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

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

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.)

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)

Finite fields 4. Definition of Field (Kn,+,x) (Kn,+) 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 1 x a = a x 1 = a,  a  Kn The operation x is distributive over + a x ( b + c ) = ( a x b ) + ( a x c ) It is communicative a x b = b x a,  a,b  Kn All non-zero elements have multiplicative inverses Galois Fields: GF(2r) For every prime power order pm, there is a unique finite field of order pm Denoted by GF(pm)

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

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 = 1 Kn\{0}={i | i = 0,1,…,2n-2}

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

Minimal polynomials 1. Root of a polynomial 2. Order of   : 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

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 m(x) is a factor of

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).

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

Minimal polynomials If f()=0, then f(2)=(f())2=0 If  is a root of f(x), so are , 2, 4,…, The degree of m(x) is |{, 2, 4,…, }|

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

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

Minimal polynomials Table 5.2: Minimal polynomials in GF(24) constructed using 1+x+x4 Element of GF(24) Minimal polynomial 1 , 2, 4, 8 3, 6, 9, 12 5, 10 7, 11, 13, 14 x 1+x 1+x+x4 1+x+x2+x3+x4 1+x+x2 1+x3+x4