# Finite Fields. Saravanan Vijayakumaran Department of Electrical Engineering Indian Institute of Technology Bombay

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1 1 / 25 Finite Fields Saravanan Vijayakumaran Department of Electrical Engineering Indian Institute of Technology Bombay September 25, 2014

2 2 / 25 Fields Definition A set F together with two binary operations + and is a field if F is an abelian group under + whose identity is called 0 F = F \ {0 is an abelian group under whose identity is called 1 For any a, b, c F a (b + c) = a b + a c Definition A finite field is a field with a finite cardinality. Example F p = {0, 1, 2,..., p 1 with mod p addition and multiplication where p is a prime. Such fields are called prime fields.

3 Some Observations Example F 5 = {0, 1, 2, 3, = 2 mod 5, 3 5 = 3 mod 5, 4 5 = 4 mod 5 All elements of F 5 are roots of x 5 x 2 2 = 4 mod 5, 2 3 = 3 mod 5, 2 4 = 1 mod 5 F 5 = {1, 2, 3, 4 is cyclic Example F = {0, 1, y, y + 1 under + and modulo y 2 + y + 1 y 4 = y mod (y 2 + y + 1), (y + 1) 4 = y + 1 mod (y 2 + y + 1) All elements of F are roots of x 4 x (y + 1) 2 = y mod (y 2 + y + 1), (y + 1) 3 = 1 mod (y 2 + y + 1) F = {1, y, y + 1 is cyclic 3 / 25

4 4 / 25 Field Isomorphism Definition Fields F and G are isomorphic if there exists a bijection φ : F G such that for all α, β F. Example F = G = φ(α + β) = φ(α) φ(β) φ(α β) = φ(α) φ(β) {a 0 + a 1 x + a 2 x 2 a i F 2 under + and modulo x 3 + x + 1 {a 0 + a 1 x + a 2 x 2 a i F 2 under + and modulo x 3 + x 2 + 1

5 Uniqueness of a Prime Field Theorem Every field F with a prime cardinality p is isomorphic to F p Proof. Let F be any field with p elements where p is prime F has a multiplicative identity 1 Consider the additive subgroup S(1) = 1 = {1, 1 + 1,... By Lagrange s theorem, S(1) divides p Since 1 0, S(1) 2 = S(1) = p = S(1) = F Every element in F is of the form {{ + 1 i times F is a field under the operations {{ + 1 i times {{ i times {{ j times {{ j times = {{ + 1 and i+j mod p times = {{ ij mod p times 5 / 25

6 6 / 25 Proof of F being Isomorphic to F p Consider the bijection φ : F F p φ {{ + 1 = i mod p i times φ 1 + {{ {{ = φ {{ i times j times i+j times = (i + j) mod p = i mod p + j mod p φ [ ] [ ] = φ {{{{{{ i times j times ij times = ij mod p = (i mod p) (j mod p)

7 7 / 25 Subfields Definition A nonempty subset S of a field F is called a subfield of F if α + β S for all α, β S α S for all α S α β S \ {0 for all nonzero α, β S α 1 S \ {0 for all nonzero α S Example F = {0, 1, x, x + 1 under + and modulo x 2 + x + 1 F 2 is a subfield of F

8 8 / 25 Characteristic of a Field Definition Let F be a field with multiplicative identity 1. The characteristic of F is the smallest integer p such that {{ = 0 p times Examples F 2 has characteristic 2 F 5 has characteristic 5 R has characteristic 0 Theorem The characteristic of a finite field is prime

9 9 / 25 Prime Subfield of a Finite Field Theorem Every finite field has a prime subfield. Examples F 2 has prime subfield F 2 F = {0, 1, x, x + 1 under + and modulo x 2 + x + 1 has prime subfield F 2 Proof. Let F be any field with q elements F has a multiplicative identity 1 Consider the additive subgroup S(1) = 1 = {1, 1 + 1,... S(1) = p where p is the characteristic of F S(1) is a subfield of F and is isomorphic to F p

10 Order of a Finite Field Theorem Any finite field has p m elements where p is a prime and m is a positive integer. Example F = {0, 1, x, x + 1 has 2 2 elements Proof. Let F be any field with q elements and characteristic p F has a subfield isomorphic to F p F is a vector space over F p F has a finite basis v 1, v 2,..., v m Every element of F can be written as α 1 v 1 + α 2 v α m v m where α i F p 10 / 25

11 11 / 25 Polynomials over a Field Definition A nonzero polynomial over a field F is an expression f (x) = f 0 + f 1 x + f 2 x f m x m where f i F and f m 0. If f m = 1, f (x) is said to be monic. Definition The set of all polynomials over a field F is denoted by F[x] Examples F 3 = {0, 1, 2, x 2 + 2x F 3 [x] and is monic x is a monic polynomial in R[x]

12 12 / 25 Divisors of Polynomials over a Field Definition A polynomial a(x) F[x] is said to be a divisor of a polynomial b(x) F[x] if b(x) = q(x)a(x) for some q(x) F[x] Example x i 5 is a divisor of x in C[x] but not in R[x] Definition Every polynomial f (x) in F[x] has trivial divisors consisting of nonzero elements in F and αf (x) where α F \ {0 Examples In F 3 [x], x 2 + 2x has trivial divisors 1,2, x 2 + 2x, 2x 2 + x In F 5 [x], x 2 + 2x has trivial divisors 1, 2, 3, 4, x 2 + 2x, 2x 2 + 4x, 3x 2 + x, 4x 2 + 3x

13 13 / 25 Prime Polynomials Definition An irreducible polynomial is a polynomial of degree 1 or more which has only trivial divisors. Examples In F 3 [x], x 2 + 2x has non-trivial divisors x, x + 2 and is not irreducible In F 3 [x], x + 2 has only trivial divisors and is irreducible In any F[x], x + α where α F is irreducible Definition A monic irreducible polynomial is called a prime polynomial.

14 14 / 25 Constructing a Field of p m Elements Choose a prime polynomial g(x) of degree m in F p [x] Consider the set of remainders when polynomials in F p [x] are divided by g(x) R Fp,m = {r 0 + r 1 x + + r m 1 x m 1 r i F p The cardinality of R Fp,m is p m R Fp,m with addition and multiplication mod g(x) is a field Examples R F2,2 = {0, 1, x, x + 1 is a field under + and modulo x 2 + x + 1 R F2,3 = {r 0 + r 1 x + r 2 x 2 r i F 2 under + and modulo x 3 + x + 1

15 15 / 25 Factorization of Polynomials Theorem Every monic polynomial f (x) F[x] can be written as a product of prime factors k f (x) = a i (x) i=1 where each a i (x) is a prime polynomial in F[x]. The factorization is unique, up to the order of the factors. Examples In F 2 [x], x = (x + 1)(x 2 + x + 1) In C[x], x = (x + i 5)(x i 5) In R[x], x is itself a prime polynomial

16 16 / 25 Roots of Polynomials Definition If f (x) F[x] has a degree 1 factor x α for some α F, then α is called a root of f (x) Examples In F 2 [x], x has 1 as a root In C[x], x has two roots ±i 5 In R[x], x has no roots Theorem In any field F, a monic polynomial f (x) F[x] of degree m can have at most m roots in F. If it does have m roots {α 1, α 2,..., α m, then the unique factorization of f (x) is f (x) = (x α 1 )(x α 2 ) (x α m ).

17 17 / 25 Multiplicative Cyclic Subgroups in a Field Theorem In any field F, the multiplicative group F of nonzero elements has at most one cyclic subgroup of any given order n. If such a subgroup exists, then its elements { 1, β, β 2,..., β n 1 satisfy x n 1 = (x 1)(x β)(x β 2 ) (x β n 1 ). Examples In R, cyclic subgroups of order 1 and 2 exist. In C, cyclic subgroups exist for every order n.

18 18 / 25 Multiplicative Cyclic Subgroups in a Field Proof of Theorem. Let S be a cyclic subgroup of F having order n. Then S = { β, β 2,..., β n 1, β n = 1 for some β S. For every α S, α n = 1 = α is a root of x n 1 = 0. Since x n 1 has at most n roots in F, S is unique. Since β i is a root, x β i is a factor of x n 1 for i = 1,..., n By the uniqueness of factorization, we have x n 1 = (x 1)(x β)(x β 2 ) (x β n 1 ).

19 Factoring x q x over a Field F q Let F q be a finite field of order q For any β F q, let S(β) = {β, β 2,..., β n = 1 be the cyclic subgroup of F q generated by β The cardinality S(β) is called the multiplicative order of β and β S(β) = 1 By Lagrange s theorem, S(β) divides F q = q 1 So for any β F q, β q 1 = 1 Theorem In a finite field F q with q elements, the nonzero elements of F q are the q 1 distinct roots of x q 1 1 x q 1 1 = (x β). β F q The elements of F q are the q distinct roots of x q x, i.e. x q x = x F q (x β) 19 / 25

20 Example F 5 = {0, 1, 2, 3, 4 Factoring x q x over a Field F q (x 1)(x 2)(x 3)(x 4) = x 4 10x x 2 50x + 24 = x 4 1 x(x 1)(x 2)(x 3)(x 4) = x 5 x Example F = {0, 1, y, y + 1 F 2 [y] under + and modulo y 2 + y + 1 (x 1)(x y)(x y 1) = x 3 x 2 (y y + 1) + x(y + y y 2 + y) y 2 y = x 3 1 x(x 1)(x y)(x y 1) = x 4 x 20 / 25

21 21 / 25 F q is Cyclic A primitive element of F q is an element α with S(α) = q 1 If α is a primitive element, then {1, α, α 2,..., α q 2 = F q To show that F q is cyclic, it is enough to show that a primitive element exists By Lagrange s theorem, the multiplicative order S(β) of every β F q divides q 1 The size d of a cyclic subgroup of F q divides q 1 The number of elements having order d in a cyclic subgroup of size d is φ(d) In F q, there is at most one cyclic group of each size d All elements in F q having same multiplicative order d have to belong to the same subgroup of order d

22 22 / 25 F q is Cyclic The number of elements in Fq having order less than q 1 is at most φ(d) d:d (q 1),d q 1 The Euler numbers satisfy q 1 = φ(d) d:d (q 1) so we have q 1 φ(d) = φ(q 1) d:d (q 1),d q 1 F q has at least φ(q 1) elements of order q 1 Since φ(q 1) 1, F q is cyclic

23 23 / 25 Summary of Results Every finite field has a prime subfield isomorphic to F p Any finite field has p m elements where p is a prime and m is a positive integer. Given an irreducible polynomial g(x) of degree m in F p [x], the set of remainders R Fp,m is a field under + and modulo g(x) The nonzero elements of a finite field F q are the q 1 distinct roots of x q 1 1 The elements of F q are the q distinct roots of x q x F q is cyclic

24 24 / 25 Some More Results Every finite field F q having characteristic p is isomorphic to a polynomial remainder field F g(x) where g(x) is an irreducible polynomial in F p [x] of degree m All finite fields of same size are isomorphic Finite fields with p m elements exist for every prime p and integer m 1

25 Questions? Takeaways? 25 / 25

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