classical propositional logic is complete with respect to the 2-elementBoolean algebra.

Proof. It is well known (e.g. [7, Example 3.4], [8, Theorem 5.5]) that a formula φ is atheorem of classical propositional logic if and only if the identity φ ≈ 1 holds in everyBoolean algebra. From Corollary 4.15 we deduce that φ ≈ 1 holds in every Booleanalgebra if and only if φ ≈ 1 holds in the 2-element Boolean algebra. Therefore φ isa theorem of classical propositional logic if and only if φ ≈ 1 holds in the 2-elementBoolean algebra.

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5 Conclusions

We started this thesis with an introduction to universal algebra. We introduced uni-versal algebras, in particular lattices and Boolean and Heyting algebras. One of theresults discussed is that every Boolean algebra is a Heyting algebra but not the otherway around.

Next we introduced varieties and gave two characterizations: one of them was proved byAlfred Tarski and stated that for any class of universal algebrasK we have V (K) = HSP (K).The other characterization was first proved by Garrett Birkhoff stating that a class ofalgebras is a variety if and only if it is an equational class.

In the last chapter we discussed the building blocks of varieties: subdirectly irreduciblealgebras. We gave a characterization of subdirectly irreducible universal algebras usingcongruences. From this we obtained a concrete characterization of subdirectly irreducibleBoolean and Heyting algebras via filters. As a result we deduced a direct characteriza-tion of the subdirectly irreducible Boolean and Heyting algebras.

It follows from this last result that in classical propositional logic is complete withwith respect to the 2-element Boolean algebra and therefore we can use the {0, 1}-truthtables to check validity of formulas in classical propositional logic.

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6 Popular summary

The natural numbers, we have all heard about them, we have all seen them and we allknow that some of them are special. Some of these natural number are prime numbers.They have the nice property that their only dividers are themselves and the number1. But there is something else about them. We can build all the other numbers fromthese prime numbers by multiplicating them with each other e.g. 36 = 2 × 2 × 3 × 3and 93 = 3× 31. The prime numbers could be called the building blocks of the naturalnumbers.The natural numbers together with the binary operation × and the nullary operation0 forms a group, which is an algebraic structure (or algebra). An algebra is a set ofelements, together with a number of operations. There are different types of algebrasevery type has its own operations, within a type the algebras may differ. For every typeand set we can define equations. When we have a class of algebras we can define byequations (an algebra is in the class if and only if all the equations hold on the algebra),then the class is called a variety.As well as the natural numbers, varieties contain building blocks, elements from whichwe can ‘build up’ all the other elements in the variety. These building blocks are calledsubdirectly irreducible algebras.When we know what these building blocks are we can build the other algebras in thevariety via the operations: H, S and P. These letters stand for Homomorphic images,Subalgebras and Products, respectively. For an algebra A to be a homomorphic imageof another algebra B means that there exists a map from A to B that preserves theoperations of the algebras and such that every element in B gets reached by the map.An algebra A is a subalgebra of B when there is a map from A to B such that every twodifferent elements in A are send to different elements in B and again the map preservesthe operations of the algebras.Whenever we have algebras Bi then A is a product of the B′is if the set of A is equal tothe product of the Bi and the operations of A are defined coordinate-wise.We can use H, S and P in any combination to get any algebra in the variety. When wefirst take all the products of the subdirectly irreducible algebras, then take all subalge-bras and then take all homomorphic images, then we get the whole variety and thereforethe subdirectly irreducible algebras are to varieties what prime numbers are to the nat-ural numbers.Now that we know this it would be nice to know which algebras in a variety are thesubdirectly irreducible ones.To give a characterization we define a binary relation on any algebra called the con-gruence. All congruences form an algebra and when we know what this algebra lookslike we can tell if the algebra on which the congruence is defined is in fact subdirectly

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irreducible. But determining all congruences on an algebra can be a lot of work and amistake can easily be made.Fortunately, in this thesis we will look at particular algebras: Boolean algebras andHeyting algebras for which finding the congruences can still be a lot of work but findingthe congruences is not necessary for there is another method: via filters. A downside isthat filters can only be defined on Boolean and Heyting algebras, but they are very easyto determine and we will show that the algebra of congruences is sort of the same algebraas the algebra we get when we have all filters on a Boolean or Heyting algebra. Fromthe characterization we had via the congruences, we can now give a characterizationvia filters and since filters are easy to work with, we can give a characterization of allsubdirectly irreducible Boolean and Heyting algebras.As a consequence ofthis we obtain that there is only one subdirectly irreducible Booleanalgebra and when we want to know if an identity holds in the variety of all Booleanalgebras, we only need to check if this identity holds in the subdirectly irreducible one.

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Bibliography

[1] S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, Springer-Verlag,1981

[2] N. Bezhanishvili, Lattices of intermediate and cylindric modal logics, Institute forLogic, Language and Computation, University of Amsterdam, 2006

[3] A. Tarski, A remark on functionally free algebras, Ann. of Math. 47, 1946

[4] G. Birkhoff, On the structure of abstract algebras, Proc. Camb. Philos. Soc. 31, 1935

[5] P.A. Grillet, Abstract algebra, Springer, 2007

[6] G. Birkhoff, Subdirect unions in universal algebra, Bull. Amer. Math. Soc. 50, 1944

[7] Y. Venema, Algebras and Coalgebras, Institute for Logic, Language and Computa-tion, University of Amsterdam, 2005

[8] P. Blackburn, M. de Rijke and Y. Venema, Modal Logic, Cambridge University Press,2001

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