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Finite Fields
Rong-Jaye Chen
Finite fields

1. Irreducible polynomial

f(x)K[x], f(x) has no proper divisors in K[x]
Eg.
f(x)=1+x+x2 is irreducible
f(x)=1+x+x2+x3=(1+x)(1+x2) is not irreducible
f(x)=1+x+x4 is irreducible
p2.
Finite fields

2. Primitive polynomial


f(x) is irreducible of degree n > 1
f(x) is not a divisor of 1+xm for any m < 2n-1
Eg.
f(x)=1+x+x2 is not a factor of 1+xm for m < 3 so
f(x) is a primitive polynomial
f(x)= 1+x+x2+x3+x4 is irreducible but
1+x5=(1+x)(1+x+x2+x3+x4) and m=5 < 24-1=15 so
f(x) is not a primitive polynomial
p3.
Finite fields

3. Definition of Kn[x]




The set of all polynomials in K[x] having degree less
than n
Each word in Kn corresponds to a polynomial in Kn[x]
Multiplication in Kn modulo h(x), with irreducible h(x)
of degree n
If we use multiplication modulo a reducible h(x), say,
1+x4 to define multiplication of words in K4, however:
(0101)(0101)(x+x3)(x+x3)
= x2+x6
= x2+x2 (mod 1+x4)
= 0  0000 (K4-{0000} is not closed under
multiplication.)
p4.
Finite fields

Furthermore each nonzero element in Kn can have an
inverse if we use irreducible h(x). But if we use
reducible h(x) then there exists nonzero element,
which has no inverse.
Why? Let f(x) is nonzero element and h(x) is
irreducible
then gcd(f(x),h(x))=1 and so exists
a(x)f(x)+b(x)h(x)=1 =>
a(x)f(x)=1 mod h(x) and so a(x) is the inverse of f(x)
p5.
Finite fields

4. Definition of Field (Kn,+,x)
n
 (K ,+) is an abelian group with identity denoted 0
 The operation x is associative
 a x ( b x c) = ( a x b ) x c
 There is a multiplicative identity denoted 1, with 10
n
 1 x a = a x 1 = a,  a  K
 The operation x is distributive over +
 a x ( b + c ) = ( a x b ) + ( a x c )
 It is communicative
n
 a x b = b x a,  a,b  K
 All non-zero elements have multiplicative inverses
r
 Galois Fields: GF(2 )
m
 For every prime power order p , there is a unique finite
field of order pm
m
 Denoted by GF(p )
p6.
Finite fields

Example

Let us consider the construction of GF(23) using the
primitive polynomial h(x)=1+x+x3 to define
multiplication. We do this by computing xi mod h(x):
word 
xi mod h(x)
100
1
010
x
001
x2
110
x3  1+x
011
x4  x+x2
111
x5  1+x+x2
101
x6  1+x2
p7.
Finite fields

5. Use a primitive polynomial to construct GF(2n)

Let   Kn represent the word corresponding to x
mod h(x)

i  xi mod h(x)

m 1 for m<2n-1

since h(x) dose not divide 1+xm for m<2n-1

Since j = i for ji iff i = j-i i
 j-i
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Kn\{0}={i | i = 0,1,…,2n-2}
=1
p8.
Finite fields

6.   GF(2r) is primitive (or a generator)

 is primitive if m 1 for 1m<2r-1

In other words, every non-zero word in GF(2r) can be
expressed as a power of 

Eg.
Construct GF(24) using the primitive polynomial
h(x)=1+x+x4. Write every vector as a power of  
x mod h(x)(see Table 5.1)
Note the 15=1.
(0110)(1101)= 5.7= 12=1111
p9.
Minimal polynomials
 1. Root of a polynomial
  : an element of F=GF(2r), p(x)F[x]
  is a root of a polynomial p(x) iff p()=0
 2. Order of 
 The smallest positive integer m such that m=1
  in GF(2r) is a primitive element if it has order 2r-1
p10.
Minimal polynomials
 3. Minimal polynomial of 
 The polynomial in K[x] of smallest degree having 
as root
 Denoted by m(x)
 m(x) is irreducible over K
 If f(x) is any polynomial over K such that
f()=0,then m(x) is a factor of f(x)
 m(x) is unique
r
 m(x) is a factor of 1  x 2 1
p11.
Minimal polynomials
 Example
Let p(x)=1+x3+x4, and let  be the primitive element
in GF(24) constructed using h(x)=1+x+x4(see Table
5.1):
p()=1+3+4=1000+0001+1100=0101=9
 is not a root of p(x). However
p(7)=1+(7)3+(7)4=1+21+28=1+6+13
=1000+0011+1011=0000=0
7 is a root of p(x).
p12.
Minimal polynomials
 4. Finding the minimal polynomial of 
 Reduce to find a linear combination of the
vectors{1, , 2,…, r}, which sums to 0
 Any set of r+1 vectors in Kr is dependent, such a
solution exists
 Represent m(x) by mi(x) where =i
 eg. Find the m(x), =3, GF(24) constructed
using h(x)=1+x+x4
p13.
Minimal polynomials
n
n
n
i 0
i 0
i 0
(  ai x i )2   ai2 ( x i ) 2   ai ( x 2 )i
 If f()=0, then f(2)=(f())2=0
r 1
 If  is a root of f(x), so are , 2, 4,…,  2
 The degree of m(x) is |{, 2, 4,…,  2r 1 }|
p14.
Minimal polynomials
 Example
 Find the m(x), =3, GF(24) constructed using
h(x)=1+x+x4
 Let m(x)= m3(x)=a0+a1x+a2x2+a3x3+a4x4 then we
must find the value for a0,a1,…,a4 {0,1}
m()=0=a01+a1+a22+a33+a44
=a00+a13+a26+a39+a412
0000=a0(1000)+a1(0001)+a2(0011)+a3(0101)+
a4(1111)
 a0=a1=a2=a3=a4=1
and m(x)=1+x+x2+x3+x4
p15.
Minimal polynomials
 Example
 Let m5(x) be the minimal polynomials of =5,
5GF(24)
Since {, 2, 4, 8}={5 , 10}, the roots of m5(x)
are 5 and 10 which means that degree (m5(x))=2.
Thus m5(x)=a0+a1x+a2x2:
0=a0+a1 5+a2 10 =a0(1000)+a1 (0110)
+a2 (1110)
Thus a0=a1=a2=1 and m5(x)=1+x+x2
p16.
Minimal polynomials
 Table 5.2: Minimal polynomials in GF(24)
constructed using 1+x+x4
Element of GF(24)
0
1
, 2, 4, 8
3, 6, 9, 12
5, 10
7, 11, 13, 14
Minimal polynomial
x
1+x
1+x+x4
1+x+x2+x3+x4
1+x+x2
1+x3+x4
p17.