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Finite Fields
By: Hector L Contreras
SSGT / USMC
Outline
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Abstract Algebra review to show that Finite
Fields exist
Proof on Cyclic Difference Sets
Proof on Quadratic Residue Difference Set
What are Finite Fields?
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Finite Fields are Fields with a finite amount of
elements (i.e. the obvious). These are special types
of sets that are algebraic systems that are closed
under addition, subtraction, multiplication and
division (except for 0).
Fields >> (come from) Integral Domains
Integral Domains >> Commutative Rings
Commutative Rings >> Groups
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In fact they contain a group
Groups
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A Group G is a nonempty set together with a binary
operation (*) such that the following three properties are
satisfied:
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Associativity  (a*b)*c = a*(b*c) For all a,b,c є G
Identity  There is an element e є G such that a*e = e*a = a For all
aєG
Inverses  For each element a є G, there is an element b є G such
that a*b = b*a = e
(Note: A binary operation on G is a function that assigns each
ordered pair of elements of G an element of G (closure))
Terminology (Groups)
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Order of a Group G is the number of
elements it contains (denoted |G| ).
Order of an element g є G is the smallest
positive integer n such that gn = e (denoted
|g|).
Here gn = g * g * … * g (n times).
In a finite group, the order of each element of
the group divides the order of the group.
Properties of Groups
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For all g є G, g0 = e
For all n,m ≥ 1, g є G,
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gn = gn-1*g
gn * gm = g n + m
(gn)-1 = g-n = (g-1)n
(gm)n = gmn
Abelian Groups
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If G is a group and for all a, b є G we have
a*b = b*a (commutativity) then G is called an
Abelian Group.
In an abelian group G, for all a, b є G,
(a * b)-1 = b-1 * a-1 = a-1 * b-1
Cyclic Groups
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A group G is called cyclic if there exists an element
g є G such that G = {gn | n є Z}.
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Such an element g is called a generator of G.
Here |g| = |G|
Z4 (group of integers modulo 4) is cyclic since
Z4 = {0, 1, 2, 3} and 3 + 0 = 3, 3 + 3 = 6 ≡ 2 (mod 4),
3 + 3 + 3 = 9 ≡ 1 (mod 4), 3 + 3 + 3 + 3 = 12 ≡ 0 (mod 4)
so 3 is a generator.
An integer k in Zn is a generator of Zn if and only if
gcd(n,k) = 1.
Ring with Unity
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A Ring R is a nonempty set with two binary
operations, addition (denoted by a + b) and
multiplication (denoted ab), shuch that for all a, b, c
є R:
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R is an abelian group under addition.
a(bc) = (ab)c
(associativity)
a(b + c) = ab + ac and (b + c)a = bc + ca
A Unity in a ring is a nonzero element that is the
identity under multiplication.
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Note that whenever we speak of rings we mean for them
to have unity.
Commutative Rings
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A Commutative Ring R is ring such that for
all a, b, c є R.
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a(b + c) = ab + ac = (b + c)a (commutativity)
A Unit is a nonzero element of a
Commutative Ring with Unity that has a
multiplicative inverse.
Integral Domains
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A Zero-Divisor is a nonzero element a є R, R
is a commutative ring, such that there is a
nonzero element b є R with ab = 0.
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An Integral Domain is a commutative Ring
with unity and no zero-divisors.
Fields
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A Field is a commutative ring with unity in
which every nonzero element is a unit.
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C, R, Q are all examples of Infinite fields
Every field is an integral domain.
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In fact a finite integral domain is a field
Zm  Ring of Integers Modulo m
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For all m > 1, m є Z,
Zm = { 0, 1, 2, …, m-2, m-1}
Addition modulo m (denoted mod m)
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Multiplication modulo m (denoted mod m)
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For all x,y є Zm, (x + y) is the remainder of (x + y)
divided by m.
For all x,y є Zm, xy is the remainder of xy divided by m.
Zm is the ring of integers under addition and
multiplication modulo m.
Zp is a Field
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For every prime p, Zp, the ring of integers
modulo p, is a field
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Commutative
Unity
Finite Integral Domain
Field
Why not Zm for all m > 1, m є Z?
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Zero - Divisors
Polynomials
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Let R[x] represent the ring of polynomials
over R.
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Here you add and multiply polynomials just like
you have always done but the coefficient
arithmetic is done over R.
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If R = Z5 then 4x + 2x = x, (3x2)(3x6) = 4x8
Let p(x) є R[x].
If p(x) = anxn + an-1xn-1 + … + a1x + a0 where
an ≠ 0, then deg(p(x)) = n.
Irreducible/Reducible Polynomials
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Let D be an integral domain. A polynomial
f(x) є D[x] where deg(f(x)) ≥ 2, is said to be
Irreducible over D if we have that when,
f(x) = g(x)h(x), with g(x), h(x) є D[x], we
have deg(g(x)) = deg(f(x)) or
deg(h(x)) = deg(f(x)).
An element of D[x] that is not irreducible
over D is called Reducible over D (Duh!!!).
Irreducible/Reducible Polynomials
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Let F be a field. If deg f(x) = 2 or 3, f(x)єF[x],
then f(x) is reducible over F if and only if f(x)
has a zero in F.
Examples
x2 - x - 1 is irreducible over Q but reducible
over R:
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x2 - x - 1 = (x – (1+√5)/2)(x – (1-√5)/2)
x = (1 + √5 )/2 is a zero.
More Examples of
Irreducible/Reducible Polynomials
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2x2 + 4 is irreducible over Q and R but
Reducible over C.
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i є C, √(2i) is a zero
x2 + x + 1 is irreducible over Z2 (in fact it is
the only irreducible quadratic over Z2):
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x2 and x2 + x both have 0 as a zero
x2 + 1 = (x + 1)(x + 1) and 1 is a zero.
Example of Irreducible polynomial
with degree > 3.
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x4 + x + 1 is irreducible over Z2
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If it were reducible then it would either have
quadratic factors or linear factors.
Linear factors have zeros.
If quadratic then the factor would have to be
x2 + x + 1 (verify that not true).
Properties of a Finite Field
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It can be shown that finite fields have order
pn, where p is a prime.
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It can be shown that for each prime p and
each positive integer n, there is, up to
isomorphism, a unique finite field of order pn.
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Let GF(pn) represent a finite field of order pn..
Construction of Finite Fields
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To construct GF(pn), first find an irreducible
polynomial I of degree n, with coefficients in
Zp.
Let GF(pn) =
{an-1xn-1 + an-2xn-2 + … + a1x + a0 | ai є Zp}
(Note that here addition is done modulo Zp
while multiplication is done modulo I)
Construction of GF(pn) (cont.)
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Example GF(16) = GF(24) we want
polynomial of degree 4 with coefficients in Z2
= {ax3 + bx2 + cx + d | a,b,c,d є Z2}
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Here addition is done as in Z2[x], while multiplication is done modulo
x4 + x + 1.
More Properties of GF(pn)
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It can be shown that for each positive integer
n there exists an irreducible polynomial of
degree n over GF(p) for any p.
It can be shown that for each divisor m of n,
GF(pn) has a unique subfield of order pm.
Moreover, these are the only subfields of
GF(pn).
Example of Computation (reducing
modulos f(x) and Zp )
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GF(16) = GF(24)
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In this context x4 + x + 1 = 0 then x4 = x + 1
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So x6 = x3 + x2, x5 = x2 + x
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(x3 + x2 + x + 1)(x3 + x) = x6 + x5 + 2x4 + 3x3 + x2 + x
≡ x6 + x5 + x3 + x2 + x
≡ x2
Primitive Element
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A nonzero element a є GF(q) is called a
Primitive Element if h1, h2, …, hq-1, are
precisely all the nonzero elements of GF(q)
(i.e. the multiplicative order of a is (q-1))
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Generator of the multiplicative group of nonzero
elements
Used to simplify multiplication
It can be shown that every GF(pn) contains a
primitive element
Example of computations of Galois
Fields using a Primitive Root
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Since GF(16) = {0, 1, x, … x14} where x15 = 1,
then x is a primitive root:
 (x6)(x12) = x18 = x3
 x10 + x7 = (x2 + x + 1) + (x3 + x + 1) = x3 + x2 = x6
 (x3 + x2 + 1)(x3 + x2 + x + 1) = x13x12 = x25 = x10
= x2 + x + 1
Transition
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Now that we know that finite fields exist and
are familiar with some of the properties that
they obey, we can now use them to perform
some constructions on difference sets. Up
ahead we have:
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If q is a prime power, there exists a cyclic
(q2 + q + 1, q + 1, 1) difference set.
Let pn = 4t + 3. Then the nonzero squares in
GF[pn] form a (4t + 3, 2t + 1, t) difference set.
Finite Projective Planes
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Recall that symmetric designs with λ = 1 have
parameters v = n2 + n + 1, k = n + 1 for some
integer n ≥ 2, and are called Finite Projective
Planes (FPP).
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Let Pn = (n2 + n + 1, n + 1, 1) design (FPP).
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We know these exists if n = p where p is prime.
Will show that exists if n = pm
Difference Set
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A (v, k , λ) Difference Set in an additive
Abelian Group G of order v is a set
D = {d1, …, dk} of distinct elements of G
such that each nonzero element g of G has
exactly λ representations as g = di – dj.
If q is a prime power, there exists a cyclic
(q2 + q + 1, q + 1, 1) difference set.
Proof
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Let q = pn and let h be a Primitive Element of
GF(q3). Then
3 - 2 q3 - 1
3
0
1
q
GF(q ) = {0, h , h , …, h
,h
}. Let
G = {0, hu, h2u, …, h(q-1)u}
where u = (q3-1)/(q-1) = q2 + q + 1 (it can be
shown that G isomorphic to GF[q]). Now for
each bi є GF(q), 1 ≤ i ≤ q, bi = hui there is a
unique ai, 0 ≤ ai ≤ q3 – 1, such that h + bi = hai.
Furthermore, ai distinct (mod u).
Cont. Proof (Show that each power is
distinct)
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Aiming for a contradiction, assume that
h + b1 = ha1 and h + b2 = ha2 where
a1 ≡ a2 (mod q2 + q + 1) 
a1 = m(q2 + q + 1) + a2 for some m є Z. So
ha1 = ha2hm(q2 + q + 1) = ha2x for some x є GF[q].
This implies that h + b1 = (h + b2)x. So
h є GF[q]  since h is a primitive element
of GF[q3]. Therefore all ai are distinct(mod u).
Cont. Proof (Show that each difference
is unique)
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Now let aq+1 = 0 and consider the set
D = {a1, a2, …, aq, aq+1} (note that all elements are
distinct since for all i≠q+1 we have ai ≠ 0). Now
show that are differences in D are different. Aiming
for a contradiction, assume that ai - aj ≡ ak - al (mod
u) ≡∕ 0, ai ≡∕ ak. So we have that ai + al ≡ ak + aj (mod
u). If none of these ay equal 0 or the identity, then hai
+ al = hak + aj + ux haihal = hakhajhux 
(h
+ bi)(h + bl) = (h + bk)(h + bj)g, g є GF[q]
Cont. Proof (Cont. Show each
difference is unique)
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(h + bi)(h + bl) = (h + bk)(h + bj)g, g є GF[q]. This give
us h2(1- f) + h(bl + bi - fbj - fbk) + (bibl - fbkbj) =0. If
g = 1, then we get bi + bl = bk + bj and bibl = bkbj. This
implies that (bi – bk)(bj – bi) = 0
 bi = bk or bi = bj . If g > 1 then  since h is a
primitive element of GF[q3]. Finally, if one of these
ay = aq+1 = 0 (say it’s ai) then we get that h + bl =
(h + bk)(h + bj)g  since h is a primitive element of
GF[q3]. Therefore each difference is unique  a cyclic
(q2 + q + 1, q + 1, 1) difference set exists QED.
Quadratic Residue
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Let p be an odd prime and gcd(a, p) = 1. If
the quadratic congruence x2 ≡ a (mod p) has a
solution, then a is said to be a Quadratic
Residue (Square) of p. Otherwise a is called
a quadratic nonresidue of p (once again
DUH!!!).
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Let p = 7, then h = 3 is a primitive element
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32 ≡ 2 (mod 7), 34 = (32)2 ≡ 4 (mod 7),
36 = (33)2 ≡ 1 (mod 7). So {1, 2, 4} is the set of
nonzero squares (quadratic residues) in 7.
Properties of Squares
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Let h be a primitive element of GF[pn] where pn is
odd. It can be shown that:
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The nonzero squares in GF[pn] are the even powers of h.
-1 is a square if pn ≡ 1 (mod4) but is not a square if pn ≡ 3
(mod 4).
If pn ≡ 1 (mod 4) then x is a square iff –x is a square.
If pn ≡ 3 (mod 4) then x is a square iff –x is not a square
Let pn ≡ 4t + 3. Then the nonzero squares in GF[pn]
form a (4t + 3, 2t + 1, t) difference set.
Proof
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Since pn ≡ 3 (mod 4) then let pn = 4t + 3 for some
integer t. Let h be a primitive element of GF[pn] and
let Q denote the set of nonzero squares. So
Q = {h2, h4, …, h4t + 2} and – Q = {h, h3, …, h4t+1}.
Let S = {(x,y)| x, y є Q, x, y are distinct squares}. If
x – y ≡ a (mod p) (a є Zp) then we say the (x,y)
represents a and denote N(a) as the number of pairs
in S that represent a. Now define a mapping fa of
ordered pairs mod p by:
fa(x,y) = {(a-1x, a-1y)
if a is a square
{(-a-1y, -a-1x)
if a is not a square
Cont. Proof (rep’s of a lead to rep’s of 1)
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If a is a square then a-1x and a-1y are both squares
and then (a-1x , a-1y) represent 1 since whenever we
have (x,y) represent a, then x – y ≡ a (mod p) 
a-1x – a-1y ≡ 1 (mod p). Now if a is not a square then
(-a-1y, -a-1x) represents 1 since whenever we have
(x,y) represent a, then -y – (-x) = x – y ≡ a (mod p)
 -a-1y – -a-1x = a-1x – a-1y ≡ 1 (mod p). So we
have that every representation of a leads to a
representation of 1. So whenever (x,y) represents a
we have fa(x,y) represents 1.
Cont. Proof (rep’s of 1 lead to rep’s of a)
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fa-1 maps (a-1x , a-1y) back to (x,y), if a is a
square, and (-a-1y, -a-1x) back to (x,y), if a is
not a square. If (w,v) represents 1 then
fa-1(w,v) represents a. For if w – v ≡ 1 (mod p)
then w – v = a-1x – a-1y for some x,y є Q 
a(w – v ) = a(a-1x – a-1y) = x – y ≡ a (mod p).
So we have that every representation of 1
leads to a representation of a.
End Proof
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We now have the N(a) = N(1) for all a є Zp (i.e. the
difference a shows up the same amount of times as
the difference 1). So Q is a difference set. We know
that we have v = pn = 4t + 3. Since
Q = {h2, h4, …, h4t + 2} then it is easy to see that
k = (4t + 2)/2 = 2t + 1. From lecture we know that
λ(v – 1) = k(k – 1)  λ(4t + 2)= 4t2 + 2t 
λ = t. Therefore the nonzero squares in GF[pn] form
a (4t + 3, 2t + 1, t) difference set. Q.E.D.