Basic galois field arithmatics required for error control codes

GALOIS FIELD ALGEBRA FOR ERROR
CONTROL CODES

SETS
 Collection of objects.
 A set with n objects is -- { s1, s2, …., sn}, where s1, s2, etc are set’s
elements.
 A new set formed using any m elements of n elements of a set is
subset of original set.
 Finite sets – finite number of elements.
 e.g. set of Decimal digits is { 0 1 2 3 4 5 6 7 8 9}
 Binary set { 0 1} is subset of decimal set.
 Infinite sets – infinite number of elements.
 e.g. non-zero positive integers, {1,2,3,….}
 Set of real numbers.
 Operations between elements within a set can be understood by
higher mathematical structure called “Groups”.

GROUPS
 A set G on which a binary operation * is defined between elements
within the set is called Group if following conditions are satisfied:-
1. The group is closed under operation *. (Closure property)
1. If a and b are elements of set, then c = a * b also belong to the set.
2. The operation * is associative.
1. a * ( b * c ) = ( a * b ) * c
3. There exists a unique identity element e within the set such that
for any element a,
1. a * e = e * a = a
4. For any element a, there exists unique inverse a’ within the set
such that,
1. a * a’ = a’ * a = e
 If a * b = b * a, group is commutative (Abelian Group). It is NOT a
necessary condition for a set to be group.

GROUP OPERATIONS
 Group operation is normally addition or multiplication.
 Gives Additive Groups or Multiplicative Groups.
 Example: Binary numbers { 0, 1 } in modulo–2 addition operation.
1. Closure property obeyed.
2. It is associative.
3. Identity element is 0.
4. Each element is its own inverse. ( Addition should result in 0.)
5. It also forms a commutative group.
 For coding theory, it is an important group.

EXAMPLE
 Find whether ordinary addition forms a group over
 A) { 0,1,2,…}
 B) {0. ±1, ±2, …}
 A) No. Addition over this set fails to generate the group.
 1. Set is closed. 3 + 4 = 7, a member.
 2. Set is associative. (3 + 4) + 7 = 3 + ( 4 + 7) = 14
 3. Identity element is 0. Element unaltered when added to 0.
 4. But set does not contain inverse of any element. E.g. inverse of 9 is -9.
 B) Yes. Addition over this group forms a commutative group.
 1. Set is closed. 3 + 4 = 7, a member.
 2. Set is associative. (3 + 4) + 7 = 3 + ( 4 + 7) = 14
 3. Identity element is 0. Element unaltered when added to 0.
 4. Inverse of each element is within the group. E.g. inverse of 9 is -9.
 5. Moreover obeys commutative rule.

EXAMPLE
 Find whether ordinary multiplication forms a group over
 A) { 0,1,2,…}
 B) {0. ±1, ±2, …}
 None form a multiplicative group as group does not contain
multiplicative inverse of the members.
 Find whether set of nonzero rational numbers form multiplicative
group?
 Yes.
 Identity element is 1.
 If c = a/b, multiplicative inverse c’ = b/a
 Multiplication is closed, associative and commutative too.
 Additive group can be created by including 0 element.

GROUP OPERATIONS – SUBTRACTION AND DIVISION
 Group operation is normally addition or multiplication.
 No second group operation definition required within additive
group.
 Subtraction can be carried out by replacing element to be
subtracted with its additive inverse.
 a – b = a + (-b)
 Division can be carried out by multiplicative group by replacing
divisor with its multiplicative inverse.
 a / b = a * ( b’)
 Associative and commutative properties applies to group elements.
 a – b = a + ( - b) = ( - b) + a = - b + a

SUB-GROUP
 Subset of elements within a group forms a subgroup.
 Addition over the set of integers ( a subgroup of rational numbers) form
additive subgroup.
 Set of positive integers ( a subset of integers) do not form additive
subgroup. Why?
 Set of positive integers do not contain additive inverses.

FINITE GROUP
 Group having finite number of elements.
 Number of members within finite group is called ‘Order” of the
group.
 A finite group can be constructed by taking integers modulo-m.
 n modulo-m is remainder r obtained after dividing n by m.
 Written as n = r modulo-m or n modulo-m = r
 5 modulo-4 = 1
 13 modulo-5 = 3

MODULO-M ADDITION
 n = r modulo-m or n modulo-m = r
 Modulo-m produces m integers 0,1,2,…,m-1.
 If n=m, m modulo-m = 0
 If n=a x m, (a x m) modulo-m = 0 ( a is an integer)
 If a and b are two integers, using modulo-m addition,--
 (a + b ) modulo-m = r
 Modulo-m addition set {0,1,2,…,m-1} is commutative group as -
 Closed,
 Associative,
 Commutative ,
 Each nonzero integer a has an additive inverse m – a
 0 is its own inverse.


EXAMPLE-1
 Construct the additive group of integers modulo-5 over the set
{0,1,2,3,4}.
 Need to find remainders when integers are added pair wise.
 (1 + 1) modulo-5 = 2
 (1 + 2) modulo-5 = 3
 (1 + 3) modulo-5 = 4
 (1 + 4) modulo-5 = 0
 (2 + 2) modulo-5 = 4
 (2 + 3) modulo-5 = 0
 (2 + 4) modulo-5 = 1
 (3 + 3) modulo-5 = 1
 (3 + 4) modulo-5 = 2
 (4 + 4) modulo-5 = 3

EXAMPLE-1
 Adding 0 to any element does not alter element.
 Remaining pairs are same as commutative equivalent.
 Additive Group of Integers Modulo-5 -

EXAMPLE-2
 Find inverse of each element in Additive Group of integers modulo-
5.
 Additive integer of a nonzero integer a is m – a.
 When added to a, it should give 0 after given operation modulo-5.
 Additive inverse of 1,2,3, and 4 are 4, 3, 2, and 1.
 Additive inverse of 0 is 0.

MODULO-M MULTIPLICATION
 Gives remainder of the product of two integers.
 Modulo-m multiplication generates groups for only prime
values of m.
 Group is over the set {1,2,3,…m - 1}.
 0 element is excluded as it does not have a multiplicative
inverse.
 If m is not a prime number, then modulo-m multiplication
over the set {1,2,3,…m - 1} does not produce a set.
 Does not give multiplicative inverse.
 Hence, two groups can be constructed over the set
{0,1,2,3,…m - 1},
 An additive group over all elements,
 A multiplicative group over set’s nonzero elements.

EXAMPLE-3
 Construct the multiplicative group over the set {1,2,3,4}using
modulo-5 multiplication.
 Obtain remainders when integers are multiplied pair wise.
 2 X 2 = 4 modulo-5
 2 X 3 = 1 modulo-5
 2 X 4 = 3 modulo-5
 3 X 3 = 4 modulo-5
 3 X 4 = 2 modulo-5
 4 X 4 = 1 modulo-5
 Product with 1 keeps integers unaltered.
 Remaining pairs are commutative equivalents.

EXAMPLE-3
 Construct the multiplicative group over the set {1,2,3,4}using
modulo-5 multiplication.
Multiplicative inverse for a is a’ such that a x a’ = 1 modulo-5.
Inverse of 2 is 3. Find other inverses.

PROOF
 Prove 1: The identity element in a group G is unique.
 Let there exists two identity elements e and e’ in G.
 Then e’ = e’ * e = e.
 This implies that e and e’ are identical.
 Hence there is one and only one identity element.
 Prove 2: The inverse of a group element is unique.
 Let there exist two inverses a’ and a’’ for group element a.
 Then a’ = a’ * e = a’ * (a * a’’) = (a’ * a )* a’’ = e * a’’ = a’’.
 This implies that a’ and a’’ are identical.
 Hence there is one and only one inverse for a.

FIELDS
 A set F on which two binary operations, addition and multiplication
are defined between elements within the set is called Field if
following conditions are satisfied:-
1. The Field is a commutative group under addition having
1. Additive identity - zero.
2. Additive inverse.
2. A set of non-zero elements in F is commutative under
multiplication.
1. Having multiplicative identity – unity.
2. Having multiplicative inverse.
3. Multiplication is distributive over addition.
1. a * ( b + c) = a*b + a*c

FIELDS
 Field consists of at least two elements, AI and MI.
 Number of elements in a field is called the ORDER of the field.
 Types – Finite field and Infinite field.
 Subtraction element b from a is by adding AI (–b) to a.
 a – b = a + (-b)
 Division of a by b is by multiplying a by MI b-1.
 a / b = a * b-1
 Hence, within a field, all following are possible-
 Addition
 Subtraction
 Multiplication
 Division

FIELDS - PROPERTIES
1. For every element a in field, a · 0 = 0 · a = 0
2. For any two nonzero elements, a and b in field, a · b ≠ 0
3. a · b = 0 and a ≠ 0 imply that b = 0
4. For any two elements, a and b in field,
- (a · b) = (- a) · b = a · (- b)
5. For a ≠ 0 , a · b = a · c implies that b = c

FIELDS – EXAMPLE –
 Set of real numbers
 Addition and multiplication form commutative groups.
 Two operations satisfy distributive rule.
 Gives REAL FIELD.
 Set of integers -- { 0, ±1, ±2, ±3, …}
 Addition forms a group.
 Multiplication FAILS to form a group.
 Product of two integers is an integer BUT Division of integer by
another integer is not an integer.
 MI not available in this set.
 Called a RING.

RING
 Set of integers -- { 0, ±1, ±2, ±3, …} is a ring.
 Addition, subtraction and multiplication are possible but division
not possible.
 Structure between Group and Field. Why?
 Group-- Either addition and subtraction (AG)or Multiplication and
division (MG).
 Field – Addition, subtraction, multiplication and division all are
possible.

CONSTRUCTION OF FIELDS
 Fields can be created from modulo-m addition and multiplication.
 Modulo-2 arithmetic forms AG over set { 0, 1} and MG over non-
zero elements of { 0, 1}.
 Generates binary field GF(2)
 Finite fields are called Galois fields.
 Important for coding theory.
 Widely used in digital computers and digital data transmissions.
modulo-2 addition modulo-2 multiplication
+ 0 1 * 1
0 0 1 1 1
1 1 0

EXERCISE: CONSTRUCT THE PRIME FIELD UNDER MODULO-7 ARITHMETIC GF(7). FIND
ADDITIVE IDENTITY AND INVERSE AND MULTIPLICATIVE IDENTITY AND INVERSE.
 (Prime fields GF(p)– larger fields generated by modulo-m addition and
multiplication over set { 0, 1, …, m-1}, p =m is prime. )
Additive inverse of a– (7 – a)
Additive identity -- 0
Multiplicative inverse of a– a X a* = 1 modulo-m
Multiplicative identity -- 1

EXERCISE: EVALUATE ((2 – 3 ) X 3 ) / 4 OVER PRIME FIELD
MODULO-M WHEN
A) M = 5
B) M = 7
 A) M = 5
 ((2 + (-3 ) X 3 ) X (1/ 4)
 Additive inverse of 3 is 2 and multiplicative inverse of 4 is 4.
 (2 + 2) X 3 X 4
 48 modulo-5 = 3 modulo-5
 =3
 B) M = 7
 AI = 4, MI = 2
 Ans = 1

EXERCISE
 Solve the equations –
 1. 3x + 2y = 2 modulo - 7
 4x + 6y = 3
 2. 2x + 3y = 1 modulo – 5
 x + 2y = 2

1. 3X + 2Y = 2 MODULO - 7
4X + 6Y = 3
 Eliminating x by multiplying 1 by 6
 18x + 12Y = 12 ≈ 4x + 5y = 5
 Solving with 2 we get
 Y = 3 – 5 = 3 + 2 = 5
 Hence-
 3x + 10 = 2
 3x + 3 = 2
 3x = 2 – 3 = 2 + 4 = 6
 x = 6/3 = 6 X 5 = 30 modulo-7 = 2
 x = 2, y = 5
 2. Ans x = 1, y = 3

PROOFS
 Prove : If a be nonzero element of a finite field GF(q), then
Prove that a
q-1
= 1
 Let b1, b2, …, bq-1 be the q-1 non-zero elements of GF(q).
 Hence q-1 elements a.b1, a.b2, …, a.bq-1 are non-zero and
distinct.
 (a.b1) . (a.b2) …(a.bq-1) = b1. b2. … bq-1 (Try multiplying a=2 with all
nonzero elements of GF(7) using mod-7)
 a
q-1
(b1 . b2 …bq-1) = b1. b2. … bq-1
 a
q-1
= 1

PROOFS
 Prove : If a be nonzero element of a finite field GF(q), and n
be the order of a then,
Prove that n divides q-1
 Let n does not divide q-1.
 Dividing q-1 by n , we obtain q-1 = kn +r where 0 < r < n
 Then, a
q-1
= a
kn + r
= a
kn
. a
r
= (a
n
)
k
. a
r
 Since a
q-1
=1 and a
n
=1 as order of a is n—
 a
r
=1
 This is impossible as 0 < r < n and n is smallest integer such
that a
n
=1.
 Hence assumption is wrong.
 n must divide q-1.

PRIME FIELDS
 For any prime p, there exists a finite field of p elements GF(p).
 For any positive integer m, it is possible to extend the prime
field GF(p) to a field of pm
elements called extension field
GF(pm
).
 Order of any prime field is the power of the prime.

BINARY FIELD ARITHMETIC
 Codes can be constructed with symbols from any Galois field
GF(p), where p is either prime or a power of p.
 Codes most widely used in data communication use p = 2.
 GF(2) or its extension GF(2m)
 GF(2) uses binary arithmetic – modulo-2 addition and
multiplication.
 1 + 1 = 0, 1 = -1 → Subtraction is same as addition.
 Binary arithmetic can be used to solve set of equation using
Cramer’s rule. As
 X + Y = 1
 X + Z = 0
 X + Y + Z = 1

CRAMER’S RULE
Given
which in matrix format is
x, y and z can be found as follows:
Answer—
X Y Z = 0 1 0

POLYNOMIAL OVER GF(2) (ONE VARIABLE)
 Degree of polynomial is largest power of X with nonzero
coefficient.
 Polynomial over GF(2) takes its coefficients from GF(2).
 Total of 2n
polynomials over GF(2) with degree n.
 If n = 1, → X and 1 + X
 If n = 2 → X2
, 1 + X2
, X + X2
, 1+ X + X2
 Polynomial over GF(2) can be added, subtracted, multiplied and
divided. Using modulo-2 addition and multiplication.

 a(X) = 1 + X + X3
+ X5
 b(X) = 1 + X2
+ X3
+ X4
+ X7
 a(X) + b(X) = ?
 X + X2
+ X4
+ X5
+ X7

 Polynomials can be multiplied over GF(2).
 When f(X) is divided by g(X), we get, using Euclid’s division algorithm,
 Degree of r(X) is less than that of g(X),

 Polynomial over GF(2) follow following conditions:

 f(X) = 1 + X + X
4
+ X
5
+ X
6
 g(X) = 1 + X + X
3
 Find f(X) · g (X) = ?
 1 + X
2
+ X
3
+ X
8
+ X
9
.
 Find f(X) / g (X) = ?
 1 + X + X
2
is remainder and quotient is X
2
+ X
3
.

ROOTS OF POLYNOMIAL OVER GF(2)
 If a is root of polynomial f(X), f(X) is divisible by (X – a).
 If f(X) = 1 + X
2
+ X
3
+ X
4
, and X = 1, f(1) = 0
 Hence 1 is root and f(X) is divisible by (X + 1).
 If a Polynomial over GF(2) has even number of terms, it is divisible by
(X + 1). Why? 1 is root.
 Polynomial over GF(2) of degree m is called Irreducible over GF(2), if
it is not divisible by any Polynomial over GF(2) of degree less than m
but greater than zero. (Divisible by only 1 and self.)
 If degree m = 2, then
 X
2
, 1 + X
2
, and X + X
2
are divisible by X or 1 + X, → roots 0 and 1.
 but 1+ X + X
2
is not divisible by any polynomial of degree 1
 1+ X + X
2
is irreducible polynomial of degree 2.
 1+ X + X
3
 1+ X + X
4

PROVE THAT
 Expanding above equation repeatedly
 Since fi
= 0 or 1, f2
i
= fi

CONSTRUCTION OF GALOIS FIELD GF(23)
 As seen before, polynomial 1+ X + X3
has no binary (0,1) roots.
 Let , not belonging to binary field GF(2) but lying within finite
field GF(23
) is one of the roots of above polynomial.
 Hence 3
+ + 1 = 0
 Elements of GF(23
) = ?
 0 and 1 form additive and multiplicative identity elements.
 +0 = · 1 =
 Additive inverse of is itself.
 + = 1 + 1 = (1 + 1) = 0 · = 0
 Subtraction and addition of are equivalent
 From above = -
 Multiplicative inverse of = -1
= 1/ .
 The other elements of field GF(23
) can be generated from .

FIELD ELEMENTS OF GALOIS FIELD GF(2M)
 0 (additive identity)
 1 (multiplicative identity)


2

3
= + 1 ( 3
+ + 1 = 0, - = , 1 = - 1)

4
= ( + 1) = 2
+

5
= ( 2
+ ) = 3
+ 2
= 2
+ + 1

6
= ( 2
+ + 1) = 3
+ 2
+ = 2
+ 1

7
= ( 2
+ 1) = 3
+ = 1

8
= 7
=

9
=
2 7
=
2
 And so on…

FIELD ELEMENTS OF GALOIS FIELD GF(2M)
 We can see higher powers as

7
= 1 ( 7 = 0 modulo-7)

8
= ( 8 = 1 modulo-7)

9
= 2
(9 = 2 modulo-7)
 And so on…

12
= 5
(12 = 5 modulo-7)

17
= 3
(17 = 3 modulo-7)
 GF(23
) has 8 basic elements- 0, 1, , 2
, 3
, 4
, 5
, and 6
 Degree of polynomial = m = 3
 Field’s characteristics = no, of elements in base field = p = 2
 Order of the field = q = p
m
.
 Coefficients of polynomials belong to base field while roots belong to
extension field.

FIELD ELEMENTS OF GALOIS FIELD GF(23)
Power Polynomial 3 – tuple
Representation Representation Representation
0 0 0 0 0
1 1 1 0 0
0 1 0
2 2
0 0 1
3
1 1 1 0
4 2
0 1 1
5
1 2
1 1 1
6
1 2
1 0 1

ADDITION AND MULTIPLICATION OVER GF(23)
 Find 1.
5
+ + 1 2.
6
+
2
+ 1
 1. 5
+ + 1 = 2
+ + 1+ + 1 = 2
 2.
6
+
2
+ 1 =
2
+ 1 +
2
+ 1 = 0
 Find 1. 5
2. 4 5
3. 5 6 4
 1. 5
= 6
 2. 4 5
= 9
= 7 2
= 2
 3. 5 6 4
= 15
= 7 7
=
 Find 1. 2
/ 5
2. 1 / 3
3. 3
/ 4. 1/
 1. 2
/ 5
= 2 -5
= 2 7-5
= 4
 2. 1 / 3
= -3
= 7-3
= 4
 3. 3
/ = 3 -1
= 3 7-1
= 3 6
= 2
 4. 1/ = -1
= 7-1
= 6

CONSTRUCTION OF GALOIS FIELD GF(24)
 As seen before, polynomial 1+ X + X
4
has no binary (0,1) roots.
 Let , not belonging to binary field GF(2), but lying within finite
field GF(24
) is one of the roots of above polynomial.
 Hence 4
+ + 1 = 0
 The field GF(24
) can be generated from .
 0 and 1 form additive and multiplicative identity elements.
 +1 = · 1 =
 Additive inverse of is itself.
 + = 1 + 1 = (1 + 1) = 0 · = 0
 Subtraction and addition of are equivalent
 From above = -
 Multiplicative inverse of = -1
= 1/ .

FIELD ELEMENTS OF GALOIS FIELD GF(24)
 0, 1, , 2
, 3

4
= + 1

5
= 2
+

6
= 3
+ 2

7
= 4
+ 3
= 3
+ + 1

8
= 2
+ 1

9
= 3
+

10
= 2
+ + 1

11
= 3
+ 2
+

12
= 3
+ 2
+ +1

13
= 3
+ 2
+ 1

14
= 3
+ 1

15
= 1 repeats
 Total 16 field elements.
 Find field elements for GF(25
)

ADDITION AND MULTIPLICATION OVER GF(24) , GF(24)
 Find for GF(2
4
)

2
+ 9

7
+
3
+
11

2 9

13 8

7 12 4

13
/ 2

3
/ 10
 / 9
 1/

PRIMITIVE FIELD ELEMENTS
 Field element that can generate all the nonzero elements of
the field are primitive elements.
 is primitive in GF(23
), GF(24
) and GF(25
).
 All elements except 0 and 1 are primitive for GF(23
). e.g…
 Show that
2
is primitive in GF(2
3
).
 Let = 2
,

2
=( 2
)2
= 4

3
=( 2
)3
= 6

4
=( 2
)4
= 8
=

5
=( 2
)5
= 10
= 3

6
=( 2
)6
= 12
= 5

7
=( 2
)7
= 14
= 1

8
=( 2
)8
= 16
= 2
And repeats.

PRIMITIVE FIELD ELEMENTS
 Show that
5
3
).
 Show that 2
).
 Show that
3
is NOT primitive in GF(2
4
).
 Show that 5
is NOT primitive in GF(24
).

ORDER OF THE ELEMENTS
 Smallest positive integer n for which
n
= 1
 Determines if is primitive or not.
 (Not same as order of the field, which is number of elements
within field.)
 In GF(23
), all field elements have same order 7. Show.
 In GF(24
), all field elements do not have same order . Show.
 Order of an element in GF(2m
) divides 2m
- 1
 For GF(24
) determine the order of 12
and 7
. Find if they are
primitive or not. What field elements they generate?
 Smallest power of
12
to give unity is 5. Not primitive as 15
elements are required. Elements – 12 9 6 3
1
 Smallest power of
7
to give unity is 15. Primitive as 15
elements are required. Elements – all

IRREDUCIBLE AND PRIMITIVE POLYNOMIAL
 Polynomials, divisible by only 1 and self are called irreducible
polynomials.
 Irreducible polynomial having a primitive field element as a root
is called a primitive polynomial.
 An irreducible polynomial p(X) of degree m is said to be primitive
if the smallest positive integer n for which p(X) divides X
n
+ 1 is
n=2m
– 1. (Not any n <2
m
– 1)
 The irreducible polynomial p(X) = X4
+ X + 1, divides (X15
+ 1)
(n=15), but does not divide any Xn
+ 1 for 1<n<15, hence p(X) is
primitive polynomial.
 The polynomial p(X) = X4
+ X3
+ X2
+ X + 1 is irreducible but not
primitive as it divides X5
+ 1 also.

CONJUGATE OF FIELD ELEMENT OVER GF(2
M
)
 In ordinary algebra, a polynomial may have complex conjugate
roots occurring in pair.
 Similarly roots of polynomial with coefficients from GF(2) also
occur In groups or sets of conjugates.
 X4
+ X3
+ 1 has no roots from GF(2), but has 4 roots from GF(24
).
 By substitution, roots are -
7
,
11
,
13
and
14
. Verify
 Then (X + 7
)(X + 11
)(X + 13
)(X + 14
) = X4
+ X3
+ 1
 If one root is known, its other conjugate roots can be found as—

M
)
 If one root is known, its other conjugate roots can be found as—
 Theorem- Let f(X) be a polynomial with coefficients from GF(2).
Let β an element in an extension field of GF(2). If β is a root of
f(X), then for any l ≥ 0, β2
l
is also a root of f(X).
 Proof: We have [f(X)] 2
l
= f(X 2
l
)
 β is Root --- [f(β)] 2
l
= f(β2
l
)
 Since f(β)= 0, f(β2
l
) = 0
 Hence β2
l
is also a root of f(X).
 The element β2
l
is called a conjugate of β.
 Hence if β from GF(2m
) is a root of f(X) over GF(2), then all
conjugates of β ( which are elements of GF(2
m
),are also roots of
f(X).

M
)
 Example – Let f(X) = 1+ X3
+ X4
+ X5
+ X6
has 4
from GF(24
), as a
root. Verify. Also find its conjugate roots.
 f( 4
) = 0.
 The conjugates of 4
are –
 ( 4
)2
= 8
,
 ( 4
) 22
= 16
=
 ( 4
) 23
= 32
= 2
. Higher powers repeat roots.
 Check.
 Find other two roots. Are they conjugates?

5 10
. They are conjugates.

THEOREM
 Theorem2.8- The 2m
- 1 non zero elements of GF(2m
) are all the
roots of (X(2m– 1)
+ 1).
 Proved earlier that -
 If β is nonzero element in the field GF(2m
),---- β2m-1 =1
 Example – For GF(24
) , β15 = 1, for β = 1, 2 3 4 5 6
…… 14
 Adding 1 on both sides,
 β2m-1 +1 = 0
 Hence β is the root of X2m-1 +1 = 0
 Hence all 2
m
-1 nonzero element are roots of X2m-1 +1 = 0
 Prove Corollary 2.8.1– The elements of GF(2m
) form all the roots
of X2m
+X
 X2m
+X = X (X2m-1 +1)
 Hence element 0 is also a root.

MINIMAL POLYNOMIAL
 Let any element β in GF(2m
) is a root of (X(2m– 1)
+ 1) over GF(2).
 β may also be a root of a polynomial over GF(2)with degree < 2m
.
 The binary polynomial of smallest degree, of which β is a root, is
called minimal polynomial of β.
 Theorem 2.9: Minimal polynomial φ(x) of a field element β is
irreducible.
 Suppose φ(x) is NOT irreducible. Then φ(x) = φ1(x) φ2(x)
 Both φ1(x) and φ2(x) have degrees > 0 and < degree of φ(x)
 As φ(β) = φ1(β) φ2(β) = 0
 Either φ1(β) = 0 or φ2(β) = 0
 This contradicts the hypothesis that φ(x) is a polynomial of
smallest degree, such that φ(β) = 0.
 Therefore φ(x) must be irreducible.

MINIMAL POLYNOMIAL
 Theorem 2.10: Let f(x) be a polynomial over GF(2). Let φ(x) be the
minimal polynomial of a field element β. If β is the root of f(x),
then f(x) is divisible by φ(x).
 Dividing f(x) by φ(x),
 f(x) = a(x) φ(x) + r(x)
 Degree of r(x) < degree of φ(x).
 If x= β ,then f(β) = a(β) φ(β) + r(β)
 As f(β) = φ(β) = 0
 r(β) = 0
 If r(x) ≠ 0, r(x) would have degree < φ(x), which has β as root.
 This contradicts fact that φ(x) is a minimal polynomial of β.
 Hence r(x) must be zero and φ(x) divides f(x).

MINIMAL POLYNOMIAL
 Theorem 2.11: Minimal polynomial φ(x) of a field element β in
GF(2m
) divides (X2m
+ X).
 All roots of φ(x) are from GF(2m
).
 Theorem 2.8 corollary 2.8.1 says field elements of GF(2m
) are
roots (X2m
+ X).
 β is root of φ(x) as well as (X2m
+ X).
 As φ(x) is minimal polynomial, from theorem 2.10, φ(x) divides
(X2m
+ X).

MINIMAL POLYNOMIAL
 Theorem 2.12: Let f(x) be an irreducible polynomial over GF(2).
Let β be an element in GF(2m
). Let φ(x) be the minimal
polynomial of element β. If f(β) = 0 then φ(x) = f(x).
 From 2.10, φ(x) divides f(x).
 Since φ(x) ≠ 0, and f(x) is irreducible, φ(x) = f(x).

MINIMAL POLYNOMIAL
 Theorem 2.13: Let β be an element in GF(2m
) and Let e be the
smallest non-negative integer such that β2e
= β. Then
 f(X) = ∏i=0
e-1
(X + β2i
) is an irreducible polynomial over GF(2).

MINIMAL POLYNOMIAL
 For 0 ≤ i ≤ e, fi = fi
2
.
 It is possible only when , fi = 0 or 1.
 Hence f(x) Has coefficients from GF(2).
 Suppose f(X) is NOT irreducible over GF(2) and f(X) = a(X) b(X).
 Since f(β) = 0, either a(β) = 0 or b(β) = 0.
Hence

MINIMAL POLYNOMIAL
 If a(β) = 0, a(X) has , β, β2, …β2e-1
as roots, so
 a(X) has degree e and
 a(X) = f(X).
 Similarly if b(β) = 0, b(X) has , β, β2, …β2e-1
as roots, so
 b(X) has degree e and
 b(X) = f(X).
 This contradicts itself.
 Hence f(X) must be irreducible.

MINIMAL POLYNOMIAL
 Theorem 2.14 : Let ¢(X) be the minimal polynomial of an element
β in GF(2m
). Let e be the smallest integer such that β2e
= β. Then
 ¢(X) = ∏i=0
e-1
(X + β2i
)
 From 2.13, f(X) is irreducible.
 ¢(X) is irreducible minimal polynomial.
 Hence ¢(X) = ∏i=0
e-1
(X + β2i
)

EXAMPLE
 For Galois Field GF(24
), β = 3
. Find conjugates of β and the
minimal polynomial of β = 3
.
 Conjugates are – 6
, 12
, 24
= 9
.
 Minimal polynomial ¢(X) =(X + 3
) (X + 6
)(X + 12
)(X + 9
).
 = 1+ X + X2
+ X3
+ X4
 For Galois Field GF(23
), find minimal polynomial of 3
.

COMPUTATION USING GALOIS FIELD GF(2M
) ARITHMETIC.
 Given linear equations over GF(24
) as
 X + 7
Y = 2

12
X + 8
Y = 4
 Multiply eq2 with 3
and add the two.
 X + 7
Y = 2
 X + 11
Y = 7
 ( 7
+ 11
) Y= 2
+ 7

8
Y= 12
 Y = 4
 X = 9
 Alternately use Cramer’s rule.

) ARITHMETIC.
 Find roots of X2
+ 12
X + 9
= 0 for GF(24
)
 Let β1 ,β2 are required roots.
 X2
+ X(β1+β2 ) + β1 β2 = 0
 β1+β2 = 12
 β1 β2 = 9
 Roots that satisfy = 2
, 7
. Check by finding minimal polynomial.
 Expand (X + 4
)2
in GF(23
).
 =X2
+ X 4
+ X 4
+ 8
= X2
+ 8
= X2
+ .
 Expand (X + 3
)5
(X + 10
) in GF(24
).
 (X + 3
)4
(X + 3
) (X + 10
)
 (X4
+ 12
) (X2
+ 12
X + 13
)
 X6
+ 12
X5
+ 13
X4
+ 12
X2
+ 9
X + 10
)

) ARITHMETIC.
 Find inverse of following in GF(24
).

2 13
 0 10

7 3
1

14
0 12

5 13
1

14 9
 Find inverse of following in GF(24
).

3 5
 1

8 12

7 10

) ARITHMETIC.
 Find solution of following in GF(24
) using Cramer’s rule.

3
X + Y +Z = 5

2
X + 6
Y +Z = 6

14
X + 7
Y + 7
Z = 1
 X =
 Y = 4
 Z = 4

RELATION WITH CYCLIC CODES
 Code word c(x) for an (n,k) cyclic code is c(x) = f(x) g(x)
 Nonsystematic CRC – f(x) is data polynomial d(x)
 Systematic CRC – f(x) is quotient q(x) after dividing d(x) xn-k
by
g(x)
 Root of g(x) is also root of c(x).
 Let for (7,4) CRC, g(x) = 1+ X + X3
and c(x) = 1+ X + X2
+ X5
 Roots , 2
and 4
of g(x) belonging to GF(23
) are roots of c(x).
 Polynomials like 1+ X + X3
generate CRC as well as construct
finite field.
 Generator polynomial can be obtained by finding minimal
polynomial using all roots and finding LCM to eliminate
common multiples.
 Find generator polynomial using , 2
and 4
in GF(23
) .

VECTOR SPACES
 Code word belonging to (n.k) block code can be interpreted as
vectors within k-dimensional vector space.
 Vector space is defined as collection of objects, called vectors,
together with operations of vector addition and scalar
multiplication satisfying following conditions :-
1. Set of vectors forms additive cumulative group satisfying
closure, associative and commutative property and has
additive inverse and additive identity vector 0.
2. Multiplication of vectors with scalar is defined. Scalar product
is a vector with distributive and associative property and has
multiplicative identity 1.

ORDERED SEQUENCES – N-TUPLE
 Vector v = (v1, v2, ….vn)
 Where vi for i = 1,2,…n, are scalars from a field and are
components of v.
 Vector addition is component wise.
 If u = (u1, u2, ….un)
 w = u +v is
 w = (u1 + v1, u2 + v2, … un + vn)
 = (w1, w2, ….wn)
 Scalar product is also component wise.
 au = (au1, au2, …aun)

EXAMPLE
 Given vectors u = (2,7,1) and v = (10,-3,8),
 Find
1. u + v
2. 4u
3. 2u – 7v
 Answer-
1. (12, 4, 9)
2. (8, 28, 4)
3. (-66, 35, -54)

LINEARLY DEPENDENT VECTORS
 A set of vectors v1, v2, ….vm are said to be linearly
dependent if there exists m scalars a1, a2, …am, not all
of them zero, such that— a1v1 + a2v2 + ….amvm = 0
 Otherwise vectors are said to be linearly independent.
 For linearly independent vectors, only scalars that
satisfy above equation are zeros.
 Find if following ser of vectors are linearly dependent.
1. u1 = (3, -8, 5), u2 = (-2, 2, 14) and u3 = (-1, 6, -19)
2. v1 = (7, 3, -3), v2 = (16, 2, -1) and v3 = (6, -12, 15)
3. w1 = (-2, 0, 0), w2 = (0, 7, 0) and w3 = (0, 0, 9)

ANSWER
1. If a1 = 1, a2 = 1 and a3 = 1
• a1u1 + a2u2 + ….amum = 0
• Linearly dependent
2. If a1 = 1, a2 = 1 and a3 = 1
• a1v1 + a2v2 + ….amvm ≠ 0
• But if a1 = 2, a2 = -1 and a3 = 1/3
• a1u1 + a2u2 + ….amum = 0
• Linearly dependent
3. No combination of scalars satisfy condition.
1. Linearly independent

IMPORTANCE OF LINEAR INDEPENDENCE OF VECTORS
 Ability to form linear combination and linear independence
simplify construction of vector spaces.
 Within a vector space, there exists a set of linearly independent
vectors, from which all other vectors can be generated.
 This set is called Basis and these vectors, Basis Vectors.
 Each vector in a vector space is unique combination of basis
vectors.
 Basis vectors are said to span the vector space.
 Number of vectors in a Basis is called Dimension of vector
space.
 In m-dimensional vector space, m linearly independent vectors
span the space.

EXAMPLE
 Three dimensional space over the real field.
 Basis Vectors are—
 i = ( 1 0 0)
 j = ( 0 1 0)
 k= ( 0 0 1)
 Also called Standard Basis because --
 All other vectors within space can be expressed using them as
–
 v = vxi + vyj + vzk
 v = vx ( 1 0 0) + vy ( 0 1 0) + vz ( 0 0 1)
 v = (vx vy vz)
 What is Standard Basis for V4?

BINARY N-TUPLE
 n bit vectors made with 0s and 1s.
 A vector space Vn is formed by 2n n-tuples using modulo-2
addition.
 Example - vector space V4 is formed by 16 binary 4-tuples.
 From (0 0 0 0) to (1 1 1 1)
 (0 0 0 0) is identity element and each vector is its own inverse.
 Scalars are 0 and 1.
 Vector addition and scalar multiplication obey required
associative, distributive and commutative laws.

VECTOR SUBSPACE
 Subset of vectors, existing within a vector space, having all the
characteristics of vector space under vector addition and scalar
multiplication.
 example – following four vectors under V4.
 (0 0 0 0), (0 1 1 1), (1 0 1 0), (1 1 0 1)
 The subset follows closure, commutative, associative and
distributive property under vector addition and scalar
multiplication .
 Has additive identity and inverse.
 Hence forms subspace of V4.
 One more subspace of V4:-
 (0 0 0 0), (0 1 1 0), (0 0 1 0), (0 1 0 0)
 Standard Basis of Vn can not be a basis for a subspace. Why?

CREATION OF VECTOR SPACE FROM MATRICES
 Given n X m matrix as-
 a1,1
a1,2
….. v1
 a2,1
a2,2
…. v2
 an,1
an,2
… vn
where ai,j
= 0 or 1.
 Let v1
, v2
, …vn
are row vectors ( or row spaces) of matrix, then
set of all linear combinations of row vectors forms vector space.
 Interchanging rows and columns also give same vector space.
 Example :- Let matrix A =
 1 0 0 1 1
 0 1 0 1 0
 0 0 1 0 1
 Find row spaces and vectors in vector spaces.
 Row spaces v1
= 1 0 0 1 1 , v2
= 0 1 0 1 0 v3
= 0 0 1 0 1

 Row spaces v1
= 1 0 0 1 1 , v2
= 0 1 0 1 0 v3
= 0 0 1 0 1
 Taking 23
linear combinations of v1
, v2
, v3
vector space created.
 0. v1
+ 0.v2
+0.v3
= 0 0 0 0 0
 0. v1
+ 0.v2
+1.v3
= 0 0 1 0 1
 0. v1
+ 1.v2
+0v3
= 0 1 0 1 0
 0. v1
+ 1.v2
+1v3
= 0 1 1 1 1
 1. v1
+ 0.v2
+0v3
= 1 0 0 1 1
 1. v1
+ 0.v2
+1v3
= 1 0 1 1 0
 1. v1
+ 1.v2
+0v3
= 1 1 0 0 1
 1. v1
+ 1.v2
+1v3
= 1 1 1 0 0


 Exercise:- Form matrix B from A by adding row 3 to row 1 and
then interchanging rows 2 and 3.
 Find row spaces and vectors in vector spaces. Show that they
are same as for A.

LINEARITY OF CODES BASED ON VECTOR SPACES
 For (n,k) code, two sets of words are defined.
 SET I : 2n
n-bit words,
 form vector space Vn,
 its elements form commutative group
 Vector addition and scalar multiplication satisfy associative,
commutative and distributive conditions.
 SET II: smaller set Ck formed by 2k
code words.
 LINEARITY requirement:-
 Code words Ck forms subspace over vector space Vn.
 Subspace over vector space is also a vector space.
 There exist at least one set of k linearly independent basis
vectors which span the vector space such that vectors in V
can be formed by linear combinations of basis vectors.

 For a linear code, rows of generator matrix G form a set of basis
vectors for the code.
 Taking all combinations of rows of G code words can be
generated.

 Example: The (6,3) code has generator matrix
1 0 0 0 1 1
0 1 0 1 0 1 = G
0 0 1 1 1 0
 Show that the codewords can be generated by taking their
linear combinations of the rows.
 Row 1 = (1 0 0 0 1 1)
 Row 2 = (0 1 0 1 0 1)
 Row 3 = (0 0 1 1 1 0)
 Row 1+ Row 2= (1 1 0 1 1 0)
 Row 1+ Row 3= (1 0 1 1 0 1)
 Row 2+ Row 3= (0 1 1 0 1 1)
 Row 1 +Row 2 +Row 3 = (1 1 1 0 0 0 )
 Along with 0 0 0 0 0 0) , 8 codewords.

CODES BASED ON VECTOR SPACES
 Example: The (6,3) code has generator matrix
g1
=1 0 0 0 1 1
g2
= 0 1 0 1 0 1 = G
g3
= 0 0 1 1 1 0
 Show that row vectors of G are linearly independent
 Check all 8 combinations of g1
,g2
,g3
to find for which
condition, following is satisfied.
 a1
g1
+ a2
g2
+ a3
g3
=0

CODES BASED ON VECTOR SPACES
 Codeword for information word d = d1d2d3d4 is expressed as
 c = dG = d1
g1
+ d2
g2
+ d3
g3
+ d4
g4
 Equivalent to adding those rows of G for which dk
= 1
 Same example – d = 1101
 Code can be obtained by multiplying matrix d with G
 Or by adding rows 1,2 and 4.
 Check.
 Example:
1 0 0 0 1 0 1
0 1 0 0 1 1 1 = G
0 0 1 0 1 1 0
0 0 0 1 0 1 1
Find code words by adding rows and check.


Basic galois field arithmatics required for error control codes

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