ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...
ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...
ABSTRACT ALGEBRAIC STRUCTURES OPERATIONS AND ...
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Anders Madsen<br />
<strong>OPERATIONS</strong><br />
<strong>AND</strong><br />
<strong>STRUCTURES</strong><br />
<strong>ABSTRACT</strong><br />
<strong>ALGEBRAIC</strong><br />
<strong>STRUCTURES</strong><br />
y z<br />
y z<br />
yz<br />
bc<br />
b c<br />
b c<br />
1
This is a booklet in a series with the title ”Abstract algebraic structures”, which I have<br />
written for the algebra courses (E1,BE2) part of the mathematics education at RUC<br />
These booklets should be seen in connection with another series with the title ”Concrete<br />
algebraic structures”. These two series should be viewed as the jaws of a pair of pinchers.<br />
It would obviously be futile (and frustrating) to teach abstract algebra without involving<br />
concrete examples and it would be poor (and devoid of perspective) only to build concrete<br />
examples without involving the underlying abstract structures.<br />
And then again, there is a certain beauty in emphasizing the abstract character by isolating it<br />
and let its top-down nature appear clearly as in done in the controversial idol ”The Elements<br />
of Mathematics” by Bourbaki. Starting with the coarsest structures and then step by step<br />
refining by adding structural elements. All results which occur in many circumstances are<br />
proved once and for ever in its most elementary form.<br />
In the same way there is some satisfaction combined with letting each concrete structure stand<br />
as simple as possible without to much ado, das Ding an sich. And the pleasure by recognizing<br />
the essentially same type of argument recurring over and over in different disguises.<br />
Besides having the aesthetic satisfaction from the pure abstraction and the pure pleasure<br />
from the concrete details, both perspectives have a great cognitive influence and contributes<br />
to the development of competencies which are essential to any mathematician.<br />
I have chosen to emphasize these to oppositely directed but coordinated perspectives which<br />
is implied in the two series.<br />
The individual concrete structures are displayed in separate expositions without mutual<br />
references. Subjects which are needed at more places are repeated at each place. But the<br />
choice of details is made in such a way as to best possible deliver stuff to be used in examples<br />
in the abstract part.<br />
The text of the abstract part is not printed but can be found at<br />
http://milne.ruc.dk/~am/algebra<br />
Anders Madsen, may 2012<br />
2
Table of contents<br />
1 Introduction<br />
2 Operations<br />
1 Background and motivation 8<br />
2 Background terminology 8<br />
3 Definition 9<br />
4 Homomorphisms 10<br />
5 Induced operations 14<br />
6 Induced operation on subset 14<br />
7 Induced operation on function spaces. 16<br />
8 Induced operation on product spaces. 16<br />
9 Induced operation on quotient. 19<br />
10 Axioms for operations. 23<br />
3 Algebraic structures.<br />
1 Definition. 26<br />
2 Homomorphisms and isomorphisms 26<br />
3 Induced structures on subset 27<br />
4 Induced structure on function spaces 29<br />
5 Induced structure on quotient 31<br />
4 Appendix: Tables.<br />
1 Symbols for sets. 32<br />
2 Tables with operations 34<br />
3 Tables with homomorphisms 35<br />
5 Examples and exercises<br />
6 Appendix : Classification<br />
7 Semigroups<br />
1 Definition 46<br />
3
2 Induced semigroups 47<br />
3 Powers 48<br />
8 Monoids<br />
1 Definitions 48<br />
2 Induced monoids 49<br />
3 Examples 51<br />
4 Monoid homomorphisms 51<br />
5 Inverse 52<br />
6 Powers 52<br />
7 Translations 53<br />
9 Groups<br />
1 Definitions 54<br />
2 Induced group structures 55<br />
3 subgroup 55<br />
4 Quotient groups 59<br />
5 Group homomorphisms 61<br />
10 Rings<br />
1 Definitions and rules 62<br />
2 Subring 62<br />
3 Induced rings 63<br />
4 Quotient rings and ideals 64<br />
5 Integral domains 66<br />
11 Fields<br />
1 Definitions 68<br />
2 Quotient ring over a maximal ideal 69<br />
3 Fields of fractions 69<br />
12 Examples and exercises<br />
13 Stikordsregister<br />
4
1: Introduction<br />
The word algebra is of arabic origin meaning take into parts and put together,<br />
in short cut and paste. An equation is cut into pieces which are gathered to<br />
a new equation. And so algebra is about expressions put together using operations,<br />
originally the simple operations addition, subtraction, multiplication<br />
and division, and build together to create equations.<br />
Algebra is about manipulating such algebraic equations, with the intention to<br />
do it in a way which makes them solvable. The typical algebraic equation is<br />
an equation of the n-th degree in one unknown or a system of equations with<br />
several unknowns. The challenge to find methods for solution has been a main<br />
driving force in the development of mathematics.<br />
The specification of what is considered to be an algebraic problem has been<br />
widened in the course of this development. It has been recognized that a lot of<br />
specific methods can be subordinated as special cases of more general points of<br />
view. And today algebra is treated as an abstract discipline, where you do not<br />
necessarily define the objects which are the building stones or the operations<br />
used to connect them. In stead you make some assumptions of their properties<br />
and develop a theory which can be applied to all mathematical objects which<br />
have these properties. It is obvious that the more assumptions you make the<br />
more results may be deduced. On the other hand more assumptions will reduce<br />
the applicability. Therefore there exists a plethora of algebraic structures, and<br />
it is part of the sport to keep account of what precisely can be achieved with<br />
a specific choice of assumptions.<br />
So we start out with very few assumptions and gradually extend the number<br />
of assumptions. In this way you construct hierarchies of algebraic structures.<br />
Here we are going to focus on a single hierarchy, starting with the simplest<br />
possible configuration and then successively adding structure. This refinement<br />
of the structure can be accomplished by taking more operations into account<br />
and demanding more conditions on the operations.<br />
And now a survey over the text, which may be appropriate to bear in<br />
mind during reading, since it highlights the system behind the text. It might<br />
otherwise be a little more difficult to extract the main lines.<br />
The introductory chapters exhibit the fundamental notions for a lot of structures<br />
defined using operations in a set. There are the following categories of<br />
structures.<br />
5
Category Operations Axioms<br />
1 semigroups a + b + is associative<br />
2 monoids 0 0 neutral wrt +<br />
3 groups −a −a inverse wrt +<br />
4 rings a ⋆ b,1<br />
⋆ is assoc.<br />
⋆ is distr wrt +,<br />
1 neutral wet ⋆<br />
5 integral domain no zero divisors wrt ⋆<br />
6 fields<br />
We have successive refinements of structures generated by adding new operations<br />
and/or sharpening the conditions on these operations. In the column<br />
with operations and the row of a category you find the operations which have<br />
been added at this stage. The same holds for the column of conditions.<br />
For each of the mentioned categories you can construct new structures within<br />
the same category by carrying the structure which you have on some set to<br />
another set, which is suitably related to the first, it may for instance be a<br />
subset. We say that such structures are induced.<br />
We Shall consider the following types of induced structures:<br />
- substructures (the structure is induced on a subset of a set on which you<br />
already have a structure.)<br />
- structures on function spaces (the structure is induced on a set of functions<br />
with images in a set in which already has a structure. So addition of real<br />
numbers induce addition of reel functions )<br />
- product structures (the structure induced on a product of sets, all of which<br />
already have a structure. For instance addition of real numbers induce addition<br />
of pairs of reals. )<br />
- quotient structure (the structure induced on a quotient of a set which already<br />
has a structure. For instance addition of integers induce addition of remainder<br />
classes of integers.)<br />
Most algebraic structures are namely induced structures, induced from some<br />
fundamental structures in each category, which can be considered to be some<br />
sort of ground building blocks<br />
So constructing new structures from old ones is one important issue. The<br />
other important issue is to study the mappings between structures in the same<br />
category which relates the structures.<br />
6
This is connected to the fact that a number of phenomena which only depend<br />
on the structure may be studied in another structure, sometimes in a more<br />
convenient way. So multiplications can be carried out by means of additions by<br />
using logarithm functions. (This was of practical importance in the days prior<br />
to electronic calculating machines).<br />
This type of mappings which respects the structures are called homomorphisms.<br />
Among them you will find isomorphisms, which completely equal the structures.<br />
These three main aspects (the categories, induced structures, homomorphism)<br />
are mirrored in the architecture of the text:<br />
First part: Operations is about single operations. After the definition of operation,<br />
homomorphisms are treated in relation to single operations.<br />
Second part: Structures is about collections of operations. After the definition<br />
of structures homomorphisms are treated in relation to structures.<br />
Third part: Categories of structures introduces successively the above mentioned<br />
categories. A category is a certain type of structure together with a set<br />
of rules (axioms) to be satisfied by the operations.<br />
For each category the definition is followed by results concerning the structures<br />
induced by structures within the category. These induced structures will often<br />
them selves be in the category, although this is not always the case. Especially<br />
when it comes to quotient structures, you have to deal with the problem of<br />
having the defining equivalence relation to be compatible with the structure.<br />
This is the motivation for some special substructures where you are dealing<br />
with what is called normal subgroups and ideals.<br />
7
2: Operations<br />
2. 1: Background and motivation<br />
In this part we have collected stuff about the operations as such, without<br />
considering them as part of some algebraic structure.<br />
An operation is a way of composing a number of operands to a result, as you<br />
know addition as an operation which takes two summands and gives their sum.<br />
When we are dealing with operations we are mainly thinking of binary operations.<br />
These have two operands- But we are also going to consider operations<br />
with any number of operations, for instance one operand. So we give a definition<br />
which covers all cases in one.<br />
We are not in this context going to handle operations with more than two<br />
operands. But operations with one operand will show to be important. And<br />
we are even going to consider operations with zero operands (to be defined<br />
next).<br />
The rationale of this more general view of operations is as usual to make it<br />
possible to handle separate cases in one sweep. This may be considered to be<br />
utterly affected and a sign of the mathematicians ubiquitous urge for formalism<br />
and abstraction. I just find it smart! And not that difficult after all, may be a<br />
little inconvenient and strange in the beginning.<br />
2. 2: Background terminology<br />
We are going to use extensively the Cartesian product of sets. Lets therefore<br />
recall that we are calling the set A1 × . . . × An the Cartesian product or simply<br />
the product of the factors A1, . . . , An. This set consists per definition of all the<br />
ordered tuples (a1, . . . , an) for which ai ∈ Ai for all i ∈ In, where In = {i ∈<br />
Z: 1 ≤ i ≤ n}. Notice that this definition includes I0 = ∅.<br />
The mapping pi : A1 × . . . × An → Ai defined by pi(a1, . . . , an) = ai is called<br />
the projection on the i-th factor. Many things concerning the product can be<br />
conveniently formulated using the projections.<br />
We have that two elements a, b ∈ B are equal, if pi(a) = pi(b) for all i ∈ In.<br />
8
Let f : B → A1 × . . . × An be a mapping into the product, then we call the<br />
mapping pi ◦f the i ′ th component of f or the component of f of index i . Then<br />
f is completely determined by the collection of its components. So to define f<br />
it suffices to define all the pi ◦ f.<br />
So given a map fi : B → Ai for each i ∈ I there exists a unique map f with<br />
pi ◦ f = fi for all i ∈ I.<br />
If all the factors are the same we shall use power notation : A n = A × . . . × A.<br />
For a map f : A → B we define the associated multi map f : A n → B n by<br />
f(a1, . . . , an) = (f(a1), . . . , f(an)).<br />
Occasionally we shall just write f for f.<br />
These definitions also are meaningful when n = 1. It even shows to be appropriate<br />
to make a convention which extends to the case n = 0, that is the case<br />
where there are no factors at all. We do this by defining the empty tuple, the<br />
ordered set with 0 elements and denote it by (). Then we let the product set<br />
of 0 factors be the set with the only element ().<br />
You should remember that this is just a convention, which is a useful way to<br />
avoid a lot of special cases in some formulations. With this convention we also<br />
have that<br />
A 0 = {()}.<br />
2. 3: Definition<br />
1. Definition: Operation<br />
Let A be a set and let n ≥ 0 be an integer. An operation ♢ in A with n<br />
operands is a mapping<br />
♢ : A n → A.<br />
An operation with n operands is also called an operation with the arity n or an<br />
n-ary operation. For n = 0, 1, 2, 3, 4 one uses the special terms 0-ary, unary,<br />
binary, ternary and quaternary.<br />
2. Remark: Constant operations<br />
9
An operation ♢ in A with 0 operands is uniquely determined by ♢ (()) that is a<br />
certain element of A, which the operation can be identified with. The operations<br />
with 0 operands are therefore sometimes called the constant operations. So we<br />
shall all the time think of an operation with 0 operands as an element of A.<br />
3. Remark: Operation symbols<br />
To underline the general aspect we are going to use operation symbols which do<br />
not already have a definite meaning such as x ♢ y, x ♡ y.<br />
4. Definition: Prefix, Infix, Postfix<br />
We have a number of alternative ways to denote an operation, other than<br />
♢ (a1, . . . , an), namely<br />
♢ a1 . . . an, a1 ♢ . . . ♢ an, a1 . . . an ♢<br />
called respectively prefix, infix and postfix notation. In some contexts we do not<br />
state which notation we use in the hope that it is otherwise easily understood<br />
from the context. For unary operations only prefix and postfix notation are<br />
relevant. Generally we use the operator notation. Infix notation is mostly used<br />
for binary operations.<br />
5. Remark: Backwards polish<br />
Postfix notation is in principle the simplest and is used in programming contexts,<br />
for instance on some pocket calculators. It is simply because you can<br />
avoid parentheses E1 X2<br />
2. 4: Homomorphisms<br />
We know from the real numbers the the rule exp(x + y) = exp(x) · exp(y). For<br />
a linear mapping L we have that L(x + y) = L(x) + L(y). If we let M(X)<br />
denote the matrix which belongs to a certain rotation X, then M(X ◦ Y ) =<br />
M(X) · M(X). In each of these three cases we are dealing with a mapping with<br />
a particularly nice behaviour with respect to a couple of operations.<br />
This may be generalized to arbitrary pairs of operations if they have the same<br />
arity:<br />
6. Definition: Homomorphism wrt a pair of operations<br />
10
Let ♢ and ♡ be operations on A and B respectively and both having n<br />
operands. Let f : A → B be a mapping, so that for all (a1, . . . , an) ∈ A n :<br />
you have that<br />
f( ♢ (a1 . . . an)) = ♡ (f(a1) . . . f(an))<br />
Then we say that f is homomorphic wrt ( ♢ , ♡ ).<br />
7. Remark: Homomorphisms wrt unary and 0-ary<br />
In infix notation the defining equation may be written<br />
f(a1 ♢ . . . ♢ an) = f(a1) ♡ . . . ♡ f(an)<br />
which in the binary case is<br />
f(a1 ♢ a2) = f(a1) ♡ f(a2)<br />
In the unary case it means that f( ♢ a) = ♡ f(a), and for n = 0 we get<br />
f( ♢ ) = ♡<br />
Loosely we can say that it doesn’t matter if we carry out the operation before<br />
using the mapping or first map all the operands and then use the operation on<br />
the result. This can be most elegantly and concentrated stated by the equation<br />
⋆ f ◦ ♢ = ♡ ◦ f<br />
which is displayed in the diagram<br />
♢<br />
A n f ✲ B n<br />
♡<br />
❄ ❄<br />
f<br />
A<br />
✲ B<br />
8. Definition: Isomorphism wrt a pair of operations<br />
Let ♢ and ♡ be operations in A and B both with n operands. Let f : A → B<br />
be a bijective mapping such that f is a homomorphism wrt ( ♢ , ♡ ) and f −1<br />
is a homomorphism wrt ( ♡ , ♢ ) Then we say that f is an isomorphism wrt<br />
( ♢ , ♡ ).<br />
11
You should realize that you have already met a number of homomorphisms by<br />
studying the following examples and exercises: E3 E4<br />
E5 E6 E7<br />
E8 X9 X10<br />
We could say that a homomorphism preserves the algebraic structure and it<br />
should then not be surprising that a composition of homomorphisms is again a<br />
homomorphism. There are other rules for combining homomorphisms. These<br />
are the subject of the next section:<br />
9. Theorem: Composition and homomorphisms<br />
Let ♢ , ♡ and ♠ be operations on A,B and C respectively, all with n operands,<br />
And let f : A → B, g : B → C be mappings with h = g ◦f, which we summarize<br />
in the diagram<br />
Then<br />
A<br />
✒ f<br />
B<br />
❅ g<br />
❅❅❘<br />
h ✲ C<br />
1) If f and g are homomorphisms then so is h.<br />
2) If f and h are homomorphisms and f is surjective then g is a homomorphism.<br />
3) If g and h are homomorphisms and g is injective then f is a homomorphism.<br />
Proof : We use the compact homomorphism equation ⋆ in D6.<br />
Then the proof of (1) takes the form<br />
h ◦ ♢ = g ◦ f ◦ ♢ = g ◦ ♡ ◦ f =<br />
♠ ◦ g ◦ f = ♠ ◦ g ◦ f = ♠ ◦ h<br />
The proof of (2) consists in combining the two equations<br />
g ◦ ♡ ◦ f = g ◦ f ♢ = h ◦ ♢ = ♠ ◦ h,<br />
♠ ◦ g ◦ f = ♠ ◦ h<br />
12
to yield g ◦ ♡ = ♠ ◦ g, using that f and so f is surjective.<br />
Regarding (3) we combine the equations<br />
g ◦ f ◦ ♢ = h ◦ ♢<br />
g ◦ ♡ ◦ f = g ◦ f ◦ ♢ = h ◦ ♢<br />
to deduce that f ◦ ♢ = ♡ ◦ f, since g is injective.<br />
Since these proofs may seem very formalistic and concentrated we repeat them<br />
for the binary case in a more simple form<br />
Proof of (1):<br />
Let a1, a2 ∈ A then<br />
h(a1 ♢ a2) =g(f(a1 ♢ a2))<br />
Proof of (2):<br />
Let b1, b2 ∈ B.<br />
=g(f(a1) ♡ f(a2)) = g(f(a1)) ♠ g(f(a2))<br />
=h(a1) ♠ h(a2)<br />
As f is surjective we can choose a1, a2 ∈ A such that f(a1) = b1, f(a2) = b2.<br />
We then have that<br />
g(b1 ♡ b2) =g(f(a1) ♡ f(a2)) = g(f(a1 ♢ a2)) = h(a1 ♢ a2)<br />
Proof of (3):<br />
=h(a1) ♠ h(a2) = g(f(a1)) ♠ g(f(a2))<br />
=g(b1) ♠ g(b2)<br />
Let a1, a2 ∈ A. First we see that<br />
g(f(a1 ♢ a2)) = h(a1 ♢ a2) = h(a1) ♠ h(a2)<br />
and next that<br />
g(f(a1) ♡ f(a2)) = g(f(a1)) ♠ g(f(a2)) = h(a1) ♠ h(a2)<br />
13
which gives that<br />
g(f(a1 ♢ a2)) = g(f(a1) ♡ f(a2))<br />
and since g is injective it then follows that<br />
f(a1 ♢ a2) = f(a1) ♡ f(a2)<br />
10. Theorem: A bijective homomorphism is automatically an isomorphism<br />
En bijective homomorphism is an isomorphism.<br />
Proof : Follows from (3) (3) applied to (f, g, h) = (f, f −1 , Id)<br />
11. Theorem: Iterated composition of homomorphisms gives a homomorphism.<br />
If we have a number of sets, each with its own operation, combined in a chain by<br />
homomorphisms then their composition is a homomorphism between the start<br />
and the end of the chain.<br />
Proof : Follows by induction from point (1) (1)<br />
2. 5: Induced operations<br />
Now follow some ways of constructing new operations from given ones, so called<br />
induced operations. These are suboperation, function space operation, product<br />
operation and quotient operation.<br />
2. 6: Induced operation on subset<br />
It will show convenient to extend an operation on a set A to act also on subsets<br />
of A in the following obvious way:<br />
12. Definition: Extension of an operation to subsets.<br />
14
Let ♢ be an operation on A with n operands and let B1, . . . , Bn be subsets of<br />
A. Then we define<br />
♢ (B1, . . . , Bn) = { ♢ (b1, . . . , bn) : b1 ∈ B1, . . . , bn ∈ Bb}<br />
which in the binary case may be written<br />
B1 ♢ B2 = {b1 ♢ b2 : b1 ∈ B1, b2 ∈ B2}<br />
If any of the subsets is a singleton, that is a set of the form {b}, then we shall<br />
write b ♢ B2 instead of {b} ♢ B2.<br />
13. Definition: Closed or invariant subset<br />
A subset B of A is called closed (or invariant) wrt the operation ♢ on A if<br />
♢ (B, . . . , B) ⊆ B.<br />
14. Remark: Closed wrt to a constant operation.<br />
B is closed wrt the constant operation ♢ if and only if ♢ ∈ B. X11<br />
X12<br />
15. Definition: Induced operation on a subset<br />
Suppose that B is a subset of A, which is closed wrt to the operation ♢ on<br />
A. Then the restriction of ♢ to B n is an operation on B with the same arity.<br />
We call it the operation induced by ♢ on B. We use the same symbol for the<br />
induced operation. When necessary we write ♢ B.<br />
If we let i : B → A denote the inclusion map i(x) = x, x ∈ B we can summarize<br />
the definition in the diagram<br />
♢B<br />
B n i ✲ A n<br />
♢<br />
❄ ❄<br />
i<br />
B<br />
✲ A<br />
which just states that ♢B is the operation on B which makes i an homomorphism.<br />
16. Definition: Translations<br />
15
Let ♢ be a binary operation on A and let a ∈ A. The mapping of A into it self<br />
defined by x ↦→ a ♢ x ([A ∋ x ↦→ a ♢ x ∈ A]) is then called the left translation<br />
by a. Right translation is defined analogously<br />
17. Remark:<br />
This is inspired by addition of vectors (fx in space).<br />
2. 7: Induced operation on function spaces.<br />
For any sets M and A we let F(M, A) denote the set of mappings from M to<br />
A. We are going to use an operation on A to define an operation on F(M, A).<br />
This is<br />
18. Definition: Induced operation on function spaces<br />
If ♢ is an operation on A with n operands and M is a set, then we can define<br />
an operation ♢M in F(M, A) by the formula<br />
♢ (f1, . . . fn)(x) = ♢ (f1(x), . . . , fn(x))<br />
and we call it the the operation on F(M, X) induced by ♢ . Usually we shall<br />
just use the same symbol for the induced operation, which hopefully will cause<br />
no serious ambiguity. If necessary we denote it ♢ M . For binary operations<br />
the definition may be written<br />
(f1 ♢ f2)(x) = f1(x) ♢ f2(x)<br />
For each x ∈ M we let δx denote the mapping from F(M, X) to A defined by<br />
δx(f) = f(x). This mapping is called evaluation at x. Then we can summarize<br />
the definition of ♢ M in the diagram<br />
F(M, A) n δx ✲ A n<br />
♢M<br />
♢<br />
❄ ❄<br />
δx F(M, A)<br />
✲ A<br />
from which we see that ♢ M is exactly that operation which makes δx a homomorphism<br />
for all x.<br />
16
2. 8: Induced operation on product spaces.<br />
On the linear space R n we have defined addition by component wise addition:<br />
(x1, . . . , xn) + (y1, . . . , yn) = (x1 + y1, . . . , xn + yy)<br />
Here we have a set R n = R × . . . × R which is a set product of factors on<br />
which an addition + is known and from these component additions we create<br />
an addition on the product.<br />
To create this component wise operation it is however not needed neither that<br />
the factors nor the operations are the same. So we generalize by allowing the<br />
different factors to have its own operation. The definition may look a little<br />
scary at first<br />
19. Definition: Product operation. Direct product<br />
Let there be given r sets Ai, i = 1, . . . r and on each of these an operation ♢ i<br />
with s operands.<br />
We then define the product operation ♢ = ♢ 1<br />
× · · · × ♢ r<br />
to be the operation uniquely defined by the formula<br />
(⋆) pi ◦ ♢ = ♢ i<br />
◦ pi<br />
on A = A1 × · · · × Ar<br />
(A map into a product is determined by its projections (as mentioned in the<br />
background terminology))<br />
which simply states that to find the component with index i of the result of the<br />
operation you just take component with index i in each of the factors and use<br />
the operation with that index on those.<br />
To see a more explicit formula we restrict to the case where the operations are<br />
binary and we get<br />
(a1, . . . , ar) ♢ (b1, . . . , br) =<br />
(<br />
a1 ♢ 1<br />
b1, . . . , ar ♢ r<br />
The definition may also be summarized in the diagram<br />
17<br />
br<br />
)<br />
.
A n pi ✲ A n i<br />
♢<br />
i<br />
♢<br />
❄ ❄<br />
A<br />
pi ✲ Ai<br />
and we see that we have defined the operation in such a way that all the projections<br />
are homomorphisms.<br />
In the special case where all the factors are the same Ai = A and also the<br />
operations are the same ♢ i<br />
= ♢ for all i we shall use the notation ♢ for<br />
♢ 1<br />
× . . . × ♢ r<br />
. So in case of binary operations we have<br />
(a1, . . . , ar) ♢ (b1, . . . , br) = (a1 ♢ b1, . . . , ar ♢ br).<br />
where we recognize vector addition as a special case, namely ♢ being + in a<br />
linear space.<br />
The case with the same factor repeated is the most common, (you can practice<br />
it in X13) though it may also be relevant with different factors like in X14<br />
Next consider the question concerning how the properties of the factor operations<br />
are reflected in the product operation. First we have:<br />
20. Theorem: Projection on a factor is a homomorphism<br />
With the notation in D19 we have that the projections pi are homomorphisms<br />
wrt the pair ( ♢ , ♢ i<br />
)<br />
Proof : This follows by comparing the definition of the product operation (⋆<br />
in D19) with the definition of homomorphism (⋆ in D6). See also the diagram<br />
in the definition.<br />
The following theorem shows that for a mapping with values in a product<br />
the question of being a homomorphism may be answered by looking at its<br />
components:<br />
21. Theorem: The components are homomorphisms and conversely<br />
18
Suppose that ♢ is an operation on A, that ♡ 1<br />
, . . . , ♡ r<br />
B1, . . . , Br, all with n operands.<br />
Further let B = B1 × · · · × Br<br />
and ♡ = ♡ 1<br />
× · · · × ♡ r<br />
.<br />
and assume that fi : A → Bi are the components of f : A → B.<br />
are operations on<br />
Then f a homomorphism wrt ♢ and ♡ if and only if fi is a homomorphism<br />
wrt to ♢ and ♡ i<br />
for all i<br />
Proof : Since fi = pi ◦ f the only if part follows from T9: 1 and the if part<br />
follows from T9:2, since pi is surjective.<br />
This is used to prove a very famous theorem, see the example E15<br />
2. 9: Induced operation on quotient.<br />
While building product sets is about how to create new objects with many properties<br />
from objects with simpler properties, then building quotients is about to<br />
forget properties, which are not relevant in the context.<br />
A radical example is to forget all properties of a permutation except its parity.<br />
We are then left with only two elements ”even” and ”odd”. When we transfer<br />
some operation on permutations to these object we must assure our selves that<br />
the operation is insensitive to all other properties of the operation.<br />
This is the case with multiplication of permutations, since the parity of the<br />
product only depends on the parity of the factors.<br />
The general framework for this is notion of a partition K of a set X into classes<br />
according to some criterion. Then K is the set of classes. Each class is non<br />
empty and all elements must be in exactly one class. If K is a class and x ∈ K<br />
we say that x is a representative of K and that K is the class of x, which is<br />
also denoted [x]K.<br />
The mapping of X into K which to x assigns its class [x]K is called the canonical<br />
projection and we denote it by kK.<br />
Any mapping f of X into some set Y which is constant on the classes can in<br />
an obvious way be considered to be defined on XK, lets call it fK. Then we<br />
19
have the diagram<br />
kK<br />
X<br />
❅ f<br />
❅❘<br />
❄<br />
✒ fK<br />
K<br />
Y<br />
The most usual way to specify a partition is based on an equivalence relation.<br />
Lets recall that an equivalence relation ∼ in a set X is characterized by the<br />
three properties<br />
1) reflexivity (x ∼ x)<br />
2) symmetry (x ∼ y ⇒ y ∼ x)<br />
3) transitivity (x ∼ y, y ∼ z ⇒ x ∼ z)<br />
For each x ∈ X we define the equivalence class of x to consist of all elements<br />
equivalent with x. This partition is denoted X∼, and we simply write [x]∼<br />
for the class of x. This partition is called the quotient of X modulo ∼. The<br />
canonical map is the denoted by k∼.<br />
Often an equivalence relation is based on a mapping f of X into some set Y<br />
by defining x1 ∼ x2 to mean f(x1) = f(x2). Then f is constant on the classes<br />
and the situation is described by the diagram<br />
k∼<br />
X<br />
❅ f<br />
❅❘<br />
❄<br />
✒ f∼<br />
X∼<br />
Y<br />
Now we return to considering operations. We have been dealing with the task<br />
of transferring a map on a set X to a quotient of X and we are interested in<br />
extending this idea to operations, that is to have an operation on the quotient.<br />
Operations for which this is possible will be considered to be compatible with<br />
the partition. If the partition is defined by some equivalence relation we shall<br />
say that the relation is compatible.<br />
The mathematical precision of this is in the following, where we know a relation<br />
defining the quotient.<br />
20
22. Definition: Compatibility<br />
Suppose that ♢ is an operation and ∼ an equivalence relation in A, such that<br />
a1 ∼ b1, . . . , an ∼ bn ⇒ ♢ (a1, . . . , an) ∼ ♢ (b1, . . . , bn)<br />
for all (a1, . . . , an) and (b1, . . . , bn). Then we say that the equivalence relation<br />
and the operation are compatible<br />
Lets for convenience also state the definition for binary operations:<br />
a1 ∼ b1, a2 ∼ b2 ⇒ a1 ♢ a2 ∼ b1 ♢ b2<br />
You have most certainly met this notion previously, for instance in the examples:<br />
X16 X17 X18 X19 More examples (which may be new to you) are in<br />
E20 E21<br />
The reason for introducing compatibility is as mentioned that it enables us to<br />
induce an operation on the equivalence classes. this is clarified in the next<br />
theorem :<br />
23. Theorem: Calculating with representatives<br />
Suppose that the operation ♢ is n-ary operation on A and ∼ is en equivalence<br />
relation on A. If they are compatible, then the class of the result by the operation<br />
is determined by the classes of the operands. More formally<br />
Let K1, . . . , Kn ∈ A∼ and let a1, . . . , an ∈ A be representatives for the respective<br />
classes. Then the equivalence class<br />
[ ♢ (a1, . . . , an)]∼.<br />
is independent of the way the representatives are chosen.<br />
21
This makes the following definition meaningful:<br />
24. Definition: The quotient operation<br />
The formula<br />
[a1] ♢ ∼<br />
. . . ♢ ∼<br />
[an] = [a1 ♢ , . . . , ♢ an]<br />
defines am operation on the set of equivalence classes. We call it the quotient<br />
operation induced by ♢ modulo ∼, and we denote it ♢ ∼.<br />
The definition is summarized by the diagram<br />
♢<br />
A n k ✲ A∼ n<br />
∼<br />
♢<br />
❄ ❄<br />
A<br />
k ✲ A∼<br />
which shows that the quotient operation is the one making the canonical projection<br />
a homomorphism.<br />
25. Theorem: The canonical projection is a homomorphism.<br />
Let ♢ be an operation on A and let ∼ be an equivalence relation compatible<br />
with ♢ and let k be canonical projection of A on A∼ associated with ∼. Then<br />
k is a homomorphism wrt to ♢ and ♢ ∼.<br />
Proof : Since k is the canonical projection, we have that k(a) = [a], the<br />
equivalence class containing a. Then the diagram above proves the theorem.<br />
Lets give a less formalistic proof in the case of a binary operation<br />
k(a1) ♢ k(a2) = [a1] ♢ [a2] = [a1 ♢ a2] = k(a1 ♢ a2)<br />
which is the diagrammatic formulation of the condition for homomorphism<br />
(D6).<br />
The previous theorem show that an equivalence relation which is compatible<br />
with an operation gives rise to a homomorphism. The next theorem shows the<br />
opposite, that any homomorphism generates a compatible equivalence relation.<br />
22
26. Theorem: Criterion for homomorphy on the quotient<br />
Let ∼ be an equivalence relation on A compatible with the operation ♢ . Let k<br />
denote the canonical projection of A on A∼. Suppose that ♡ is an operation<br />
on B and f is a mapping of A∼ into B. Then f is a homomorphism wrt<br />
( ♢ ∼, ♡ ) if and only if f ◦ k is a homomorphism wrt ( ♢ , ♡ )<br />
Proof : Consider the diagram<br />
A<br />
✒ k<br />
A∼<br />
❅ f<br />
❅❅❘<br />
f◦k ✲ B<br />
The if part follows from T9:2 (since k is surjective) and the only if part follows<br />
from T9:1<br />
27. Theorem: Homomorphism and compatibility.<br />
Suppose that f : A → B is a homomorphism wrt ♢ and ♡ . Let ∼ be the<br />
equivalence relation induced by f, (that means a ∼ b ⇔ f(a) = f(b)). Then ∼<br />
will be compatible with ♢ .<br />
Proof : Suppose that a1 ∼ a ′ 1, . . . , an ∼ a ′ n. Then<br />
and so<br />
f( ♢ (a1, . . . , an)) = ♡ (f(a1), . . . , f(an))<br />
= ♡ (f(a ′ 1), . . . , f(a ′ n)) = f( ♢ (a ′ 1, . . . , a ′ n))<br />
♢ (a1, . . . , an) ∼ ♢ (a ′ 1, . . . , an) ′<br />
2. 10: Axioms for operations.<br />
28. Definition: Associative operation<br />
A binary operation ♢ on A is said to be associative if<br />
a ♢ (b ♢ c) = (a ♢ b) ♢ c<br />
23
holds for all a, b, c ∈ A<br />
29. Theorem: Parentheses are redundant for associative operations<br />
Easy to understand what it means but clumsy to formulate precisely.<br />
30. Definition: Commutative operation<br />
A binary operation ♢ on A is said to be commutative if<br />
a ♢ b = b ♢ a<br />
holds generally, that is for all a, b ∈ A<br />
31. Definition: Distributive operation<br />
En binary operation ♢ is said to be distributive wrt a binary operation ♡ if<br />
a ♢ (b1 ♡ b2) = (a ♢ b1) ♡ (a ♢ b2)<br />
holds generally, that is for all a, b ∈ A<br />
32. Theorem: Associativity, commutativity and distributivity is hereditary<br />
If an operation ♢ on A is associative, the same holds for the operations induced<br />
on subsets, function spaces, product spaces and quotient spaces. The same holds<br />
commutativity and distributivity.<br />
Proof : An easy, may be tedious exercise.<br />
33. Theorem: Associativity, commutativity and distributivity is preserved<br />
by homomorphisms<br />
Let ♢ be an operation in A and ♡ an operation in B and let f be a homomorphism<br />
of A into B. If f is surjective and ♢ is associative then ♡<br />
is associative. If f is injective and ♡ is associative then ♢ is associative.<br />
Obvious analogues hold for commutativity and distributivity.<br />
24
Proof : Assume f surjective and ♢ associative. Let b1, b2, b3 ∈ B. We can<br />
choose a1, a2, a3 ∈ A such that f(a1) = b1, f(a2) = b2, f(a3) = b3. Then<br />
b1 ♡ (b2 ♡ b3) = f(a1) ♡ (f(a2) ♡ f(a3)) = f(a1 ♢ (a2 ♢ a3))<br />
(b1 ♡ (b2)) ♡ b3 = (f(a1) ♡ f(a2)) ♡ f(a3) = f((a1 ♢ a2) ♢ a3)<br />
Since the two right ends of the lines are equal (by associativity of ♢ ) also the<br />
two left ends are equal, showing that ♡ is associative.<br />
The remaining part of the proof is left as an exercise.<br />
25
3: Algebraic structures.<br />
3. 1: Definition.<br />
34. Definition: Algebraic structure<br />
An algebraic structure is a set equipped with operations in the set. If the set<br />
itself is A and if the operations are the ordered tuple ( ♢ 1<br />
, . . . , ♢ n<br />
) we shall use<br />
(A, ♢ 1<br />
, . . . , ♢ n<br />
) to denote the structure. We call A the underlying set of the<br />
structure. Often it will cause no confusion to simply let A denote the structure.<br />
35. Remark: Infinitely many operations<br />
In general you will often allow more than finitely many operations. We do not<br />
do so here for notational convenience.<br />
36. Definition: Type of structure<br />
By the type of an algebraic structure we understand the ordered tuple (i1, . . . , ik)<br />
of the arities of the operations.<br />
We are now going to extend all the notions we have for a single operation to<br />
structures. It simply consists in letting claims and constructions apply to all<br />
the operations of the structure. for instance a mapping is a homomorphism<br />
for a pair of structures if it is a homomorphism for all the pairs of operations<br />
defining the structures. A product of structures consists of the products of all<br />
the defining operations and so for all the other notions. This might be enough<br />
to know but we are carrying out the details for completeness and reference.<br />
3. 2: Homomorphisms and isomorphisms<br />
37. Definition: Homomorphism<br />
Let (A, ♢ 1<br />
, . . . , ♢ n<br />
) and (B, ♡ 1<br />
, . . . , ♡ n<br />
) be algebraic structures of the same<br />
type and let f : A → B be a mapping, which is a homomorphism wrt ( ♢ i<br />
, ♡ i<br />
)<br />
for all i = 1, . . . , n. Then we say that f is a homomorphism wrt the pair of<br />
structures A and B.<br />
38. Definition: Isomorphism<br />
26
Let (A, ♢ 1<br />
, . . . , ♢ n<br />
) and (B, ♡ 1<br />
, . . . , ♡ n<br />
) be algebraic structures of same type<br />
and let f : A → B be a mapping, which is an isomorphism wrt ( ♢ i<br />
, ♡ i<br />
i = 1, . . . , n. Then f is an isomorphism wrt the pair of structures.<br />
) for all<br />
39. Theorem: Composition of homomorphisms<br />
Let (A, ♢ 1<br />
, . . . , ♢ n<br />
), (B, ♡ 1<br />
, . . . , ♡ n<br />
) and (C, ♠ 1<br />
, . . . , ♠ n<br />
) be algebraic structures<br />
of same type. And let f : A → B, g : B → C be mappings with h = g ◦ f.<br />
Then we have that<br />
1) If f and g are homomorphisms then so is h<br />
2) If f and h are homomorphisms and f is surjective then also g is a<br />
homomorphism<br />
3) If g and h are homomorphisms and g is injective then f is a homomorphism<br />
Proof : Follows from the related theorem T9 about mappings which are homomorphisms<br />
wrt a single pair of operations.<br />
40. Theorem: Bijectivity implies automatically isomorphism<br />
A bijective homomorphism is an isomorphism.<br />
Proof : Follows from T39 (3) used on the scenario (f, g, h) = (f, f −1 , Id)<br />
41. Theorem: Composition of homomorphisms is homomorphism.<br />
Analogously for isomorphisms<br />
If we have a chain of structures linked in by homomorphisms, then the composition<br />
of these homomorphisms is itself an homomorphism<br />
Proof : Follows by induction from the previous theorem<br />
3. 3: Induced structures on subset<br />
42. Definition: Substructure<br />
By a substructure of a structure is meant a (non empty) subset which is invariant<br />
wrt to all of the operations of the structure.<br />
27
43. Theorem: Substructure is a structure of same type<br />
Let B be a substructure of A. When B is equipped with the induced operations<br />
you get a structure of same type as A. We shall always let B also denote this<br />
structure if no other structure is explicitly specified.<br />
Substructures occur in a lot of situations.<br />
44. Theorem: The generated substructure<br />
Let (A, ♢ 1<br />
, . . . , ♢ n<br />
) be a algebraic structure and let D ⊂ A. Then there exists<br />
a substructure B that contains D and is the least one to do so, which means<br />
that it is contained in any other substructure containing D.<br />
It is the intersection of all substructures containing D and can be explicitly<br />
written as<br />
B = ∩ {C ∈ A : D ⊆ C}<br />
where A denotes the set of substructures of (A, ♢ 1<br />
, . . . , ♢ n<br />
Proof : We define B by the formula and starts by showing that B is in fact a<br />
substructure. (Remark that A ∈ A and so A is not empty). To do so we must<br />
show that B is invariant wrt any operation.<br />
So let ♢ be any operation of the structure, say with arity k. Let (b1, . . . , bk) ∈<br />
B k . We must show that ♢ (b1, . . . , bk) ∈ B. So let C be any substructure<br />
containing D. Since C is a substructure we have that ♢ (b1, . . . , bk) ∈ C.<br />
Since this is true for any C we have that ♢ (b1, . . . , bk) must also be in the<br />
intersection of all these C which is B.<br />
By construction B is contained in any substructure containing C<br />
45. Definition: The generated substructure, the algebraic closure<br />
The structure B in the preceding theorem is said to be the subalgebra generated<br />
by D, or the algebraic closure of D. It is denoted by ⟨D⟩. If D = {d1, . . . , dn}<br />
we shall also write ⟨D⟩ = ⟨d1, . . . , dn⟩<br />
46. Theorem: A characterization of the algebraic closure<br />
The algebraic closure ⟨D⟩ consists of all elements which can be obtained by<br />
successive application of the operations on elements of D<br />
28<br />
).
Proof : It follows from the definitions that any substructure C containing D<br />
also must contain the elements constructed as above, since C must be invariant.<br />
So all of these elements must be in ⟨D⟩. On the other hand these elements will<br />
constitute a substructure.<br />
47. Remark: What is meant by generation<br />
The characterization in the theorem more clearly expresses the meaning of generation.<br />
And we could as well have taken this as a definition.<br />
48. Theorem: The image by a homomorphism is a substructure<br />
Let f be a homomorphism wrt (A, ♢ 1<br />
C be a substructure of (A, ♢ 1<br />
, . . . , ♢ n<br />
Then f(C) a substructure of (B, ♡ 1<br />
Proof : Obvious.<br />
, . . . , ♢ n<br />
).<br />
, . . . , ♡ n<br />
).<br />
) and (B, ♡ 1<br />
, . . . , ♡ n<br />
), and let<br />
49. Theorem: The inverse image by a homomorphism is a substructure<br />
Let f be a homomorphism wrt (A, ♢ 1<br />
, . . . , ♢ n<br />
) and (B, ♡ 1<br />
, . . . , ♡ n<br />
), and let<br />
C be a substructure of (B, ♡ 1<br />
, . . . , ♡ n<br />
). Then f −1 (C) is a substructure of<br />
(A, ♢ 1<br />
, . . . , ♢ n<br />
).<br />
Proof : Let ( ♢ , ♡ ) be a pair of coupled operations with n operands. Let<br />
a1, . . . , an ∈ f −1 (C), which means that f(a1), . . . , f(an) ∈ C. It is to be shown<br />
that ♢ (a1 . . . , an) ∈ f −1 (C), which means that f( ♢ (a1, . . . , an)) ∈ C. But<br />
since f is a homomorphism we have that f( ♢ (a1, . . . , an)) = ♡ (f(a1), . . . , f(an)),<br />
and this expression is in C, since f(a1), . . . , f(an) are so per assumption, and<br />
C as a substructure is closed wrt ♡ .<br />
3. 4: Induced structure on function spaces<br />
For any sets M and A we will use F(M, A) to denote the set of all mappings<br />
form M to A. For each x ∈ M we define the map ex from F(M, A) to A by<br />
ex(f) = f(x) and we call this map the evaluation at x. It so to speak changes<br />
the roles of map and argument.<br />
29
We are now going to move any algebraic structure on A to be a structure<br />
on F(M, A) (See the introduction to D 18), which we shall call the induced<br />
operation:<br />
50. Definition: Induced structure on function spaces<br />
Antag at (A, ♢ 1<br />
, . . . , ♢ n<br />
) is a algebraic structure and at M is a set. Then kalder<br />
vi den algebraic structure on F(M, A) der som operations har de inducerede<br />
operations for den inducerede structure. We tænker os altid, uden at behøve at<br />
fremhæve det, at F(M, A) is equipped with denne structure.<br />
51. Theorem: The structure on the function space is of same type<br />
Using the notation from the previous theorem we have that F(M, A) is of the<br />
same type asA<br />
Proof : Oplagt.<br />
3.4.1: Induced structure on product.<br />
Since we can construct the product of a number of operations we can also<br />
construct the product of structures by taking the product of the involved operations.<br />
This is made precise in the following<br />
52. Definition: Direct product of structures<br />
We take the case of two factors first: Suppose that (A, ♢ 1<br />
, . . . , ♢ n<br />
) and (B, ♡ 1<br />
, . . . , ♡ n<br />
are two structures of same type. Then we define their product as the structure<br />
(A × B, ♢ 1<br />
×♡ 1<br />
, . . . , ♢ n<br />
× ♡ n<br />
)<br />
Next lets take m factors each of which have m operations:<br />
Suppose that (Ai , ♢ i 1, . . . , ♢ i n) for i = 1, . . . , m are algebraic structures all of<br />
same type. Then we define the product structure as (A, ♢ 1<br />
, . . . , ♢ n<br />
), where<br />
A = A1 × · · · × An ♢ i<br />
= ♢ i 1 × · · · × ♢ i m.<br />
53. Theorem: The product structure is of same type<br />
The product structure has the same type as the factors.<br />
30<br />
)
Proof : The factors have per definition same type, lets say with aryties (k1, . . . , kn).<br />
Men each of product operations has the same arity as its factors. Therefore we<br />
have that the aryties for the product structure also be (k1, . . . , kn).<br />
3. 5: Induced structure on quotient<br />
54. Definition: Quotient structure modulo an equivalence relation.<br />
Canonical projection.<br />
Let there be given an algebraic structure (A, ♢ 1<br />
, . . . , ♢ n<br />
), where the underlying<br />
set A is equipped with an equivalence relation ∼ compatible with all the<br />
operations.<br />
We then define an algebraic structure, called the quotient structure of (A, ♢ 1<br />
modulo ∼, to be the algebraic structure (A∼, ♢ 1<br />
55. Theorem: Quotient structure type.<br />
The quotient structure is of same type.<br />
Proof : Obvious .<br />
∼, . . . , ♢ n<br />
56. Theorem: The canonical projection is a homomorphism.<br />
Den canonical projection is a homomorphism.<br />
Proof : Obvious.<br />
57. Theorem: Homomorphism gives equivalence.<br />
∼).<br />
, . . . , ♢ n<br />
Assume that f is a homomorphism of the algebraic structure (A, ♢ 1<br />
, . . . , ♢ n<br />
)<br />
into the algebraic structure (B, ♡ 1<br />
, . . . , ♡ n<br />
), and let the relation ∼ in A be the<br />
equivalence relation induced by f (that is: a ∼ b ⇔ f(a) = f(b)). Then we<br />
have that ∼ is compatible with the structure on A.<br />
31<br />
)
58. Definition: Quotient induced by a homomorphism.<br />
The equivalence relation mentioned in the preceding theorem is said to be induced<br />
by f and we denote it by ∼f . The quotient structure induced by this<br />
relation will be called the quotient structure induced by f and will be denoted<br />
by A/f<br />
4: Appendix: Tables.<br />
4. 1: Symbols for sets.<br />
In the table below X and Y are sets G, H are groups, M monoids, R rings,<br />
L Fields and V, W linear spaces. If needed in the context we let X and Y be<br />
equipped with topologies and we let V, W have inner products.<br />
Notation Set<br />
1 N natural numbers<br />
2 Z integers<br />
3 Q rational numbers<br />
4 R real numbers<br />
5 C complex numbers<br />
6 M ∗ invertible elements in M<br />
7 Zn classes of remainders Z/nZ<br />
8 Un invertible elements in Zn, (Z ∗ n)<br />
9 Lin(V, W ) linear mappings of V into W .<br />
10 Iso(V, W ) isometries of V into W<br />
11 Rot(2) rotations in the plane<br />
12 Rot(3) rotations in space<br />
13 Rfl(2) reflections in the plane<br />
14 Rfl(3) reflections in space<br />
15 T classes of remainders R/Z<br />
16 Mat(m, n, L) m × n matrices over L<br />
17 GL(n, L) invertible elements i Mat(n, n, L)<br />
18 SL(n, L) {A ∈ GL(n, L)| det(A) = 1}<br />
19 En unity matrix of dimension n<br />
20 O(n) orthogonal matrices<br />
21 SO(n) special orthogonal matrices<br />
22 U(n) unitary matrices<br />
32
23 SU(n) special unitary matrices<br />
24 Pern(R) permutation matrices over R<br />
25 F(X, Y ) mappings from X into Y<br />
26 C(X, Y ) continuous mappings from X into Y<br />
27 D(X) subsets of X.<br />
28 O(X) open subsets of X.<br />
29 F(X, Y ) ∗ bijective mappings from X into Y<br />
30 C(X, Y ) ∗ homeomorphisms from X into Y<br />
31 ⟨a, b, c, . . .⟩ words over {a, b, c, . . .}<br />
32 Sn permutations of n objects<br />
33 An even permutations of n objects<br />
34 Gx elements in G with x as fixed point<br />
35 R[X] polynomials in X over the ring R<br />
36 R(X) the canonical field extension of R[X]<br />
37 L(α) minimal field extension of L, containing α<br />
38 Fix(G, X) {g ∈ G|g(X) ⊆ X}<br />
39 Mb Möbius transformations<br />
40 Rot(X) rotations invariant for X<br />
41 V Klein’s Vierer-Gruppe (Z2 × Z2)<br />
42 T tetrahedron group<br />
43 H hexahedron group<br />
44 I icosahedron group<br />
45 S n unit hyper sphere i R n+1<br />
33
4. 2: Tables with operations<br />
.<br />
Operations with 0 operands<br />
Set operation Closed subsets<br />
1 N 0 kN<br />
2 C 0 R, Q, Z<br />
3 C 1 R, Q, Z, N<br />
4 C i<br />
5 Mn(L) En, E SLn(L), On(L), Pern(L)<br />
6 Mn(C) En, E Mn(R), Mn(Q), Mn(Z)<br />
7 D(X) ∅<br />
8 F(X, X) I<br />
9 ⟨a, b, c, . . .⟩ (), empty word<br />
Unary operations<br />
Set operand operation comment closed subsets<br />
1 C a −a Opposite R, Q, Z, N<br />
2 C ⋆ a a −1 Reciprocal<br />
3 C a ā conjugate<br />
4 L n x −x Opposite<br />
5 Mn(L) A A † ,A t ,A ⊤ transposed<br />
6 Mn(C) A A ⋆ adjoint<br />
7 F(X, X) f f −1 inverse<br />
8 D(X) A X \ A complement<br />
Binary operations<br />
Set operands operation name closed subsets<br />
1 C a, b a + b R, Q, Z, N<br />
2 C ⋆ a, b ab<br />
3 Mn(L) A, B A + B<br />
4 Sn s, t st<br />
5 D(X) A,B A ∩ B<br />
6 D(X) A,B A ∪ B<br />
34
4. 3: Tables with homomorphisms<br />
In the following tables A, X, S, M, G, R, L, V are denoting an arbitrary structure<br />
of the following types respectively general structure, set, semigroup, monoid,<br />
group, ring, field.<br />
If we have more structures of the same type we shall use indexed letters in the<br />
usual way, so that M1, M2, . . . etc denote sets. We let M ′ denote a subset of<br />
M and so on.<br />
The set of homomorphisms of a structure A into a structure B is denoted by<br />
Hom(A, B).<br />
Homomorphisms wrt binary operations.<br />
x ♢ y A x f(x) B u ♡ v symbo<br />
1 x + y Rn x Ax Rm u + v<br />
2 m + n Z n an R uv exp. m<br />
3 A + B Mat(m, n, L) A L(A, V, W) Lin(V, W ) u + v ass. lin. m<br />
4 L + M Lin(V, W ) L M(L, V, W) Mat(m, n, L) u + v ass. ma<br />
5 LM Lin(V, W ) L M(L, V, W) Mat(m, n, L) u ◦ v ass. ma<br />
6 P + Q L[X] P P (L) Lin(V, V ) u + v<br />
7 x + y Rn x Ax Rm u + v mult. m.<br />
8 xy N x σ(x) N0 N<br />
u + v prime spec<br />
9 xy OrdA x |x| N u + v word len<br />
10 x × y Finite sets x |x| N0 uv ant. ele<br />
11 x ◦ y Rot(3) x ˆx Mb u ◦ v ass. Möbius tran<br />
12 x ◦ y Mb x ˆx Rot(3) u ◦ v ass. ro<br />
13 xy SU(2) x ˆx SO(3) uv ass. ro<br />
14<br />
5: Examples and exercises<br />
Example 1: Different ways of notation. (see nr 5)<br />
We have that +(a, ·(b, c)) = +a(·bc) = a + b · c = abc · +<br />
Exercise 2: Backward polish. (see nr 5)<br />
10 2 · 5 + √ 2 · 10 : = 1<br />
Example 3: Linear mappings are homomorphisms (see nr 8)<br />
35
Let L be a linear mapping from the vectors pace V to the vector space W .<br />
Then L is a homomorphism wrt the pair (+, +).<br />
Let Λ denote multiplication with the scalar λ, that is Λ(x) = λx for any vector<br />
x. Then Λ is a unary operator and L is a homomorphism wrt the pair (Λ, Λ)<br />
Example 4: Presenting a linear mapping by a matrix is homomorphic (see nr<br />
8)<br />
If L is a linear mapping from the linear space V to the linear space W and<br />
bases are chosen, then there exists a unique matrix A such that if w = L(v)<br />
then y = Ax where x are the coordinates of v and y are the coordinates of w.<br />
We say that A represents L (or is associated with or belongs to L).<br />
The mapping that sends L to A is a homomorphism wrt the pair of binary operations<br />
(◦, ·) consisting of composition of maps and multiplication of matrices.<br />
If we restrict to invertible mappings and invertible matrices then it is also<br />
a homomorphism wrt the pair of unary operations (inversion of mappings,<br />
inversion of matrices).<br />
Example 5: Associating a Möbius transformation with a matrix is homomorphic<br />
(see nr 8)<br />
(<br />
a<br />
To any invertible complex matrix<br />
c<br />
hA given by<br />
)<br />
b<br />
we associate the complex mapping<br />
d<br />
hA(z) =<br />
az + b<br />
cz + d ,<br />
The mapping that takes A to hA is a homomorphism wrt to the pair (·, ◦) of<br />
matrix multiplication and composition of mappings.<br />
Example 6: The dependence of the power on the exponent is a homomorphy<br />
(see nr 8)<br />
Let x be a real number. Then the mapping N ∋ n ↦→ x n ∈ R is a homomorphism<br />
wrt the pair (+, ·).<br />
Example 7: The dependence of the matrix power on the exponent is a homomorphy<br />
(see nr 8)<br />
Let A be a square matrix. Then the mapping N ∋ n ↦→ A n is a homomorphism<br />
wrt the pair (+, ·).<br />
Example 8: Homomorphisms for unary operations (see nr 8)<br />
36
We consider the unary operation − on R and want to determine those mappings<br />
f of R into it self, which are homomorphisms. the condition is that<br />
f(−x) = −f(x),<br />
which is the condition for f to be an odd function<br />
This result can be generalized to general linear spaces.<br />
Exercise 9: Homomorphisms for unary operations (see nr 8)<br />
Write the condition f : R → R to be a homomorphism wrt the operations<br />
x ↦→ −x and x ↦→ x −1 , and find such a homomorphism. (Hint: f(x) = a x )<br />
Exercise 10: The four fundamental isomorphisms concerning the algebraic operations.<br />
(see nr 8)<br />
Let a be a real constant. Study the four function definitions ax, a x , log a(x), x a<br />
and explain that they define homomorphisms for suitable choices of domain<br />
and codomain. Notice that all combinations of addition and multiplication are<br />
possible.<br />
Exercise 11: Closed wrt to addition. (see nr 14)<br />
Construct an increasing chain of subsets of C which are closed wrt addition in<br />
C<br />
Exercise 12: Show that the set {x+y √ 3 : x, y ∈ Q} is closed both wrt addition<br />
and multiplication .<br />
Show that the same is not true for {x + y 3√ 2 : x, y ∈ Q}. (see nr 14)<br />
Exercise 13: Product of equal factors (see nr 19)<br />
Show that Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}. Show that the product operation<br />
+ × + partially is given in the following table and complete:<br />
(+,+) (0,0) (0,1) (1,0) (1,1)<br />
(0,0) (0,0) (0,1) (1,0) (1,1)<br />
(0,1) (0,1) (0,0) (1,1) (1,0)<br />
(1,0) (1,0) (1,1)<br />
(1,1) (1,1) (0,1)<br />
(·, ·) (0,0) (0,1) (1,0) (1,1)<br />
(0,0) (0,0) (0,0) (0,0) (0,0)<br />
(0,1) (0,0) (0,1) (0,0) (0,1)<br />
(1,0) (0,0) (0,0)<br />
(1,1) (0,0) (1,0)<br />
Exercise 14: Product of different factors (see nr 19)<br />
Make tables for (Z2, +) × (Z3, +) and (Z2, ·) × (Z3, ·), in the same way as in<br />
X13<br />
Example 15: Chinese remainder theorem (see nr 21)<br />
37
Assume that p divides n. Then the mapping Zn ∋ x ↦→ x mod p ∈ Zp is a<br />
homomorphism (with obvious choices of operations). Prove it<br />
Assume that n = pq. The mapping Zn ∋ x ↦→ (x mod p, x mod q) ∈ Zp × Zq is<br />
then according to T21 a homomorphism. If p and q are coprime then this is<br />
even an isomorphism. The latter claim is a way of formulating the so called<br />
Chinese remainder theorem. The most common way of formulating it is to say<br />
that the system of equations<br />
x mod p = u x mod q = v<br />
always has solutions and that they together constitute a class of remainders<br />
modulo n. This formulation corresponds to the claim that the above mapping<br />
is surjective (since the domain and the codomain has the same number of<br />
elements.<br />
From this you can see that Z2 ×Z3 is isomorphic with Z6 with obvious modular<br />
additions. compare with the results in X13 and X14<br />
The Chinese remainder theorem is proved as follows :<br />
Since p and q are coprime we can use the extended Euclidean algorithm to find<br />
integers k and l such that kp + lq = 1, and then u − v = (kp + lq)(u − v). This<br />
gives<br />
u + k(v − u)p = v + l(u − v)q.<br />
And so x = u + k(v − u)p is a solution. Prove this.<br />
Exercise 16: Permutations with the same sign (see nr 22)<br />
Show that the relation ”x and y has the same sign is an equivalence relation<br />
on Sn. Find the equivalence classes.<br />
Show that this relation is compatible with product of permutations.<br />
Show that the sign considered as a mapping is a homomorphism wrt a suitable<br />
pair of operations.<br />
Exercise 17: Orthogonal matrices with the same determinant (see nr 22)<br />
Show that the relation ”the determinant of x and y is the same” is an equivalence<br />
relation on On. Find the equivalence classes.<br />
Show that this relation is compatible with product of matrices.<br />
38
Show that the determinant considered as a mapping is a homomorphism wrt<br />
to suitable choices of operations.<br />
Exercise 18: Similar matrices (see nr 22)<br />
By definition two square matrices A and B are similar if they are matrix (<br />
each with respect to some bases of its own) for the same linear mapping. Show<br />
that this relation is an equivalence relation which is compatible with product<br />
of matrices. This is easily proven by using the definition given here, but may<br />
also be proven by using the characterization that A and B are similar if and<br />
only if there exists an invertible matrix S such that AS = SB.<br />
Exercise 19: Integers with same remainder (see nr 22)<br />
Show that the relation ”tje principal remainder modulo n of x and y is the<br />
same” is an equivalence relation on Z. Find the equivalence classes.<br />
Show that this relation is compatible with addition.<br />
Show that x mod n is a homomorphism wrt suitable choices of operations.<br />
Example 20: Colour compatibility (see nr 22)<br />
A cube is coulored with three colours same colour on opposite faces. We define<br />
two rotations of the cube to be colour equivalent if they have the same action<br />
if you only consider how the colours are placed after the rotation<br />
Show that this relation is a equivalence relation on the set of rotations of the<br />
cube and is compatible with composition<br />
Example 21: Möbius equivalence (see nr 22)<br />
Two 2 × 2-matrices over C which induces the same Möbius transformation are<br />
said to be Möbius equivalent. This relation is compatible with multiplication<br />
of matrices.<br />
39
6: Appendix : Classification<br />
Mathematicians are very fond off classifying and have developed a terminology<br />
which is well suited to describe the act of classifying.<br />
A given universe is partitioned into classes in such a way that any element is<br />
in a class and any class contains an element.<br />
Formally a partition is a set X (the universe) together with a set K of subsets<br />
of X. The sets which are members of K are said to be classes and an element<br />
in a class is said to be a representative of this class.<br />
It shall be so that each element in X is a representative for exactly one class<br />
and each class has at least one representative. In other words the classes are<br />
mutually disjoint and non empty and their union is all of X. The class to which<br />
x belongs is denoted [x]K.<br />
We shall use two examples throughout :<br />
Example 1: The set K = {3Z, 1 + 3Z, 2 + 3Z} is a partition of Z with [x]K =<br />
x + 3Z. These are the remainder classes modulo 3.<br />
Example2: The set of lines that are parallel with the y-axes is a partition of the<br />
plane.<br />
Up to know we have been dealing with the result of a partition. The process of<br />
partitioning must be based on some specific properties of the members of X.<br />
One typical method is to have a rule which tells you if two members go into<br />
the same class or not.<br />
Formally this is takes the form of a relation x ∼ y which tells that x and y<br />
are in the same class. For such a relation to lead to a partition it is necessary<br />
and sufficient that it is an equivalence relation. On the basis of such a relation<br />
the classes are constructed in the following (obvious way): For each x ∈ X we<br />
define the class Kx = {y: y ∼ x}. Then we get the set K = {Kx: x ∈ X} of<br />
classes, which we can check actually to be a partition of X, (if ∼ happens to<br />
be an equivalence relation, and we have that at [x]K = Kx.<br />
We shall say that this partition is generated by (or induced by or associated<br />
with ) ∼ and call it the quotient of X modulo ∼ and denote it X/∼ or simply<br />
X∼. We shall return to the reason for calling it quotient.<br />
We are going to use [x]∼ to denote the class for which x is a representative.<br />
40
Lets take a look on the two examples<br />
If we define x to be congruent with y modulo 3 if x − y is et multiple of 3,<br />
then we have an equivalence relation ≡ which is the one which generates the<br />
partition into remainder classes.<br />
For the next example we shall interpret the coordinates (x, y) to record that<br />
some specific event takes place at timex at the location y. Then the relation<br />
”simultaneously” is an equivalence relation i the set of events and will generate<br />
the partition into vertical lines.<br />
Another method of classification is based on some property of the elements<br />
which determines its class. Formally we can model a property as a mapping f<br />
of X into some set I, such that f(x) is the instance of the property characterizes<br />
x. Then for each possible property, that is each member of i ∈ I, we can assign<br />
the class K i = {x: f(x) = i}.<br />
We shall say that this partition has f as its indicator. And we shall speak of<br />
the quotient of X modulo f and denote it X/f or simply Xf . and [x]f the class<br />
of x.<br />
Lets take another look at the examples:<br />
We get the partition into remainder classes modulo 3 by using I = {0, 1, 2} and<br />
f(x) = x mod 3.<br />
We get the partition into vertical lines for each of the following choices of<br />
indicator function:<br />
1) f is the projection on the x-axis<br />
2) f(x, y) = x<br />
3) f is the time of the event<br />
Having an indicator f there is an obvious equivalence relation which tells you<br />
if two elements belong to the same class for this indicator. We just say that<br />
x ∼ y if f(x) = f(y). We shall say that ∼ is generated by f. It is then obvious<br />
that X/∼= X/f and that [x]∼ = [x]f .<br />
To any partition you can assign an equivalence relation ∼ which generates it by<br />
defining x ∼ y to mean that x and y are in the same class, that is [x]K = [y]K .<br />
We can also f ind a function which is the indicator for the partition, namely<br />
the so called canonical indicator or canonical projection k, by letting I = K<br />
and k(x) = [x]K. To check that k is indicator for K you have to show that all<br />
41
classes in K have the form {x: k(x) = K} for some K ∈ K. But this is clear<br />
from the definitions.<br />
In our examples we see that<br />
x ↦→ [x]r is the canonical indicator for the partition into remainder classes<br />
(x, y) ↦→ the vertical line(x, y) is the canonical indicator for the partition into<br />
vertical lines<br />
Many indicators are made by selecting a special well suited representative for<br />
each class to represent all the members of the class, so to speak some sort of<br />
principal representative .<br />
Some examples:<br />
1) The principal remainder modulo some integer<br />
2) The projection on the x-axis (that is (x, 0))<br />
3) Rn in the class of linear spaces of dimension n in the partioning of<br />
the set of finite dimensional linear spaces with the dimension as the<br />
indicator.<br />
4) the principal argument of a complex number. To be more explicit:<br />
In the real numbers you have an equivalence relation which makes to<br />
real numbers equivalent if theire difference is a multiple of 2π. As<br />
5)<br />
the indicator of a class you often choose the (unique) member in the<br />
interval [0, 2π[.<br />
The principal value of arcsin by the partition which has sin as its<br />
indicator function. (the pricipal value of arcsin y is that solution x to<br />
the equation sin x = y which is in the interval [ − π<br />
]<br />
π<br />
2 , 2 . This is the<br />
one which computer programmes chooses if not otherwise instructed.<br />
For mathematicians arcsin is the equivalence class itself, since this<br />
leads to the simplest rules for calculation.<br />
We promised to motivate, explain, excuse the terminology quotient. We do it<br />
by an example: Let Z = X × Y , and let pX and pY denote the projections on<br />
the factors. If we use pX as the indicator function we get a partition into classes<br />
of the form K x = {(x, y): y ∈ Y } and the mapping X ∋ x ↦→ K x ∈ X/pA is a<br />
bijection. We use this to identify X/f with X and so Z = K × Y . This could<br />
be expressed that K is Z divided by Y .
Anders Madsen<br />
CATEGORIES<br />
OF<br />
<strong>STRUCTURES</strong><br />
<strong>ABSTRACT</strong><br />
<strong>ALGEBRAIC</strong><br />
<strong>STRUCTURES</strong><br />
y z<br />
y z<br />
yz<br />
bc<br />
b c<br />
b c<br />
2
This is a booklet in a series with the title ”Abstract algebraic structures”, which I have<br />
written for the algebra courses (E1,BE2) part of the mathematics education at RUC<br />
These booklets should be seen in connection with another series with the title ”Concrete<br />
algebraic structures”. These two series should be viewed as the jaws of a pair of pinchers.<br />
It would obviously be futile (and frustrating) to teach abstract algebra without involving<br />
concrete examples and it would be poor (and devoid of perspective) only to build concrete<br />
examples without involving the underlying abstract structures.<br />
And then again, there is a certain beauty in emphasizing the abstract character by isolating it<br />
and let its top-down nature appear clearly as in done in the controversial idol ”The Elements<br />
of Mathematics” by Bourbaki. Starting with the coarsest structures and then step by step<br />
refining by adding structural elements. All results which occur in many circumstances are<br />
proved once and for ever in its most elementary form.<br />
In the same way there is some satisfaction combined with letting each concrete structure stand<br />
as simple as possible without to much ado, das Ding an sich. And the pleasure by recognizing<br />
the essentially same type of argument recurring over and over in different disguises.<br />
Besides having the aesthetic satisfaction from the pure abstraction and the pure pleasure<br />
from the concrete details, both perspectives have a great cognitive influence and contributes<br />
to the development of competencies which are essential to any mathematician.<br />
I have chosen to emphasize these to oppositely directed but coordinated perspectives which<br />
is implied in the two series.<br />
The individual concrete structures are displayed in separate expositions without mutual<br />
references. Subjects which are needed at more places are repeated at each place. But the<br />
choice of details is made in such a way as to best possible deliver stuff to be used in examples<br />
in the abstract part.<br />
The text of the abstract part is not printed but can be found at<br />
http://milne.ruc.dk/~am/algebra<br />
Anders Madsen, may 2012<br />
44
Use the table of contents in the first part Go<br />
45
In the previous part the subject was general algebraic structures. We made no<br />
assumptions on the number or the character of the operations. Nor did we use<br />
any laws of calculation. Now we turn to specific algebraic structures. These<br />
are defined by specifying which types of operations are in play and which rules<br />
(also known as axioms) they must obey. For each such specification there will<br />
be many concrete algebraic structures answering the specification. We shall<br />
speak about a category of algebraic structures when we consider the collection<br />
of all structures determined by a given specification.<br />
We shall advance gradually through different categories, in each step adding<br />
some extra operation or some extra axiom.<br />
Consequently we start out with the category of semigroups, where there is only<br />
one operation and only one axiom, associativity.<br />
In the next step we pick out an element (considered to be a 0-ary operation)<br />
and claim as an axiom that this element is a neutral element. This is the<br />
category of monoids.<br />
We continue this way adding more operations and more axioms through the<br />
categories of groups, rings and fields.<br />
For each of these categories we study the induced structures and check if they<br />
them selves belong to the category. Since they have the same type it is enough<br />
to check if the axioms are satisfied.<br />
Then we study homomorphisms between structures of the same category .<br />
7: Semigroups<br />
7. 1: Definition<br />
59. Definition: Semigroup<br />
An algebraic structure (A, ♢ ), where ♢ is binary and associative is said to be<br />
a semigroup.<br />
60. Remark: Semigroups are half groups<br />
The name semigroup refers to the fact that many semigroups make up half (so<br />
to speak) of some group, for instance the natural numbers with addition, which<br />
is (roughly) half of the group of integers with addition. As a matter of fact it<br />
46
is always possible to extend (double) a semigroup to a genuine group, but we<br />
are not going to show how.<br />
Almost all the binary operations appearing in algebraic operations are associative.<br />
Semigroups thus are in the fundament of most algebraic structures.<br />
Accordingly most examples of semigroups can be embedded in more refined<br />
structure. In the sequel we shall meet monoids and groups as examples of this<br />
aspect. But semigroups which are essentially semigroups exist, and are treated<br />
in E23 and X28 and in X33 .<br />
7. 2: Induced semigroups<br />
61. Definition: Subsemigroup.<br />
A substructure of en semigroup is called a subsemigroup if the induced structure<br />
is a semigroup.<br />
62. Theorem: All substructures of a semigroup are subsemigroups<br />
A substructure of a semigroup is a subsemigroup<br />
Proof : Associativity is inherited (T32).<br />
This may be stated more operationally:<br />
63. Theorem: Criterion for subsemigroup<br />
A subset B of a semigroup (A, ♢ ) is a subsemigroup if and only if it is closed<br />
wrt to the operation (B ♢ B ⊆ B),<br />
64. Theorem: Structures induced by a semigroup are semigroups.<br />
If (S, ♢ ) is a semigroup and M is a set, then F(M, S) is a semigroup with the<br />
induced operation.<br />
If S1, . . . , Sn are semigroups then S1 × . . . × Sn is a semigroup.<br />
If ∼ is an equivalence relation compatible with the semigroup S then the quotient<br />
structure S∼ is a semigroup.<br />
47
Proof : Associativity is inherited by the induced structures, and no more is<br />
needed.<br />
7. 3: Powers<br />
The notion of power with positive integer exponent can be defined and the<br />
power rules proved at this very primitive level. And then we have dealt with<br />
this aspect once and for always. Notice that the definitions and the proofs are<br />
the ones you learned very early in your life.<br />
65. Definition: Powers with positive exponent.<br />
In a semigroup parentheses are superfluous and so powers with positive integer<br />
exponent of an element a may be immediately defined by the following recursive<br />
procedure: a 1 = a, a n+1 = aa n<br />
66. Theorem: Rules for calculating with Powers<br />
a n a m = a n+m , (a n ) m = a mn for all a and all n, m > 0.<br />
If ab = ba then (ab) n = a n b n .<br />
Proof : Induction. Example. Let m be fixed. We show the formula for this m<br />
and all n by induction after n.<br />
For n = 1 we have u n u m = uu m = u m+1 per definition of power.<br />
Suppose then that you have proved the formula for n = k. Then we have that<br />
u k+1 u m = uu k u m = uu m+k = u m+k+1 , and so the formula is also proved for<br />
n = k + 1.<br />
8: Monoids<br />
8. 1: Definitions<br />
A very large part of the binary operations, used in algebraic structures, have a<br />
neutral element, an element without action when used by the operation.<br />
To study the pure effect of introducing a neutral element we introduce the<br />
notion of a monoid, a structure which differs from a semigroup only by admitting<br />
a neutral element. Actually there are a lot of interesting monoids without<br />
further structure. For the sake of completeness we take the following<br />
48
67. Definition: Neutral element for a binary operation<br />
The element e is said to be a neutral element wrt den binary operation ♢ on<br />
A, if for all a ∈ A we have that a ♢ e = e ♢ a = a. We identify e with the<br />
constant operation (with 0 operands) for which e is the constant value.<br />
The prototype of a ”genuine” monoid, one without further ado is the set<br />
F(X, X) of all mappings of X into itself with composition as binary operation<br />
and the identity mapping as the neutral element. Other examples are seen<br />
in X33<br />
68. Theorem: Uniqueness of neutral element<br />
Any operation has at most one neutral element.<br />
Proof : If e and e ′ are neutral elements, then e = e ♢ e ′ = e ′<br />
69. Definition: Monoid<br />
A monoid is an algebraic structure (A, ♢ , e) for which (A, ♢ ) is a semigroup<br />
(the underlying semigroup) and e is a neutral element wrt ♢ . Since the neutral<br />
element is uniquely determined we shall also let (A, ♢ ) denote the monoid.<br />
8. 2: Induced monoids<br />
70. Definition: Submonoid<br />
A substructure of a monoid which is itself a monoid with the induced structure,<br />
is said to be a submonoid.<br />
71. Theorem: Each substructure of a monoid is a submonoid<br />
Substructure of monoid is submonoid<br />
49
Proof : The first thing to show is that the substructure is a semigroup. This<br />
follows from T62. Next we show that there is a neutral element. A substructure<br />
is per definition closed wrt each operation in the structure, and the neutral<br />
element is (considered to be ) an operation. The neutral element will therefore<br />
be a member of the substructure.<br />
A somewhat more operational way is<br />
72. Theorem: Criterion for submonoid<br />
A subset B of en monoid (A, ♢ , e) is a submonoid if and kun if B is closed<br />
wrt the operation (B ♢ B ⊆ B) and e ∈ B.<br />
Even though the following result may seem somewhat meager it is useful to<br />
have it coined for later reference (T77). But logically it belongs right here:<br />
73. Theorem: The trivial submonoid<br />
The singleton {e} is a submonoid<br />
Proof : It is obviously a substructure and therefore a submonoid.<br />
74. Theorem: Structures induced by a monoid.<br />
If (A, ♢ , e) is a monoid and M is a set, then F(M, A) is a monoid with the<br />
induced operations. The neutral element is constant mapping M ∋ x ↦→ e.<br />
Product structures of monoids (Ai, ♢ i<br />
e = (e1, . . . , en).<br />
, ei) is a monoid. The neutral element is<br />
A quotient structure (A, ♢ , e) modulo ∼ of a monoid is a monoid. The neutral<br />
element is [e]∼.<br />
Proof : We only need to check the axioms: In all three cases we know that the<br />
underlying structure is a semigroup, which is the first axiom. Which remains<br />
is to check the neutral element. This is straightforward in all the cases. Lets<br />
do it.<br />
For F(M, A) the induced operation is the constant mapping defined on M with<br />
the constant value e. This is seen to be neutral by simple inspection..<br />
50
For product structures we must inspect e1 × . . . × en, which is the constant<br />
operation (e1, . . . , en), and is easily seen to be neutral :<br />
(a1, . . . , an) ♢ 1<br />
×· · ·× ♢ n<br />
(e1, . . . , en) = (a1 ♢ 1<br />
e1, . . . , an ♢ n<br />
en) = (a1, . . . , an).<br />
For quotient structures we must inspect e/ ∼, which is the constant operation<br />
[e]. To show it to be neutral we have the following simple calculation:<br />
[a][e] = [ae] = [a]<br />
Examples of function spaces are X31,X32. Example quotient monoids are found<br />
in: E22<br />
8. 3: Examples<br />
Deal with the following examples and exercises. E23 E24 E25 E26 X27<br />
X28 X29 X30 X31<br />
8. 4: Monoid homomorphisms<br />
75. Definition: Monoid homomorphism<br />
A homomorphism between monoids is said to be a monoid homomorphism.<br />
X31,X33.<br />
X32<br />
76. Definition: The kernel for a monoid homomorphism<br />
By the kernel for a monoid homomorphism we mean the inverse image of the<br />
neutral element. So if (A, ♢ , eA) and (B, ♡ , eB) are monoids and f : A → B<br />
is a monoid homomorphism, then f −1 ({eB}) is called the kernel for f (wrt to<br />
the structures). It is denoted by ker f<br />
77. Theorem: The kernel of a monoid homomorphism is a submonoid<br />
Let (A, ♢ , eA) and (B, ♡ , eB) be monoids and f : A → B a monoid homomorphism,<br />
Then ker f is a submonoid.<br />
51
Proof : The kernel is the inverse image of a substructure, since the neutral<br />
element (as a singleton) is a submonoid (T73). A general result (T49) states<br />
that the inverse image of a structure is substructure.<br />
You can meet more monoids in X33 E34 E35 E36<br />
8. 5: Inverse<br />
78. Definition: Invertible element. Inverse.<br />
Suppose that (A, ♢ , e) is a monoid, and let a ∈ A. We shall say that a is<br />
invertible if there exists b ∈ A such that a ♢ b = e and b ♢ a = e. Then b is<br />
said to be an inverse element to a wrt ♢ .<br />
79. Theorem: Inverse is unique.<br />
An invertible element has only one inverse element.<br />
Proof : Let b1 and b2 be inverses of a. Then by associativity b1 = b1 ♢ e =<br />
b1 ♢ (a ♢ b2) = (b1 ♢ a) ♢ b2 = e ♢ b2 = b2.<br />
80. Definition: The inverse element.<br />
The unique inverse is said to be the inverse element for a.<br />
If you insist on having a general way of denoting the inverse you may write<br />
a ♢ −1 . But usually the context can tell you what operation is involved and then<br />
we simply write a −1 .<br />
8. 6: Powers<br />
81. Definition: Powers with negative exponents in monoids<br />
Let a be an invertible element. For n ∈ N we define a −n to be (a −1 ) n , and a 0<br />
to be e (the operation being known from context)<br />
82. Theorem: Power rules<br />
For any invertible element a and for all m, n ∈ Z we have that at a m ♢ a n =<br />
a m+n and (a m ) n = a mn .<br />
52
Proof : The theorem is already proved for n, m > 0 when dealing with semigroups<br />
,T66. It is easily seen also to hold for n = 0. And by using that we<br />
have a n a −n = a n (a −1 ) n = (aa −1 ) n = e for all m ≥ n ≥ 0 you then get<br />
a m a −n = a m−n+n−n a n a −n = a m−n a n a −n = a m−n . It is left to the reader to<br />
fill in the remaining details of the proof<br />
The following theorem is known in a lot of special cases, such as functions and<br />
matrices<br />
83. Theorem: Inverse of combinations<br />
If a and b are invertible elements of monoid (M, ♢ ) then so is a ♢ b and a −1 .<br />
Their inverses can be computed as (a ♢ b) −1 = b −1 ♢ a −1 and (a −1 ) −1 = a<br />
Proof : Exercise<br />
8. 7: Translations<br />
84. Theorem: The monoids of left translations<br />
Let La denote left translation by a in the monoid (M, ♢ ). Let LM denote<br />
the set of all such left translations. Then the mapping a ↦→ La is a monoid<br />
homomorphism from M to LM wrt to ♢ and composition of functions, that is<br />
La ♢ b = La ◦ Lb.<br />
Proof : Exercise<br />
85. Theorem: A translation with invertible element is invertible,<br />
that is a bijection<br />
If a is an invertible element in a monoid, then left translation with a is a<br />
bijection of the monoid on itself.<br />
Proof : Using the terminology from T84 we have that La ◦ L a − 1 = L a ♢ a −1 =<br />
Le = I (and similarly from the right) and therefore La is a bijection with<br />
inverse L a −1<br />
This has the following corollary:<br />
86. Theorem: Cancelation in monoids<br />
53
If a is invertible then for all x, y in the monoid you have that<br />
ax = ay ⇒ x = y<br />
meaning that you can cancel multiplication with invertible elements<br />
Proof : Exercise<br />
87. Theorem: A monoid homomorphism respects inversion<br />
For a homomorphism between monoids you have that if a is invertible then so<br />
is f(a) and its inverse is given by f(a) −1 = f(a −1 )<br />
Proof : Just check that the candidate for an inverse is ok: f(a −1 )f(a) =<br />
f(a −1 a) = f(eA) = eB.<br />
9: Groups<br />
And now we take one step up in the hierarchy by assuming all elements to have<br />
inverses which is equivalent to adding a new operation namely taking inverses.<br />
Therefore<br />
9. 1: Definitions<br />
88. Definition: Group<br />
An algebraic structure (G, ♢ , ⋆, e) is said to be a group if (G, ♢ , e) is a monoid<br />
( the underlying monoid) and if all a ∈ G is invertible wrt ♢ with ⋆a as inverse<br />
element. So the axioms are<br />
1) x ♢ (y ♢ z) = (x ♢ y) ♢ (x ♢ z)<br />
2) e ♢ x = x ♢ e = x<br />
3) a ♢ (⋆a) = (⋆a) ♢ a = e<br />
89. Remark: Short notation<br />
Since the neutral element and all inverses are uniquely determined by the binary<br />
operation ♢ it is common practice to let the group be denoted by the short form<br />
(G, ♢ )<br />
90. Remark: Multiplicative notation<br />
54
When dealing with abstract groups one often chooses to use the common multiplication<br />
symbol (the invisible dot) to denote the binary operation, that is ab.<br />
Similarly we use a −1 for ⋆a and 1 for e. If we say that G is a group without<br />
further specification of the operations this notation is understood. If there is<br />
more than one group in the context it may be relevant to denote the neutral<br />
element of G by 1G.<br />
9. 2: Induced group structures<br />
91. Theorem: The induced structures are group structures<br />
The structures that a group induces on subsets, products, function spaces and<br />
quotients are groups<br />
Proof : The only things left over from the analogue theorem for monoids is to<br />
check for invertibility, which is left to the patient reader.<br />
9. 3: subgroup<br />
92. Definition: Subgroup<br />
A substructure of a group is called a subgroup if it is a group when it is equipped<br />
with the induced operations<br />
It is easy to check for subgroups by the following characterization<br />
93. Theorem: Any substructure of a group is a subgroup<br />
Let H be a substructure of the group G. Then H is a subgroup .<br />
Proof : Since H is a substructure, H will be closed wrt all the operations,<br />
especially the operations in the underlying monoid. So H is a submonoid, and<br />
so a monoid.<br />
The operation of taking inverses in G induces an operation on H, which is<br />
obviously the operation of taking inverses in H.<br />
The next theorem is a more operative formulation of the same fact<br />
94. Theorem: Criterion for subgroup .<br />
55
A subset H of a group G is a subgroup if and only if HH ⊆ H, H −1 ⊆ H and<br />
1 ∈ H.<br />
95. Definition: Group homomorphism<br />
A homomorphism between two groups is said to be a group homomorphism<br />
96. Theorem: Images and inverse images of subgroups<br />
Let f : G1 → G2 be a group homomorphism and let H1 be a subgroup of G1.<br />
Put H2 = f(H1).Then H2 is a subgroup of G2.<br />
Let H2 be a subgroup of G2. Put H1 = f −1 (H2). Then H1 a subgroup of G1.<br />
Proof : A general result (T48) states that the image of a substructure is itself<br />
a substructure, so H2 is a substructure of G2, and therefore a subgroup. The<br />
proof for inverse image is similar.<br />
97. Definition: The Kernel<br />
The kernel for a group homomorphism f is the kernel for the underlying monoid<br />
homomorphism, and is denoted ker(f)<br />
98. Theorem: The trivial subgroup<br />
{1} is a subgroup (called the trivial subgroup ).<br />
Proof : It is obvious that {1} is closed wrt all the operations, therefore a substructure<br />
and therefore a subgroup .<br />
99. Theorem: The kernel for a group homomorphism is a subgroup<br />
Let f : A → B be a homomorphism. Then ker(f) is a Subgroup of A.<br />
Proof : The set {1B} is a subgroup . Its inverse image therefore also is<br />
subgroup . But this is per definition the kernel.<br />
The following theorem is useful for proving that a homomorphism is injective.<br />
The technique is to move questions to the neutral element by translation. It is<br />
then sufficient to have injectivity at the neutral element.<br />
100. Theorem: Characterization of injective homomorphisms<br />
56
A homomorphism f : A ↦→ B is injective precisely when its kernel is trivial (it<br />
consists of the neutral element alone<br />
Proof : Necessity is obvious. To show the sufficiency suppose that the only<br />
element in the kernel is 1A.. Then if f(a) = f(b) it follows that f(a −1 b) =<br />
f(a) −1 f(b) = 1B from which it follows that a −1 b = 1A, and therefore a = b.<br />
This proves the injectivity.<br />
A well known special case of this theorem is about linear space, and is recalled<br />
in E37. Another useful special case E38 is about matrices.<br />
The preceding examples are most probably already well known to you from<br />
past experience. May be X39 presents a new example.<br />
Now we are ready for the important subject of quotient groups. So we must<br />
consider equivalence relations compatible with the the group structure.<br />
Our first step is to introduce a way of creating such equivalence relations by<br />
using a subgroup .<br />
101. Theorem: A subgroup induces an equivalence relation<br />
Let H be a subgroup of the group G. The relation ∼ defined by<br />
a ∼ b ⇔ b ∈ aH<br />
is an equivalence relation.<br />
Proof : Reflexivity:<br />
a ∼ a, since a = a1 and 1 ∈ H, and so a ∈ aH.<br />
Symmetry:<br />
If a ∼ b then b ∈ aH per definition. Therefore there exists h ∈ H with b = ah.<br />
From this you can deduce that a = bh −1 and since h −1 ∈ H you have that<br />
a ∈ bH. Therefore b ∼ a<br />
Transitivity :<br />
Let x ∼ y and y ∼ z, which per definition means that y ∈ xH and z ∈ yH.<br />
Therefore there exists h1 and h2 with y = xh1 and z = yh2, and therefore<br />
z = yh2 = xh1h2 ∈ xH, and consequently x ∼ z.<br />
57
102. Definition: The equivalence relation induced by a subgroup<br />
The equivalence relation established in the preceding theorem is said to be the<br />
left equivalence relation induced by H. The equivalence classes are said to be<br />
the left cosets for H.)<br />
It is obvious to define right equivalence relation and right cosets analogously.<br />
Lets give some interpretation of what is going on here. Imagine that the elements<br />
in G have some property (call it their color). You should think of the<br />
elements h in the subgroup H to be so that their action doesn’t change the<br />
colour, by what we mean that ah has the same colour as a. Then the coset aH<br />
consists of the elements with the same colour. If H consists of all the ”neutral”<br />
elements then each coset represents a specific colour. Here are two examples<br />
X40 and X41<br />
103. Theorem: A Left coset for a subgroup is the image of H by a<br />
bijection.<br />
The left cosets have the form aH and are the result of a left translation of H.<br />
The mapping h ↦→ ah is a bijection of H on aH.<br />
Similar for right cosets.<br />
Proof : Let ∼ denote the left equivalence corresponding to H. This means<br />
per definition that b ∼ a ⇔ b ∈ aH. This shows that [a] = aH. Let La denote<br />
left translation with a that means that La(x) = ax. Then La(H) = aH. Left<br />
translation with an invertible element in a monoid is injective (T85). And in<br />
a group all elements are invertible .<br />
You have maybe met this result in the situation in X42. X43<br />
104. Definition: Index for subgroup<br />
If H is a subgroup in G then we define the index for H to be the number<br />
(possibly ∞) of left cosets and we denote it by [G : H]. Notice that we would<br />
have the same number if we had used the right cosets<br />
X44 X45<br />
105. Theorem: Lagrange’s theorem<br />
58
If H is a subgroup of the finite group G then |G| = [G : H]|H| and consequently<br />
|H| is a divisor in |G|.<br />
Proof : Each coset has the same number of members as the subgroup, since<br />
the coset is the result of left translation of subgroup and the left translation is<br />
bijective. Since G is the disjoint union of the cosets the result follows.<br />
A useful application of this theorem is exhibited in E46.<br />
9. 4: Quotient groups<br />
As seen a subgroup will give rise to a partition into classes, and the resulting<br />
quotient offers interesting information on the action of the subgroup on the<br />
group structure.<br />
However it is not always possible to transfer the group structure to the quotient<br />
since multiplication of two cosets does not always produce another coset, X41.<br />
To ensure the possibility of that, more is needed. Exactly what is the content<br />
of the following notions and results.<br />
106. Definition: Normal subgroup<br />
Let N be a subgroup of the group G. We shall say that N is normal if the<br />
induced equivalence relations are compatible with the group structure.<br />
We have the following convenient way to test if we have a normal subgroup<br />
107. Theorem: Criteria for normality<br />
The following conditions are equivalent<br />
(1) N is normal<br />
(2) (aN)(bN) = (ab)N and (Na)(Nb) = N(ab)<br />
(3) x −1 Nx = N for all x ∈ G<br />
(4) xN = Nx for all x ∈ G<br />
59
Proof :<br />
(1) ⇒ (2): If N is normal you have per definition that the associated left equivalence<br />
relation is compatible with the group operation and as a result of this<br />
calculation with the classes can be carried out by calculation with representatives,<br />
which is what is stated in (2).<br />
(2) ⇒ (3) Let x be an arbitrary member of G. Then by (2) we have that<br />
(xN)(x −1 N) = (xx −1 )N = N. For an arbitrary member h ∈ N we then have<br />
that xhx −1 e ∈ N. Therefore also xhx −1 ∈ N, and since h was arbitray it<br />
follows that xNx −1 ⊆ N. The opposite inclusion is obvious.<br />
(3) ⇒ (4): Exercise.<br />
(4) ⇒ (1): Assuming (4) we must show that the left equivalence relation is<br />
compatible with the group operation. Let a ∼ b and c ∼ d. Per definition this<br />
means that b ∈ aN and d ∈ cN. So we may choose h, k ∈ N with b = ah and<br />
d = ck. Since hc ∈ Nc and since Nc = cN per assumption it is possible to find<br />
p ∈ N with hc = cp. Therefore bd = ahck = acpk and pk ∈ N. This shows that<br />
bd ∈ acN which means that bd ∼ ac, establishing the compatibility. To show<br />
that also inversion is compatible let a ∼ b, that is a ∈ bH = Hb. So you may<br />
find h ∈ N with a = hb. Then a −1 = b −1 h −1 , which shows that a −1 ∈ b −1 H<br />
and so a −1 ∼ b −1 which establishes the compatibility. These criteria are used<br />
in X47 X48<br />
108. Theorem: Commutativity guaranties normality<br />
Any subgroup of a commutative group is normal.<br />
Proof : Trivial.<br />
109. Theorem: Same left and right coset in case of normality<br />
If N is a normal subgroup , then the right equivalence relation is identical with<br />
the left and so this relation is just said to be the equivalence relation induced<br />
by N and is denoted by ∼N .<br />
110. Theorem: Quotient group.<br />
Let N be a normal subgroup of the group G. The induced quotient structure is<br />
a group structure.<br />
60
Proof : This is just T91<br />
111. Definition: Quotient group.<br />
Let N be a normal subgroup of the group G. The induced quotient structure<br />
is said to be the quotient G and N and is denoted G/N.<br />
G∼N<br />
9. 5: Group homomorphisms<br />
112. Theorem: The kernel of a group homomorphism is a normal<br />
subgroup .<br />
The kernel of a group homomorphism f : G1 → G2 is a normal subgroup and<br />
the coset are the level sets of f. The equivalence relation induced by f is denoted<br />
∼f and is identical with the one induced by the kernel ∼ ker(f).<br />
Proof : Let N denote the kernel of f. We use one of the criteria for normality<br />
(T 107), namely x −1 Nx ⊆ N. Let h ∈ N and x ∈ G be arbitrary. Then<br />
f(x −1 hx) = f(x −1 )f(h)f(x) = f(x) −1 1Bf(x) = 1B, and so x −1 hx ∈ N.<br />
Let now aN be an arbitrary coset. Then we have that<br />
x ∈ aN ⇔ a −1 x ∈ N ⇔ f(a −1 x) = 1B ⇔ f(a) −1 f(x) = 1B ⇔ f(x) = f(a),<br />
so that aN = {x|f(x) = f(a)}. From this it also follows that the cosets of N<br />
are the same as the equivalence classes for f<br />
113. Definition: The homomorphism induced by a quotient group.<br />
The quotient group induced by the kernel of some homomorphism f is also<br />
called the quotient group induced by f and we can denote it by G/f.<br />
A normal subgroup can thus be constructed from a homomorphism. A little<br />
more surprising is it may be than any normal subgroup may be constructed<br />
this way. But that is what is said in<br />
114. Theorem: Any normal subgroup is the kernel of some homomorphism<br />
(for instance the canonical projection)<br />
Let N be a normal subgroup i G. The canonical projection kN of G on G/N is<br />
a group homomorphism with kernel N.<br />
61
Proof : Exercise<br />
These results are used in X49, X50, E51<br />
X52<br />
10: Rings<br />
10. 1: Definitions and rules<br />
115. Definition: Ring<br />
An algebraic structure (R, +, −, 0, ∗, 1) is said to be a ring if (R, +, −, 0) is a<br />
commutative group and (R, ∗, 1) is a monoid and if ∗ is both left and right<br />
distributive wrt +. Furthermore 0 ̸= 1. If ∗ is commutative the ring is called<br />
commutative. We speak in a obvious way about the underlying additive group<br />
and the underlying multiplicative monoid.<br />
There are very many sorts of rings, see for instance X53 X54 X55<br />
There are a lot of obvious rules that nevertheless needs proof :<br />
116. Theorem: Rules for calculation in rings<br />
(1) a ∗ 0 = 0 ∗ a = 0<br />
(2) (−a) ∗ b = a ∗ (−b) = −(a ∗ b)<br />
(3) (−a) ∗ (−b) = a ∗ b<br />
(4) (−1) ∗ a = −a<br />
Proof : Lets start recalling that −(−a) = a follows by considering the underlying<br />
group.<br />
(1) We have that a = 1 ∗ a = (1 + 0) ∗ a = 1 ∗ a + 0 ∗ a = a + 0 ∗ a, and so<br />
a = a + 0 ∗ a and then 0 ∗ a = 0.<br />
(2) We also have that (−a) ∗ b + a ∗ b = ((−a) + a) ∗ b = 0 ∗ b = 0, and so<br />
(−a) ∗ b + a ∗ b = 0 and therefore (−a) ∗ b = −(a ∗ b).<br />
(3) We have, using (2) twice, that (−a)∗(−b) = −(a∗(−b)) = −(−(a∗b)) = a∗b<br />
and therefore (−a) ∗ (−b) = a ∗ b.<br />
(4) We have that (−1) ∗ a = −(1 ∗ a) = −a.<br />
62
10. 2: Subring<br />
117. Definition: Delring<br />
A substructure of a ring which is a ring in its own right is said to be a subring.<br />
118. Theorem: Any substructure of ring is a subring<br />
Let D be a substructure of the ring R. Then D is a subring.<br />
Proof : When D is a substructure as a ring then it is also a substructure of<br />
the underlying additive group and so (D, +, 0, −) is itself a group, obviously<br />
commutative. In a similar way we see that (D, ∗, 1) is a monoid. Distributivity<br />
follows since this propery is hereditary.<br />
This may be formulated as the following criterion<br />
119. Theorem: Subring criterion<br />
A subset D of a ring (R, +, −, 0, ∗, 1) is a subring if and only if D + D ⊆<br />
D,D ∗ D ⊆ D,−D ⊆ D and 1 ∈ D.<br />
Proof : The conditions are clearly necessary. And they explicitly guaranties<br />
the closure of all the operations except 0. To see that 0 =∈ D notice that<br />
0 = 1 − 1.<br />
10. 3: Induced rings<br />
120. Theorem: Function spaces<br />
Suppose that A is a ring and that M is a set. The algebraic structure on<br />
F(M, A) induced by A is a ring.<br />
Proof : (F(M, A), +, −, 0) is a commutative group since it is induced by the<br />
underlying additive group. In the same way we see that (F(M, A), ∗, 1) is a<br />
monoid. Distributivity is inherited.<br />
The preceding theorem furnishes many important examples of rings, the rings<br />
of functions: X56, with some important subrings, rings of polynomials E57.<br />
121. Theorem: The product of rings is a ring<br />
63
Suppose that A and B are rings then A×B is a ring. Analogously for a product<br />
with several factors.<br />
Proof : (A × B, + × +, 0 × 0, − × −) is the product of de underlying additive<br />
groups and is therefore a group , and obviously commutative. In a similar way<br />
we have that (A×B, ∗×∗, 1×1) is a monoid. Again distributivity is inherited.<br />
10. 4: Quotient rings and ideals<br />
122. Theorem: Quotient of a ring is a ring.<br />
The quotient of en ring wrt to a compatible equivalence relation a ring<br />
Proof : The underlying additive group in the quotient is the quotient of the<br />
underlying additive group and therefore a group, obviously commutative. The<br />
underlying multiplicative monoid of the quotient is the quotient of the underlying<br />
multiplicative monoid and therefore a monoid. Distributivity is inherited<br />
by quotients.<br />
123. Theorem: The canonical projection is a homomorphism<br />
Let ∼ be an equivalence relation on a ring R, which is compatible with the<br />
structure. The canonical projection k∼ of R onto the quotient R∼ is a ring<br />
homomorphism<br />
Proof : This is an immediate consequence of the general result for quotient<br />
structures (T56).<br />
124. Definition: Kernel of a ring homomorphism<br />
The kernel of homomorphism f between rings is the kernel of f considered as<br />
a homomorphism of the underlying additive groups<br />
We are now going to introduce the analogy to a normal subgroup . A normal<br />
subgroup is a subgroup which induces an equivalence relation, which is compatible<br />
with the underlying additive group. But when we want to transfer the<br />
operations to the quotient ring we also need the relation to be compatible with<br />
multiplication. This special type of substructure is called an ideal. An ideal<br />
will always be the kernel of a homomorphism of the ring structures.<br />
125. Definition: Ideal.<br />
64
A subset I is said to be et ideal of the ring R, if I is a subgroup of the underlying<br />
additive structure and is closed wrt to multiplication by any element in the ring.<br />
This can be expressed by<br />
I + I = I<br />
RI = I<br />
IR = I<br />
126. Theorem: The kernel is an ideal.<br />
Let f be a homomorphism of the ring R1 into the ring R2. Then the kernel of<br />
f is an ideal<br />
Proof : Let K denote the kernel of f. Then f a homomorphism between the<br />
underlying groups with kernel K, which therefore is a subgroup of the underlying<br />
group in R1.<br />
This means that the first condition in the definition of ideals is fulfilled.<br />
The second condition: Let k ∈ K, r ∈ R then f(kr) = f(k)f(r) = 0f(r) = 0.<br />
Consequently kr ∈ I. Analogously for rk ∈ I.<br />
127. Theorem: The quotient wrt to an ideal is a ring<br />
Let I be an ideal in the ring R. Then I is a normal subgroup of the underlying<br />
additive group R. Let R/I denote the quotient group. Then the equivalence<br />
relation associated with I is compatible with the ring structure on R and the<br />
quotient structure R/I is a ring.<br />
Proof : We must show that the multiplication is compatible with the equivalence<br />
relation associated with I. Therefore let x1 ∼ y1 and x2 ∼ y2. This<br />
means per definition that x1 − y1 ∈ I and that x2 − y2 ∈ I. We use the good<br />
old trick to express a difference between two products by the difference of the<br />
factors to get<br />
x1x2−y1y2 = x1x2−x1y2+x1y2−y1y2 = x1(x2−y2)+(x1−y1)y2 ∈ IR+RI<br />
and by the properties of an ideal we have that RI + IR = I + I = I and<br />
so we have proved that x1x2 ∼ y1y2. Therefore the quotient structure is well<br />
defined and the product is given by the formula [x][y] = [xy] or in other words<br />
(x + I)(y + I) = (xy) + I. It is easy to check that this product makes the<br />
quotient a monoid with [1] as neutral element for multiplication. To check that<br />
this multiplication is distributive wrt addition is again routine. So the quotient<br />
is a ring.<br />
65
128. Remark: Motivation for ideals.<br />
The preceding result is the announced motivation for notion of an ideal.<br />
129. Definition: Quotient ring wrt to an ideal<br />
The ring constructed in the previous theorem is said to be the quotient ring wrt<br />
to the ideal ideal and is denoted R/I<br />
E58<br />
130. Theorem: An ideal is the kernel of a homomorphism.<br />
Let I be an ideal in the ring R and let kI denote the canonical projection of R<br />
onto the quotient ring R/I. Then kI is a homomorphism with the ideal as its<br />
kernel.<br />
Proof : The projection kI on a quotient is a homomorphism in any algebraic<br />
structure. So it only remains to show that the kernel of k is I. For all a ∈ R<br />
we have that kI(a) = [a]I = a + I. The 0-element in R/I is kI(0) = 0 + I = I.<br />
It therefore holds that a ∈ ker(kI) ⇔ kI(a) = I ⇔ a ∈ I. And so I = ker(kI).<br />
10. 5: Integral domains<br />
Our next goal is to study fields which are rings where division is always possible<br />
except by 0. There is however an interesting and import intermediary step. In<br />
fields you can always cancel by division by a non zero factor. But there exists<br />
rings that are not fields and where cancelation is possible. Such rings are called<br />
integral domains and they are important stepping stones for constructing fields<br />
from rings. We are going to see this in use in two cases.<br />
The cancelation property is prepared by a couple of definitions.<br />
131. Definition: Proper element.<br />
An element in a ring is called proper if it is not the zero element.<br />
132. Definition: Zero divisor<br />
A proper element a in a ring is said to be a zero divisor, if 0 is a non trivial<br />
multiple of a, that is if there exists a proper element k such that ka = 0 or<br />
ak = 0<br />
66
133. Definition: Integral domain<br />
A ring without zero divisors is said to be an integral domain<br />
This is obviously equivalent to<br />
134. Theorem: Equivalent definition<br />
Integral domains are the rings in which the set of proper elements is closed wrt<br />
multiplication.<br />
135. Theorem: Cancelation in integral domains<br />
In an integral domain you may cancel proper elements: If a is proper and<br />
ax = ay then x = y.<br />
Proof : If ax = ay then a(x − y) = 0 and since a is proper we must have that<br />
x − y = 0.<br />
We are not going to study structures induced by integral domains.<br />
We are next going to consider those ideals for which the quotient ring is a field:<br />
136. Definition: Proper ideal<br />
An ideal which is not a proper subset is said to be a proper ideal<br />
137. Theorem: The ideal closure.<br />
Let D be a subset of the ring R. Then the intersection of all ideals containing<br />
D is an ideal, called the ideal closure of D.<br />
Proof : Lets first show that the mentioned intersection I is again an ideal.<br />
So let r ∈ R, i ∈ I. Let I1 be an arbitrary ideal containing D. Then ri ∈ I1.<br />
Since this holds for all I1 then ri ∈ I.<br />
Next I must be the least ideal containing D, since it is per construction part of<br />
all others.<br />
138. Definition: The ideal closure<br />
The ideal constructed in the previous theorem is said to be the ideal generated<br />
by D or the ideal closure of D.<br />
67
139. Theorem: Description of the ideal closure.<br />
Let D be en subset of the ring R. Then the ideal closure of D is the set<br />
{x1a1 + . . . + xnan : x1, . . . , xn ∈ R, a1, . . . , an ∈ D}<br />
Proof : Let B denote the set above. It is readily checked that its an ideal<br />
according to the definition of ideals. It also contains D since any element<br />
a ∈ D may be written in the form 1a, where 1 denotes the 1-element of the<br />
ring. And any ideal containing D must contain all elements of B. And so B<br />
must be the least ideal containing D.<br />
140. Definition: Maximal ideal<br />
An ideal, which is maximal among all proper ideals, is said to be a maximal<br />
ideal<br />
E59<br />
Now we are ready for<br />
11: Fields<br />
11. 1: Definitions<br />
141. Definition: Field<br />
En ring, with commutative multiplication, for which any proper element is invertible<br />
wrt multiplication is called a field.<br />
This means an algebraic structure (L, +, −, 0, ·, ⋆, 1) satisfying the following<br />
axioms<br />
1) · is commutative<br />
2) (L, +, −, 0, ·, 1) is a ring<br />
3) (L ∗ , ·, 1, ⋆) is a group , where L ∗ = L \ {0}.<br />
142. Definition: Subfield<br />
En substructure of et field, being itself a field, is said to be et subfield.<br />
68
143. Theorem: Any substructure of a field is a subfield<br />
Any substructure of a field is a subfield<br />
Proof : You have to check that the substructure satisfies the axioms, which<br />
results from the analogous results for subrings and subgroups (T118 and T93.<br />
11. 2: Quotient ring over a maximal ideal<br />
Now we are prepared for one of the main results in algebra:<br />
144. Theorem: The Quotient ring wrt a maximal ideal is a field<br />
Let I be a maximal ideal in the ring R. Then the quotient ring R/I is a field<br />
Proof : Let L denote R/I. We know then that L is a ring (T 127). What<br />
remains is then to show that any proper element x in L is invertible. Lets<br />
choose a ∈ R with x = [a] and so x = a + I. Since x is proper we have that<br />
a /∈ I. Now let J be the ideal generated by I ∪ {a}. Since I is a proper subset<br />
of J it is not possible for J to be a proper ideal, therefore J = R, and so<br />
1 ∈ J. This means according to T139 that you can find a1, . . . , an ∈ I ∪ {a}<br />
and r1, . . . , rn ∈ R such that<br />
1 = r1a1 + . . . + rnan,<br />
which after some change of order also may be written<br />
1 = r1a1 + . . . + rpap + rp+1a + . . . + rna = ra + r ′ ,<br />
where we have put r = rp+1 + . . . + rn and r ′ = r1a1 + . . . + rnan. This implies<br />
since r ′ ∈ I, that [ar] = [1]. Putting now y = [r] then yields xy = [a][r] =<br />
[ar] = [1], which says that x has y as its inverse. E60<br />
11. 3: Fields of fractions<br />
Now we come to yet another way to construct fields from rings. We shall show<br />
how to construct all the fractions.<br />
The rational numbers constitutes a field, which moreover is the least to contain<br />
the integers Z as a subring. Any integral domain like Z, can in the same<br />
way be considered to be situated inside some field. Next follows a way to<br />
construct such a field. You can also consider this to be a way to construct the<br />
69
ationals from the integers. This make it possible to take the stand that you<br />
can define the rationals if only you take the integers for granted. This fits into<br />
a program where you want to have all concepts defined. This includes a way of<br />
constructing the integers from the natural numbers. And so on till you reach<br />
the foundations of mathematic based on set theory. But this is quite another<br />
story to be told in another afsnit.<br />
It is them important to say that the construction can be used in a lot of other<br />
cases, where we don’t have the result of the construction already.<br />
First we have to clarify the notions with the following<br />
145. Definition: Field of fractions<br />
Let R be a ring and let L be et field. Suppose there exists an injective homomorphism<br />
i : R → L where L is the underlying ring with the property that any<br />
x ∈ L may be written in the form i(p)/i(q) where p and q are in R. Now we<br />
use i to identify R with i(R), that is we consider p and i(p) to be the same.<br />
Then R is a subring of L and any element in L can be written in the form p/q,<br />
where p and q belongs to R. We shall say that L via i is a field of fractions for<br />
R. We shall call i the associated embedding of R into L<br />
Now follows the construction, which has the form of a<br />
146. Theorem: Existence of field of fractions<br />
There exists a field of fractions for the ring R if and only if R is an integral<br />
domain<br />
Proof : The necessity of the condition follows from the fact that L as a field<br />
is also an integral domain and so the i(R) as a substructure of an integral<br />
domain is itself an integral domain, and then of course this also holds for R by<br />
isomorphism.<br />
To prove the sufficiency assume that R is an integral domain.<br />
We define the set S = {(p, q) ∈ R × R : q ̸= 0}. On S we define an addition<br />
+ by the assignment (a, b) + (c, d) = (ad + bc, bd) and a multiplication by<br />
the assignment (a, b)(c, d) = (ac, bd). That this in fact defines operations rely<br />
heavily on R being an integral domain, since the non existence of zero divisors<br />
ensures that bd ̸= 0. We also define the opposite operation −(a, b) = (−a, b)<br />
and a reciprocal operation (a, b) −1 = (b, a). Finally we define a zero element<br />
(0, 1) and et unit-element (1, 1).<br />
70
Then it is routine to check that S equipped with these operations is a ring.<br />
Now the set I consisting of all elements of the form (0, b) can be checked to be<br />
an ideal.<br />
We let L denote the quotient ring of S wrt this ideal. Let ∼ denote the associated<br />
equivalence relation, then the following calculation<br />
(x, y) ∼ (u, v) ⇔ (x, y) − (u, v) ∈ I ⇔ (xv − yu, yv) ∈ I ⇔ xv − yu = 0<br />
shows that<br />
(x, y) ∼ (u, v) ⇔ xv = yu<br />
In what follows we shall use [p/q] to denote the equivalence class containing<br />
(p, q) ∈ S.<br />
We notice that [1/1] is the class which contains the unit element (1, 1) in the<br />
ring S, and so is the unit element, 1, in the quotient L. Furthermore this class<br />
also contains all the elements (a, a) because (a, a) ∼ (1, 1) since a · 1 = 1 · a.<br />
Analogously you see that [0/1] is the zero element,0.<br />
If [a/b] ̸= 0 the a ̸= 0, and then (b, a) ∈ S, and [b/a][a/b] = [ab/ab] = 1, and<br />
this shows that [a/b] is invertible with [a/b] −1 = [b/a] . And the L is a field.<br />
The mapping defined by h(a) = (a, 1) is a homomorphism of R into S, since<br />
h(a) + h(b) = (a, 1) + (b, 1) = (a · 1 + b · 1, 1 · 1) = (a + b, 1) = h(a + b)<br />
h(a)h(b) = (a, 1)(b, 1) = (ab, 1) = h(ab)<br />
The mapping i : R → L defined by i(a) = [h(a)] = [a/1] is a composition of<br />
two homomorphisms and therefore itself a homomorphism. To show that it is<br />
injective lets assume that i(a) = i(b). Using the definitions we see that<br />
i(a) = i(b) ⇔ [a/1] = [b/1] ⇔ (a, 1) ∼ (b, 1) ⇔ a · 1 = 1 · b ⇔ a = b.<br />
which proves injectivity<br />
As a result of the identification of a with i(a) we have justified the the following<br />
147. Definition: Canonical field of fractions for an integral domain<br />
The field constructed in the previous theorem is said to be the canonical field of<br />
fractions for R. The mapping i is said to be the canonical embedding of R.<br />
71
The canonical field of fractions L for an integral domain R has the property<br />
that it is a field which is an extension of R. Furthermore it is minimal wrt<br />
to this property in the sense that it does not contain more than absolutely<br />
necessary, namely the fractions made from elements in R.<br />
In some sense it is also the only one possible. This is stated in the following<br />
148. Theorem: ”Uniqueness” of field of fractions<br />
Assume that L, M are fields of fractions for the ring R via i, j. Then there<br />
exists an isomorphism F between the fields L and M such that i = F ◦ j, which<br />
in particular means that the fields are isomorphic.<br />
Proof : It is enough to show the theorem when L is the canonical fields of<br />
fractions for R. We use the notation from the proof of the existence part.<br />
So let M be any field of fractions for R and let j be the embedding of R into<br />
M. We define the map f of S into M by<br />
f(p, q) = j(p)/j(q)<br />
The first step is to show that f is a homomorphism (between rings). This is<br />
straight forward using that j is a homomorphism and L is a field:<br />
f ((p, q) + (r, s)) =<br />
j(ps + qr)<br />
j(qs)<br />
= f(p, q) + f(r, s)<br />
= j(p)j(s) + j(q)j(r)<br />
j(q)j(s)<br />
= j(p) j(r)<br />
+<br />
j(q) j(s)<br />
Next we note that the equivalence relation ∼ is generated by f in the sense that<br />
(p, q) ∼ (r, s) ⇔ f(p, q) = f(r, s)<br />
This means that we can define a function F on S/I by F ([p/q]) = f(p, q)<br />
since the value does not depend on the chosen representative of the class. Then<br />
f = F ◦k and therefore F is a homomorphism by a general theorem, T392, since<br />
f is surjective as a result of the definition D145 and since k is a homomorphism<br />
( T56)<br />
Next we show that F is injective. So assume that F (a) = F (b) with a =<br />
[p/q], b = [r/s]. Then f(p, q) = F (a) = F (b) = f(r, s) and so (p, q) ∼ (r, s)<br />
which means that a = b. Since f is surjective, so is F . Now we have shown<br />
that F is an isomorphism between fields<br />
72
The prototype example of field extension is the extension from Z to Q (see<br />
E61)<br />
A very important field of fractions is the one resulting from the ring of polynomials<br />
: E62<br />
12: Examples and exercises<br />
Example 22: Words of equal length (see nr 74)<br />
Th relation x and y have the same length is an equivalence relation on the<br />
monoid of words and it is compatible with concatenation En equivalence class<br />
consists of all words of a certain length. This classification is induced by the<br />
homomorphism which is defined to be the length of the word.<br />
Example 23: The semigroup of words (see nr 74)<br />
The set S of words over a given alphabet with concatenation as operation is a<br />
semigroup.<br />
Example 24: The semigroup of words is not a monoid, but ... (see nr 74)<br />
If the semigroup of words is extended with the empty word you will get a<br />
monoid with the empty word as the neutral element.<br />
Example 25: The set of naturals with addition is a semigroup (see nr 74)<br />
The set of naturals with addition is a semigroup<br />
Example 26: The word length is a homomorphism (see nr 74)<br />
The mapping x ↦→ |x| which to a word assigns its length (number of letters) is a<br />
semigroup homomorphism. It can in an obvious way be extended to a monoid<br />
homomorphism<br />
Exercise 27: Greatest common divisor gives a semigroup (see nr 74)<br />
On the integers you have the binary operation x ⊓ y which to x and y assigns<br />
the greatest common divisor of x and y.<br />
Show that Z with this operation is a semigroup.<br />
Exercise 28: Greatest common divisor does not yield a monoid (see nr 74)<br />
Show that the semigroup i X27 is not (the underlying semigroup of) a monoid<br />
Exercise 29: Minimum gives a semigroup (see nr 74)<br />
We define x ∧ y as the minimum of x and y.<br />
73
Show that ∧ is a binary operation on Z, making it a semigroup and that N is<br />
a subsemigroup.<br />
Exercise 30: Minimum does not give a monoid (see nr 74)<br />
Show that the semigroup in X29 is not (the underlying semigroup of) a monoid.<br />
Exercise 31: Minimum induces an operation on sequences (see nr 74)<br />
The set of sequences of integers may be identified with F(N, Z).<br />
We let ∧ (also) denote the operation F(N, Z) which is induced by ∧. This<br />
operation is the component wise minimum,that is (a ∧ b)(n) = a(n) ∧ b(n).<br />
Show that F(N, Z) with this operation is a semigroup.<br />
We shall say about a sequence of integers that it has finite support if only<br />
finitely many of its elements are different from zero. We denote the set of<br />
sequences with finite support by Z ∞ 0 .<br />
Show that Z ∞ 0 is a subsemigroup.<br />
Exercise 32: The prime spectrum as a homomorphism (see nr 75)<br />
Continuation of X27 and X31. To any natural number n we assign the sequence<br />
e with finite support, where e(i) is defined by the property that<br />
n = p(1) e(1) p(2) e(2) · · ·<br />
is the factorization of n into primes. Here p(i) is the prime nr i. So e(i) = 0<br />
if the prime p(i) do not appear in the factorization. We denote this sequence<br />
with σ(n) and call it the prime spectrum of n.<br />
Show that σ is a homomorphism from N into Z ∞ 0 for a suitable choice of<br />
operations.<br />
Show also that σ is a monoid homomorphism, when the operation on N is<br />
multiplication and the operation on Z ∞ 0 is componentwise addition.<br />
Exercise 33: Operations on subsets (see nr 77)<br />
Let X be the set of subsets of X. Show that (X , ∩, X) and (X , ∪, ∅) are<br />
monoids. Let Xe denote the set of finite subsets of X and Xc denote the set of<br />
subsets with finite complement.<br />
Show that (A, ♢ ) is a semigroup when A is Xe or Xc and ♢ is ∩ or ∪.<br />
Some of these semigroups may be extended to monoids. Decide which.<br />
74
Consider the possibility of the mapping A ↦→ X \ A being a homomorphism or<br />
an isomorphism.<br />
Example 34: Remainder classes with multiplication is a monoid (see nr 77)<br />
The relation x ≡ y on Z defined by demanding x and y to be in the same<br />
remainder class modulo n is an equivalence relation which is compatible with<br />
multiplication. Since (Z, ·, 1) is a monoid so is (Z/≡, ·, 1), where 1 also is used<br />
to denote the remainder class containing 1.<br />
Example 35: The monoid of remainder classes is isomorphic with Zn (see nr<br />
77)<br />
The relation x ≡ y i E 34 is induced by the mapping f(x) = x mod n with<br />
values in Zn = {1, . . . , n − 1}. By defining a multiplication ⊙ on Zn by the<br />
formula x ⊙ y = xy mod n, you see that f becomes an isomorphism between<br />
the monoids (Z/≡, ·, 1) and (Zn, ⊙, 1). You often see Zn defined as Z/≡.<br />
Example 36: Condition for existence of modular reciprocal (see nr 77)<br />
Continuation of E34. You can prove by using the extended Euclidean algorithm<br />
that the remainder class [x] is invertible if and only if x and n are coprime<br />
Example 37: Criterion for injectivity of (see nr 100)<br />
A linear mapping is injective if its kernel (null space) only contains the null<br />
vector. You can consider a linear space as a group with u + v, 0 og −u as<br />
operations.<br />
Example 38: The matrix representing a linear mapping with respect to given<br />
bases is uniquely determined (see nr 100)<br />
The mapping A ↦→ LA, which to a matrix A assigns the linear mapping LA<br />
represented by A is bijective. The mapping is obviously linear. Its kernel<br />
consists of the matrices which represents the zero mapping. This means that<br />
each column is coordinate vector for the 0 vector and therefore itself the 0<br />
vector. Therefore the matrix is the 0 matrix which therefore is the only element<br />
of the kernel.<br />
Exercise 39: Short proof of the Chinese remainder theorem (see nr 100)<br />
Use that the mapping in E15 is a group homomorphism to show it is injective.<br />
This gives another proof of the Chinese remainder theorem (but not a way of<br />
finding the solution).<br />
Exercise 40: A subgroup of the group of rotations of the tetrahedron (see nr<br />
102)<br />
Let G be the group of of rotations of a regular tetrahedron with corners A, B,<br />
C and D. Let H be the set of rotations with axis through A. It can be seen<br />
75
that H is a subgroup with the members I, a and a −1 , where a is a rotation<br />
of 120 ◦ about the axis through A. Determine the partition in left and right<br />
cosets.<br />
Information: a = (BCD) b = (ADC) c = (DAB) d = (ACB).<br />
Exercise 41: A subgroup of the group of isometries of the triangle (see nr<br />
102)<br />
Let G be the group of rotations and reflections of a regular triangle with corners<br />
A, B, C. Let d denote rotation 120 ◦ about the midpoint of the triangle and<br />
let a, b, c denote the reflections in the lines through A, B, and C, respectively.<br />
Let H be the set {I, a}. It may be seen that H is a subgroup. By a direct<br />
calculation (for instance by calculation with permutations of the corners ) you<br />
get that bH = {b, d} and cH = {c, d}. But (bH)(cH) = {b, c, d, d −1 }. This is<br />
an example that the product of two left cosets not necessarily is again a left<br />
coset<br />
Exercise 42: There are the same number of even and odd permutations (see<br />
nr 103)<br />
Show that the set of even permutations constitutes a subgroup with exactly<br />
one more coset, namely the set of odd permutations. Use this prove that there<br />
is the same number of even and odd permutations.<br />
Exercise 43: Any half subgroup is normal (see nr 103)<br />
Show that any subgroup which contains half of the elements is normal. Use it<br />
to solve the preceding exercise<br />
Exercise 44: Index (see nr 104)<br />
Determine the index for subgroup H in X40<br />
Exercise 45: Index for the remainder class (see nr 104)<br />
Determine the index for 3Z as a subgroup of (Z, +).<br />
Example 46: The cosets of the stabilizer. A counting formula (see nr 105)<br />
Let X be a set and F a set of bijections of X into itself. Assume further that<br />
F is a group with composition of mappings as operation. Then we say that we<br />
have a group of transformations acting on X.<br />
For x ∈ X we define the stabilizer of x to be the subset of F consisting of those<br />
members that fix x. We denote it by Fx. To summarize:<br />
h ∈ Fx ⇔ h(x) = x<br />
76
Now it is easily checked that Fx is a subgroup of F. Let f ∈ F and let us<br />
determine the left coset fFx, (we use multiplicative notation for composition).<br />
From the definition of coset we get that<br />
g ∈ fFx ⇔ f −1 g ∈ Fx ⇔ f −1 g(x) = x ⇔ g(x) = f(x)<br />
Letting y = f(x) we see that the coset<br />
fFx = {g : g(x) = y},<br />
lets denote it by Fx→y. Therefore we have a coset for each y which is f(x) for<br />
some f ∈ F, that is for each y in the set {f(x) : f ∈ F}. This set is usually<br />
called the orbit of x wrt F and may be denoted be Fx. Therefore the index of<br />
Fx is |Fx|, the number of elements in Fx.<br />
Then it follows from Lagranges formula that<br />
|F| = |Fx||Fx|,<br />
a formula that may serve to determine the number of elements in a group of<br />
transformations if you have a suitable stabilizer.<br />
Exercise 47: A non normal subgroup of the tetrahedron group (see nr 107)<br />
Show that the subgroup H in X40 not is normal<br />
Exercise 48: A normal subgroup of the tetrahedron group (see nr 107)<br />
Let G be the group of rotations of a regular tetrahedron. Let H be the set<br />
consisting of the three edge preserving rotations (that is having axis through<br />
midpoints of edges) and the identity mapping. Show that H is a normal subgroup.<br />
Determine the cosets an set up a table for calculation with the classes.<br />
Exercise 49: Same argument (see nr 114)<br />
Let C ∗ denote the set of invertible complex numbers, that is all complex numbers<br />
except 0. Then C ∗ is a group with complex multiplication. Let ∼ be the<br />
relation z1 ∼ z2 defined by z1 = rz2 for some positive real number r. Show that<br />
this is a compatible equivalence relation. Determine the equivalence classes,<br />
and the quotient group. Find a function f of C ∗ into a group that generates<br />
this relation. Show that it is a homomorphism and find its kernel.<br />
Exercise 50: Same modulus (see nr 114)<br />
Let C ∗ denote the set of invertible complex numbers, that is all complex numbers<br />
except 0. Then C ∗ is a group with complex multiplication. Let ∼ be<br />
the relation z1 ∼ z2 defined by z1 = uz2 for some complex number u on the<br />
77
unit circle. Show that this is a compatible equivalence relation. Determine the<br />
equivalence classes, and the quotient group. Find a function f of C ∗ into a<br />
group that generates this relation. Show that it is a homomorphism and find<br />
its kernel.<br />
Example 51: Definition of argument (see nr 114)<br />
We let Rot denote the set of rotations of the plane around the origin.<br />
We let SO2 denote the set of orthogonal orientation preserving matrices of<br />
order 2.<br />
For any t ∈ R we let R(t) denote member of Rot with angle t.<br />
( )<br />
cos t − sin t<br />
For any t ∈ R we let A(t) denote the matrix<br />
sin t cos t<br />
For any t ∈ R we let E(t) denote the number e it .<br />
For any u ∈ U we let L(u) denote the mapping of C into itself defined by<br />
L(u)(z) = uz. Then L(u) ∈ Rot.<br />
( )<br />
a −b<br />
For any u = a + ib ∈ U we let U(u) denote the matrix<br />
b a<br />
For any 2×2 real matrix A we let M(A) be the mapping of the plane into itself<br />
given by M(A)(x) = Mx.<br />
Show that the following diagram reflects these definitions and argue that all<br />
the arrows are homomorphisms. Tell which are isomorphisms and find the<br />
quotients modulo the kernels for those which are not.<br />
(R, +)<br />
E<br />
❄<br />
(U, ·)<br />
❅<br />
❅❅A<br />
R ✲ (Rot, ◦)<br />
✻<br />
M<br />
❅❘<br />
U ✲ (SO2, ·)<br />
(R, +)<br />
R ✲ (Rot, ◦)<br />
E<br />
<br />
❄ <br />
✒<br />
L<br />
✻<br />
M<br />
(U, ·)<br />
U ✲ (SO2, ·)<br />
Exercise 52: The stabilizer is usually not normal (see nr 114)<br />
Continuation of E46. When is the stabilizer normal. Not very often. We are<br />
going to use the criterion (3) in T107. We have seen that g ∈ fFx if and only<br />
if g(x) = f(x). In a similar way we can show that g ∈ Fxf if and only if<br />
g −1 (x) = f −1 (x) or equivalently g(f −1 (x) = x.<br />
78
We shall use this to illustrate an important example:<br />
Let G be the group of rotations of the unit sphere and let H be the subgroup<br />
of rotations having the z-axis as its rotation axis. Then H is the stabilizer<br />
of the north pole (0, 0, 1). Lets choose f to be the rotation 90 degrees about<br />
the y-axis then g ∈ fH if and only if g(0, 0, 1) = f((0, 0, 1) = (1, 0, 0) and<br />
g ∈ Hf if and only if g −1 (0, 0, 1) = f −1 (0, 0, 1) = (−1, 0, 0) which means that<br />
g(−1, 0, 0) = (1, 0, 0). It is an interesting exercise to show that this implies<br />
that g = f (g and f coincides in two points and so everywhere!) and so fH<br />
and Hf only have f in common and so cannot coincide.<br />
Lets take a deeper look at this example. Each coset is determined by the value<br />
at the the north pole which is common for all its members. So one class could<br />
consist of all the rotations which rotates the north pole to (1, 0, 0), independent<br />
of axis and angle. Any point on the sphere is a possible image (the orbit of<br />
the north pole is the whole sphere). So the quotient can be identified with the<br />
sphere. With the customary notation this is often written as the formula:<br />
SO3/SO2 = S 2<br />
Exercise 53: Z is the prototype ring (see nr 115)<br />
Convince yourself that Z with obvious operations is a ring<br />
Exercise 54: Ring of matrices (see nr 115)<br />
Convince yourself that the set of n × n matrices is a ring with the operations<br />
matrix addition, matrix subtraction, zero matrix, opposite matrix, matrix multiplication<br />
and identity matrix.<br />
Show that this also can be given meaning and holds when the elements of the<br />
matrices are taken from an arbitrary ring.<br />
Exercise 55: Ring of endomorphisms (see nr 115)<br />
This is a generalization of X 54. Let V be et linear space and let R denote<br />
Lin(V, V ), the set of linear mappings of V into itself. Show that it is possible<br />
to equip R as a ring with addition being the usual addition of functions with<br />
values in a linear space and multiplication is composition of functions.<br />
Exercise 56: Function rings (see nr 120)<br />
Show that the following sets (with obvious operations) are rings:<br />
1) F(X, R), the set of real functions on an arbitrary set X<br />
2) F(X, Z), the set of functions with integer values on a set X<br />
79
Example 57: Polynomials with real coefficients (see nr 120)<br />
The set of real functions on R is a ring.<br />
The set of polynomials is a subring and so a ring. This ring is usually denoted<br />
by R[X]. In the sequel we shall denote it P.<br />
In P we have a lot of results which are analogous to results valid in the ring of<br />
integers when we replace the order of the integers by the order of polynomials<br />
according to degree: one polynomial is considered to be less than another one<br />
if its degree is less :<br />
You can carry out division with remainder. A well known algorithm learns you<br />
how to. The resulting remainder will be less than the divisor. And so you can<br />
use a Euclidean algorithm to find a greatest common divisor. And you can<br />
define an extended algorithm: to any two given polynomials a and b you can<br />
express their greatest common divisor c in the form xa + yb, where x and y are<br />
polynomials.<br />
Let p be a polynomial. Put I = pP, then I is an ideal consisting of all multiples<br />
of p. Therefore P/I is a ring, the ring of remainder classes modulo p.<br />
Lets see how multiplication works:<br />
[a + bx][c + dx] = [(a + bx)(c + dx)] = [ac + (bc + ad)x + bdx 2 ] = [ac − bd + (bc + ad)x +<br />
[ac − bd + (ad − bc]x] + [bd(1 + x 2 )] = [ac − bd + (ad − bc]x]<br />
From which we see that the mapping (a+ib) → [a+bx] is a homomorphism from<br />
the complex numbers with multiplication. It is actually a field isomorphism.<br />
If q divides p then I = pP must be a subideal of J = qP. If moreover p not<br />
divides q then the subideal is a proper subideal.<br />
The polynomial p is said to be irreducible if it can not be factored into two<br />
polynomials, unless one of the factors is a constant. This is the polynomial<br />
analogue of a prime number.<br />
If p is irreducible then I is a maximal ideal (show it !) and the quotient ring<br />
consequently a field.<br />
The polynomial p(x) = 1 + x 2 is irreducible and the associated field is isomorphic<br />
with the field of complex numbers.<br />
80
What is said above may be generalized to polynomials, whos coefficients are<br />
taken from an arbitrary ring. This may lead to finite fields.<br />
Example 58: Rings of remainder classes are the prototypes of a quotient ring.<br />
(see nr 129)<br />
Let n ∈ N with n > 1.<br />
Let I = nZ, and notice that I is an ideal in the ring (Z, +, ·). Two elements<br />
x and y are equivalent if y ∈ x + I, that is when y − x ∈ nZ. Let k be the<br />
canonical projection of Z on Z/I, defined by k(x) = [x] = x + I = x + nZ,<br />
which means that k(x) is the remainder class modulo n which contains x.<br />
We shall here see an alternative way to construct this ring:<br />
Let Zn = {0, . . . , n − 1}. We define an addition on Zn by letting u ⊕ v =<br />
(u + v) mod n. Analogously we define subtraction and multiplication. There<br />
are also obvious candidates for 0-element and 1-element. We are going to show<br />
that this defines a ring isomorphic with Z/nZ.<br />
It is seen that the mapping f : Z → Zn, defined by f(z) = z mod n, is a<br />
homomorphism with respect to the underlying algebraic structure. To check<br />
the axioms we us that f is surjective to see that addition and multiplication<br />
are associative. The remaining axioms are straightforward.<br />
The mapping g : Zn → Z/I defined by g(x) = [x], is en bijection. Its inverse<br />
h = g −1 is given by h([x]) = x mod n.<br />
Direct check gives that f = h ◦ k. Then f and k are homomorphisms and since<br />
f is surjective it follows that h is a homomorphism and consequently an isomorphism.<br />
OBS: indsÆt om muligt reference !!!It may be noticed that instead<br />
of using the classes we are using the principal representatives and in stead of<br />
calculating with classes we are doing calculation with the representatives. You<br />
can see an example of multiplication tables on the figure: In praxis you wont<br />
look in the table for each operation but instead carry out the operations in the<br />
usual way and then take the remainder at last or at certain practical moments.<br />
0 1 2 3<br />
0 0 1 2 3<br />
1 1 2 3 0<br />
2 2 3 0 1<br />
3 3 0 1 2<br />
81<br />
0 1 2 3<br />
0 0 0 0 0<br />
1 0 1 2 3<br />
2 0 2 0 2<br />
3 0 3 2 1
Tabel 1. Addition table (⊕) and multiplication table (⊗) i Z4)<br />
Example 59: Prototype maximal ideal (see nr 140)<br />
The ideal nZ in the ring Z is maximal if and only if n is prime.<br />
If n is not prime, n = pq, then nZ is a proper subset of pZ and then of course<br />
not maximal. This proves the only if part.<br />
To prove the if part assume that n is prime and I is an ideal containing nZ as<br />
a proper subset. Then I must contain a number m which is coprime with n.<br />
By the properties of ideals I must then also contain all numbers of the form<br />
xm + yn. By the extended Euclidean algorithm then also the greatest common<br />
divisor of n and m must be in I. Since m and n are coprime I then contains<br />
1. And so I = R. This proves that nZ is maximal.<br />
Example 60: A finite field with 4 elements (see nr 144)<br />
Let P be the ring of polynomials over Z2. Let p be the polynomial x 2 + x + 1,<br />
which is irreducible. The ideal I = pP is therefore maximal, and the quotient<br />
ring P/I a field.<br />
By using the Euclidean algorithm you can choose a polynomial of degree less<br />
than 2 in each remainder class. There are only four such polynomials , namely<br />
0, 1, x and x + 1. These must be in different classes and so the field has four<br />
element. As an example of multiplication of classes we compute [x][x + 1],<br />
which gives [x 2 + x] = [x 2 + x + 1 − 1] = [x 2 + x + 1] + [−1] = [1]. And so the<br />
product is [1]. We can summarize the operations in the following tables, where<br />
i is the class contains x and j is the class which contains x + 1.<br />
+ 0 1 i j<br />
0 0 1 i j<br />
1 1 0 j i<br />
i i j 0 1<br />
j j i 1 0<br />
· 0 1 i j<br />
0 0 0 0 0<br />
1 0 1 i j<br />
i 0 i j 1<br />
j 0 j 1 i<br />
Example 61: Extension from Z to Q (see nr 148)<br />
We take the stand that the good Lord has offered us the integers as a gift. If<br />
we furthermore thinks that he also gave os the rationals, then we can observe<br />
that the rationals is a field of fractions for the integers. If God had been<br />
less generous we could have constructed the rationals as the canonical field of<br />
fractions for Z.<br />
Example 62: Power series as fractions (see nr 148)<br />
82
The field of fractions for the ring of polynomials may be identified with a set<br />
of formal power series with finitely many negative exponents.<br />
We define a formal power series to be a sequence a ∈ F(Z, R), that is a doubly<br />
infinite sequence for which there exists an integer N such that an = 0 for all<br />
n < N. We write the series as ∑<br />
n∈Z aixi , but this is only formal or symbolical,<br />
we are not thinking of it as a proper sum, and no notion of convergence is<br />
involved.<br />
A polynomial a0 + a1X + a2X 2 + . . . + aMx M may be identified with its sequence<br />
of coefficients . . . , 0, 0, a0, a1, . . . , aM , 0, 0, . . . which may be considered<br />
as a formal power series.<br />
We define addition and multiplication by<br />
where<br />
(<br />
∑<br />
aix i<br />
) (<br />
∑<br />
+ bix i<br />
)<br />
= ∑<br />
cix i<br />
(<br />
∑<br />
aix i<br />
) (<br />
∑<br />
bix i<br />
)<br />
= ∑<br />
dix i<br />
n∈Z<br />
n∈Z<br />
cn = ∑<br />
ai + bidn = ∑<br />
i∈Z<br />
i∈Z<br />
n∈Z<br />
aibn−i<br />
It is straight forward to check that this are operations in the set of formal<br />
power series and that you get a ring. When restricted to polynomials we have<br />
the usual operations on polynomials.<br />
And it is actually a field! This takes some effort to show. Moreover it is a field<br />
of fractions for the ring of polynomials. This is harder to show.<br />
You should check that<br />
1<br />
1 − x = 1 + x + x2 + . . .<br />
83<br />
n∈Z<br />
n∈Z<br />
n∈Z
13: Stikordsregister<br />
84