cc 2 algebra

8/2/2019 CC 2 Algebra

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Chapter 2

Introduction to Algebra


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2

OutlineGroups

Fields

Binary Field Arithmetic

Construction of Galois Field

Basic Properties of Galois FieldComputations Using Galois Field Arithmetic

Vector Spaces


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Groups


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GroupsLet Gbe a set () of elements ().A binary operation * on G is a rule that assign to each pair of

elements a and b a uniquely defined third element c=a*b in G.Definition 2.1: Agroup is a set G with a binary operator * that

satisfies the following conditions :

Closure ():Associative ():Identity ():

This element e is called an identity elementofG.Inverse ():The element ais called an inverse ofa.

,a b G a b G ( ) ( ), , * * * *a b c G a b c a b c =

s.t. , * *e G a G a e e a a = =

, ' , s.t. * ' '*a G a G a a a a e = =


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GroupsTheorem 2.1: The identity element in a group G is unique.

pf : Suppose it is not unique (e and e ):

e = e*e = e*e = e (from definition)

Theorem 2.2: The inverse of a group element is unique.

pf : Suppose it is not unique (a and a ):a = a*e (identity)

= a*(a*a) (inverse)

= (a*a)*a (associative)= e*a (inverse)

= a (identity)


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GroupsRemarks:

A group G is said to be commutative (abelian group)

if a, b G, a*b = b*aOrder: The number of elements in a group. We denote it |G|.

Finite group: A group of finite order.

Examples of commutative groups:integers under +

rational numbers under

{0,1} under

2*2 real-valued matrix under +{0, 1, 2, ., m-1} under modulo-m addition is commutative (seeexample 2.2)

{1, 2, 3.p-1} under modulo-p multiplicator (p is a prime) is alsocommutative (see example 2.3)


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GroupsExample 2.1Consider the set of two integers, G = {0, 1}. Let us define a

binary operation, denoted by , on G as follows :0 0 = 0 0 1 = 1 1 0 = 1 1 1 = 0

This binary operation is called modulo-2 addition.

The set G = {0, 1} is a group under modulo-2 addition.It follows from the definition of modulo-2 addition that G is

close under and 0 is the identity element.

The inverse of 0 is itself and the inverse of 1 is also itself.

It is easy to show that is associative.

Thus, G together with is a commutative group.


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GroupsExample 2.2Let mbe a positive integer. Consider the set of integerG = {0, 1,

2, , m-1}. Let + denote real addition. Define a binary operationon G as follows:

For any integers i andj in G, i j = r , where ris the

remainder resulting from dividing i +jby m. The remainderr

is an integer between 0 and m-1 (Euclids division algorithm)

and is therefore in G. Hence G is closed under the binary

operation , called modulo-m addition.

First we see that 0 is the identity element.For 0 < i < m, i and mi are both in G. Since

i + (m i) = (m i) + i = m


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GroupsIt follows from the definition of modulo-m addition that

i (m i) = (m i) i = 0

Therefore, i and m-i are inverses to each other with respectto .

It is also clear that the inverse of 0 is itself.

Since real addition is commutative, it follows from thedefinition of modulo-m addition that, for any integers i andj in

G, i j = j i . Therefore modulo-m addition is commutative.

Next we show that modulo-m addition is also associative. Let i,

j, and kbe three integers in G. Since real addition is

associative, we have

i + j + k = (i + j) + k = i + (j + k)


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GroupsDividing i + j + kby m, we obtain

i + j + k = qm + r,

where q and rare the quotient and the remainder, respectively,and . Now, dividing i + jby m, we have

i + j = q1m + r1, with (2.1)

Therefore, i j = r1. Dividing r1 + kby m, we obtain

r1 + k = q2m + r2 with (2.2)Hence r1 k = r2 and (i j) k= r2.

Combining (2.1) and (2.2), we have

i + j + k = (q1

+q2

)m + r2

,

This implies that r2 is also the remainder when i + j + k is

divided by m. Since the remainder resulting from dividing an

integer by another integer is unique, we must have r2 = r.

mr


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GroupsAs a result, we have

(i j) k= r.

Similarly, we can show thati (j k) = r.

Therefore (i j) k= i (j k) and modulo-m addition is

associative.

This concludes our proof that the set G = {0, 1, 2, , m-1} isa group under modulo-m addition. We shall call this group an

additive group.


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GroupsExample 2.3:Letpbe a prime (e.g.p =2, 3, 5, 7, 11,). Consider the set ofintegers, G = {0, 1, 2, ,p-1}. Let denote real

multiplication.

Define a binary operation $ on G as follows: Fori andj in G,i $j = r, where ris remainder resulting from dividing i jbyp.

The set G = {0, 1, 2, ,p-1} is a group under modulo-pmultiplication.

First we note that i j is not divisible byp. Hence 0 < r


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GroupsLet ibe an element in G. Sincep is a prime and i


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GroupsTherefore a is the inverse ofi. However, ifa is not in G, we

divide abyp ,

a = q p + r.Since a andp are relatively prime, the remainderrcannot be 0

and it must be between 1 andp-1. Therefore ris in G. Now

combining (2.4) and (2.5), we obtain

r i = - (b + qi)p + 1.

Therefore r$ i = i $ r= 1 and ris the inverse ofi. Hence any

element i in G has an inverse with respect to modulo-p

multiplication. The group G = {0, 1, 2, ,p-1} under modulo-pmultiplication is called a multiplicative group.

Def: LetH G &H (empty set), thenHis said to be a

subgroup ofG ifHis a group.


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SubgroupTheorem 2.3: Let Gbe a group under the binary operation *. Let

Hbe a nonempty subset ofG. ThenHis asubgroup ofG if the

following conditions hold:(i)His closed under the binary operation *.

(ii) For any element a inH, the inverse ofa is also inH.

Proof:Condition (ii) says that every element ofHhas an inverse inH.

Condition (i) & (ii) ensure that the identity element ofG is

also inH. (a*a=e is an element ofH)

Because the elements inHare elements in G, the associative

condition on * holds automatically.

Hsatisfies all the conditions of a group.


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CosetDefinition 2.2: LetHbe a subgroup of a group G with binary

operation *. Let a be an element ofG. Then the set of elements

is called a left coset ofH; the set ofelements is called a right coset ofH.

If the group G is commutative, then every left coset is identical

to every right coset.

Example: Consider the additive group G={1,2,3,,15} under

modulo-16 addition. SubgroupH={0,4,8,12}. The coset

{ }* * :a H a h h H { }* * :H a h a h H

3 H

{ }

{ }

3 3 0,3 4,3 8,3 12

3,7,11,15

H =

=

{ }

{ }

7 7 0,7 4,7 8,7 12

7,11,15,3

H =

=


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CosetThere are only four distinct cosets of H:

The four distinct cosets ofHare disjoint, and their union formsthe entire group G.

Theorem 2.4: LetHbe a subgroup of a group G with binary

operation *. No two elements in a coset ofHare identical.The proof is based on the fact that all the elements in the

subgroupHare distinct.

Consider the coset with

{ }

{ }

{ }

0 0,4,8,12

1 1,5,9,13

2 2,6,10,14

H

H

H

=

=

=

{ }* * :a H a h h H = .a G


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CosetSuppose two elements, say a*h and a*h, in a*Hare

identical, where h and h are two distinct elements inH.

Let a-1 denote the inverse ofa with respect to the binaryoperation *. Then

a-1 *(a*h)= a-1 *(a*h),

(a-1

*a)*h=(a-1

*a)*h,e *h=e *h,

h=h.

This result is a contradiction to the fact that all the elementsofHare distinct.

Therefore, no two elements in a coset are identical.


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Coset2.5: No two elements in two different cosets of a subgroupHof a

group G are identical.

Proof: Let a*Hand b*Hbe two distinct cosets ofH, with a andb in G.

Let a*h and b*h be two elements in a*Hand b*H,

respectively.

Suppose a*h=b*h.

Let h-1be the inverse ofh.

(a*h)* h-1 =(b*h)* h-1

a*(h* h-1) =b*(h* h-1)

a*e=b*h

where (h= h* h-1) is an element in H.


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CosetThe equality a=b*h implies that

This result says that a*Hand b*Hare identical, which is a

contradiction to the given condition that a*Hand b*Hare twodistinct cosets ofH.

Therefore, no two elements in two distinct cosets ofHare

identical.

( )

( ){ }

( ){ }

{ }

* * '' *

* '' * :

* ''* :

* ''' : '''*

a H b h H

b h h h H

b h h h H

b h h H b H

=

=

=

= =


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CosetFrom Theorem 2.5 and 2.5, we obtain the following properties of

cosets of a subgroupHof a group G:

Every element in G appears in one and only one coset ofH;All the distinct cosets ofHare disjoint;

The union of all the distinct cosets ofHforms the group G.

All the distinct cosets of a subgroupHof a group G form a

partition ofG, denoted by G/H.


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Lagranges TheoremLagranges Theorem: Let Gbe a group of ordern, and letHbe a

subgroup of orderm. Then m divides n, and the partition G/H

consists ofn/m cosets ofH.Proof:

Every coset consists ofm elements ofG.

Let ibe the number of distinct cosets ofH.Since n=im, m divides n and i=n/m.


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Fields


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FieldsRoughly speaking, afieldis a set of elements in which we can do

addition, subtraction, multiplication, and division without leaving

the set. Addition and multiplication must satisfy the commutative,associative, and distributive laws.

Definition 2.3: LetFbe a set of elements on which two binary

operations, called addition + and multiplication , are defined.

The setFtogether with the two binary operations + and is afieldif the following conditions are satisfied:

Fis a commutative group under addition +.

The identity element with respect to addition is called thezeroelementor the additive identity ofFand is denoted by 0.


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FieldsThe set of nonzero elements inFis a commutative group

under multiplication .

The identity element with respect to multiplication is called theunit elementor the multiplicative identity ofFand is denoted

by 1.

Multiplication is distributive over addition; that is, for any

three elements a, b, and c inF,

a (b+c)=a b+a c


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FieldsA field consists of at least two elements, the additive identity and

the multiplicative identity.

The number of elements in a field is called the orderof the field.A field with finite number of elements is called afinite field.

In a field, the additive inverse of an element a is denoted by a

and the multiplicative inverse ofa is denoted by a

-1

provided thata0.

Subtracting a field element b from another field element a is

defined as adding the additive inverse b ofb to a. [a-ba+(-b)].

Ifb is a nonzero element, dividing aby b is defined as

multiplying aby the multiplicative inverse b-1 ofb. [a/ba b-1].


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FieldsProperty I. For every element a in a field, a 0=0 a=0.

Proof. a=a1=a (1+0)=a+a 0

Adding a to both sides of the equality above, we have:-a+a= -a+a+a 0 0=0+a 0 0=a 0

Similarly, we can show that 0 a=0. Therefore, we obtaina 0=0 a=0.

Property II. For any two nonzero elements a and b in afield, a b0.

Proof. From definition, nonzero elements of a field are closed

under multiplication.


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Fields

Property III. ab=0 and a0 imply that b=0.

This is a direct consequence of Property II.

Property IV. For any two elements a and b in a field, -(ab)=(-a)b=a(-b).

0=0b=(a+(-a))b=ab+(-a)b

(-a)b must be the additive inverse ofab and (ab)=(-a)b.

Similarly, we can prove that (ab)=a(-b).

Property V. Fora0, ab=ac implies that b=c.Since a is a nonzero element in the field, it has a multiplicative inverse a-1.

Multiplying both side ofab=acby a-1, we obtaina-1(ab)= a-1(ac)

(a-1a)b= (a-1a ) c

1b=1c => b=c


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Fields

Some examples:

R (real number set)

C(complex number)

Q (Rational number)

GF(q) exists ifq = ,p is a prime

Ex.Binary field

GF(2) with modulo-2 addition

infinite fields

mp

+

0

1

0 1

0

1

1

0

modulo-2 multiplication0

1

0 1

0

0

0

1


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Fields

Ex: GF(p) ,p is a prime. (Prime Field){0, 1, 2, .p-1} is an abelian group under modulo-p addition.

{1, 2, p-1} is an abeian group under modulo-p multiplication.Fact: real number multiplication is distributive over real numberaddition. This implies that modulo-p multiplication isdistributive over modulo-p addition.

{0, 1, 2, .p-1} is a field of orderp under modulo-p additionand multiplication.

In fact, for any positive integerm, it is possible to extend the prime

field GF(p) to a field ofpm

elements called an extension fieldofGF(p) and is denoted by GF(pm).

Furthermore, the order of any finite field is a power of a prime.

Finite fields are also called Galois field.


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Fields

Def: Characteristic ofGF(q)Consider the unit element 1 in GF(q); a smallest positive

integer s.t. , then is called the characteristic ofGF(q)EX.

The characteristic ofGF(2) is 2

The characteristic ofGF(p) isp

011 =+

= =

1 01i

1 1

1 0 for 1 , 1 0pk

i i

k k p= =

= < =


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Fields

Theorem 2.7: The characteristic ofGF(q) is prime.(pf)

(contradicts the definition of )

Fact: For any two distinct positive integerk, m


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Fields

Remarks:

1, 1+1, 1+1+1, , 1+1++1=0

are distinct elements in GF(q), which form asubfield

GF( ) ofGF(q)

If , then q is a power of (proven in later)

Def: (Orderof field element a)Let a , a smallest positive integern s.t. .

n is called the orderof the field element.

q

)(qGF ,0a 1=na


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Fields

Remarks:a, =a a, =a a a, , , =1 are all distinct, which form

a group under the multiplication ofGF(q).(pf):

Closure

Inverse For is the multiplicative inverse ofSince the powers ofa are nonzero elements in GF(q), they

satisfy the associative and communitative laws.

2a 3a1na na

, i j i jif i j n a a a ++ =

rnjinjiif +=+>+ havewe,

0where nri j i j n r r a a a a a a+ = = =

in

ani


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Fields

Def: (Cyclic)A Group is said to be cyclic, if there exists an element in the

group whose powers constitute the whole group.Theorem 2.8:

(pf): Let b1, b2, bq-1be the q-1 nonzero element ofGF(q),

then a b1, a b2, a bq-1 are also nonzero and distinct.Thus,

(a b1) (a b2)..(a bq-1) = b1 b2bq-1

aq-1(b1 b2bq-1) = b1 b2bq-1

aq-1=1Theorem 2.9: , n is the order ofa then

(pf) : If not, q-1 = kn + r 0


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Def: (Primitive), a is said to be primitive if the order ofa is q-1

Remarks:The powers of a primitive element generate all the nonzero

elements ofGF(q)

Every finite field has a primitive element.Example. GF(5)

2 is a primitive elements.

primitive elements are useful for constructing fields.

Example. GF(7)

3 is a primitive element, the order of 4 is 3, which divides 6.

( )& 0a GF q a

22,12,32,42,22 5432' =====

Fields


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Binary Arithmetic

In general, we can construct codes with symbols from any Galois

field GF(q), where q is either a primep or a power ofp; however,

codes with symbols from the binary field GF(2) or its extension

GF(2m) are most widely used in digital data transmission and

storage systems.

In this text, we are concerned only with binary codes and codes

with symbol from the field GF(2m).Most of the results presented in this text can be generalized to

codes with symbols from any finite field GF(q) with q=2 or 2m.

In binary arithmetic, we use modulo-2 addition and multiplication,which are defined by Tables 2.3 and 2.4, respectively.


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Sets of equationse.g.X+Y=1,X+Z=0,X+Y+Z=1

Solved by Gramers rule

11

010

11

111

11

101

111

101

011

+==

1100111 =+=

01

0111

100

011

==

=x 11

1111

101

011

==

=y 01

0111

001

111

==

=z


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Polynomials overGF(2). We denote it GF(2)[X].

Def:

iffn=1, deg[f(x)] = nif , deg[f(x)] = 0

Remarks:

Polynomials overGF(2) with degree = 1ex :x , 1+x

Polynomials overGF(2) with degree = 2

ex :In general, with degree = n we have polynomials.

1,0....01

==== fffn

n

nxfxffxf +++= ...)(

10(2)if GF

2222

1,,1, xxxxxx++++

n2


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Added (or subtracted)

Multiplied

Ifg(x) = 0, thenf(x) 0 = 0

Commutative

n

n

m

m

m

mm

xfxfxgfxgfgfxgxf

++++++++=+ +

+

1

1

1100

)()()()()(

)(...)( 10 nmxgxggxgm

m +++=

0 1

0 1 1

( ) ( ) ......

n m

n m

i i i i o

f x g x c c x c xc f g f g f g

+

+

= + + += + + + ),(

000 mnmngfcgfc ==

+

)()()()(

)()()()(

xfxgxgxf

xfxgxgxf

=

+=+



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Associative

f(x) + [g(x) + h(x)] = [f(x) +g(x)] + h(x)

f(x) [g(x) h(x)] = [f(x) g(x)] h(x)

Distributive

f(x) [g(x) + h(x)]

=[f(x) g(x)] + [f(x) h(x)]

Euclids division algorithmSuppose deg[g(x)] 0,

s.t.f(x)=q(x)g(x) + r(x), where deg[r(x)] < deg[g(x)]

q(x) : quotient, r(x) : remainder

e.g.

Ifr(x)=0,f(x) is divisible byg(x). [g(x) dividesf(x),g(x)|f(x)]

( ), ( ) (2)[ ]q x r x GF x

1)1)(()1( 2323654 ++++++=++++ xxxxxxxxxx



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Root

, thenf(x) is divisible by (x - )

e.g.f(1)= 1+1+1+1=0, f(x) is divisible byx+1

Def: (Irreducible)p(x) GF(2) [x] with deg[p(x)]=m is said to be irreducible over

GF(2) ifp(x) is not divisible by any polynomial overGF(2) ofdegree less than mbut greater than zero.

(2), ( ) 0GF if f = )..( +xei432

1)( xxxxf +++=( ))1)(1()1( 3432 +++=+++ xxxxxx


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e.g.

is an irreducible polynomial with degree 2.is also an irreducible poly with degree 3.

In general, for any , there exists an irreducible polynomial

of degree m.Theorem 2.10: Any irreducible polynomial overGF(2) ofdegree m divides .

e.g.

1only,1,,1,among 22222 ++++++ xxxxxxxx

13 ++ xx

1m

112 +m

x

1x1|)1( 71233

+=+++ xxx


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Def: (Primitive)An irreducible polynomialp(x) of degree m is said to be primitiveif the smallest positive integern for whichp(x) divides is

.

e.g.butp(x)!| for primitive

it can also not primitive

Remarks:

For a given m>0, there may be more than one primitivepolynomials of degree n.

Lists of primitive polynomials (see p.42 Table 2.7)

1+nx12 =

mn

11)( 154 +++= xxxxp

1+nx


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In this section, we present a metnod for constructing the Galois

field of 2m elements (m>1) from the binary field GF(2).

Consider 0,1 in GF(2) and a new symbol

.Define as follows:

2 3

0 0 0

0 1 1 0 01 1 1

0 0 0

1 1

, ,j

=

= = =

= =

= == = =

j times


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0 0 0

1 1

j j

j j j

i j j i i j

+

= =

= =

= =

{ }0

0,1, ,... ,...

with 1 be sometime denoted by

jF

=


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Let a primitive polynomial

With deg [p(x)]=m &

Since

Therefore, under the condition that

Fis finite i.e.

From is closed under

])[2()( xGFxp

0)(assume =p

)()(11)( 1212 xpxqxxxp mm =++

2 1 1 ( ) ( )m

q p + = 00)( == q

1

12 = m

0)( =p

{ }222 ,...,,,1,0* == mFF *, F


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FACT: The nonzero elements of form a commutative group

with order under

Now define an additive operation + on F* s.t.F* forms

a commutative group under +

For overGF(2)

s.t.

where

*F

12 m

)(&)(,120 xaxgi iim


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FACT:

For

( x andp(x) are relative prime is not divisible byp(x) )For

(pf): If (i


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C f G l F ld


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Since zero element 0 in may be represented by the zero poly.

elements in are represented by distinct poly.

of overGF(2) with degree m-1 or less and are regarded as

distinct elements.

Define + as follows:

1. 0+0=02. for

So,

*Fm2 *F m2

m2

iiimji =+=+


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FACT: is a commutative group under +

FACT: is a Galois field of elements.

(pf): is a commutative group under +

is a commutative group under

*F

additive identity

additive inverse

commutative

associative

{ }22* ,......,,1,0 = mF m2

*

{ }0* F


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C t ti f G l i Fi ld


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Example: (2)overpoly.primitiveais1)(44 GFxxxpm ++==

(1001)1

(1011)1

(1111)1(0111)

(1110)1

(0101)

(1010)1

(1101)1

1

)(

(0011))(

(0110))(1

(1100)101)p(Set

314

3213

3212

3211

210

39

28

3

3

433267

32256

245

44

+=

++=

+++=++=

++=

+=

+=

++=

++=

+=+==

+=+==

+=+==

+==++=

C st ti f G l is Fi ld


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0)(1)(11

1)(1)(

)1(,

22105

13323275

72210125

12

71215412

4

154197121275

=+++++=++

=++=++++=+

===

==

====

aa

mAnother useful representation of field elements in GF(2 )

),......,( 110 maaatuplem1

110 ...:

+++m

maaa

C nstructi n f Gal is Field


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Ex. Construct GF(4) from GF(2) with

{ } { }110101

2

2

+

++=

,,,,,,

xx)x(p

+

01

2

0

01

1+

1

10

1+

+1

0

1

2

1+

1

0

01

1+

0

00

1

01

1+

0

0

0

1+

1+

01+

1

1


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Basic Properties of Galois Field


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In ordinary algebra, we often see that a polynomial with real

coefficients has roots not from the field of real numbers but from

the field of complex numbers that contains the field of real

numbers as a subfield.

This situation is also true for polynomials with coefficients from

GF(2). In this case, a polynomial with coefficients from GF(2)

may not have roots from GF(2) but has roots from an extensionfield ofGF(2).

For example,X4+X3+1 is irreducible overGF(2) and therefore it

does not have roots from GF(2); however, it has four roots from

the field GF(24).



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)(2over1xx)(ofroottheis 4347 GFxp ++=

)(2over)(ofrootsare,,,

1

)x()x()x(x

))(x(x

])x(][x)x([x))(x)(x)(x(x

41413117

34

155202310123284

1222382

2714132181172

1413117

GFxp

xx

xx

++=++++++++=

++++=

++++++=++++

A polynomial with coefficients from GF(2) may not have roots

from GF(2), but has roots from an extension field of GF(2).



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p(x)

If, an element in GF(2m), is a root off(X), the polynomialf(X)

may have other roots form GF(2m). What are these roots?

Theorem 2.11:

(pf):

4 3

4

x x 1 is irredeucible over (2) i.e. it doesn't have any

root over (2). However, it has four roots over (2 )

GF

GF GF

= + +

rootsallarethenroot,aisifi.e.

00)(then0,)(If

2

2

== ff

00)]([)(

)()]([

222

22

===

=

ff

xfxf

)2(offieldextensionan],)[2()( GFxGFxf

From eq. 2.10.



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Remarks:

The

e.g.

FACT:

2 m 2

Let ( ) (2)[x], if (2 ) s.t. ( ) 0, then

(2 ) 0 and ( ) 0

m

f x GF GF f

GF f

=

=

ofconjugateacalledis2

2.8)Tablebygiven),2((where0...)(Then

)2(1)(

444

446543

GFf

GFxxxxxf

==

++++=

( )

( )

4 8 2 5 10The conjugates of : , , are all roots of 0, besides &

are roots of 0

f x

f x

=

=m2 1If GF(2 ) and 0, then is a root of x 1m +

m2 1(pf): 1 (from Theorem 2.8) =



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Remarks:

All nonzero elements ofGF(2m) form all the roots of

All elements ofGF(2m

) form all the roots ofDef: (minimal polynomial)

Remarks:

)1(122

+=+mm

xxxx

112 +m

x

.ofpoly.minimalthecalledis0)(

s.t.(2)overdegreesmallestof)(poly.the),(2Let

=

GFxGF m

mxpxGFxp

xxm

2)](degree[with])[2()(ofrootabemay

ofrootais 2


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Theorem 2.13:

(pf): If not

not?or1isofpoly.minimalThe:Q1isofpoly.minimalThe

1is1ofpoly.minimalThe

is0ofpoly.minimalThe

65434

347

xxxxxx

x

x

++++++

+

The minimal poly. (x) of a field element

is irreducible.

(x)(x)(x) 21 =

21,for(x)]deg[(x)]deg[0when =


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Theorem 2.14: Let be the minimal poly.of .

(pf):

0)()(

0)(

21 =

=

0)(or0)( 21 ==

poly.minimalanotis(x)

].(2)[)( xGFxf ( )x)((x)then0,)(If xff =

0)(0)()( === rf

( )

If (x) 0, then ( ) is a polynomial of lower degree than

, which has as a root. This is a contradiction.

r r x

x

)](deg[)](deg[,)()()()( xxrxrxxaxf


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Theorem 2.15: The minimal polynomial(x) of

Remark:(according to Corollary 2.12.1)

Thm 2.16: andf(x) is an irreducible poly.Let and be the minimal poly.

of .

(pf): from Theorem 2.14

2(2 ) dividesm

mGF x x +

)(2fromare(x)ofrootstheall mGF

])[2()( xGFxf )(2mGF (x)

)()(then0,)(If xfxf ==

)()( xfx

f(x)(x) = eirreduciblis(x)&1(x) f

(Thm 2.12.1 and 2.14)



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Remarks:

This fact implies that an irreducible poly.f(x) with root is

the minimal poly.From previous result (Thm 2.11)

are roots of Let ebe the smallest

integer s.t.

are all the distinct conjugates of (see prob. 2.15)

.of(x)

,,,,, 2222

(x).12 2222 ,,then,

=ee



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Bas c Propert es of Galo s F eld

Thm 2.17: s.t. , then

is an irreducible poly overGF(2)

(pf):

prove that by first prove

smallestthebe&)(2 eGF m =e2

)()( 21

0

i

xxfe

i+=

=

[ ]xGFxf (2))( )()]([ 22 xfxf =

221

0

2

21

0

2 )()()]([ii

xxxfe

i

e

i+=

+=

=

=

))((122221

0

+

+++= =

iii

xxe

i

)(122

1

0

+

+=

=

i

xe

i



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71

p f

Let

)( 221

i

xe

i+=

=

)( 221

0

i

xe

i+=

=

)(

2

xf=

))((

22221

1

ei

xx

e

i ++=

=

e

exfxffxf +++= ...)( 10 Expand1where =ef

[ ]

( )

2 2

0 1

2 2 2 2

0 0 0 0

( ) ( )

1 1i i

e

e

e e e ei j

i i j i

i i j ii j

f x f f x f x

f x f f x f x+

= = = =

= + + +

= + + =

=e2



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p

This holds only when or 1

prove thatf(x) is irreducible overGF(2)

&from eiff ii = 02

= =

...)(0

22e

i

i

i

xfxf

[ ]xGFxf (2))( 0

=if

has)(0,)(if0,)(or0)( xaaba === if not, ( ) ( ) ( )& ( ) ( ) ( ) 0f x a x b x f a b = = =

e2 2 1roots , , . (Theorem 2.11)

[ ] )()(and)(deg xfxaexa ==)()(s.t.reasonsamethe0,)(if xfxbb ==

Therefore,f(x) must be irreducible.



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73

p

Thm 2.18:

, Then

(following from Thm 2.16 & 2.17)

e.g.

)(2ofpoly.minimalthebeLet mGF(x)

smallestthebeand e s.t.integere2 =

)()(

1

0

2

=+=

e

i

i

xx

2.8Tablebygiven)(243

GF =924212262 32 ,, ====

isofpoly.minimalThe 3=

=++++= ))(x)()(()(12963

xxxx

1234 ++++= xxxx


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75

p

All the minimal poly. of elements in GF(24). (See p.52 Table2.9)

Thm2.19: Let. Then ebe the smallest integer s.t.

(direct from Thm2.18&2.19)

Remarks:

f(x) be the minimal poly. of

then (proof is omitted)

101 3210 ==== , aa, aa431)( xxx ++=

(x) be the minimal poly. of )2(mGF

ex =)](deg[& me = Moreover.2

ef(xGF m = )]deg[with)2(me |



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Minimal poly. of m = 2 to 10 see Appendix B

In the construction of the Galois field GF(2m), we use aprimitive polynomialp(X) of degree m and require that

the elementbe a root ofp(X).

Because the powers of generate all the nonzeroelements ofGF(2m), is a primitive element.

In fact, all the conjugates of are primitive elements of

GF(2m).

)(2 mGF



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77

Thm2.20: ofGF(2m),then all its conjugates are also primitive

element ofGF(2m)

(pf): Let nbe the order of , then .

from Thm2.9.

elementprimitiveaisif),2( GF m

,222,

0for2 >

1)(22 ==

nn

)1(2mn

)(2ofelementprimitiveais mGF

12isorderits m

21)-(21,For 2 =

n mn

prime,relativeare2&1-2 m



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Example: given by Table 2.8

Clearly, is a primitive element ofGF(24)

are all primitive

elements ofGF(24)

nm 1)-(2

12,from = mn element.primitiveaalsois2

)(247 GF =

6213142710 1 ,, , =====

== 13284

11051589814 ==== ,

7=112132142 22 ,, ===



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Thm2.21: then all theconjugate have the same ordern. (See prob.2.15)

,orderhasand)2( nGF m


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Computations Using Galois FieldArithmetic

Computations Using Galois Field Arithmetic


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EX1: Consider overGF(24)

=+

=+4812

27

Y

Yx

x

2 7

4 8 1 0 1 1 3 1 49

8 1 9 2 57

1 2 8

1X

1

+ += = = = =

+ +

45

9

2

3

198

164

812

7

412

2

1

1

Y ==+ +=+ +=

=

Computations Using Galois Field Arithmetic


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EX2: Solve over GF(24) given byTable2.8 (try and error)

072 =++= xxf(x)

0)(

13126 =++=f

0)( 172010 =++= f

106 ,=x

0))(()( 106 =++= xxxf


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Vector Spaces

Vector Spaces


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Def: Vbe a set of elements with a binary operation + is defined.Fbe a field. A multiplication operator between

and is also defined. The Vis called a vector space

over the fieldFif:

Vis a commutative group under +

distributive law

Associative Law

.

Vv Fa

VvV av&Fa

Fa,bVv,u andvauavua +=+ )(

vbvavba +=+ )(

v)bavba = ()(

vv =1


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Vector Spaces


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86

Property III. For any scalarc inFand any vectorv in V,

(-c) v = c (- v ) = -(c v)i.e., (-c) v orc (- v ) is the additive inverse of the vectorc v.

(Left as an exercise)

Consider an ordered sequence ofn components, ,

where each component ai is an element from the binary field GF(2)(i.e., ai = 0 or 1). This sequence is called an n-tuple overGF(2).

Since there are two choices for each ai, we can construct

distinct n-tuples. Let denote this set. Now we define an

addition + on as following : For any u = and

v = in , u + v =

(2.7)

),...,,( 110 nuuu

n2

nVnV

),...,,( 110 naaa

),...,,( 110 nvvv nV ),...,,( 111100 +++ nn vuvuvu

Vector Spaces


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where ui+vi is carried out in modulo-2 addition. Clearly, u + v is

also an n-tuple overGF(2). Hence is closed under the addition.

We can readily verify that is a commutative group under the

addition defined by (2.27) .(1) we see that allzero n-tuple 0 = (0, 0, ,0) is the additive

identity. For any v in ,

v + v = = (0, 0, ,0) = 0

Hence, the additive inverse of each n-tuples in is itself.

Since modulo-2 addition is commutative and associative, the

addition is also commutative and associative.

Therefore, is a commutative group under the addition.

(2) we defined scalar multiplication of an n-tuple v in

nV

nV

nV

),...,,(111100

+++nn

vvvvvv

nV

nV

nV

Vector Spaces


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by an element a from GF(2) as follows :

a = (2.28)

where avi is carried out in modulo-2 multiplication.Clearly, a is also an n-tuple in .

Ifa = 1,

1 =

=By (2.27) and (2.28), the set of all n-tuples overGF(2) forms a

vector space overGF(2)

),...,,( 110 nvvv ),...,,( 110 nvavava

),...,,( 110 nvvv nV

),...,,( 110 nvvv )1,...,1,1( 110 nvvv

),...,,( 110 nvvvnV

Vector Spaces


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ExampleLet n=2. The vector space V2 of all 2-tuples overGF(2) consists of

the following 4 vectors :

(0 0) (0 1) (1 0) (1 1)The vector sum of (0 0) and (0 1) is

(0 0) + (0 1) = (0 + 0 0+1) = (0 1)

Using the rule of scalar multiplication defined by (2.28), we get0 (1 0) = (01 00) = (0 0)

1 (1 1) = (11 11) = (1 1)

Vbeing a vector space of all n-tuples over any fieldF, it mayhappen that a subset SofVis also a vector space overF. Such a

subset is called asubspace ofV.

Vector Spaces


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Theorem 2.18Let Sbe a nonempty subset of a vector space Vover a fieldF.

Then Sis a subspace ofVif the following conditions are satisfied :

(1) For any two vectors u and v in S, u + v is also a vector in S.(2) For an element a inFand any vector u in S, a u is also in S.

(pf). Conditions (1) and (2) say simply that Sis closed under

vector addition and scalar multiplication ofV. Condition (2)

ensures that, for any vectorv in S, its additive inverse (-1) v

is also in S. Then, v + (-1)v = 0 is also in S. Therefore, Sis a

subgroup ofV. Since the vectors ofSare also vectors ofV,the associative and distributive laws must hold forS. Hence,

Sis a vector space overFand is a subspace ofV.

Vector Spaces


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Let v1, v2, ,vkbe kvectors in a vector space Vover a fieldF.Let a1, a2, , akbe kscalars fromF. The sum

a1v1 + a2v2 + + akvk

is called a linear combination ofv1

, v2

, ,vk

.Clearly, the sum of two linear combinations ofv1, v2, ,vk ,(a1v1 + a2v2 + + akvk ) + (b1v1 + b2v2 + + bkvk )= (a1+b1 )v1 + (a2+b2 )v2 ++ (ak+bk)vk

is also a linear combination ofv1

,v2

, ,vk

, and the product of ascalarc inFand a linear combination ofv1, v2, ,vk ,c(a1v1 + a2v2 + + akvk) =is also a linear combination ofv1, v2, ,vk

Theorem 2.19Let v1, v2, ,vkbe kvectors in a vector space Vover a fieldF.The set of all linear combinations ofv1, v2, ,vk forms a subspaceofV.

k2211 v)(...v)(v)( kacacac +++

Vector Spaces


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A set of vectors v1, v2, ,vk in a vector space Vover a fieldFis

said to be linearly dependentif and only if there exit kscalars a1,

a2, , ak fromF, not all zeros, such that

a1v1 + a2v2 + + akvk = 0A set of vectors v1, v2, ,vk is said to be linearly independentif it

is not linearly dependent. That is, ifv1, v2, ,vk are linearly

independent, then a1v1 + a2v2 + + akvk 0

unless a1 = a2 = = ak = 0.

EX. The vectors (1 0 1 1 0), (0 1 0 0 1), and (1 1 1 1 1) arelinearly dependent since

1(1 0 1 1 0) + 1(0 1 0 0 1) + 1(1 1 1 1 1) = (0 0 0 0 0)

Vector Spaces


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However, (1 0 1 1 0), (0 1 0 0 1), and (1 1 1 1 1) are linearly

independent.

A set of vectors is said tospan a vector space Vif every vector in

Vis a linear combination of the vectors in the set.

In any vector space or subspace there exits at least one setB of

linearly independent vectors which span the space. This set is

called a basis (orbase) of the vector space.The number of vectors in a basis of a vector space is called the

dimension of the vector space. (Note that the number of vectors in

any two bases are the same.)

Vector Spaces

C id h f ll l GF(2) LV


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Consider the vector space of all n-tuples overGF(2). Let us

form the following n n-tuples :

where the n-tuple ei has only nonzero component at ith position.

Then every n-tuple in can be expressed as alinear combination ofe0, e1,,en-1 as follows :

1),0...0000(

0)0...0010(

0)0...0001(

1

1

0

=

=

=

n-e

e

e

),...,,( 110 naaa

nV

nV

111100110 ...),...,,( +++= nnn eaeaeaaaa

Vector Spaces

Th f th t f ll t lV


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Therefore, e0, e1,,en-1 span the vector space of all n-tuples

overGF(2). We also see that e0, e1,,en-1 are linearly independent.

Let u = and v = be two n-tuples in

. We define the inner product(ordot product) ofu and v as

where uivi and uivi + ui+1vi+1 are carried out in modulo-2

multiplication and addition. Hence the inner product uv is a

scalar in GF(2). Ifuv = 0, u and v are said to be orthogonalto

each other.

The inner product has the following properties :

uv = vu

u(v+w) = uv + uw

(au)v = a(uv)

nV

),...,,( 110 nuuu ),...,,( 110 nvvv

nV,...vu 111100 +++= nn vuvuvu

Vector Spaces

Let S be a k dimension subspace of and let S be the set ofV


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Let Sbe a k-dimension subspace of and let Sdbe the set ofvectors in such that, for any u in Sand v in Sd, uv = 0. The setSd contains at least the all-zero n-tuple 0 = (0, 0, , 0), since for

anyu

in S,0u

= 0. Thus, Sd

is nonempty. For any element a inGF(2) and any v in Sd,

Therefore, av is also in Sd. Let v and wbe any two vectors in Sd .For any vectoru in S, u(v+w) = uv + uw = 0 + 0 = 0. This saysthat ifv and w are orthogonal to u, the vector sum v + w is alsoorthogonal to u. Consequently, v + w is a vector in Sd. It followsfrom Theorem 2.18 that Sd is also a subspace of . This subspaceis called the null(ordual) space ofS. Conversely, Sis also thenull space ofSd.

nV

nV

1ifv

0if0{v

=

==

a

aa

nV

Vector Spaces

Th 2 20


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Theorem 2.20Let Sbe a k-dimension subspace of the vector space Vn of all

n-tuples overGF(2). The dimension of its null space Sd is n-k. In

other words, dim(S) + dim(Sd)= n.

cc 2 algebra

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