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INTRODUCTION TO ABSTRACT ALGEBRA

G. JANELIDZE

Department of Mathematics and Applied Mathematics

University of Cape Town

Rondebosch 7701, Cape Town, South Africa

Last updated on 20 September 2013

This is an elementary course aiming to introduce very first concepts of abstract

algebra for third year students interested in pure mathematics. It is divided into the

following (two chapters and) sections:

I. First algebraic and related structures

1. Algebraic operations..........................................2 2. Magmas and unitary magmas.....................................3 3. Semigroups....................................................4 4. Monoids.......................................................5 5. Closure operators.............................................5 6. Equivalence relations.........................................7 7. Order relations...............................................8 8. Categories....................................................9 9. Isomorphism..................................................11 10. Initial and terminal objects.................................12 11. Algebras, homomorphisms, isomorphisms........................13 12. Subalgebras..................................................15 13. Products.....................................................16 14. Quotient algebras............................................16 15. Canonical factorization of homomorphisms.....................17 16. Classical algebraic structures...............................18 17. Quotient groups, rings, and modules..........................20

II. Lattices, semirings, number systems, and foundation of linear algebra

1. Commutativity................................................25

2. Semilattices.................................................26

3. Lattices.....................................................29

4. Distributivity and complements...............................31

5. Complete lattices............................................33

6. Boolean algebras.............................................34

7. Semirings and semimodules....................................37

8. The semiring ℕ of natural numbers............................38

9. Number systems, ℤ, ℚ, ℝ, and ℂ as rings, and their modules...40 10. Pointed categories...........................................46 11. Products and coproducts......................................47 12. Direct sums..................................................51 13. Free algebras and free semimodules...........................52 14. Vector spaces................................................55

We shall refer to these sections as follows: say, “Section 10” means “Section 10 in

this chapter”, while “Section I.10” means “Section 10 in Chapter I”, when we refer to

it in Chapter II.

I am grateful to Dr. Amartya Goswami for a number of misprint corrections.

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I. FIRST ALGEBRAIC AND RELATED STRUCTURES

1. Algebraic operations

For a natural number n and a set A, the set of all maps from {1,…,n} to A will be

denoted by An; this includes the case n = 0, where {1,…,n} becomes the empty set and

therefore An becomes a one-element set. For n = 1, the set A

n will be identified with A,

and for n = 2, 3,…, the elements of An will be written as n-member sequences

(a1,…,an) of elements in A.

Definition 1.1. For n = 0, 1, 2,…, an n-ary operation on a set A is a map from An to A.

We will also use special terms for small n’s:

0-ary nullary

1-ary unary

2-ary binary

3-ary ternary

Remarks and Conventions 1.2. (a) Since to give a nullary operation on A is the same

as to pick up an element in A, we will simply identify them; accordingly the nullary

operations are sometimes called constants.

(b) If is an n-ary operation with n = 2, 3,…, we will write (a1,…,an) instead of

((a1,…,an)). Furthermore, for n = 2, instead of (a1,a2) we usually write a1a2, or

simply a1a2 – especially when we think of as a kind of multiplication.

Binary operations play an especially important role. When the ground set A has a

small number of elements, it is convenient to define binary operations on it with

tables, such as:

a b

a a a

b a b

where A = {a,b} and aa = ab = ba = a, bb = b, or

a b

a b b

b b b

where A = {a,b} again, but now aa = ab = ba = bb = b. That is, when A has n

elements, the table contains n 1 rows and n 1 columns with the following entries

in their cells:

the first cell is blank, and the rest of the first row lists the elements of A in any

order;

the rest of the first column lists the elements of A (preferably) in the same order;

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for i and j both greater than 1, the cell on the intersection of i-row and j-column has

the element ab in it – where a stays in the i-th cell of the first column and b stays in

the j-th cell of the first row.

Definition 1.3. Let be a binary operation on a set A, and let us write (a,b) = ab, as

above. The operation is said to be

(a) associative, if a(bc) = (ab)c for all a, b, c in A;

(b) commutative, if ab = ba for all a, b in A;

(c) idempotent, if aa = a for all a in A.

2. Magmas and unitary magmas

Definition 2.1. A magma1 is a pair (M,m), where M is a set and m a binary operation

on M.

Convention 2.2. Whenever only one magma (M,m) is considered, we will always

write m(a,b) = ab and often write just M instead of (M,m).

Definition 2.3. An element e in a magma M is said to be:

(a) an idempotent, if ee = e;

(b) a left identity (or a left unit), if ea = a for every a in M;

(c) a right identity (or a right unit), if ae = a for every a in M;

(d) an identity2 (or a unit), if it is a left and a right identity at the same time, i.e. if ea =

a = ae for every a in M.

Theorem 2.3. In an arbitrary magma:

(a) every left identity and every right identity is an idempotent;

(b) if e is a left identity and e' is a right identity, then e = e'.

Proof. (a) is trivial. (b): e = ee' = e'.

Remark 2.4. For an arbitrary set M one can define a binary operation on it by ab = b

for all a and b in M. In the resulting magma, every element is a left identity. And of

course one can do the same with the right identities by putting put ab = a. On the

other hand Theorem 2.3(b) tells us that the existence of at least one left identity and at

least one right identity in the same magma immediately implies that all of them are

equal, yielding a unique identity.

Definition 2.5. A unitary magma is a triple (M,e,m), where (M,m) be a magma and e

its identity.

1 In old literature magmas were sometimes called groupoids; this has been changed in order to avoid

confusions with groupoids in category theory. 2 As follows from 2.3(b) below (see also Remark 2.4), we could also say “the identity”.

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Convention 2.6. Whenever only one unitary magma (M,e,m) is considered, we will

(still use Convention 2.2 and) write 1 instead of e.

Remark 2.7. As we see from Remark 2.4, it is a triviality that every unitary magma

has exactly one left identity, exactly one right identity, and exactly one identity (all

the same).

3. Semigroups

Definition 3.1. A magma (M,m) is called a semigroup if m is associative.

Theorem 3.2. If (M,m) is a semigroup, then there exists a unique sequence of

operations m1, m2, … on M such that:

(a) mn is an n-ary operation (n = 1, 2,…);

(b) m1(a) = a for every a in M;

(c) mp(a1,…,ap)mq(b1,…,bq) = mp+q(a1,…,ap,b1,…,bq) for all p = 1, 2,…; q = 1, 2,…;

and a1,…, ap, b1,…, bq in M.

Proof. Existence. Let us define m1, m2, … inductively by (b) for m1 and by

mn+1(a1,…,an1) = mn(a1,…,an)an1

for m2, m3, …, i.e. for n = 1, 2,…. After this we will prove (c) by induction in q as

follows:

for q = 1 we have mp(a1,…,ap)mq(b1) = mp(a1,…,ap)b1 = mp1(a1,…,ap,b) for each

p;

for q > 1, using the inductive assumption and associativity, we then obtain

mp(a1,…,ap)mq(b1,…,bq) = mp(a1,…,ap)(mq1(b1,…,bq1)bq) =

(mp(a1,…,ap)mq1(b1,…,bq1))bq = mpq1(a1,…,ap,b1,…,bq1)bq =

mp+q(a1,…,ap,b1,…,bq),

as desired.

Uniqueness. From our assumptions on the sequence m1, m2, … we obtain

mp+1(a1,…,ap1) = mp(a1,…,ap)m1(ap1) = mp(a1,…,ap)ap1

and so our inductive definition was a consequence of those assumptions.

According to Convention 2.2, it is also convenient to avoid writing the letter m for the

operations introduced in Theorem 3.2. That is, one writes

n

mn(a1,…,an) = a1…an = ai = n

i=1ai (3.1)

i=1

and, in this notation, the formula given in 3.2(c) becomes

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(a1…ap)(ap1…apq) = a1…apq, (3.2)

or, equivalently

(p

i=1ai)(p

i

=

q

p1ai) = p

i

=

q

1ai, (3.3)

Furthermore, for a1 = … = an = a, one writes an instead of a1…an, and (3.2) becomes

apa

q = a

pq. (3.4)

4. Monoids

Definition 4.1. A monoid is a unitary magma (M,e,m), in which m is associative.

In other words, a monoid is a unitary magma (M,e,m), in which (M,m), is a

semigroup. Theorem 3.2 reformulates for monoids as follows:

Theorem 4.2. If (M,e,m) is a monoid, then there exists a unique sequence of

operations m0, m1, … on M such that:

(a) mn is an n-ary operation (n = 0, 1,…);

(b) m1(a) = a for every a in M;

(c) mp(a1,…,ap)mq(b1,…,bq) = mp+q(a1,…,ap,b1,…,bq) for all p = 0, 1,…; q = 0, 1,…;

and a1,…, ap, b1,…, bq in M.

Proof. As follows from Theorem 3.2, all we need is to take care of m0. That is, given

a sequence m1, m2, … as in Theorem 3.2, we need to prove that there exists a unique

element m0 in M satisfying

m0m0 = m0, mp(a1,…,ap)m0 = mp(a1,…,ap), and m0mq(b1,…,bq) = mq(b1,…,bq)

for all p = 1, 2,…; q = 1, 2,…; and a1,…, ap, b1,…, bq in M. This is the same as to

prove that there exists a unique element m0 in M satisfying am0 = a = m0a for all a in

M, i.e. the same as to prove that the magma (M,m) has a unique identity element –

which follows from the fact that (M,e,m) is a monoid (see Remark 2.7 for the

uniqueness).

According to Theorem 4.2, we will use the notation introduced in Section 3 also when

one or more of the indices involved is 0. Hence

m0 = 1 = a0 for all a in M; (4.1)

Note also that (4.1) makes (3.4) valid for p = 0 and/or q = 0.

5. Closure operators

Definition 5.1. A closure operator on a set X is a unary operation c on the power set

P(X) with:

(a) A B c(A) c(B),

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(b) A c(A),

(c) c(c(A)) = c(A),

for all A and B in P(X). For A P(X), the set c(A) is called the closure of A; if c(A) =

A, we say that A is closed.

Theorem 5.2. For a given closure operator c on a set X and A P(X), the following

sets are (well defined and) equal:

(a) c(A);

(b) the smallest closed subset in X containing A;

(c) the intersection of all closed subsets in X containing A.

Proof. Since c(A) is closed (by 5.1(c)) and since it contains A (by 5.1(b)), all we need

to prove is the following implication

(C is closed and A C) (c(A) C). (5.1)

But this is obvious, since: A C implies we have c(A) c(C) by 5.1(a), and c(C) = C

when C is closed.

Consider the map

CO(X) PP(X) (5.2)

from the set CO(X) of all closure operators on X to the set PP(X) of all subsets in P(X)

sending the closure operators to the corresponding sets of closed subsets, i.e. sending

a c to {A P(X) c(A) = A}. As follows from Theorem 5.2, the set of closed subsets

“remember” the closure operator used to define them; that is, Theorem 5.2 gives:

Corollary 5.3. The map (5.2) is injective.

After this, we are interested to know, what is the image of the map (5.2)? In other

words, we are interested to know which subsets in P(X) can be described as sets of all

closed subsets in X with respect to some closure operator on X ? The answer is:

Theorem 5.4. A subset C in P(X) admits a closure operator c with

C = {A P(X) c(A) = A} if and only if it is closed under intersections, i.e. satisfies

the property A C A C.

Proof. “If”: For A P(X), put c(A) = {C C A C}. The map c defined in this

way will obviously satisfy 5.1(a) and 5.1(b). Moreover, the required property

A C A C will then tell us that c(A) is always in C and so

c(A) = {C C c(A) C}.

Since the right-hand side of this equality is c(c(A)), this proves 5.1(c). That is, the

map c we defined is a closure operator. We will also have

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A C A = {C C A C} c(A) = A,

and so C = {A P(X) c(A) = A}.

“Only if”: We have to prove that if c is a closure operator on X, then the set

C = {A P(X) c(A) = A} has the property A C A C. That is, we have to

prove that if a subset A in P(X) has c(A) = A for all A in A, then c(A) = A. And

since c is a closure operator, proving c(A) = A of course means just proving

c(A) A. For every A in A, since A A, we have c(A) c(A) = A, and so

c(A) A, as desired.

Putting Corollary 5.3 and Theorem 5.4 together, we obtain:

Conclusion 5.5. The map (5.2) induces a bijection between the set of all closure

operators on X, and the set of all subsets C in P(X) that are closed under intersections.

6. Equivalence relations

For a binary relation E on a set X and elements x and y in X, we might write xEy

instead of (x,y) E.

Definition 6.1. A binary relation E on a set X is said to be an equivalence relation if

for all x, y, z in X we have:

(a) xEx (reflexivity);

(b) xEy yEx (symmetry);

(c) xEy & yEz xEz (transitivity).

In this situation, for any x in X, the set {y X xEy} = {y X yEx} is called the

equivalence class of x, and we will denote it by [x]E, or simply by [x]. The set

{[x] x X} is called the quotient set (sometimes factor set) of X by E, and it is

denoted by X/E.

The map X X/E sending x to [x] is usually called canonical. We will use this term

whenever there is no confusion with other types of canonical maps. The following

theorem is obvious:

Theorem 6.2. Let E be an equivalence relation on a set X, and x and y elements in X.

The following conditions are equivalent:

(a) xEy;

(b) [x] = [y];

(c) x and y have the same image under the canonical map X X/E;

(d) [x] [y] .

Remark 6.3. In the situation above, the sets of the form [x] form a partition of X,

which means that they are disjoint (by 6.2(b)(d)) and their union coincides with X

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(since x [x] by reflexivity). Moreover, this passage from an equivalence relation to a

partition obviously determines a bijection between the set of all equivalence relations

on X and all partitions of X. Theorem 6.4 below is closely related to this fact.

Theorem 6.4. The following conditions on a binary relation E on a set X are

equivalent:

(a) E is an equivalent relation;

(b) there exists a map f : X Y with xEx' f(x) = f(x').

Proof. (a)(b): Take f to be the canonical map X X/E and apply 6.2(a)(c).

(b)(a): Under the condition xEx' f(x) = f(x'), the desired properties of E (see

Definition 6.1) immediately follow from the similar properties of the equality relation.

7. Order relations

Definition 7.1. A binary relation R on a set X is said to be an order relation if for all x,

y, z in X we have:

(a) xRx (reflexivity);

(b) xRy & yRx x = y (antisymmetry);

(c) xRy & yRz xRz (transitivity).

A pair (X,R), where R is an order relation on X is called an ordered set, or simply an

order. We shall also write X = (X,R) for simplicity.

Remark 7.2. (a) In old literature the order relations are often required to be linear,

that is, in the notation above, having the property

either xRy or yRx holds for all x and y.

When this property is included in the definition of an order relation, what we called

“order” would be called “partial order”. Accordingly, ordered sets are sometimes

called posets, where “po” is an abbreviation of “partial order”.

(b) When only one fixed order relation R is considered, it is usual to write instead of

R and x y (or y x) instead of xRy. If so, then x < y (or y > x) means (x y)&(x y).

This so-called strict order relation < is irreflexive (i.e. there is no x with x < x),

transitive, and has the property that x < y and y < x exclude each other. Conversely,

every binary relation S on a set X satisfying these three properties is the strict order

determined by a unique order, namely by S{(x,x) x X}.

(c) Removing antisymmetry from Definition 7.1, we obtain the definition of a

preorder; that is, both equivalence and order relations are special cases of a preorder

relation. One also considers preorders (X,), and repeat various constructions and

arguments used for orders, sometimes with suitable modifications of course.

Definition 7.3. Let (X,) be an ordered set. An element x in X is said to be:

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(a) smallest, if x y for all y in X;

(b) largest, if y x for all y in X;

(c) minimal, if y x y = x in X;

(d) maximal, if x y y = x in X.

Theorem 7.4. Let (X,) be an ordered set and x an element in X. Then:

(a) if x is (the) smallest element in X, then it is a unique minimal element in X;

(b) if x is (the) largest element in X, then it is a unique maximal element in X.

8. Categories

All maps from a fixed set S to itself (also called endomaps of S) form a monoid

End(S):

for f and g in End(S), we take fg to be the composite of f and g; that is, fg is defined

by fg(s) = f(g(s)), and therefore:

the 1 in End(S) is then to be defined as the identity map of S.

Now, a category is structure that has the collection of all sets3 and all maps between

them as an example of it in the same way as End(S) is an example of a monoid. The

formal definition is:

Definition 8.1. A category is a system C = (C0,C1,d,c,e,m), in which:

(a) C0 is a set, whose elements are called objects in C;

(b) C1 is a set, whose elements are called morphisms in C;

(c) d and c are maps from C1 to C0, called domain and codomain respectively; when

d(f) = A and c(f) = B, we write f : A B, or f homC(A,B), and say that f is a

morphism from A to B;

(d) e is a map from C0 to C1, called the identity and written as e(A) = 1A; it has d(1A) =

A = c(1A);

(e) m is a map from the set {(f,g) C1C1 d(f) = c(g)} to C1, called composition,

written as m(f,g) = fg, and satisfying the equalities

d(fg) = d(g), c(fg) = c(f), f1d(f) = f = 1c(f)f, f(gh) = (fg)h,

whenever d(f) = c(g) and d(g) = c(h).

3 The collection of all sets does not form a set; this problem is what one usually refers to as the problem

of size. We will simply ignore it here – assuming that in our entire considerations one can always

replace “the collection of all” with “a collection of sufficiently many”. A proper treatment of the

problem of size would involve a considerable amount of material from mathematical logic, e.g. as

much as needed to speak about models of set theory.

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Displaying the elements of C1 as arrows (as in 8.1(c)) sometimes makes convenient to

think of objects in C as points. For instance it is useful to display the first two

equalities in 8.1(e) as

fg

g f

,

which actually explains where do they come from. Accordingly, when d(f) = c(g), we

will say that (f,g) is a composable pair of morphisms and that fg is their composite.

Example 8.2. The category C = Sets of sets has:

(a) C0 = the class of all sets;

(b) C1 = the class of all maps between sets;

(c) f : A B if and only if f is a map from A to B in the usual sense;

(d) fg the composite of f and g in the usual sense; therefore

(e) for a set S, the identity morphism 1S : S S is defined as the identity map of S in

the usual sense.

Example 8.3. A monoid M determines a one-object category C as follows:

(a) C0 is any one-element set, say {M};

(b) C1 = M;

(c) since C0 has only one element, the maps d and c are uniquely determined;

(d) fg in C is defined as fg in M; therefore

(e) 1M : M M is the identity element of M 4.

Example 8.4. An ordered (or, more generally, a preordered) set X determines a

category C as follows:

(a) C0 = X;

(b) C1 = is the order relation itself;

(c) the maps d and c are defined by d(x,y) = x and c(x,y) = y respectively;

(d) the composition in C is therefore defined by (y,z)(x,y) =(x,z);

(e) and therefore 1x = (x,x).

Example 8.5. Starting from a category C and a subset S in C0, we can form a new

category D as follows:

(a) D0 = S;

4 Not to be confused with the identity map of the set M!

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(b) D1 = the class of all f : A B in C with A and B in S, and then f : A B in D is

the same as f : A B in C;

(c) identity morphisms and composition in D are defined as in C.

A category D of this form is called a full subcategory in C. More generally, instead of

taking the class of all morphisms f : A B in C with A and B in S we could take any

subclass of it closed under composition and containing all identities of objects from

D; then we will drop “full” and talk about (just) a subcategory.

Convention 8.6. (a) Since we have f(gh) = (fg)h in 8.1(e), we will write simply fgh,

etc. – when appropriate (compare this with Theorem 3.2!).

(b) We will often omit the indices of identity morphisms, i.e. write 1 instead of, say,

1A.

(c) We might describe morphisms f : A B in some particular categories in such a

way that the same f could also be a morphism A' B' for A' A and/or B' B –

which formally contradicts to our notation since f : A B means d(f) = A and c(f) =

B. But in all such cases the original f can be replaced with, say, the triple (A,f,B). This

is the same convention as used in set theory, where a map f from a set A to a set B is

sometimes defined as a triple (A,Gf,B), where Gf is a subset in the cartesian product

AB satisfying suitable conditions and called the graph of f.

9. Isomorphism

Definition 9.1. Let f : A B be a morphism in a category C. A morphism g : B A

is said to be

(a) a left inverse of f if gf = 1A;

(b) a right inverse of f if fg = 1B;

(c) an5 inverse of f if it is a left inverse of f and a right inverse of f at the same time,

i.e. if gf = 1A and fg = 1B.

If f is invertible, that is, a g as in (c) does exist, we say that f is an isomorphism, call

the objects A and B isomorphic, and write A B.

Theorem 9.2. (a) Every identity morphism is the unique inverse of itself;

(b) if g and g' are left inverses of f and f ' respectively and (f,f ') is a composable pair

of morphisms, the composite g'g is a left inverse of the composite ff ';

(c) if g and g' are right inverses of f and f ' respectively and (f,f ') is a composable pair

of morphisms, the composite g'g is a right inverse of the composite ff ';

(d) if g is a left inverse of f and h is a right inverse of f, then g = h.

Proof. (a) is obvious; (b): g'gff ' = g'1f ' = g'f ' = 1; (c) is similar to (b); (d): g = g1 =

g(fh) = (gf)h = 1h = h.

5 Just as for 2.3(d), “an” can be replaced with “the”.

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As follows from 9.2(d), every isomorphism (=a morphism that admits an inverse), has

a unique inverse. The inverse of an isomorphism f will be denoted by f1

; from

Theorem 9.2 we obtain:

Corollary 9.3. (a) Every identity morphism is an isomorphism, and 1A1

= 1A;

(b) if (f,g) is a composable pair of isomorphisms, then fg is an isomorphism, and

(fg) 1

= g1

f 1

;

(c) if f is an isomorphism, then so is its inverse, and (f 1

) 1

= f.

Corollary 9.4. The isomorphism relation is an equivalence relation, i.e. for every

three objects A, B, C in a given category, we have:

(a) A A,

(b) A B implies B A,

(c) A B and B C imply A C.

Remark 9.5. Using Definition 9.1 in Example 8.3 we obtain usual notions of left

inverse, right inverse, inverse, and invertibility of elements in a monoid.

10. Initial and terminal objects

Definition 10.1. An object Z in a category C is said to be

(a) initial, if for every object A in C there exists a unique morphism from Z to A,

which we will denote by !A : Z A (when Z is fixed);

(b) terminal, if for every object A in C there exists a unique morphism from A to Z,

which we will denote by !A : A Z.

The following table provides the list of expressions used by various authors instead of

“initial” and “terminal”:

initial terminal

initial final

universal couniversal

left universal right universal

left zero right zero

zero one

0 1

However some authors say “zero” and write “0” only for an object that is initial and

terminal at the same time.

It is useful to see “initial” and “terminal” as generalizations of “smallest” and

“largest” using Example 8.4. In particular Theorems 10.2 and 10.3 below imply that

an ordered set can have at most one smallest and at most one largest element, as we

already know from Theorem 7.4. However, the story of initial/terminal is not at all a

simple copy of the story of smallest/largest. For instance 0 = 1 (=the existence of an

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object that is initial and terminal at the same time) holds in many important categories

of classical algebra.

Theorem 10.2. Let Z be an initial object and A an arbitrary object in the same

category. The following conditions are equivalent:

(a) A is initial;

(b) Z A;

(c) the morphism !A : Z A is an isomorphism.

Proof. (a)(b): When both Z and A are initial, there is a unique morphism from Z to

A, and a unique morphism from A to Z. We then need to show that the composites

Z A Z and A Z A of these morphisms are identity morphisms of Z and of A

respectively. This follows from the fact that since Z and A are initial, there is only one

morphism Z Z and only one morphism A A.

(b)(c) is trivial since there is exactly one morphism Z A.

(c)(a): Assuming that !A is an isomorphism, we have to prove that for every object

B in the given category, there exists a unique morphism A B.

Existence: Take the composite !B(!A)1

of !B with the inverse of !A.

Uniqueness: Let f and g be two morphisms A B. Since every two morphisms

Z B are equal, so are f!A and g!A, and we have f = f1A = f!A(!A)1

= g!A(!A)1

= g1A =

g.

Let us repeat, omitting the proof, the similar theorem for terminal objects:

Theorem 10.3. Let Z be a terminal object and A an arbitrary object in the same

category. The following conditions are equivalent:

(a) A is terminal;

(b) A Z;

(c) the morphism !A : A Z is an isomorphism.

11. Algebras, homomorphisms, isomorphisms

Definition 11.1. Let be a set equipped with a map l : {0,1,2,…}, and let

0 = l1

(0), 1 = l1

(1), 2 = l1

(2),…; such a pair (,l) is called a signature. An

(,l)-algebra is a pair (A,v), in which A is a set, and v =

(v : Al()

A)

is a family of operations on A (with arities as displayed, i.e. for each in n, v must

be an n-ary operation). We will also say that A is the underlying set of (A,v), and that

v is the algebra structure of (A,v), or that v is an algebra structure on A. (,l)-algebras

are also called universal algebras or algebraic structures.

Convention and Remark 11.2. (a) We will use the following simplified notation:

14

(,l)-algebra -algebra

(A,v) A

v(a1,…,an) (a1,…,an)

and moreover, we will sometimes write = (,l) and A = (A,v) in order to express the

fact that we write just and A, but we also remember l and v. Thus, whenever no

confusion is expected, we will not distinguish between an algebra (A,v) and its

underlying set A.

(b) The abbreviations (a1,…,an) = v(a1,…,an) and A = (A,v) actually suggest

removing and v from the notation completely. For example: a monoid M is an

-algebra with

{e} for n = 0,

n = {m} for n = 2,

for 0 n 2

(satisfying the conditions we required in Definition 4.1), but we simply write M =

(M,e,m) instead of (M,v) and in fact consider e and m not as elements of , but as

their images under v. We will use such simplified notation for all particular classes of

algebras that we will consider.

Definition 11.3. Let A and B be -algebras. A homomorphism f : A B is a map f

from the set A to the set B with

f((a1,…,an)) = (f(a1),…,f(an))

for every n = 0, 1, 2, …, for every in n, and for every a1, …, an in A.

Theorem 11.4. Let A, B, C be -algebras. Then:

(a) the identity map 1A : A A is a homomorphism;

(b) if f : A B and g : B C are homomorphisms, then so is their composite

gf : A C.

That is, all -algebras and their homomorphisms form a category in the same way as

all sets and all maps between sets do.

The category of all -algebras and their homomorphisms will be denoted by -Alg.

Theorem 11.5. A morphism f : A B of -algebras is an isomorphism (in -Alg) if

and only if the map f is bijective.

Proof. Being bijective, the map f has a unique inverse, say g. And all we need to

prove is that g is a homomorphism. For every n = 0, 1, 2, …, for every in n, and

for every b1, …, bn in B we have

g((b1,…,bn)) = g((fg(b1),…,fg(bn))) = g(f((g(b1),…,g(bn)))) =

15

(g(b1),…,g(bn))

as desired.

In addition to this characterization of isomorphisms, it is useful to know that we can

always transport an algebra structure along a bijection that will then become an

isomorphism:

Theorem 11.6. Let A = (A,v) be an -algebra, B a set, and f : A B a bijection. Then

there exists a unique -algebra structure w on B, such that f : (A,v) (B,w) is an

isomorphism.

Proof. Just observe that when f is a bijection, the equalities f((a1,…,an)) =

(f(a1),…,f(an)) (for all a1, …, an in A) are equivalent to the equalities (b1,…,bn) =

f((g(b1),…,g(bn))) (for all b1, …, bn in B), where g is the inverse of f.

12. Subalgebras

Theorem and Definition 12.1. Let A = (A,v) be an -algebra and X a subset in A.

The following conditions are equivalent:

(a) X is closed in A under all operations of the -algebra structure, i.e.

a1, …, an X (a1,…,an) X

for every n = 0, 1, 2, …, for every in n, and for every a1, …, an in A;

(b) there exists an -algebra structure w on X such that the inclusion map X A is a

homomorphism from (X,w) to (A,v);

(c) there exists a unique -algebra structure w on X such that the inclusion map

X A is a homomorphism from (X,w) to (A,v).

If these equivalent conditions hold, we will say that X, equipped with the w above, is a

subalgebra in A (or a subalgebra of A), and that w is the induced structure on X. In

other words, (X,w) is a subalgebra of (A,v) if and only if X is a subset in A and the

inclusion map X A is a homomorphism from (X,w) to (A,v). Considering X as just a

set, i.e. without w, we will express the equivalent conditions (a)–(c) by saying that X

forms a subalgebra. We will say “-subalgebra” instead of “subalgebra” when the

signature = (,l) is not fixed.

Remark and Definition 12.2. Let A be an -algebra. The set of subsets in P(A) that

form subalgebras is obviously closed under intersections. Therefore this set

determines a closure operator c on the set A, which will be written as c(X) = X. The

same symbol X will be used for the subalgebra in A with the underlying set X; that

is

X = the smallest subalgebra in A containing X

= the intersection of all subalgebras in A containing X.

16

We will say that X is the subalgebra in A generated by X.

Theorem 12.3. Let A be an -algebra and X a subset in A. Then

X = X0 X1 X2 …,

where X0, X1, X2, … is the increasing sequence of subsets in A defined inductively as

follows:

(a) X0 = X;

(b) Xn1 = Xn {(a1,…,am) m = 0, 1, 2, …; m; a1, …, am Xn}.

Proof. On the one hand the union X0 X1 X2 … obviously forms a subalgebra in

A containing X. On the other hand every subalgebra in A must contain Xn1 as soon as

it contains Xn, and so every subalgebra in A containing X = X0 also contains the union

X0 X1 X2 … .

One sometimes expresses this result by saying that the elements of X are (iterated)

combinations of elements of X (with operators from ).

13. Products

Definition 13.1. Let (Ai)iI be a family of -algebras. The (cartesian) product

Ai iI

of the family (Ai)iI is defined as the product of the underlying sets, with the

-algebra structure defined component-wise, i.e. defined by

((a1i)iI,…,(ani)iI) = ((a1i,…,ani))iI. We will also write

n

Ai = A1…An i=1

when the family is presented as a sequence A1, …, An, etc.

In particular, for the product AB of two algebras A and B we have

((a1,b1),…,(an,bn)) = ((a1,…,an),(b1,…,bn)). (13.1)

14. Quotient algebras

Theorem and Definition 14.1. Let A = (A,v) be an -algebra and E an equivalence

relation on A. The following conditions are equivalent:

(a) E forms a subalgebra in AA, i.e.

a1Eb1, …, anEbn (a1,…,an)E(b1,…,bn)

for every n = 0, 1, 2, …, for every in n, and for every a1, …, an, b1, …, bn in A;

17

(b) there exists an -algebra structure w on A/E such that the canonical map A A/E

is a homomorphism from (A,v) to (A/E,w);

(c) there exists a unique -algebra structure w on A/E such that the canonical map

A A/E is a homomorphism from (A,v) to (A/E,w).

If these equivalent conditions hold, we will say that:

E is a congruence on A;

A/E equipped with the w above, is a quotient algebra of A, and that

w is the induced structure on A/E.

In other words, A/E = (A/E,w) is a quotient algebra of A = (A,v) if and only if the

canonical map A A/E is a homomorphism from (A,v) to (A/E,w).

Remark and Definition 14.2. Let A be an -algebra. The set of subsets in P(AA)

that are congruences is obviously closed under intersections. Therefore this set

determines a closure operator c on the set AA. A binary relation R on A being a

subset in AA has its closure

c(R) = the smallest congruence on A containing R

= the intersection of all congruences on A containing R.

We will say that c(R) is the congruence closure of R, or the congruence generated by

R.

15. Canonical factorization of homomorphisms

Theorem 15.1. Every homomorphism f : A B of -algebras factors as the

composite

A

canonical homomorphism

A/{(a,b) f(a) = f(b)}

isomorphism sending [a] to f(a)

f(A)

inclusion homomorphism

B

of -algebra homomorphisms; in particular {(a,b) f(a) = f(b)} is a congruence on A

and f(A) is a subalgebra in B. In this composite:

(a) A A/{(a,b) f(a) = f(b)} is an isomorphism if and only if f is injective, and this

the case if and only if {(a,b) f(a) = f(b)} is the equality relation;

(b) f(A) B is an isomorphism if and only if f is surjective, and this the case if and

only if f(A) = B;

18

(c) in particular f is an isomorphism if and only if {(a,b) f(a) = f(b)} is the equality

relation and f(A) = B.

16. Classical algebraic structures

In this section we simply list some of especially important algebraic structures, not

asking the readers to remember the list – since only a few of these structures are

mentioned above in these notes:

(a) Sets can be considered as -algebras with = .

(b) Pointed sets are pairs (A,a), where A is a set, and a an element in A.

(c) Magmas were defined in Section 2.

(d) Semigroups were defined in Section 3.

(e) Unitary magmas were defined in Section 2.

(f) Monoids were defined in Section 4.

(g) Groups: a group is a system (A,e,m,i) in which (A,e,m) is a monoid and i is a unary

operation on A with m(i(a),a) = e (= m(a,i(a)). One usually writes either (A,e,m,i) =

(A,1,,1

) and

m(a,b) = ab = ab, e = 1, and i(a) = a1

,

or (A,e,m,i) = (A,0,,) and

m(a,b) = a b, e = 0, and i(a) = a,

i.e. uses either multiplicative or additive notation; the additive notation is preferable

when m is commutative, in which case the group is called commutative, or abelian.

The same applies also to monoids, yielding two other important types of algebraic

structures:

(h) Abelian monoids = commutative monoids.

(i) Abelian groups = commutative groups.

(j) Semirings = rigs: a semiring is a system (A,0,,1,) in which (A,0,) is a

commutative monoid and (A,1,) is a monoid with 0a = 0 = a0, a(b c) = ab ac, and

(a b)c = ab ac (“distributivity”, or “distributive laws” for multiplication with

respect to addition and 0) for all a, b, and c in A.

(k) Rings: a ring is a system (A,0,,,1,) in which (A,0,,1,) is a semiring and

(A,0,,) is a group. Sometimes, however, the associativity requirement for the

multiplication is dropped and/or there is no 1 (i.e. the structure becomes (A,0,,,)

in the notation above). We would then say “non-associative ring” and/or “ring without

1”. The requirements 0a = 0 = a0 are dropped in any case, since they can deduced

from others (e.g. 0a = 0a 0a (0a) = (0 0)a (0a) = 0a (0a) = 0).

(l) Commutative Rings: as above, but the multiplication is required to be

commutative.

19

(m) Fields: a field is a system (A,0,,,1,,1

) in which (A,0,,,1,) is a commutative

ring and (A\{0},1,,1

) is a group. Since A\{0} is involved here, this definition as

stated does not present a field as an algebraic structure. However, since the operation 1

on the set A\{0} is uniquely determined (when it exists), a field can be considered

as a commutative ring satisfying an additional condition.

(n) M-sets: let M = (M,1,) be a monoid (or, more generally, a unitary magma) – then

an M-set (or an (M,1,)-set) is a pair (A,), where A is a set and : MA A a map

written (usually) as (m,a) = ma and satisfying 1a = a and m(m'a) = (mm')a for all a

in A and m, m' in M. This definition as stated does not present an M-set as an algebraic

structure since it involves a map, namely , that is not an operation in the sense of

Definition 1.1. However, this map can equivalently be described an M-indexed family

(m)mM of unary operations on A, where m(a) = (m,a). Therefore an M-set can be

considered as an (,l)-algebra with = M and l(m) = 1 for each m in M.

(o) R-semimodules: for a semiring R = (R,0,,1,), an R-semimodule, or a semimodule

over R, is a system (A,0,,) in which (A,0,) is a commutative monoid and (A,) an

(R,1,)-set with r0 = 0 = 0a, r(a b) = ra rb, and (r s)a = ra sa for all r and s in

R and a and b in A.

(p) R-modules: for a ring R = (R,0,,,1,), an R-module, or a module over R, is the

same as a semimodule over (R,0,,1,). Here again, the requirements r0 = 0 = 0a are

dropped since they follow from others (cf. (k)).

(q) K-vector spaces: for a field K = (K,0,,,1,,1

), a K-vector space, or a vector space

over K, is the same as a module over (K,0,,,1,).

(r) R-algebras: for a commutative ring R = (R,0,,,1,), an R-algebra, or an algebra

over R, is a system A = (A,0,,,), where (A,0,,) is a ring without 1 and (A,0,,) an

R-module with r(ab) = (ra)b = a(rb) for all r in R and a and b in A. Similarly to rings

one can introduce “non-associative” and “non-commutative” algebras.

(s) Unitary R-algebras: for a commutative ring R = (R,0,,,1,), an unitary

R-algebra is a system A = (A,0,,,1,), where (A,0,,,) is an R-algebra and (A,1,) is

a unitary magma. And again, similarly to rings one can introduce “non-associative”

and “non-commutative” unitary algebras.

(t) Lie R-algebras: for a commutative ring R, a non-associative non-commutative

R-algebra A is said to be a Lie R-algebra, or a Lie algebra over R if ab = ba and

a(bc) b(ca) c(ab) = 0 for all a, b, and c in A. Lie R-algebras are usually considered

for R being a field.

(u) Semilattices: a commutative monoid (A,1,) is said to be a semilattice if it is

idempotent, i.e. a2 = aa = a for all a in A. For a semilattice A, we also write

A = (A,1,) and then call it a -semilattice, or write A = (A,0,) and then call it a

-semilattice.

(v) Lattices: a lattice is a system A = (A,0,,1,), in which (A,0,) and (A,1,) are

semilattices with a (a b) = a = a (a b) for all a and b in A.

(w) Modular lattices: a lattice A = (A,0,,1,) is said to be modular if

a (b (c a)) = (a b) (c a) for all a, b, and c in A.

(x) Distributive lattices: a lattice A = (A,0,,1,) is said to be distributive if

a (b c) = (a b) (a c) for all a, b, and c in A.

20

(y) Heyting algebras: a Heyting algebra is a system A = (A,0,,1,,) in which

(A,0,,1,) is a distributive lattice and a binary operation on A with

a (b (a b)) = a = a (b (b a)) for all a, b, and c in A.

(z) Boolean algebras: a Boolean algebra is a system A = (A,0,,1,,) in which

(A,0,,1,) is a distributive lattice and is a unary operation on A with a a = 0

and a a = 1 for all a in A.

17. Quotient groups, rings, and modules

Quotient algebras of the form A/E, where E is a congruence on A, were already

described in Section 14. However, for many classical algebraic structures one can

make a next step, namely to prove that all congruences are determined by their single

classes that are special subsets of the ground algebra. We shall give details for the

three types of algebras mentioned in the title of this section. Their subalgebras will be

called subgroups, subrings, and submodules, respectively.

Theorem 17.1. Let A = (A,1,,1

) be a group. For every subgroup S of A, the sets

{(a,b) AA ab1

S} = {(a,b) AA ba1

S} (17.1)

and

{(a,b) AA a1

b S} = {(a,b) AA b1

a S} (17.2)

are equivalence relations on A.

Proof. First note that the equalities of (17.1) and (17.2) indeed hold, since

(ab1

)1

= ba1

and (a1

b)1

= b1

a. To prove that the set (17.1) is an equivalence

relation we observe:

Since, for every a in A, we have aa1

= 1 S, it is reflexive.

Since, for ab1

S, we have ba1

= (ab1

)1

S (cf. 9.3(b)), it is symmetric (and

the equality of (17.1) holds).

Since, for ab1

, bc1

S, we have ac1

= (ab1

)(bc1

) S, it is transitive.

The fact the set (17.2) is an equivalence relation can be proved similarly.

For any monoid A, the set P(A) of all subsets of A has a natural monoid structure

defined by

UV = {uv u U, v V}; (17.3)

the identity element in this monoid is the one-element set {1}, and, moreover, sending

u to {u} determines a monoid homomorphism A P(A). Whenever there is no

danger of confusion, we shall identify A with its image in P(A), and write e.g. aV

instead of {a}V.

21

When A is a group, and S a subgroup of A, the sets of the form aS and Sa (for a A)

are called right and left6 cosets (of S), respectively; when left and right cosets coincide

they are simply called cosets.

Proposition 17.2. For A and S as in Theorem 17.1, the left and right cosets of S are

the equivalence classes with respect to the equivalence relations (17.1) and (17.2).

Proof. Just note that ab1

S a Sb and b1

a S a bS.

Moreover, we have:

Theorem 17.3. For A and S as in Theorem 17.1, the following conditions are

equivalent:

(a) the sets (17.1) and (17.2) coincide;

(b) aS Sa for every a A;

(c) Sa aS for every a A;

(d) aS = Sa for every a A;

(e) aSa1

S for every a A;

(f) S aSa1

for every a A;

(g) aSa1

= S for every a A.

Proof. (a)(d) follows from Proposition 17.2.

(b)(e): aS Sa aSa1

Saa1

= S.

(e)(b): aSa1

S aSa1

a Sa, but aSa1

a = aS.

(c)(f): Sa aS Saa1

aSa1

, but Saa1

= S.

(f)(c): S aSa1

Sa aSa1

a = aS.

(b)(f): if aS Sa for every a A, then a1

S Sa1

for every a A, and so

aa1

S aSa1

for every a A; but aa1

S = S.

(f)(b): if S aSa1

for every a A, then S a1

Sa for every a A, and so

aS aa1

Sa = Sa for every a A.

The last six implications show that conditions (b), (c), (e), and (f) are equivalent to

each other. Since the conjunction of (b) and (c) is equivalent to (d), and the

conjunction of (e) and (f) is equivalent to (g), we can conclude that all conditions

(b)-(g) are equivalent to each other. Together with the equivalence (a)(d) this

completes the proof.

Definition 17.4. A subgroup S of a group A is said to be normal if it satisfies the

equivalent conditions of Theorem 17.3.

Lemma 17.5. If S is a normal subgroup of a group A, then sending each a A to aS

determines a semigroup homomorphism A P(A).

6 Some authors would say “left” for aS and “right” for Sa.

22

Proof. Just note that we can write (aS)(bS) = aSbS = abSS = abS = (ab)S.

Theorem 17.6. For a subset S of a group A, the following conditions are equivalent:

(a) S is a normal subgroup of A;

(b) at least one of the sets (17.1) and (17.2) is a congruence on A;

(c) the sets (17.1) and (17.2) coincide with each other and it is a congruence on A;

(d) there exists a congruence E on A with [1]E = S;

(e) there exists a unique congruence E on A with [1]E = S;

(f) there exists a group homomorphism f : A B with Ker(f) = S, where the kernel

Ker(f) of f is defined as Ker(f) = f 1

(1).

Proof. Suppose S is normal and consider the set A/S of cosets of S. As easily follows

from Lemma 17.5, it has a group structure whose multiplication is defined by

(aS)(bS) = (ab)S.

By Proposition 17.2 this means that the quotient set A/E, where E is the equivalence

relation (17.1) (or (17.2), which is the same since S is normal), has a group structure

whose multiplication is defined by

[a]E[b]E = [ab]E.

Therefore (a) implies (c). The implications (c)(b) and (e)(d) are trivial, and we

also have:

(b) implies (d), since, for E being either the set (17.1) or the set (17.2), we have

(a,1) E a S;

(d) implies (e) since (a,b) E (a,b)(b1

,b1

) E (ab1

,1) E;

(d) implies (f) since we can use the canonical homomorphism A A/E as f, and

then f 1

(1) = [1]E = S;

(f) implies (a) since, for f : A B as in (f) and every a A and s S we have

f(asa1

) = f(a)f(s)f(a)1

= f(a)1f(a)1

= f(a)f(a)1

= 1, which gives aSa1

S.

Since every ring has its additive group structure, the results above will help us to

establish similar results for rings. However:

Considering rings instead of groups we have to use the additive notation, in which

we will have e.g. Ker(f) = f 1

(0) = [0]E, (a,b) E a b S (where a b is

defined as a (b)), and a S instead of aS (in the situations above).

When a group A is abelian, which is always the case for the additive group of ring,

every subgroup of A is normal. On the other hand, an arbitrary subgroup S of the

additive group of a ring A will make A/S a group, but not a ring in general. For

example, if S contains the identity element 1 of A, then it determines a congruence

on the ring A only in the trivial case S = A.

There is no counterpart of Theorem 17.3 for rings.

23

Accordingly the story of quotient rings is short: it consists of one definition and a

counterpart of Theorem 17.6, whose substantial part actually follows from Theorem

17.6.

Definition 17.7. A subgroup S of the additive group of a ring A is said to be an ideal if

as and sa are in S for every a in A and every s in S.

Remark 17.8. A ring A can always be considered as a module over itself, that is, as

an A-module; submodules of this module are called left ideals of A. On the other

hand, every ring A has its opposite ring Aop

defined as the same abelian group with the

opposite multiplicative structure, making ab in Aop

the same as ba in A. The left ideals

of Aop

are called the right ideals of A. That is:

(a) a subgroup S of the additive group a ring A is a left ideal of A if and only if as is in

S for every a in A and every s in S;

(b) a subgroup S of the additive group a ring A is a right ideal of A if and only if sa is

in S for every a in A and every s in S;

(c) a subset S of a ring A is an ideal of A if and only if it is a left and a right ideal of A

at the same time.

Ideals are therefore also sometimes called two-sided ideals, and, as next theorem

shows, the play the same role in ring theory as normal subgroups do in group theory.

Theorem 17.9. For a subset S of a ring A, the following conditions are equivalent:

(a) S is an ideal of A;

(b) the set {(a,b) AA a b S} is a congruence on A;

(c) there exists a congruence E on A with [0]E = S;

(d) there exists a unique congruence E on A with [0]E = S;

(e) there exists a ring homomorphism f : A B with Ker(f) = S, where the kernel

Ker(f) of f is defined as Ker(f) = f 1

(0).

Proof. Having in mind Theorem 17.6, we only need to prove that (a) implies that

{(a,b) AA a b S} is closed in AA under the multiplication, and that (e)

implies (a). The first of these assertions follows from the simple formula

ac bd = (a b)c b(c d),

since a b S makes (a b)c S and c c S makes b(c d) S. The second one

follows from two other simple formulas, namely a0 = 0 and 0a = 0 (for all a A).

Similarly, but even easier, for a module over a ring we have:

Theorem 17.10. For a subset S of an R-module A, the following conditions are

equivalent:

(a) S is a submodule of A;

24

(b) the set {(a,b) AA a b S} is a congruence on A;

(c) there exists a congruence E on A with [0]E = S;

(d) there exists a unique congruence E on A with [0]E = S;

(e) there exists an R-module homomorphism f : A B with Ker(f) = S, where the

kernel Ker(f) of f is defined as Ker(f) = f 1

(0).

These results allow talking about quotients with respect to:

normal subgroups in the case of groups,

ideals in the case of rings,

submodules in the case of modules,

instead of quotients with respect to congruences. Accordingly we shall write A/S

instead of A/{(a,b) AA ab1

S} in the situation of Theorem 17.6, and instead of

A/{(a,b) AA a b S} in the situations of Theorems 17.9 and 17.10. We say that

S is trivial if it has only one element, making A/S canonically isomorphic to A.

Theorem 15.1 gives:

Theorem 17.11. Let f : A B be a homomorphism of either groups, or rings, or

R-modules; f factors as the composite

A

canonical homomorphism

A/Ker(f)

isomorphism sending [a] to f(a)

f(A)

inclusion homomorphism

B

of groups, or rings, or R-modules respectively. In this composite:

(a) A A/Ker(f) is an isomorphism if and only if f is injective, and this the case if and

only if Ker(f) is trivial;

(b) f(A) B is an isomorphism if and only if f is surjective, and this the case if and

only if f(A) = B.

(c) in particular f is an isomorphism if and only if Ker(f) is trivial and f(A) = B.

25

II. LATTICES, SEMIRINGS, FOUNDATION OF LINEAR

ALGEBRA, NUMBERS AND DIVISIBILITY

1. Commutativity

In this section we will use additive notation for semigroups and monoids; accordingly,

in the situation of Section I.3, instead of

n

mn(a1,…,an) = a1…an = ai = n

i=1ai

i=1

we will write

n

mn(a1,…,an) = a1 … an = ai = n

i=1ai (1.1)

i=1

Theorem 1.1. Let (a1, …, an) be an n-tuple of elements a commutative semigroup S

and f : {1, …, n} {1, …, n} a bijection. Then

n

i=1ai = n

i=1af(i)

in S.

Proof. Using mathematical induction and fact that the statement of the theorem is

(trivially) true for n = 1, we can assume that, for every bijection

g : {1, …, n1} {1, …, n1}, we have

n

i =

1

1ai = n

i =

1

1ag(i),

which we will call here our inductive assumption. We will use this equality for g

defined by

f(i) if i < f1

(n),

g(i) =

f(i1) if i f1

(n);

it is easy to see that the image of {1, …, n1} under g coincides with

{1, …, n}\{n} = {1, …, n}, and so g is a bijection indeed. We have

n

i=1ai = (n

i =

1

1ai) an (using (the additively written version of) (3.3))

= (n

i =

1

1ag(i)) an (by our inductive assumption)

= (n

i =

1

1ag(i)) af(m) (where m = f1

(n))

= ((m

i =

1

1ag(i)) (n

i =

1

mag(i))) af(m) (by (3.3) again)

= ((m

i =

1

1af(i)) (n

i =

1

maf(i1))) af(m) (by the definition of g)

= (m

i =

1

1af(i)) ((n

i =

1

maf(i1)) af(m)) (by associativity)

= (m

i =

1

1af(i)) (af(m) (n

i =

1

maf(i1))) (by commutativity)

= (m

i =

1

1af(i)) (af(m) (n

i =m1af(i))) (using obvious re-indexing)

26

= (m

i =

1

1af(i)) (n

i =maf(i)) (by (3.3))

= n

i=1af(i) (by (3.3)).

This theorem tells us that performing callucations in a commutative semigroup, not

only omitting parentheses but even arbitrary permutations of summands are allowed.

In particular we can write

(a b) (c d) = (a c) (b d), (1.2)

which in fact expresses the fact that the addition map SS S is a homomorphism

whenever S is a commutative semigroup. Furthermore, it characterizes commutative

monoids as follows:

Theorem 1.2. The following conditions on a unitary magma (M,0,) are equivalent:

(a) (M,0,) is a commutative monoid;

(b) the equality (1.2) holds for all a, b, c, and d in M.

Proof. (a)(b) follows from Theorem 1.1 (but can also easily be proved directly of

course). In order to prove (b)(a), we need to show that (b) implies associativity and

commutativity of . We have:

a (b c) = (a 0) (b c) = (a b) (0 c) = (a b) c,

and

a b = (0 a) (b 0) = (0 b) (a 0) = b a,

as desired.

2. Semilattices

Definition 2.1. Let (X,) be an ordered set, S a subset of X and a an element in X. We

say that a is the infimum, or meet of S, and write

a = inf S = S, (2.1)

if the following conditions hold:

(a) a s for every s S;

(b) if an element x in X is such that x s for every s S, then x a.

Briefly, (2.1) means that a is the largest element in X with the property 2.1(a). When S

is a finite set, say S = {s1,…,sn}, we shall also write

a = n

i=1si = s1 … sn, (2.2)

and, in particular, this defines the binary meet operation

27

: XX X, (2.3)

whenever every two-element subset of X has a meet. This operation, whenever it

exists, is associative and commutative, and so Theorem 1.1 applies to it. Moreover, it

is obviously idempotent, and in fact this establishes a bijection between the set of

order relations with binary meets on any given set X and the set of idempotent

commutative semigroup structures on X. Let us explain:

For a fixed set X consider the diagram

f

(2.4)

g

h

in which:

the map f is defined by f(m) = {(x,y) XX x = m(x,y)};

the map g is defined by the restriction of f, which requires to show that when

m : XX X is associative, commutative, and idempotent, the corresponding

relation f(m) is an order relation – see Lemma 2.2 below;

the map h sends order relations to the corresponding meet operations as

described above.

Lemma 2.2. Let m be a binary operation on a set X, and

R = f(m) = {(x,y) XX x = m(x,y)} the corresponding binary relation on X, as

described above. Then:

(a) if m is idempotent, then R is reflexive;

(b) if m is commutative, then R is antisymmetric;

(c) if m is associative, then R is transitive.

Therefore, if m is idempotent, commutative, and associative, then R is an order

relation.

Proof. Let us write xy instead of m(x,y).

(a): Just note that (x,x) R x = xx.

(b): ((x,y) R & (y,x) R) (x = xy & y = yx) (x = xy = yx = y).

(c): ((x,y) R & (y,z) R) (x = xy & y = yz) (x = xy = x(yz) = (xy)z = xz)

(x,z) R.

The following theorem completes the desired explanation:

The set of order

relations on X with

binary meets

The set of binary

relations on X The set of binary

operations on X

The set of idempotent

commutative semigroup

structures on X

28

Theorem 2.3. The maps g and h in diagram (2.4) are inverse to each other, and

therefore they establish a bijection between the set of order relations with binary

meets on any given set X and the set of idempotent commutative semigroup structures

on X.

Proof. To prove that hg is the identity map is to prove that, given an idempotent,

commutative, and associative binary operation m on X we have

xy = x y, (2.5)

for every x, y X, where xy = m(x,y), while x y is the meet of {x,y} with respect to

the order defined by

x y x = xy. (2.6)

And to prove (2.5) is to prove that xy is the largest element z in X with z x and z y

with respect to the same order. Indeed, we have:

(xy)x = (yx)x = y(xx) = yx = xy = x and (xy)y = x(yy) = xy = x, which shows that

xy x and xy y;

if z x and z y, then z = zx and z = zy; this gives z(xy) = (zx)y = zy = z and so

z xy.

To prove that gh is the identity map is to prove that, for a given order relation on X,

we have

x y x = x y, (2.7)

for every x, y X, where x y is the meet of {x,y} with respect to the order . But

this easily follows from the definition of meet.

Example 2.4. An order on X is said to linear if, for every x, y X we have either

x y or y x. It is easy to see that the following conditions on an order and the

corresponding binary meet operation are equivalent:

(a) is a linear order;

(b) x y {x,y} for every x, y X;

(c) every subset of X is a subsemigroup of (X,).

Example 2.5. Let X be a set, and the inclusion order on the power set P(X). Then

= , that is, the corresponding meet operation is nothing but the intersection

operation. Formally this applies to binary meets, but we also have S = S for an

arbitrary set S of subsets in X of course.

Example 2.6. Let X be a set, and (X) the set of equivalence classes of assertions on

elements of X; not going into details of the corresponding logic, let us assume that two

such assertions and are identified if and only if

{x X (x)} = {x X (x)}. (2.8)

29

Then the sets (X) and P(X) are bijective (again, under suitable assumptions on the

logic involved, which we are not discussing here); under this bijection an assertion

corresponds to the subset {x X (x)} of X, while a subset S corresponds to the

assertion x S. We have:

(a) S T if and only if x S implies x T;

(b) x S T if and only if (x S)&(x T).

That is, under the bijection above, the inclusion order on P(X) corresponds to the

implication order on (X), while the corresponding meet operation, which the

intersection operation, corresponds to the conjunction operation on (X).

Remark 2.7. In fact 2.6(a) and 2.6(b) are used in mathematics to define inclusion and

intersection respectively. Moreover, conjunction is in fact used to define meets.

Indeed, the definition of, say, the binary meet can be formulated as

x a b (x a)&(x b) (2.9)

(here we use the fact that an element c in an ordered set X is completely determined

by the set {x X x c}). Furthermore, in fact the binary meet operation is

associative, commutative, and idempotent because so is the logical conjunction.

Indeed,

x a (b c) (x a)&(x b c) (x a)&((x b)&(x c))

((x a)&(x b))&(x c) (x a b)&(x c) x ((a b) c),

x a b (x a)&(x b) (x b)&(x a) x b a,

x a a (x a)&(x a) x a x a.

Nevertheless it is true of course that the conjunction operation is a very special case of

the meet operation (we should say “up to logic”; see Example 2.6).

Having finite meets is of course equivalent to have binary meets and empty meet

(=the meet of the empty subset). And the empty meet is nothing but the largest

element 1 in the given ordered set; note also that x 1 x = x1. Therefore we obtain:

Theorem 2.8. The bijection described in Theorem 2.3 induces a bijection between the

set of order relations with finite meets on X and the set of idempotent commutative

monoid structures on X.

Recall from paragraph (u) of Section I.16 that idempotent commutative monoids are

called semilattices.

3. Lattices

Definition 3.1. Let (X,) be an ordered set, S a subset of X and a an element in X. We

say that a is the supremum, or join of S, and write

30

a = sup S = S, (3.1)

if the following conditions hold:

(a) s a for every s S;

(b) if an element x in X is such that s x for every s S, then a x.

Briefly, (3.1) means that a is the smallest element in X with the property 3.1(a). When

S is a finite set, say S = {s1,…,sn}, we shall also write

a = n

i=1si = s1 … sn, (3.2)

and, in particular, this defines the binary join operation

: XX X, (3.3)

whenever every two-element subset of X has a join.

Every order relation on a set X determines the reverse (or opposite) order on X

defined by

x y y x, (3.4)

and we have

= , (3.5)

which tells us that the passage from orders to their reverse orders is a bijection inverse

to itself. Under this bijection joins correspond to meets and meets correspond to joins,

that is

(a = S under ) (a = S under ), (3.6)

(a = S under ) (a = S under ). (3.7)

Therefore all constructions and results of previous sections have their obvious

counterparts, in which meets are replaced with joins. Having this in mind, the

following theorem immediately follows from the results of previous section:

Theorem 3.2. Let X be a set. There exist a bijection between all systems (X,0,,1,),

in which (X,0,) and (X,1,) are semilattices, and systems (X,0,1), in which 0 and

1 are order relations with finite joins and finite meets respectively, under which

x 0 y y = x y and z 1 t z = z t, (3.8)

for all x, y, z, t X.

This theorem suggests to find out conditions on (X,0,,1,) under which the two

orders 0 and 1 coincide, and we have:

31

Theorem 3.3. Let (X,0,,1,) be as in Theorem 3.2, and let 0 and 1 be the

corresponding orders on X defined via (3.8). Then the following conditions are

equivalent:

(a) 0 = 1;

(b) (X,0,,1,) is a lattice (see (v) in Section I.16), that is ((X,0,) and (X,1,) are

semilattices and) the absorption law x (x y) = x = x (x y) holds for all x and y

in X.

Proof. (a)(b): Assuming (a) and writing instead of 0 and 1, we have

x y x x y, (3.9)

and so x (x y) = x = x (x y) as desired.

(b)(a): Assuming (b) and x 0 y, we have

x y = x (x y) = x,

and so x 1 y. Similarly (or applying the same argument to the reverse order) we can

show that, assuming (b), x 1 y implies x 0 y.

Corollary 3.4. The bijection described in Theorem 3.2 induces a bijection the set of

lattice structures on X and the set of order relations on X with finite joins and finite

meets.

4. Distributivity and complements

Theorem 4.1. The following conditions on a lattice L = (L,0,,1,) are equivalent:

(a) x (y z) = (x y) (x z) for all x, y, z L;

(b) x (y z) = (x y) (x z) for all x, y, z L.

Proof. It suffices to show that (a)(b) – since (a)(b) applied to the opposite order

gives (b)(a). Assuming (a), we have:

(x y) (x z) = ((x y) x) ((x y) z) (by (a))

= ((x (x y)) (z (x y)) (by commutativity of )

= x ((z x) (z y)) (by absorption law and (a))

= (x (z x)) (z y) (by associativity of )

= x (z y) (by absorption law)

= x (y z) (by absorption law),

as desired.

That is, the distributive laws 4.1(a) and 4.1(b) are equivalent to each other, and each

of them can be used to define a distributive lattice (see (x) in Section I.16).

32

Definition 4.2. Let a be an element in a lattice L = (L,0,,1,). An element b in L is

said to be a complement of a, if a b = 0 and a b = 1 in L. In particular 0 and 1 are

unique complements to each other.

Theorem 4.3. If L is a distributive lattice, a an element in L, and b a complement of a

in L, then:

(a) b is the largest element among the elements c in L with a c = 0;

(b) b is the smallest element among the elements c in L with a c = 1.

Proof. Since (a) applied to the opposite order becomes (b), it suffices to prove (say)

(a). If a c = 0, we have

b = b 0 = b (a c) = (b a) (b c) = 1 (b c) = b c,

and so c b, as desired.

Corollary 4.4. If L is a distributive lattice and b and c are complements of a in L, then

b = c.

Example 4.5. Examples 2.5 and 2.6 in fact describe distributive lattices

P(X) = (P(X),,,X,) and (X) = ((X), false, disjunction, true, conjunction),

in which every element has a (unique) complement, namely:

(a) for A X, the complement of A in P(X) is the usual set-theoretic complement

{x X x A};

(b) for in (X), the complement of is its negation .

Example 4.6. A linear order is a lattice if and only if it has the smallest element 0 and

the largest element 1. Every such lattice is distributive, but its only complemented

elements (that is, elements that have complements) are 0 and 1.

Example 4.7. When a lattice is finite and, moreover, has a small number of elements

it is convenient to display it as a set of points on a plane connected by lines – in such a

way that x y if and only if x and y are connected by a line and x is “below” y. In

particular, consider the following four lattices Li (i = 1, 2, 3, 4):

1 1 1 1

L1 L2 L3 L4 b

a a b a b c a

c

0 0 0 0,

and observe that the element a has no complement in L1, has a unique complement in

L2 (namely b), and has two complements in L3 and two complements in L4 (which are

33

b and c in each case). Therefore among these lattices only L1 and L2 can be

distributive, and, indeed, they are – since for L1 is a (or has the) linear order, while L2

is isomorphic to (P(X),,,X,), when X is a two-element set.

5. Complete lattices

Theorem 5.1. The following conditions on an ordered set X = (X,) are equivalent:

(a) every subset of X has a meet;

(b) every subset of X has a join.

Proof. Just observe that

S = {x X sS s x} and S = {x X sS x s}, (5.1)

for every subset S of X; here equality signs mean “the left-hand side exists if and only

if the right-hand does, and they are equal”.

That is replacing finite meets and joins with infinite ones, makes meets and joins

“better related to each other”. On the other hand, in contrast to Theorem 4.1, the

infinite versions of the distributive laws 4.1(a) and 4.1(b), namely

x (S) = {x s s S} and x (S) = {x s s S}, (5.2)

are not equivalent to each other. For example the lattice of open subsets of the real

line satisfies the first of them, but not the second one (we omit the details).

Definition 5.2. A lattice is said to be complete if it admits all meets and joins, that is,

has meets and joints of each of its subsets.

Remark 5.3. From Theorem 5.1 it immediately follows that:

(a) In Definition 5.2, it suffices either to require (just) the existence of meets or to

require the existence of joins.

(b) Every finite semilattice (X,1,) has a unique (complete) lattice structure

(X,0,,1,), under which (X,1,) = (X,1,), and unique lattice structure (X,0,,1,),

under which (X,1,) = (X,0,). In order to avoid confusion between these two

structures, one usually begins with a lattice (X,0,,1,) and calls (X,1,) and (X,0,)

the corresponding meet-semilattice (or -semilattice) and join-semilattice (or

-semilattice) respectively.

Let L be a complete lattice, a an element in L and

S = {x L a x = 0}.

Then, assuming the first equality in (5.2), we would conclude that S is the largest

element among the elements c in L with a c = 0. Such element is called the

pseudo-complement of a. It is obviously uniquely determined and, according to

Theorem 4.3(a) coincides with the complement of a whenever a has a complement.

34

However, it might happen that a has no complement but still has a pseudo-

complement. For example, if L is a linear order, then 0 is the pseudo-complement of

every element in L except 0, while (as already mentioned) only 0 and 1 have

complements.

Remark 5.4. Some authors, especially old ones, prefer to avoid using empty meets

and joins. In their terminology:

(a) semilattices are idempotent commutative semigroups;

(b) hence, defined via order relations, their semigroups have just binary meets (or

joins), and, equivalently, finite non-empty ones;

(c) accordingly, their lattices have binary, or, equivalently, finite non-empty, meets

and joins – while those with 0 and 1 are called bounded lattices;

(d) there are -complete semilattices that are not lattices, and -complete semilattices

that are not lattices; the formulas (5.1) cannot be used there since empty meets/joins

are not defined. In that terminology, Theorem 5.1 would be reformulated as: every

-complete semilattice having the largest element has arbitrary joins, and every

-complete semilattice having the smallest element has arbitrary meets.

6. Boolean algebras

As defined at the end of Section I.16, a Boolean algebra is a system (A,0,,1,,), in

which (A,0,,1,) is a distributive lattice and is a unary operation on A with

a (a) = 0 and a (a) = 1 for all a in A. As we see now, a Boolean algebra is

nothing but a distributive lattice in which every element has a (unique) complement.

In particular, P(X) and (X) of Example 4.5 are in fact Boolean algebras, and the

symbol agrees with the same symbol used for the negation operation on (X).

Theorem 6.1. If (A,0,,1,,) is a Boolean algebra, then so is (A,1,,0,,), and the

map : A A is inverse to itself and determines isomorphisms

(A,0,,1,,) (A,1,,0,,) and (A,1,,0,,) (A,0,,1,,).

Proof. We already know that complements are uniquely determined in any

distributive lattice (whenever they exist), and since the binary relation

{(x,y) A x is a complement of y}

is obviously symmetric, it follows that the map : A A is inverse to itself. After

that it suffices to prove that

x y y x. (6.1)

By Theorem 4.3(a), x is the largest element among the elements a in A with

x a = 0. Therefore, all we need to prove is the implication x y x y = 0. For

x y we have

35

t x y (t x & t y) (t y & t y) t y y

t = 0,

and so x y = 0, as desired.

Corollary 6.2. In every Boolean algebra A, we have

(x y) = (x) (y), (6.2)

de Morgan laws

(x y) = (x) (y), (6.3)

x = x, (6.4)

for every x, y A.

These results of course simplify calculations of algebraic expressions in a Boolean

algebra, but there are further simplifications. We shall present here one of them,

which transforms Boolean algebras into commutative rings (see Section I.16) with

idempotent multiplication:

Theorem 6.3. (a) Let A = (A,0,,1,,) be a Boolean algebra. Putting

x y = (x y) (y x), x = x, and xy = x y (6.5)

determines a commutative ring structure on A with the same 0 and 1.

(b) Let (A,0,,,1,) be a commutative ring with idempotent multiplication. Then

x = x for every x A, and putting

x y = x y xy, x y = xy, and x = 1 x (6.6)

determines a Boolean algebra structure on A with the same 0 and 1.

(c) the constructions above determine inverse to each other bijections between the

Boolean algebra structures on a given set A and the commutative ring structures on A

with idempotent multiplication.

Proof. (a): Allowing ourselves to omit some simple details, we have

x (y z) = (x ((y z) (z y))) (((y z) (z y)) x)

= (x (y z) (z y)) (y z x) (z y x)

= (x (y z) (z y)) (y z x) (z y x)

= (x y z) (x z y) (y z x) (z y x)

= (x y z) (x y z) (x y z) (x y z),

and similarly (x y) z can be expressed in the same way; that is, is associative.

We also have

x 0 = (x 0) (0 x) = (x 1) 0 = x 0 = x,

36

x x = (x x) (x x) = 0 0 = 0,

and is obviously commutative. That is, (A,0,,) is an abelian group with x = x for

every x A. The fact that (A,1,) is an idempotent commutative monoid immediately

follows from definition of a Boolean algebra. It remains to show that

x(y z) = xy xz for every for every x, y, z A (recall from I.16(k) that 0x = 0 holds

in every ring). We have

x(y z) = x ((y z) (z y)) = (x y z) (x z y),

xy xz = ((x y) (x z)) ((x z) (x y))

= ((x y) (x z)) ((x z) (x y))

= (x y z) (x z y),

and so x(y z) = xy xz.

(b): The equality x = x follows from

x x x x = xx xx xx xx = (x x)(x x) = x x.

Next, we have

x (y z) = x (y z yz) x(y z yz) = x y z xy xz yz xyz

= (x y xy) z (x y xy)z = (x y) z,

x 0 = x 0 x0 = x,

x y = x y xy = x y (since and are commutative),

x x = x x xx = 0 x = x,

and so (A,0,) is a semilattice. The same is true for (A,1,) = (A,1,) by our

assumptions. We also have

x (x y) = x xy xxy = x xy xy = x,

x (x y) = x(x y xy) = xx xy xxy = x xy xy = x,

x (y z) = x(y z yz) = xy xz xyz = xy xz xyxz = (x y) (x z),

x x = x(1 x) = x xx = x x = 0,

x x = x 1 x x(1 x) = 1 + x x x x = 1 0 = 1,

and so (A,0,,1,,) is a Boolean algebra.

(c): Let (A,0,,1,,) be a Boolean algebra, (A,0,,,1,) the corresponding ring

defined as in (a), and then (A,0',',1',',') the Boolean algebra obtained from it as in

(b). We have

37

x ' y = xy = x y,

for all x, y A, and so the orders on A determined by the two Boolean algebra

structures coincide. This immediately tells us that (A,0,,1,,) = (A,0',',1',',').

Conversely, let (A,0,,,1,) be a commutative ring with idempotent multiplication,

(A,0,,1,,) the corresponding Boolean algebra defined as in (b), and then

(A,0',',',1',') the ring obtained from it as in (a). We have

x ' y = (x y) (y x) = x(1 y) y(1 x) xy(1 y)(1 x)

= x xy y xy 0 = x y,

x'y = x y = xy (=xy),

which completes the proof.

7. Semirings and semimodules

Semirings and semimodules are defined in paragraphs (j) and (o) of Section I.16

respectively. If R = (R,0,,1,) is a semiring and A = (A,0,,) an R-semimodule, we

shall say that (A,0,) is the underlying (commutative) monoid of (A,0,,), and the

map

: RA A, written as (r,a) = ra, (7.1)

is the scalar multiplication of (A,0,,).

Example 7.1. Let (A,0,) be a commutative monoid. The set End(A) of monoid

endomorphisms of A (that is, monoid homomorphisms A A) has a semiring

structure End(A) = (End(A),0,,1,) determined by

(f g)(a) = f(a) g(a) and (fg)(a) = f(g(a)) for all f, g End(A) and a A, (7.2)

which implies that 0 End(A) is the zero map sending all elements of A to 0 of A, and

1 End(A) is the identity map of A. We omit the proof, which is a straightforward

calculation.

The next theorem will show the importance of this simple example. In order to

formulate it, for given semiring R = (R,0,,1,) and commutative monoid (A,0,),

consider the diagram

38

inclusion inclusion (7.3)

'

in which is the canonical bijection defined by

()(r)(a) = (r,a), (7.4)

for each map : RA A, each r R and each a A; and we are going to show that

induces ' displayed as the dotted arrow, which itself is a bijection. That is, we are

going to prove:

Theorem 7.2. Let R = (R,0,,1,) is a semiring and A = (A,0,) a commutative

monoid. The canonical map of (7.3) induces a bijection between the set of maps

: RA A making (A,0,,) an R-semimodule, and the set of semiring

homomorphisms R End(A).

Proof. We have to prove that a map : RA A makes (A,0,,) an R-semimodule

if and only if the map : R End(A) defined by (r)(a) = (r,a) is a semiring

homomorphism. In other words, writing (r,a) as ra, we have to show that

1a = a, r(sa) = (rs)a, r0 = 0 = 0a, r(a b) = ra rb, and (r s)a = ra sa, (7.5)

for all r, s R and a, b A if and only if

(r)(0) = 0, (r)(a b) = (r)(a) (r)(b), (0) = 0, (r s) = (r) (s),

(1) = 0, and (rs) = (r)(s), (7.6)

also for all r, s R and a, b A. However, by the definition of , the equalities (7.6)

express the same properties as those of (7.5), up to the order of equalities.

8. The semiring ℕ of natural numbers

The structure ℕ = (ℕ,0,s) of natural numbers can be introduced as an initial object in

the category C of triples (X,e,f), in which X is a set, and e and f are, respectively, a

nullary and a unary operation on X; the morphisms of triples are defined as usually for

-algebras (see Section I.11). That is, ℕ is a set, 0 an element in ℕ, and s : ℕ ℕ a

map, such that, for every triple (X,e,f) as above, there exists a unique map u : ℕ X

with u(0) = e and us = fu. Using this definition we can prove so-called Peano axioms:

The set of all

maps RA A

The set of all maps from R

to the set of all maps A A

The set of all maps

: RA A making

(A,0,,) an

R-semimodule

The set of all semiring

homomorphisms

R End(A)

39

Theorem 8.1. (a) s : ℕ ℕ induces a bijection ℕ ℕ\{0}.

(b) (Induction Principle) If M is a subset in ℕ with

0 M and nℕ (n ℕ s(n) ℕ), (8.1)

then M = ℕ.

Proof. (a): Consider a triple (ℕ {e},e,f), in which e ℕ and f is defined by

f(e) = 0 and f(n) = s(n) for each n ℕ. (8.2)

The map v : ℕ {e} ℕ, defined by v(x) = f(x) for all x in ℕ {e}, is obviously a

morphism (ℕ {e},e,f) (ℕ,0,s) in the category C above, of triples. On the other

hand, since (ℕ,0,s) is an initial object in C, there is a morphism

u : (ℕ,0,s) (ℕ {e},e,f) in C, with vu = 1ℕ.

Since v(e) = 0, it only remains to show that uv(n) = n for each n ℕ. For, consider the

map ℕ ℕ {e} defined by n uv(n), and the inclusion map between the same

sets: since both of them are obviously morphisms from (ℕ,0,s) to (ℕ {e},e,f), they

must coincide.

(b): By the assumption on M, we can form the object (M,0,f) in C, in which

f(m) = s(m) for every m in M. The inclusion map i : M ℕ is then obviously a

morphism (M,0,f) (ℕ,0,s) in C. Again, since (ℕ,0,s) is an initial object in C, there

exists a morphism u : (ℕ,0,s) (M,0,f) in C, and iu = 1ℕ. This makes i surjective, and

so M = ℕ, as desired.

After that, introducing addition and multiplication on ℕ as usually, we make a

semiring ℕ = (ℕ,0,,1,), in which s(n) = n 1 for each n ℕ; we omit the details

since the semiring axioms are nothing but familiar laws of arithmetic here. Moreover,

we have:

Theorem 8.2. The semiring ℕ = (ℕ,0,,1,) of natural numbers is the initial object in

the category of semirings.

Proof. Let R = (R,0,,1,) be an arbitrary semiring. We can form the triple (R,0,f), in

which f : R R is defined by f(r) = r 1, and this uniquely determines a morphism

u : (ℕ,0,s) (R,0,f) in the category C considered above. According to the definition

to such a morphism, we have:

u(0) = 0 and u(n 1) = u(n) 1, (8.3)

for all n ℕ, and since this implies u(1) = u(0 1) = u(0) 1 = 0 1 = 1, we can

equivalently write

u(0) = 0, u(1) = 1, and u(n 1) = u(n) u(1). (8.4)

Since every semiring homomorphism satisfies (8.4), it only remains to prove that

40

u(x y) = u(x) u(y) and u(xy) = u(x)u(y), (8.5)

for all x, y ℕ.

We shall use the Induction Principle in the form presented in Theorem 8.1(b). For any

x ℕ, put Mx = {y ℕ u(x y) = u(x) u(y)}. Then 0 is in Mx, and for y Mx, we

have

u(x (y 1)) = u((x y) 1)) (by the associativity of addition in ℕ)

= u(x y) 1 (by the second equality in (8.3))

= (u(x) u(y)) 1 (since y is in Mx)

= u(x) (u(y) 1) (by the associativity of addition in R)

= u(x) u(y 1) (by the second equality in (8.3)),

which proves that y 1 is in Mx. By the Induction Principle, Mx = ℕ, and so the first

equality in (8.5) holds for all x, y ℕ. Next, put Mx' = {y ℕ u(xy) = u(x)u(y)}.

Then again, 0 is in Mx', and for y Mx', we have

u(x(y 1)) = u(xy x) (since ℕ is a semiring)

= u(xy) u(x) (by the first equality in (8.5), which is already proved)

= u(x)u(y) u(x) (since y Mx')

= u(x)(u(y) 1) (since R is a semiring)

= u(x)(u(y 1)) (by the second equality in (8.3)),

which proves that y 1 is in Mx'. By the Induction Principle, Mx' = ℕ, and so the

second equality in (8.5) holds for all x, y ℕ.

Let us compare this result with Theorem 7.2. By Theorem 8.2, for any commutative

monoid A, there exist a unique semiring homomorphism ℕ End(A), and on the

other hand, by Theorem 7.2, to give such a homomorphism is the same as to make A

an ℕ-semimodule. Therefore we have:

Corollary 8.3. For every a commutative monoid A = (A,0,), there exist a unique map

: RA A making (A,0,,) an ℕ-semimodule.

9. Number systems, ℤ, ℚ, ℝ, and ℂ as rings, and their modules

There are five most important classical number systems:

ℕ, the system of natural numbers. It is the set of natural numbers (where we

agreed to include 0) equipped with the familiar operations that make it a

commutative semiring, as mentioned in the previous section.

ℤ, the system of integers. The familiar operations make it a commutative ring.

Note that a commutative ring can be defined as a system (A,0,,,1,) in which

(A,0,,1,) is a commutative semiring and (A,0,,) is a group (cf. I.16(k)-(l)).

The systems ℚ, ℝ, and ℂ of rational, real, and complex numbers respectively.

The familiar operations make them fields; for the definition of a field see

Section I.16(m).

41

Assuming all this to be well known, we will present ℤ, ℚ, ℝ, and ℂ as initial objects

in certain categories. For the system of natural numbers this was done in two very

ways, very different from each other in the previous section: as initial triple (ℕ,0,s)

and as initial semiring (ℕ,0,,1,). In fact there many more ways to do this, for ℕ and

for other number systems – but we shall make only one presentation for each of them.

9.1. Integers. Let C be the category of pairs (A,f), where A is an abelian group, and f

is a monoid homomorphism from the additive monoid of natural numbers to A; a

morphism : (A',f ') (A,f) in C is a group homomorphism from A' to A with

f ' = f. We are going to prove that the pair (ℤ,), where : ℕ ℤ is the inclusion

map, is an initial object in C. For, we have to show that for every object (A,f) in C,

there exists a unique group homomorphism : ℤ A with

(n) = f(n) for every natural number n.

Existence: We define : ℤ A by

(n) = f(n) and (n) = f(n) (9.1)

for any natural number n, and we only need to show that

(z t) = (z) (t) (9.2)

for every two integers z and t. Since f is a monoid homomorphism, and (n) = f(n) for

every natural n, (9.2) certainly holds for natural z and t. After that, and using this fact,

we calculate:

For z and t both negative:

(z t) = (((z) (t))) = f((z) (t)) (since (z) (t) is natural and by (9.1))

= (f(z) f(t)) (since f is a monoid homomorphism)

= (f(z)) (f(z)) = ((z)) ((t)) (since (z) and (t) are natural and by

(9.1))

= (z) (t).

For z natural, t negative, and z t:

(z t) (z) = (((z t))) (z) = f((z t)) f(z) (since (z t) and z are

natural and by (9.1))

= (f((z t)) f(z)) = (f((z t) z)) (since f is a monoid homomorphism)

= (f(t)) = ((((t))) (since (t) is natural and by (9.1))

= (t), and so (z t) = (z) (t).

Similarly the equality (9.2) can be shown also for all other cases, and so it holds for

every two integers z and t, as desired.

Uniqueness: We need to prove that whenever a group homomorphism : ℤ A

satisfies the first equality in (9.1), it should also satisfy the second one. Indeed, we

have:

42

(n) f(n) = (n) (n) = (n n) = (0) = 0,

and so (n) = f(n), as desired.

9.2. Rational numbers. An element a of a ring A is said to be invertible if

ab = 1 = ba for some b in A. We take C to be the category of pairs (A,f), where A is a

commutative ring, and f is a ring homomorphism from the ring of integers to A,

sending all non-zero integers to invertible elements in A; a morphism

: (A',f ') (A,f) in C is a ring homomorphism from A' to A with f ' = f. We are

going to prove that the pair (ℚ,), where : ℤ ℚ is the inclusion map, is an initial

object in C. For, we have to show that for every object (A,f) in C, there exists a unique

ring homomorphism : ℤ A with

(z) = f(z) for every integer z.

Existence: We define : ℚ A by

(z/t) = f(z)(f(t))1

, (9.3)

for any integers z and t with non-zero t, and we need to show that

z/t = u/v f(z)(f(t))1

= f(u)(f(v))1

(to show that (9.3) indeed defines a map from ℚ to A), that

(z/t u/v) = (z/t) (u/v),

((z/t)(u/v)) = (z/t)(u/v),

(1) = 1,

(to show that is a ring homomorphism), and that = f. We have:

z/t = u/v zv = ut f(zv) = f(ut) f(z)f(v) = f(u)f(t) f(z)(f(t))1

= f(u)(f(v))1

,

(z/t u/v) = ((zv ut)/(tv)) = f(zv ut)(f(tv))1

= (f(zv) f(ut))(f(tv))1

= f(zv)(f(tv))1

f(ut)(f(tv))1

= f(z)f(v)(f(v))1

(f(t))1

f(u)f(t)(f(v))1

(f(t))1

= f(z)(f(t))1

f(u)(f(v))1

= (z/t) (u/v),

((z/t)(u/v)) = ((zu/tv)) = f(zu)(f(tv))1

= f(z)f(u)(f(t)(f(v))1

= f(z)(f(t)) 1

f(u)(f(v))1

= (z/t)(u/v),

(1) = f(1)(f(1))1

= 1,

(z) = (z) = (z/1) = f(z)(f(1))1

= f(z).

Uniqueness: We need to show that whenever a ring homomorphism : ℚ A has

(z) = f(z) for every integer z, it should satisfy (9.3). We have

43

(z/t)f(t) = (z/t)(t) = ((z/t)t) = (z),

and so (z/t) = f(z)(f(t))1

, as desired.

9.3. Real numbers. We shall need a notion of completeness different from the one

introduced in Section 5: A linear order (X,) (see Definition I.7.1 and Remark I.7.2) is

said to be complete, if for every non-empty subset S of X, we have

(xX sS x s) (S exists), (9.4)

that is, if there exist an element x X with x s for all s S, then there exist the

largest x X with this property. One also says: if S admits a lower bound, then it also

admits the largest lower bound. Note that, similarly to Theorem 5.1, our requirement

on (X,) is equivalent to the dual one, that is, instead of (9.4) we could equivalently

require

(xX sS s x) (S exists), (9.5)

for every non-empty subset S of X.

Here, assuming familiarity with the usual linear order on ℚ, we take C to be the

category of pairs (C,f), where C is a complete linear order and f : ℚ C a map that

preserves all (existing) meets, that is, f(S) = f(S) for every subset S in ℚ that admits

a meet in ℚ (with respect to the standard order in ℚ); a morphism : (C ',f ') (C,f)

in C is a meet-preserving map from C ' to C with f ' = f. We are going to prove that

the pair (ℝ,), where : ℚ ℝ is the inclusion map, is an initial object in C. For, we

have to prove that for every object (C,f) in C, there exists a unique meet-preserving

map : ℝ C with

(x) = f(x) for every rational number x.

Existence: We define : ℝ C by

(x) = {f(y) x y ℚ},

for every x in ℝ, and we need to prove that preserves meets (that exist), and that

= f. Let us begin with the second assertion. For each rational number x, f(x) belongs

to the set {f(y) x y ℚ}; moreover, since f preserving existing meets certainly

preserves order, f(x) is the smallest element in {f(y) x y ℚ}. Therefore, when x is

rational, we have:

(x) = (x) = {f(y) x y ℚ} = f(x),

as desired. Now let us prove the first assertion. Preservation of (existing) meets by

means that, for every subset S of ℝ that has a lower bound in ℝ, we have

(S) = (S).

44

In order to prove this equality, we observe:

If x x', then the set {f(y) x y ℚ} contains the set {f(y) x' y ℚ}, and

so (x) = {f(y) x y ℚ} {f(y) x' y ℚ} = (x'). That is, the map

preserves order.

Since preserves order, we have (S) (s), for every is in S, and so

(S) (S). It remains to prove (S) (S).

Since (S) = {f(y) S y ℚ}, to prove (S) (S) is to prove that

(S) is a lower bound of the set {f(y) S y ℚ}.

That is, we have to prove that (S) f(y) whenever S y ℚ. If S is strictly less

than y, then s y for some s S, and then (s) (y) = f(y), which immediately gives

(S) f(y). Therefore we can assume that S = y, and, in particular, that S is

rational. Next, we observe:

Let us say that subsets A and B of an ordered set P are equivalent if

aAbB b p and bBaA a b.

Then obviously, if A and B are equivalent, then A has meet if and only if B

has, and if this is the case, then A = B. Note also that if u : P Q is an

order-preserving map, and A and B are equivalent to each other, then so are

u(A) and u(B).

When S is rational, the sets S and T = {t ℚ sS s t} are equivalent.

And, using these observations and assuming S = y, we conclude: (S) = (T)

= f(T) = f(T) = f(S) = f(y), which completes the proof.

Uniqueness follows from the following three facts:

preserves (existing) meets;

restricted to ℚ coincides with f;

every real number can be presented as meet of a set consisting of rational

numbers.

9.4. Complex numbers. Now we take C to be the category of triples (A,,f), where A

is a commutative ring (with 1), is an element in A with 2 = 1, and f is a ring

homomorphism from the ring of real numbers to A; a morphism

: (A',',f ') (A,,f) in C is a ring homomorphism from A' to A with (') = and

f ' = f. We are going to prove that the pair (ℂ,i,), where : ℝ ℂ is the inclusion

map, is an initial object in C. For, we have to show that for every object (A,,f) in C,

there exists a unique ring homomorphism : ℂ A with

(i) = and (x) = f(x) for every real number x. (9.6)

Existence: We define : ℂ A by

(a bi) = f(a) f(b); (9.7)

45

then checking that is a ring homomorphism is a straightforward calculation, and

(9.6) obviously holds.

Uniqueness: If is a ring homomorphism satisfying (9.6), then

(a bi) = (a) (bi) = (a) (b)(i) = f(a) f(b),

that is, must satisfy (9.7), which determines it uniquely.

9.5. Modules. As I.16(p) says, for a ring R = (R,0,,,1,), an R-module, or a module

over R, is the same as a semimodule over (R,0,,1,). Note, however, that:

(a) If A = (A,0,,) is an R-module, then the monoid (A,0,) is a group (that is it

admits a group structure (A,0,,) with the same (A,0,) and) with a = (1)a for each

a A. Indeed, we have

a (1)a = 1a (1)a = (1 (1))a = 0a = 0.

On the other hand, when a monoid (commutative or not) admits a group structure,

such a structure is unique, since, using the additive notation again, we have

(a b = 0 = c a) (b = b a c = c).

That is, the formula a = (1)a above gives the unique way to make (A,0,) a group.

(b) Defining an R-module, we could actually omit three requirements from the

definition of a semimodule, since they can be deduced from others. They are the

identities a b = b a, 0a = 0, and r0 = 0 (for a, b A and r R). Indeed, we have:

a b = (a) a a b b (b) = (a) (1 1)a (1 1)b (b)

= (a) (1 1)(a b) (b) = (a) a b a b (b) = b a,

0a = 0a 0a ((0a)) = (0 0)a ((0a)) = 0a ((0a)) = 0,

r0 = r0 r0 ((r0)) = r(0 0) ((r0)) = r0 ((r0)) = 0.

(c) For every r R and a A we obviously have

r(a) = (ra) = (r)a, (9.8)

and so we shall write simply ra for each of these expressions. We shall also write

x y instead of x (y) when x and y are either both in R or both in A, and we have

(r s)a = ra sa and r(a b) = ra rb, (9.9)

for all r, s R and a, b A.

(d) Using Theorem 8.2, and the initiality of (ℤ,) established in 9.1, it is easy to show

that the ring ℤ of integers is the initial object in the category of rings. Then, by

46

observing that when a commutative monoid (A,0,) is a group structure, the

endomorphism semiring End(A) becomes a ring, we conclude that every abelian

group admits a unique ℤ-module structure. We should only note that this conclusion

uses that obvious fact that every semiring homomorphism between rings is a ring

homomorphism.

(e) Using (d) and the initiality of (ℚ,) established in 9.2, it is easy to show any

abelian group A admits at most one ℚ-module (=ℚ-vector space) structure, and that it

admits it if and only if, for every nonzero n ℕ, the map A A defined by a na is

bijective (which will make it an invertible element in End(A)).

Finally, we shall not describe abelian groups that admit ℝ- or ℂ-vector space

structure. Let us only mention that the homomorphisms ℚ ℝ ℂ allow

considering any ℂ-vector space as an ℝ-vector space, and considering any ℝ-vector

space as a ℚ-vector space.

10. Pointed categories

Definition 10.1. A pointed category is a category C equipped with a family

(0A,B)(A,B)C0C0

of morphisms in C indexed by pairs of objects in C, with 0A,B homC(A,B) and

g0A,Bf = 0X,Y for every f : X A and g : B Y.

Example 10.2. Let C be a full subcategory (see Example I.8.5) of a category of the

form -Alg, as defined in Section I.11. Then C is pointed whenever every -algebra

that is in C has a unique one-element subalgebra. Out of the examples of classes of

algebras given in Section I.16 this is the case for the items (b), (e)-(i), (o)-(r), (t), (u),

and for “rings without 1” mentioned in (k).

Theorem 10.3. Any category admits at most one structure of a pointed category.

Proof. Let (0A,B)(A,B)C0C0 and (0'A,B)(A,B)C0C0 be two such structures on a category C.

then 0'A,B = 0'A,B0A,A = 0A,B, where the first equality follows from the fact that

(0'A,B)(A,B)C0C0 determines a pointed category structure on C, and the second one

follows from the fact that (0A,B)(A,B)C0C0 does.

Theorem 10.4. Let C be a category that admits either initial or terminal object. Then

the following conditions are equivalent:

(a) C is pointed (i.e. it admits a pointed category structure, which is unique by

Theorem 10.3);

(b) 0 = 1 in C (that is, C has an object that is initial and terminal at the same time);

(c) C has 0 and 1 with homC(1,0) .

Proof. (a)(c) is trivial. (c)(b): Having a morphism 1 0 makes 0 1 since the

composites 0 1 0 and 1 0 1 must be identity morphisms. (b)(a): For

arbitrary objects A and B in C, just define 0A,B as the composite A 1 = 0 B. To

47

show the equality g0A,Bf = 0X,Y of Definition 10.1 follows from the commutativity of

the diagram

X f A

0A,B B

g Y

1 0

where the unlabeled arrows are the suitable uniquely determined morphisms into 1 or

from 0.

11. Products and coproducts

Definition 11.1. Let C be a category and (Ai)iI be a family of objects in C. Then:

(a) A cone over (Ai)iI is a family of the form (fi : C Ai)iI of morphisms in C. A

morphism from a cone (fi : C Ai)iI to a cone (gi : D Ai)iI is a morphism

h : C D with gih = fi for each i I.

(b) The (cartesian) product

Ai = (Ai,(i : Ai Ai)iI) = (i : Ai Ai)iI (11.1) iI iI iI iI

of the family (Ai)iI is defined as a (the) terminal object in the category of cones over

the family (Ai)iI; the morphisms i involved are called product projections. We will

also write

n

Ai = A1…An i=1

when the family is presented as a sequence A1, …, An, etc.

That is, in contrast to the usual definition of cartesian product, the categorical

definition presents it not just as an object that is product, but as a cone, that is, as an

object equipped with morphisms to all members of the given family. Furthermore,

being defined as a terminal object, it is unique only up to isomorphism of cones. This

easily gives:

Theorem 11.2. Let (gi : D Ai)iI be a product of the family (Ai)iI and

(fi : C Ai)iI be another cone over the same family. Then following conditions are

equivalent:

(a) (fi : C Ai)iI is a product of the same family (Ai)iI;

(b) the unique morphism h : C D, with gih = fi for each i I, is an isomorphism;

(c) there exists an isomorphism h : C D, with gih = fi for each i I.

48

Thinking of isomorphic objects informally as “essentially equal”, this suggests the

following convention in fact already used in the formulation of Definition 11.1(b):

Convention 11.3. Whenever a cone (gi : D Ai)iI is terminal, that is, has the

property:

for every cone (fi : C Ai)iI, over the same family,

there exists a unique morphism h : C D with (11.2)

gih = fi for each i I,

we shall say that (gi : D Ai)iI is the product of the family (Ai)iI, and write

D = Ai. iI

Example 11.4. Let C = -Alg be the category of all -algebras and their

homomorphisms, as defined in Section I.11. Then the cartesian product

Ai, defined as in Section I.13, and equipped with the family (i : Ai Ai)iI iI iI

of (ordinary) projections defined by i((ai)iI) = ai, is the product of the family (Ai)iI

in the sense of Definition 11.1(b). Indeed, given a cone (fi : C Ai)iI, to give a

morphism

h : C Ai with ih = fi for each i I, iI

is to give

h : C Ai with h(c)i = fi(c) for each c C and i I, iI

or, equivalently, is to give

h : C Ai with h(c) = (fi(c))iI for each c C. iI

And, since requiring h(c) = (fi(c))iI for each c C actually defines h as a map, all we

need to check is that h is a homomorphism of -algebras. For every n = 0, 1, 2, …,

for every in n, and for every c1, …, cn in C, we have

h((c1,…,cn)) = (fi((c1,…,cn)))iI (by the definition of h)

= ((fi(c1),…,fi(cn)))iI (since each fi is a homomorphism of -algebras)

= ((fi(c1))iI,…,(fi(cn))iI) (by Definition I.13.1)

= (h(c1),…,h(cn)),

as desired.

Remark 11.5. In particular, the construction above can be used in the case = ,

that is, for the category of sets. In fact it would be more natural to begin with the

category of sets, and then simply say that the products of algebras are constructed as

for sets, but with suitable algebraic structure.

49

Coproducts are defined dually to products, that is, by reversing all arrows in

Definition 11.1:

Definition 11.6. Let C be a category and (Ai)iI be a family of objects in C. Then:

(a) A cocone over (Ai)iI is a family of the form (fi : Ai C)iI of morphisms in C. A

morphism from a cocone (fi : Ai C)iI to a cocone (gi : Ai D)iI is a morphism

h : C D with gi = hfi for each i I.

(b) The coproduct

Ai = (Ai,(i : Ai Ai)iI) = (i : Ai Ai)iI (11.3) iI iI iI iI

of the family (Ai)iI is defined as a (the) initial object in the category of cocones over

the family (Ai)iI; the morphisms i involved are called coproduct injections. We will

also write

n

Ai = A1 … An i=1

when the family is presented as a sequence A1, …, An, etc.

We shall use convention similar to Convention 11.3.

Example 11.7. In the category of sets coproducts are disjoint unions. More precisely,

we can put

Ai = {i}Ai with i : Ai Ai defined by i(a) = (i,a). (11.4) iI iI iI

Indeed, given a cocone (fi : Ai C)iI, to give a map

h : Ai C with hi = fi for each i I, iI

is to give

h : Ai C with h(i,a) = fi(a) for each i I and a Ai, iI

which simply defines h.

Example 11.8. Let R be a semiring, and C the category of R-semimodules. Then we

can put

Ai = {(ai)iI Ai {i I ai 0} is finite} with the R-semimodule (11.5) iI iI

structure making Ai a R-subsemimodule of Ai, and with i : Ai Ai iI iI

defined by

50

a, if i = j,

(i(a))j = (11.6)

0, if i j.

To prove this we need to prove that, given a cocone (fi : Ai C)iI, there exists a

unique R-semimodule homomorphism

h : Ai C with hi = fi for each i I. (11.7) iI

For, first we observe that, for each (ai)iI Ai, we have iI

(ai)iI = i(ai), (11.8) iI

where the possibly infinite sum as the sum of the non-zero summands involved; this

formula easily follows from (11.6). Next, if h is an R-semimodule homomorphism

satisfying (11.7), then using (11.8) we obtain

h((ai)iI) = h(i(ai)) = hi(ai) = fi(ai), iI iI iI

and so whenever h exists, it is uniquely determined and must be defined by

h((ai)iI) = fi(ai). (11.9) iI

It remains to show that h, defined by (11.9), is an R-semimodule homomorphism with

hi = fi for each i I. We have:

h((ai)iI (bi)iI) = h((ai bi)iI) (by the definition of addition in Ai) iI

= fi(ai bi) (by (11.9)) iI

= (fi(ai) fi(bi)) (since each fi is an R-semimodule homomorphism) iI

= fi(ai) fi(bi) (since the addition in C is associative and commutative) iI iI

= h((ai)iI) h((bi)iI) (by (11.9)),

h(r(ai)iI) = h((rai)iI) (by the definition of scalar multiplication in Ai) iI

= fi(rai) (by (11.9)) iI

= rfi(ai) (since each fi is an R-semimodule homomorphism) iI

= rfi(ai) (by the suitable distributivity) iI

= r(h((ai)iI)) (by (11.9)),

hi(a) = fj((i(a))j) (by (11.9)) jI

= fi(ai) (by (11.6) and since fj(0) = 0 for each j),

51

as desired.

12. Direct sums

As the Examples 11.4 and 11.8 show, if a family (Ai)iI of R-semimodules has the

index set I finite, we have

Ai = Ai, (12.1) iI iI

as semimodules, that is, if we ignore the product projections and the coproduct

injections. On the other hand, the product projections and the coproduct injections

chosen exactly as in Examples 11.4 and 11.8 respectively, satisfy the equalities

1, if i = j,

ii =

0, if i j,

(12.2)

ii = 1, iI

where 1 and 0 denote suitable identity and zero morphisms respectively. This is a

general phenomenon, which we will describe below, for simplicity, in the case of a

two-element index set.

Definition 12.1. A linear category is a category C equipped with (additive)

commutative monoid structures on all homC(A,B) (A, B C0), such that for every

f : X A and g : B Y in C, the map

homC(A,B) homC(X,Y), defined by u guf

is a monoid homomorphism.

Remark 12.2. Every linear category is obviously pointed with 0A,B homC(A,B)

defined as the zero element of the additive monoid homC(A,B).

Definition 12.3. Let C be a linear category. A diagram

1 2

A AB B (12.3) 1 2

in C is said to be a direct sum diagram, if the equalities (12.2) hold, that is, if

11 = 1A, 22 = 1B, 12 = 0B,A, 21 = 0A,B, 11 22 = 1C. (12.4)

Theorem 12.4. Suppose a diagram of the form (12.3) in a linear category satisfies

only the first four equalities of (12.4). For such a diagram the following conditions are

equivalent:

(a) it is a direct sum diagram, that is, the fifth equality of (12.4) also holds;

52

(b) the pair (1,2) makes AB the product of A and B (that is, of the family, whose

members are A and B);

(c) the pair (1,2) makes AB the coproduct of A and B.

Proof. The implications (a)(b) and (a)(c) are dual to each other as well as the

implications (b)(a) and (c)(a); that is, having proved, say, (a)(b) and (b)(a),

one can transform the arguments into proofs of (a)(c) and (c)(a) simply reversing

the arrows. We shall therefore prove only (a)(b) and (b)(a).

(a)(b): Assuming that (12.3) is a direct sum diagram, we have to prove that given

arbitrary two morphisms f : C A and g : C B with the same domain, there exists

a unique morphism h : C AB with 1h = f and 2h = g.

Existence: We take h = 1f 2g, and then

1h = 1(1f 2g) = 11f 12g = 1Af 0B,Ag = f

and similarly 2h = g.

Uniqueness: If 1h = f and 2h = g, then

1f 2g = 11h 22h = (11 22)h = 1Ch = h.

(b)(a): Assuming that the pair (1,2) forms a product diagram, we have to prove

the fifth equality of (12.4). This simply means to prove

1(11 22) = 1 and 2(11 22) = 2,

which is a straightforward calculation: 1(11 22) = 111 122 = 1A1 0B,A2

= 1 and similarly for 2.

Example 12.5. For any semiring R, the category C of R-semimodules is, of course

linear. The additive monoid structure on each homC(A,B) is given by

(f g)(a) = f(a) g(a),

as it was done in a special case in Example 7.1. For this category the direct sum

diagram (12.3) can be described as

AB = AB, 1(a,b) = a, 2(a,b) = b, 1(a) = (a,0), 2(b) = (0,b). (12.5)

13. Free algebras and free semimodules

Let C be as in Example 10.2, that is, a full subcategory (see Example I.8.5) of a

category of the form -Alg, as defined in Section I.11. The objects in C will be called

C-algebras.

Definition 13.1. Let X be a set, and C[X] the category of pairs (A,f), in which A is a

C-algebra and f a map from X to (the underlying set of) A; a morphism

53

h : (A,f) (B,g) in C[X] is a homomorphism h : A B with hf = g. A (the) free

C-algebra on X is an (the) initial object in C[X].

Remark 13.2. As follows from this definition, the free C-algebra on the empty set is

nothing but the initial object in C – or, to be absolutely precise, it is the pair

(initial object in C, empty map into it).

On the other hand, the category C[X] can itself be considered as a full subcategory of

the category of [X]-algebras, where [X] is defined by

[X]0 is the disjoint union of 0 and X, and [X]n = n for n = 1, 2, …

Therefore changing allows considering any free algebra as the free algebra on the

empty set.

Example 13.3. Let R be a semiring and C the category of R-semimodules. Then the

free C-algebra (A,f) on a set X, called the free R-semimodule on X, can be constructed

as follows:

A is the set R(X)

of all maps u : X R such that the set {x X u(x) 0} is finite.

We could equivalently say of course, that it is the set of all families (ux)xX of

elements in R such that the set {x X u(x) 0} is finite, which would be more

convenient in order to describe the relationship with coproducts of semimodules

described in Example 11.8. But using the language of maps is more convenient

here.

The R-semimodule structure on A is defined accordingly; that is

(u v)(x) = u(x) v(x) and (ru)(x) = r(u(x)) (13.1)

for u and v in A, x in X, and r in R.

The map f : X A is defined by

1, if x = y,

f(x)(y) = (13.2)

0, if x y.

To prove this we need to prove that, given an object (B,g) in C[X], there exists a

unique R-semimodule homomorphism

h : R(X)

B with hf = g. (13.3)

For, first we observe that, for each u R(X)

, we have

u = u(x)f(x); (13.4) xX

indeed, for any y X, we have

(u(x)f(x))(y) = u(x)(f(x)(y)) (by the second equality in (13.1) xX xX

54

= u(y) (by (13.2)).

Next, if h is an R-semimodule homomorphism satisfying (13.3), then using (13.4) we

obtain

h(u) = h(u(x)f(x)) = u(x)hf(x) = u(x)g(x), xX xX xX

and so whenever h exists, it is uniquely determined and must be defined by

h(u) = u(x)g(x). (13.5) xX

It remains to show that h, defined by (13.5), is an R-semimodule homomorphism with

hf = g. We have:

h(u v) = ((u v)(x))g(x) (by (13.5)) xX

= (u(x) v(x))g(x) (by the first equality in (13.1)) xX

= (u(x)g(x) v(x)g(x)) (by the suitable distributivity) xX

= u(x)g(x) u(x)g(x) (since the addition in C is associative and commutative) xX xX

= h(u) h(v) (by (13.5)),

h(ru) = (ru)(x)g(x) ((13.5)) xX

= r(u(x))g(x) (by the second equality in (13.1)) xX

= r(u(x)g(x)) (by the suitable distributivity) xX = r(h(u)) (by (13.5)),

hf(y) = (f(x)(y))g(x) (by (13.5)) xX = g(y) (by (13.2) and since 0g(x) = 0 for each x),

as desired.

Remark 13.4. Obvious similarity between the proofs in Examples 11.8 and 13.3 can

be explained by the fact that to construct the free C-algebra on X is the same as to

construct the X-indexed coproduct of C-algebras free on one-element subsets on X.

Remark 13.5. Consider the special case of R = {0,1} with 1 1 = 1. In this case

R-semimodules are nothing but semilattices, since:

for every R-semimodule A and every a A, we have a a = 1a 1a = (1 1)a =

1a = a;

conversely, if A is a semilattice, that is, an idempotent commutative monoid, then

1 1 = 1 holding in A also holds in the semiring End(A); this easily implies that

sending 0 to 0 and 1 to 1 determines a (unique) semiring homomorphism

R End(A), hence making A an R-semimodule.

55

In this case the free R-semimodule on a set X can be identified with the semilattice

Pfin(X) = (Pfin(X),,) of finite subsets in X. Under this identification a map

u : X {0,1} in R(X)

corresponds to the finite subset u1

(1) of X, and obviously the

addition operation in R(X)

corresponds to the union operation in Pfin(X).

14. Vector spaces

In this section we will recall Zorn Lemma and use it to prove that every vector space

is free. In order to formulate Zorn Lemma we need:

Definition 14.1. A subset S of an ordered set P is said to be:

(a) bounded, if there exists p P with s p for every s S;

(b) a chain, if the order of P induces a linear order on S, that is, if for every s, s' S,

either s s' or s' s in P.

Theorem 14.2 (Zorn Lemma). Every ordered set, in which every chain is bounded,

has a maximal element.

Let us also recall that, depending on axioms of set theory we are using, Zorn Lemma

can either be put as one of axioms, or deduced from others; under certain “easy”

axioms it is equivalent to the so-called Axiom of Choice.

In the rest of this section K denotes a fixed field and V a K-vector space. For any set

X, we define the canonical map X : X K(X)

by (13.2), that is, by

1, if x = y,

X(x)(y) =

0, if x y.

Recall from the previous section that this makes (K(X)

,X) free, in the sense that for

every map g : X V there exists a unique linear map (that is, a morphism of K-vector

spaces) h : K(X)

V with hX = g; we shall briefly call h the linear map induced by g.

This map h can be explicitly defined by (13.5).

Definition 14.2. (a) A map g : X V is said to be linearly independent if the induced

linear map h : K(X)

V is injective.

(b) A subset S of V is said to be linearly independent if so is the inclusion map S V.

Since a linear map is injective if and only if it has a trivial kernel (see Theorem

I.17.11), formula (13.5) tells us that g : X V is linearly independent if and only if,

for any u K(X)

, we have

u(x)g(x) = 0 xX u(x) = 0. (14.1) xX

In particular, a subset S of V is linearly independent if and only if, for any u K(S)

, we

have

56

u(s)s = 0 sS u(s) = 0. (14.2) sS

Both (14.1) and (14.2) (required for u K(X)

and for u K(S)

, respectively) can be

called classical definitions of linear independence. Another classical definition is that

of a basis: one usually says that a subset S of V is a basis of V if it is linearly

independent and generates V. In the situation 14.2(a), g(X) generates V if and only if

h : K(X)

V is surjective. However, a linear map is surjective and injective at the

same time if and only if it is an isomorphism. Therefore a subset S of V is a basis of V

if and only if the pair

(S, inclusion map S V)

is a free K-vector space on S (which is nothing but another name for a free

K-semimodule on S). Therefore to say that every K-vector space is free is to say that

every K-vector space has a basis, and, again, to prove this fact is the purpose of this

section.

Remark 14.4. It is easy to see that:

(a) the empty map and the empty set are always linearly independent;

(b) the vector space {0} has a basis, which is the empty set;

(c) when X has only one element x, g : X V is linearly independent if and only if

g(x) 0; in particular a one-element subset set {s} of V is linearly independent if and

only if s 0.

Lemma 14.5. If S is a maximal linearly independent subset in V, then V is generated

by S.

Proof. Let x be an element in V that is not in the set S, which we can assume non-

empty by Remark 14.4. Then S{x} is not linearly independent, and so there exist a

map u : S{x} K, for which the set Y = {y S{x} u(y) 0} is finite and non-

empty and

u(y)y = 0. yY

Since S is linearly independent, u(x) 0; indeed, u(x) = 0 would imply

u(s)s = u(y)y = 0, sS yY

and so make the restriction of u on S a map whose existence contradicts to the linear

independence of S. Since u(x) 0, u(x) is an invertible element in K, and we can write

x = (u(x))1

u(x)x = (u(x))1

(u(x)x u(s)s u(s)s) sS sS

= (u(x))1

(u(y)y u(s)s) = (u(x))1

(u(s)s) = (u(x))1

u(s)s, yY sS sS sS

which shows that x belongs to the subspace of V generated by S.

57

Theorem 14.6. Every vector space has a basis.

Proof. Using the notation above, for a K-vector space V, let S be the set of all linearly

independent subsets of V. Using Zorn Lemma and Lemma 14.5, all we need to show

is that every chain in S is bounded. We will show more, namely, that if S' is a chain in

S, then its union S' is linearly independent (that is, is in S). For, we have to show

that, for u K(S')

with

u(s)s = 0, sS

the finite set X = {x S' u(x) 0} is actually empty. Since X is a finite subset of

S' and S' is a chain, X is included in some S S. Therefore X = {x S u(x) 0},

and

u(s)s = u(s)s = 0. sS sS

Since S is linearly independent, using the restriction of u on S, we conclude that X is

empty.

Remark 14.7. (a) Theorem 14.6 confirms that every vector space is free, as desired

(see the discussion above Remark 14.4).

(b) When V is finitely generated, the proof of Theorem 14.6 can avoid Zorn Lemma

by considering any minimal generating subset S of V, which obviously exists for a

finitely generated V. Indeed, it is easy to show that the minimality of S implies its

linear independence.