Algebraic Structures
Monoids, Groups, Rings, Fields
Algebraic Structures 2
Monoid
For a set G and an operator : G × G → G, a pair (G, ·) is a monoid iff the following properties are satisfied:IdentityThere is e ∈ G such that for all a ∈ G, a · e = a.AssociativityFor all a, b, c ∈ G, a · (b · c)=(a · b) · c.
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Algebraic Structures 32301233
Monoid
Closure Associativity Identity
Algebraic Structures 4
ExampleLet N be the set of non-negative integers.
(N, +) is a monoid because:• For any a and b in N, a + b is in N.• For any a, b and c in N, (a + b) + c = a + (b + c).• There is 0 such that for any a in N, a + 0 = a.
(N, ) is a monoid because:• For any a and b in N, a b is in N.• For any a, b and c in N, (a b) c = a (b c).• There is 1 such that for any a in N, a 1 = a.
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Algebraic Structures 5
ExampleLet N be the set of of non-negative integers.
(N, -) is not a monoid because:• There are a and b in N such that a - b is in not N.• There are a, b and c in N such that (a - b) - c a -
(b - c).
(N, ) is not a monoid because:• There are a and b in N, such that a b is in not
N.• There are a, b and c in N such that (a b) c a
(b c).2301233
Algebraic Structures 6
GroupA monoid (G, ·) is a group iff for all a ∈ G, there exists an element b ∈ G such that a · b = e.
Let I be the set of integers.(I, +) is a group because:• For any a and b in I, a + b is in I.• For any a, b and c in I, (a + b) + c = a + (b +
c).• There is 0 such that for any a in I, a + 0 = a.• For any a in I, there is a-1 = -a such that a + a-1
= 0.
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Algebraic Structures 7
GroupA monoid (G, ·) is a group iff for all a ∈ G, there exists an element b ∈ G such that a · b = e.
(I, ) is not a group because:• For any a and b in I, a b is in I.• For any a, b and c in I, (a b) c = a (b
c).• There is 1 such that for any a in I, a 1 = a.• For some a in I, there is no a-1 such that a
a-1 = 1.2301233
Algebraic Structures 82301233
Group
closure associativity identity inverse
Algebraic Structures 9
Commutative GroupA group (G, ·) is commutative or Abelian iff for all a, b ∈ G, a · b = b · a.
Let I be the set of integers.(I, +) is a commutative group because:• it is a group.• For any a and b in I, a + b = b + a.
(I, ) is not a commutative group because:• it is not a group.• For any a and b in I, a b = b a.
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Algebraic Structures 102301233
Commutative Group
closure
associative identity invers
ecommutativ
e
Algebraic Structures 11
Relationship
Monoid
group
Commutative group
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Algebraic Structures 12
RingFor a set R and binary operators · and + over R, the triple (R, +, ·) is a ring iff the following properties are satisfied:Commutative addition (R, +) is an Abelian group with identity element 0.Multiplication (R, ·) is a monoid with identity element 1.DistributivityFor all a, b, c ∈ R, a · (b + c) = a · b + a · c.
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Algebraic Structures 13
Field• A non-empty set F with two binary
operation + (addition) and (multiplication) is called a field if
• (F, +) is a commutative (additive) group, and
• (F – {0}, ) is a commutative (multiplicative) group.
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Algebraic Structures 14
Cryptography and Finite Fields• Cryptography focuses on finite fields.• For any prime integer p and any integer
n greater than or equal to 1, there is a unique field, called Galios field, with pn elements in it, denoted by GF(pn).
• “Unique” means that any two fields with the same number of elements must be essentially the same, except perhaps for giving the elements of the field different names.
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Algebraic Structures 15
Galois Fields in Cryptography
GF(p1) : ({0,1,2,…,p-1}, +, *) for integers modulo p.Example Let p = 7. Z7 = {0,1,2,3,4,5,6}.GF(7) = (Z7 , +, *).
(Z7, +) is a commutative group with identity 0, and the inverse of a is 7-a.(Z7, *) is a commutative group with identity 1, and the inverse of a is x such that ax 1 mod 7.
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Algebraic Structures 16
Galois Fields in AESGF(28) : (Z256, +, *) where Z256 = {0,1,…,255}.
Each element b=b7 b6 b5 b4 b3 b2 b1 b0in Z256
is a polynomial b7 x7 + b6x6 + b5x5 + b4x4 + b3x3 + b2x2 + b1x + b0.
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Algebraic Structures 17
AES SpecificationsInput & output block length: 128 bits.State: 128 bits, arranged in a 4-by-4 matrix of bytes.
Each byte is viewed as an element in a field.
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A0,0 A0,1 A0,2 A0,3
A1,0 A1,1 A1,2 A1,3
A2,0 A2,1 A2,2 A2,3
A3,0 A3,1 A3,2 A3,3
Algebraic Structures 18
Addition in GF(28)a7 a6 a5 a4 a3 a2 a1 a0
b7 b6 b5 b4 b3 b2 b1 b0
a7 x7 + a6x6 +…+ a1x+ a0
b7 x7 + b6x6 +…+ b1x+ b0
(a7+b7)x7+ (a6+b6)x6+ …+ (a1+b1)x+ (a0+b0)
All additions of polynomial coefficient are modulo 2. 1 + 1 =0 1 – 1 = 0 1 1 = 01 + 0 = 1 1 – 0 = 1 1 0 = 00 + 1 = 1 0 – 1 = 1 0 1 = 00 + 0 = 0 0 – 0 = 0 0 0 = 0
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Algebraic Structures 19
Multiplication in GF(28) a7 x7 + a6x6 +…+ a1x+ a0
b7 x7 + b6x6 +…+ b1x+ b0
--------------------------------------------------------------------
(a7 b0) x7 + (a6b0) x6+ …+ (a1b0) x+ (a0b0)
(a7 b1) x8 + (a6b1) x7 + (a5b1) x6+ …+ (a0b1)x
(a7 b2)x9 +(a6b2) x8 +(a5b2) x7+ (a4b2)x6 +…
…------------------------------------------------------------------------
…
(ai bj) xi+j . i=0,…,7 j=0,…,7
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Algebraic Structures 20
Multiplication in GF(28)The result can be a degree k polynomial, where k 14.Divide the result by a degree 8 polynomial .AES uses x8 + x4 + x3 + x +1.
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Algebraic Structures 21
Examplex7 + x5 + x4 + x2 + x =>
(75421) x6 + x4 + x + 1 =>
(6410)
(7 5 4 2 1) * (6 4 1 0)(7 5 4 2 1) * (6) = (13 11 10 8 7)(7 5 4 2 1) * (4) = ( 11 9 8 6 5)(7 5 4 2 1) * (1) = ( 8 6 5 3 2)(7 5 4 2 1) * (0) = + 7 5 4 2 1) ---------------------------- (13 10 9 8 5 4 3 1)
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Algebraic Structures 22
Example(x13 + x10 + x9 + x8+ x5 + x4 + x3 + x )/ (x8 + x4 + x3
+ x +1)
=> (13 10 9 8 5 4 3 1)/(8 4 3 1 0)
(13 10 9 8 5 4 3 1)(8 4 3 1 0) * (5) = (13 9 8 6 5) -------------------------
(10 6 4 3 1)(8 4 3 1 0) * (2) = (10 6 5 3 2)
-------------------------the remainder (5 4 2 1)
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