1 mathematical structures: groups, rings, and fields roger house scientific buzz café coffee catz...

1

Mathematical Structures: Groups, Rings, and Fields

Roger HouseScientific Buzz Café

Coffee CatzSebastopol, CA

2008 September 18Copyright © 2008 Roger House

2

Why haven't I ever heard of this?

“Nothing which is taught at a secondary school was discovered later than the year 1800. ... Even among physicists, apart from those who work in quantum theory or relativity, I believe that those who do experimental work use hardly any more mathematics than was known to Maxwell in 1860.”

Jean Dieudonné (1906-1992)

Mathematics – The Music of Reason

3

6 – Our first group

• We begin with a set of six numbers

6 = { 0, 1, 2, 3, 4, 5 }

• We define an operation on the set which we call addition and denote by +

• This is similar to, but not really the same as, the usual operation of addition

• To see this, let's add 1 to each element of 6

4

Add 1 to each element of 6

0 + 1 = 1

1 + 1 = 2

2 + 1 = 3

3 + 1 = 4

4 + 1 = 5

5 + 1 = ?

• What is 5 + 1?

• Usually it's 6, but there is no 6 in our set

• What to do?

5

Getting nothing from something

• Let's be bold and define

5 + 1 = 0

• This might seem unreasonable, but consider what time it is 3 hours after 11am:

11 + 3 = 2

• Out of context this seems unreasonable too, but when doing "clock arithmetic" it makes sense

• Remember we said + is not exactly addition

6

Add 2 to each element of 6

0 + 2 = 2

1 + 2 = 3

2 + 2 = 4

3 + 2 = 5

4 + 2 = 0

5 + 2 = 1

• Now we seem to have two strange results, but keep thinking about clock arithmetic

7

Addition table for 6

0 1 2 3 4 5

1 2 3 4 5 0

2 3 4 5 0 1

3 4 5 0 1 2

4 5 0 1 2 3

5 0 1 2 3 4 Pick an element x in the left column and an

element y in the top row Where the row and column meet is x + y

8

Addition made easy

• Given any two elements x and y of 6 the

addition table shows us how to add them together to get the sum x + y

• There's another way to do it: Just add x plus y in the usual way as if they were ordinary whole numbers to get a sum z

If z < 6, then z is the sum

If z ≥ 6, then z – 6 is the sum

9

Examples of addition

3 + 2 = 5

3 + 3 = 6, so 6 - 6 = 0 is the sum

3 + 4 = 7, so 7 - 6 = 1 is the sum

4 + 1 = 5

4 + 4 = 8, so 8 - 6 = 2 is the sum

5 + 5 = 10, so 10 - 6 = 4 is the sum

Note how similar this is to clock arithmetic

10

0 is an identity

0 + 0 = 0

1 + 0 = 1

2 + 0 = 2

3 + 0 = 3

4 + 0 = 4

5 + 0 = 5 So x + 0 = x for every x We say 0 is an identity element for +

11

Every element has an inverse

0 + 0 = 0

1 + 5 = 0

2 + 4 = 0

3 + 3 = 0

4 + 2 = 0

5 + 1 = 0 Every element x has an inverse element x'

such that

x + x' = 0

12

Be associative

• How to compute 5 + 4 + 2 ?

(5 + 4) + 2 = (3) + 2 = 5

5 + (4 + 2) = 5 + (0) = 5

• For any three elements x, y, z of 6 we have

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

• Another way to say this: Addition is associative

13

Can we prove + is associative?

• There are 6∙6∙6 = 216 different triples of numbers from

6 and + appears 4 times in

(x + y) + z = x + (y + z) So by using the addition table 4∙216 = 864

times, we can prove that + is associative This is a proof by “main strength and

awkwardness” There are better ways

14

Where are we?

We have a set of six elements

6 = { 0, 1, 2, 3, 4, 5 }

We have an operation + defined on the set which takes any two elements of the set and combines them to produce another element of the set

The operation is associative, so for any three elements x, y, and z we have

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

15

Still in Kansas?

• An identity element 0 exists so that for any element x we have

x + 0 = x

• Every element x has an inverse element x' so that

x + x' = 0

• In short, we have a group

16

Groups

• Note that the word “group” is being used here in a highly technical sense

• A group is a non-empty set with a binary operation defined on it such that the operation is associative, an identity element exists, and every element has an inverse

• The group is one of the most fundamental structures in mathematics

• Groups sprout like weeds in modern math

17

Examples of groups

• Let n = { 0, 1, 2, ..., n-1 } for any integer n >

0, and define + in analogy with + for 6

• nis a group

• nis called the cyclic group of n elements

• Note that for every positive integer n there exists at least one group with n elements

• So we don't have to worry about running out of groups

18

Some groups go on forever ...

• Now consider = { ..., -2, -1, 0, 1, 2, ... }, the set of all integers, and let + be the usual addition of integers

• Is <,+> a group?

• Is + a binary operation on ?• Is + associative?

• Which element is the identity?

• For x an integer, what is its inverse?

19

is indeed a group

• + on integers is a binary operation

• It is associative

• 0 is the identity element

• For any integer x, its inverse is -x

• is an infinite group, the first such we have seen

20

Is <,*> a group?

• Now consider and the operation *, the usual multiplication of integers

• Is <,*> a group?

• Is * a binary operation on ?• Is * associative?


• For x an integer, what is its inverse?

21

<,*> is NOT a group

• * on integers is a binary operation



• For any integer x, its inverse is 1/x

• Whoops! 1/x is not an integer (except when x = -1 or x = +1), so the vast majority of integers have no multiplicative inverses

22

Is <+,*> a group?

• Let + denote the set of all positive real numbers, and let * be the usual multiplication of real numbers

• Is <+,*> a group?

• Is * a binary operation on +?

• Is * associative?


• For x a positive real, what is its inverse?

23

<+,*> is a group

• * on real numbers is a binary operation



• For any real x, its inverse is 1/x

• <+,*> is an infinite group, the second such we have seen

• [It's way more infinite than <,+>, but that's the subject of another talk.]

24

Time for some geometry

• All the groups we have seen so far are arithmetic in nature

• It's time to consider a geometrical example

• We start with a regular hexagon

• hexa- means six (Greek)

• -gon means angled (Greek)

• regular means all sides are equal and all angles are equal

25

A regular hexagon

0

12

3

4 5

26

Transformations of the hexagon

• We are going to consider transformations of the hexagon which map vertices to vertices

• We will restrict our transformations to counterclockwise rotations by multiples of 60º

• Any two rotations which map vertex 0 to the same vertex v are considered identical

27

CCW rotations of the hexagon

• There are six possible rotations:

R0 – rotate the hexagon CCW by 0º






28

H - A set of rotations

• Let's define a set whose members are the six rotations of the hexagon:

H = { R0, R

1, R

2, R

3, R

4, R

5 }

• We define an operation on the set called composition and denote it by *

• The composition Ri * R

j of two rotations is

simply the rotation obtained by first doing rotation R

i and then doing rotation R

j

29

Examples of composition

R1 * R

2 = R

3

R2 * R

2 = R

4

R3 * R

4 = R

1

R5 * R

4 = R

3

R3 * R

3 = R

0

R2 * R

5 = R

1

30

R0 is an identity rotation

R0 * R

0 = R

0

R1 * R

0 = R

1

R2 * R

0 = R

2

R3 * R

0 = R

3

R4 * R

0 = R

4

R5 * R

0 = R

5

So x * R0 = x for every x

R0 is an identity element for *

31

Every rotation has an inverse

R0 * R

0 = R

0

R1 * R

5 = R

0

R2 * R

4 = R

0

R3 * R

3 = R

0

R4 * R

2 = R

0

R5 * R

1 = R

0

Every element x has an inverse element x' x * x' = R

0

32

Composition is associative

• How to determine R5 * R

4 * R

2 ?

(R5 * R

4) * R

2 = (R

3) * R

2 = R

5

R5 * (R

4 * R

2) = R

5 * (R

0) = R

5

• For any three elements x, y, z of H we have

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

• So composition is associative (but we haven't proved this)

33

So, what have we got?


H = { R0, R

1, R

2, R

3, R

4, R

5 }

We have an operation * defined on the set which takes any two elements of the set and combines them to produce another element of the set


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

34

H is a group!

• An identity element R0 exists so that for

any element x we have

x * R0 = x


x * x' = R0

• In short, H is a group!

35

Composition table for H

R0 R

1 R

2 R

3 R

4 R

5

R1 R

2 R

3 R

4 R

5 R

0

R2 R

3 R

4 R

5 R

0 R

1

R3 R

4 R

5 R

0 R

1 R

2

R4 R

5 R

0 R

1 R

2 R

3

R5 R

0 R

1 R

2 R

3 R

4

36

Is H starting to look familiar?

• Consider this mapping:

0 ↔ R0

1 ↔ R1

2 ↔ R2

3 ↔ R3

4 ↔ R4

5 ↔ R5

37

Some correspondences

1 + 2 = 3 ↔ R1 * R

2 = R

3

2 + 2 = 4 ↔ R2 * R

2 = R

4

3 + 4 = 1 ↔ R3 * R

4 = R

1

5 + 4 = 3 ↔ R5 * R

4 = R

3

3 + 3 = 0 ↔ R3 * R

3 = R

0

2 + 5 = 1 ↔ R2 * R

5 = R

1

38

It's déjà vu all over again

• Given a true statement about 6 replace

each digit i by Ri and replace + by * and we

have a true statement about H

• And vice versa

• We say that 6 and H are isomorphic

<6,+> <H,*>

iso- means same (Greek) -morph means shape or form (Greek)

39

<6,+> is isomorphic to <H,*>

• When two mathematical objects are isomorphic, they have the same structure

• When a statement is proved true about one structure, the equivalent statement is true about the other structure

• Sometimes one structure is easier to work with than another

• Addition in 6 may be easier than rotating

cardboard hexagons

40

More geometry

• Let's consider another geometrical example

• This time we'll work with a regular triangle, also called an equilateral triangle

• tri- means three (Latin)

• regular means all sides are equal and all angles are equal, so each angle is 60º

41

An equilateral triangle

A

B

C

42

Transformations of the triangle

• We're going to consider transformations of the triangle which map vertices to vertices

• One kind of transformation is a counterclockwise rotation by a multiple of 120º

• Another kind of transformation is a flip around an axis through a vertex, perpendicular to the side opposite

43

Rotations and flips

There are six such transformations:

R0 – rotate CCW by 0º



F0 – flip around the axis defined by A

F1 – flip around the axis defined by B

F2 – flip around the axis defined by C

44

Permutations, anyone?

Another view of the transformations:

R0 ≡ ABC

R1 ≡ CAB

R2 ≡ BCA

F0 ≡ ACB

F1 ≡ CBA

F2 ≡ BAC

45

S3 – Triangle transformations

• Let's define a set whose members are the six transformations of the triangle:

S3 = { R

0, R

1, R

2, F

0, F

1, F

2 }

• We define an operation on the set called composition and denote it by *

• The composition t*s of two transformations is simply the transformation obtained by first doing transformation t followed by transformation s

46

Examples of composition

R1 * R

2 = R

0

R2 * R

2 = R

1

F0 * R

1 = F

2

F2 * F

1 = R

1

R1 * F

2 = F

0

F2 * R

1 = F

1

47

R0 is an identity

R0 * R

0 = R

0

R1 * R

0 = R

1

R2 * R

0 = R

2

F0 * R

0 = F

0

F1 * R

0 = F

1

F2 * R

0 = F

2

So x * R0 = x for every x

R0 is an identity element for *

48

Every element has an inverse

R0 * R

0 = R

0

R1 * R

2 = R

0

R2 * R

1 = R

0

F0 * F

0 = R

0

F1 * F

1 = R

0

F2 * F

2 = R

0

Every element x has an inverse element x' x * x' = R

0

49

Composition is associative

• How to determine F2 * R

2 * F

0 ?

(F2 * R

2) * F

0 = (F

0) * F

0 = R

0

F2 * (R

2 * F

0) = F

2 * (F

2) = R

0

• For any three elements x, y, z of H we have

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

• So composition is associative (but we haven't proved this)

50

We've been here before


S3 = { R

0, R

1, R

2, F

0, F

1, F

2 }

We have an operation * defined on the set which takes any two elements of the set and combines them to produce another element of the set


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

51

S3 is a group

• An identity element R0 exists so that for

any element x we have

x * R0 = x


x * x' = R0

• In short, S3 is a group!

52

Composition table for S3

R0 R

1 R

2 F

0 F

1 F

2

R1 R

2 R

0 F

1 F

2 F

0

R2 R

0 R

1 F

2 F

0 F

1

F0 F

2 F

1 R

0 R

2 R

1

F1 F

0 F

2 R

1 R

0 R

2

F2 F

1 F

0 R

2 R

1 R

0

53

Something a bit odd

You may have noticed something of interest in earlier examples of composition:

R1 * F

2 = F

0

F2 * R

1 = F

1

What is a bit strange about this?

What is going on here?

54

S3 is noncommutative

• Many operations in mathematics are commutative, i.e., the order of the operands does not matter:

15 + 6 = 6 + 15 = 21

15 * 6 = 6 * 15 = 90

• But composition on S3 is noncommutative:

R1 * F

2 ≠

F

2 * R

1

Be careful not to assume commutativity

55

S3 is nonabelian

• When a group is commutative, it is said to be abelian

• When a group is noncommutative, it is said to be nonabelian

• The term abelian is derived from the name of the Norwegian mathematician, Niels Henrik Abel (1802-1829), who did significant mathematics before dying at a young age of tuberculosis

56

Is S3 really a new group?

• We have now seen three groups of order six:

6, H, and S

3

• We saw that 6 and H are isomorphic, so

they are essentially the same

• What about S3?

• Is S3 isomorphic to

6 or H?

• If S3 is isomorphic to one of

6 or H, it's

isomorphic to the other also

57

Let's try a correspondence

If S3 is isomorphic to

6 then there is a

correspondence like this one

0 ↔ R0

1 ↔ R1

2 ↔ R2

3 ↔ F0

4 ↔ F1

5 ↔ F2

58

It doesn't work out like we want

• All true statements in one group map to true statements in the other group

• We know 1 + 2 = 3 in 6

• The corresponding statement in S3 is

R1 + R

2 = F

0

• But this is false because

R1 + R

2 = R

0

And R0 corresponds to 0 in

6, not to 3

59

Are we done yet?

• Does this mean S3 is not isomorphic to

6?

• No, not at all; we tried one correspondence between elements of S

3 and

6 and it didn't

work out, but there are a bunch more possible correspondences to try

• How many?

• There are 6! = 6∙5∙4∙3∙2∙1 = 720 ways to order six things, so we only have 719 to go

60

Flips are their own inverses

Let's not use main strength and awkwardness Elements of S

3 which are their own inverses:

R0 * R

0 = R

0

F0 * F

0 = R

0

F1 * F

1 = R

0

F2 * F

2 = R

0

Elements of 6 which are their own inverses:

0 + 0 = 0

61

S3

6

• So S3 has four elements which are their

own inverses, and 6 has only one such

element

• Thus there is no possible correspondence between elements of the two groups which will preserve true statements

• S3 and

6 are NOT isomorphic

62

How many groups are there?

• So, now we have two essentially different groups of order six: S

3 and

6

• Are there other groups of order six?

• No, there are only these two

• This raises an interesting question: How many groups are there of

order 1?

order 2?

...

63

No. of groups of a given order

Ord # Ord # Ord #

1 1 8 5 15 1

2 1 9 2 16 14

3 1 10 2 17 1

4 2 11 1 18 5

5 1 12 5 19 1

6 2 13 1 20 5

7 1 14 2 21 2

64

How many groups did you say?

• For every positive integer n there is at least one group, namely the cyclic group

n

• If p is prime there is only one group of order p, namely,

p

• The general case is very complex:

n = 510 = 2∙3∙5∙17: 8 groups

n = 511 = 7∙73: 1 group

n = 512 = ?: 10,494,213 groups

n = 513 = 3∙3∙3∙19: 15 groups

65

What about multiplication?

• A group has only one operation, usually called addition (but, as we have seen, it is not necessarily the usual addition we are used to)

• In many cases of interest there are two operations: addition and multiplication

• Let's return to our favorite structure, 6, and

see if we can define multiplication for it

66

Multiplication made easy

• Denote multiplication by *

• For x and y in , we define x * y like this:

• Multiply x times y in the usual way as if they were ordinary whole numbers to get a product z

If z < 6, then z is the product If z ≥ 6, then subtract 6 from z repeatedly until a number < 6 results; it is the product

67

Examples of multiplication

2 * 2 = 4

2 * 3 = 6, so 6 - 6 = 0 is the product

3 * 3 = 9, so 9 - 6 = 3 is the product

3 * 4 = 12, so 12 - 6 - 6 = 0 is the product

1 * 5 = 5

4 * 4 = 16, so 16 - 6 - 6 = 4 is the product

5 * 5 = 25, so 25 - 6 - 6 - 6 - 6 = 1 is the product

68

Multiplication table for 6

* | 0 1 2 3 4 5

0 | 0 0 0 0 0 0

1 | 0 1 2 3 4 5

2 | 0 2 4 0 2 4

3 | 0 3 0 3 0 3

4 | 0 4 2 0 4 2

5 | 0 5 4 3 2 1

69

1 is a unity

0 * 1 = 0

1 * 1 = 1

2 * 1 = 2

3 * 1 = 3

4 * 1 = 4

5 * 1 = 5 So x * 1 = x for every x We say 1 is a unity for *

70

Keep on associating

• How to compute 5 * 4 * 2 ?

(5 * 4) * 2 = (2) * 2 = 4

5 * (4 * 2) = 5 * (2) = 4


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

• Another way to say this: Multiplication is associative (we have not proved this)

71

Better commute too

• How do 5 * 4 and 4 * 5 relate?

5 * 4 = 2

4 * 5 = 2

• For any two elements x, y of 6 we have

x * y = y * x

• Another way to say this: Multiplication is commutative (we have not proved this)

72

The distributive law

• How to compute 5 * (4 + 3) ?

5 * (4 + 3) = (5 * 4) + (5 * 3)

= 2 + 3 = 5


x * (y + z) = (x * y) + (x * z)

• Another way to say this: Multiplication distributes over addition (we have not proved this)

73

Definition of a ring

• A ring <R,+,*> is a non-empty set R together with two operations + and *, called addition and multiplication, such that

• <R,+> is an abelian group

• Multiplication is associative

• Multiplication distributes over addition

• A commutative ring with unity is a ring in which multiplication is commutative and there exists a unity element

74

<6,+,*> is a commutative ring with unity

• We have seen that <6,+> is an abelian

group (remember abelian means + is commutative)

• * is associative

• * distributes over +

• * is commutative

• 1 is a unity

• <6,+,*> is a commutative ring with unity

75

Examples of Rings

• Let n = { 0, 1, 2, ..., n-1 } for any integer n >

0, and define + and * in analogy with + and * for

6

• <n,+,*>

is a commutative ring with unity

• So there are an infinite number of such rings

• There are also infinite rings, e.g., with the usual addition and multiplication

76

Matrices

• Let Mn be the set of all nn matrices of real

numbers

• Here are some typical elements of M2:

1 4 3 2 0 2 2 1

We can define addition and multiplication of these elements in such a way that M

2 is a

ring with unity

77

More about matrices

• The identity element for addition is

0 0 0 0

The unity element for multiplication is

1 0 0 1

Matrix multiplication is NOT commutative

78

What about division?

• In a group we can add (not necessarily the kind of addition we use in daily life)

• We can also subtract, because every element has an inverse, and adding the inverse of x is the same as subtracting x

• In a ring, we can multiply

• But can we divide?

79

Be careful!

• Looking back at the multiplication table for

6, we see that

2 * 3 = 0

• Let's divide both sides by 3 to get

2 = 0

• NO! NO! NO! NO! NO! NO! NO! NO!

• What's the problem?

80

A ring divided ...

• In order to divide by an element x, we must have an element x' such that x * x' = 1, i.e., we need a multiplicative inverse for x

• Then, to divide by x, we simply multiply by the multiplicative inverse x'

• Looking back at the multiplication table for

6, we see that only 1 and 5 have

multiplicative inverses

• So division in 6 is just not going to work

81

If 6 doesn't work, try

5

• Let's look at <5,+,*>, a commutative ring

with unity consisting of 5 elements 0, 1, 2, 3, and 4

• We can add and multiply in this ring much like we did in

6: When a value is greater

than or equal to 5, we subtract 5 repeatedly until we get 0, 1, 2, 3, or 4

• Let's look at the multiplication table for 5

82

Multiplication table for 5

* | 0 1 2 3 4

0 | 0 0 0 0 0

1 | 0 1 2 3 4

2 | 0 2 4 1 3

3 | 0 3 1 4 2

4 | 0 4 3 2 1

How is this different from the multiplication table for

6?

83

Unity at last

• In every row of the table except the first, we see that the multiplicative unity 1 appears

• This means that every element except 0 has a multiplicative inverse:

1 * 1 = 1

2 * 3 = 1

3 * 2 = 1

4 * 4 = 1

84

We can divide in 5

To divide by 1: Multiply by 1




To divide by 0: Multiply by ???

85

Don't divide by nada

• Don't ever, ever, ever divide by zero

• Division is only defined when the divisor is non-zero

• So dividing by zero is meaningless

• [But can't we define x/0 = ?]

• [Yes, but that opens up another can of worms which we are not dealing with today]

86

What's your field?

• Given a non-zero element x of a ring, if there exists an element x-1 such that x * x-1 = 1, then x-1 is the multiplicative inverse of x

• A field is a commutative ring with unity in which every non-zero element has a multiplicative inverse

• In a field we can not only add, subtract, and multiply, we can also divide

87

Examples of fields

• 5 is a field

• p is a field for every prime p, so there are

an infinite number of finite fields

• Is , the set of all integers, a field?

• Is , the set of all rational numbers, a field?

• Is , the set of all real numbers, a field?

• Is , the set of all complex numbers, a field?

88

A subset of the real numbers

• Consider the set S = { a+b√2 | a,b ∈ }• Some elements of S: 2+√2, ½ +3√2, 5

• Note that S is bigger than and smaller than : ⊂S⊂

• Define addition and multiplication on S as the usual operations on the real numbers

• Is S a field?

• What are the requirements on a set in order for it to be a field?

89

We'll just do the fun ones

• Because is a field and S is a subset of , most of the requirements for S to be a field are easily shown

• There are three things, however, which we must carefully check:

1. S is closed under + 2. S is closed under * 3. Every non-zero element of S has a multiplicative inverse

90

You do the third one

• 1. (a+b√2) + (c+d√2) = (a+c)+(b+d)√2

• 2. (a+b√2) * (c+d√2) = ac + ad√2 + bc√2 + bd√2√2 = (ac+2bd) + (ad+bc)√2

• 3. The multiplicative inverse of a+b√2 is

a/d - (b/d)√2, where d = a2 - 2b2

91

Fields, fields, everywhere

• So S is a subfield of • Are there other subfields of ?• Yes, bijillions of them

• Are there other fields besides the ones we've talked about?

• Yes, bijillions of them

92

What have we seen?

• Groups – structures in which we can add and subtract

• Rings – structures in which we can add, subtract, and multiply

• Fields – structures in which we can add, subtract, multiply, and divide

• Some are finite, some are infinite

• They are abstractions of concrete number systems like , , , and

93

Who cares?

“… Monsieur Fourier was of the opinion that the principal aim of Mathematics is to serve mankind and to explain natural phenomna; but a philosopher such as he ought to have known that the sole aim of science is the fulfillment of the human spirit, and that, accordingly, a question about numbers has as much significance as a question about the workings of the world.”

C.G.J.Jacobi (1804-1882)

94

In closing …

“In mathematics you don't understand things. You just get used to them.”

John von Neumann (1903-1957)

“It's so complicated …”

Mick Jagger (1943-)

1 mathematical structures: groups, rings, and fields roger house scientific buzz café coffee catz...

Documents

element y

elements x

x y zstill

x y zso

inverse element x

x yaddition

x y zanother way

havex y z