Algebra 2. Rings and fields. Finite fields. A.M. Cohen, H. Cuypers, H. Sterk. Algebra Interactive

2 Rings and fields A.M. Cohen, H. Cuypers, H. Sterk A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 1 / 20

For p a prime number and f an irreducible polynomial of degree n in (Z/pZ)[X ], the quotient ring (Z/pZ)[X ]/(f ) is a field with p n elements. We will see that any field is essentially of this form. Let F be a finite field of order q. By a previous result [], we know that q=p a, the power of a prime number p. We need another (more general) version of Fermat s little theorem. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 2 / 20

Theorem Fermat s little theorem Each x F satisfies the equation x q =x. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 3 / 20

Example Fields of order 9 Rings and fields Each element of a field of order 9 is a zero of the polynomial The polynomial X 9 X (Z/3Z)[X ]. The elements 0, 1, and 2 of Z/3Z are zeros of this polynomial and correspond to the linear factors X, X 1, X 2. Dividing out these factors, we find a polynomial of degree 6 that factors into a product of three quadratic polynomials as follows. (X 2 + X + 2) (X 2 + 2 X + 2) (X 2 + 1). Each of these factors can be used to define a field of order 9. In the next theorem [] we shall see that they all lead to the same field up to isomorphism. That means that the fields (Z/3Z)[X ]/ ( X 2 + X + 2 ), (Z/3Z)[X ]/ ( X 2 + 2 X + 2 ) (Z/3Z)[X ], and (Z/3Z)[X ]/ ( X 2 + 1 ) (Z/3Z)[X ] are isomorphic to each other. On the other hand, Fermat s little theorem says A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 4 / 20

Example In a previous proposition, we saw that, for any power r=p b of p, the subset {x F x r =x } is a subfield of F. Apparently, for r=q, the subfield coincides with F ; the subfield only depends on the value of rem(b, a), where q=p a. Note that x q 1 =1 for nonzero x in F. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 5 / 20

Here are some more properties of finite fields. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 6 / 20

Lemma Let F be a finite field of order q. 1 X q X = x F (X x). 2 For every prime power r=p b with b a, the subset {x F x r =x } is a subfield of F of order r. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 7 / 20

Example Constructing a field of order 16 The polynomial f =X 4 + X + 1 in (Z/2Z)[X ] is irreducible. (Verify!) Put K =(Z/pZ)[X ]/(f ) and write x=x + f. We shall prove that K is a field by showing that x is invertible and establishing that, as a set, K={(), X 1, X 2,..., X 15 }. This will suffice as it implies that K has 15 invertible elements (powers of x being invertible if x is). The element x is invertible as x 3 +1 is its inverse. This observation is immediate from a rewrite of f (x)=0 to x=x 3 1. In order to establish that all powers of x up to x 15 are distinct, notice that both x 5 =x 2 + x and x 3 are distinct from 1 and that x 15 =1. The last equation implies that x has order a divisor of 15 and the two previous equations imply that the order is not a divisor of 5 or 3. Therefore, the order of x is exactly 15, which settles that the subgroup of the multiplicative group of K generated by x has order 15. In particular, A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 8 / 20

We use Fermat s little theorem [] to determine the structure of finite fields. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 9 / 20

Theorem Characterization of finite fields Let F be a finite field of order q=p a and let f (Z/pZ)[X ] be an irreducible polynomial of degree a. 1 The field F is isomorphic to (Z/pZ)[X ]/(f ). 2 The polynomial f divides X q X. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 10 / 20

Later, we shall see that, for every prime power q, there exists a field of order q. The two assertions imply that if f is an irreducible polynomial in (Z/pZ)[X ] of degree a, it factors into linear terms in F [X ]. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 11 / 20

Example A field of order 16 Rings and fields The polynomial f =X 4 + X + 1 in (Z/2Z)[X ] is irreducible. (Verify!) Put K=(Z/pZ)[X ]/f (Z/pZ)[X ]. and x=x + f. This is the field of order 16 that we shall work with. The elements of K can be arranged according to the irreducible divisors of X 16 X of which they are a zero. element(s) zero(s) of 0 X 1 1 + X x, x 2, x 4, x 8 X 4 + X + 1 x 3, x 6, x 12, x 9 X 4 + X 3 + X 2 + X + 1 x 7, x 14, x 13, x 11 X 4 + X 2 + 1 The multiplicative group of K is cyclic of order 15, with generator x. The elements of order 3 belong to the subfield {0, 1, x 5, x 10 } of order 4, isomorphic to (Z/2Z)[X ]/ ( X 2 + X + 1 ) (Z/2Z)[X ]. The elements of order 5 can be recognized by their exponents (having gcd with 15 equal to 3), but A.M. Cohen, H. Cuypers, alsoh. Sterk by the corresponding polynomial, 2 which di- September 25, 2006 12 / 20

Remark Implicit in Part 1 is the fact that any two irreducible polynomials in (Z/pZ)[X ] of the same degree, say f and g, lead to isomorphic finite fields. The theorem does not give any information on how to construct the isomorphism. A way to proceed is to look for a zero y of g in (Z/pZ)[X ]/f (Z/pZ)[X ], and to construct the isomorphism as the map (Z/pZ)[X ] (Z/pZ)[X ]/f (Z/pZ)[X ] sending X + g to y. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 13 / 20

We use this observation to prove the following result, announced before. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 14 / 20

Theorem The multiplicative group of a finite field of order q is cyclic of order q 1. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 15 / 20

By the theorem, there are always primitive elements in finite fields. If g is a primitive element of the finite field F, then the elements can be easily enumerated by their exponents with respect to g : F ={0} { g i i {0,..., q 2} }. When written in this form, multiplication on the nonzero elements of F is given by modular arithmetic, with modulus q 1. This is very efficient, but addition is less convenient. Thus, we have the opposite to the usual form, where addition is a minor effort, but multiplication is harder. In terminology introduced before, the theorem says that any field has a primitive element. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 16 / 20

Example Suppose that K is a field of order 32. Then K is a group of order 31. Each element distinct from 1 in K has order 31, as its order is a divisor of 31 and distinct from 1, see a previous theorem []. Consider the polynomial f =X 31 1. In (Z/2Z)[X ], the polynomial f factors into f =(1 + X ) (1 + X 2 + X 5 ) (1 + X 3 + X 5 ) (1 + X + X 2 + X 3 + X 5 ) (1 Let a be an element of K which is a zero of 1 + X + X 2 + X 3 + X 5. Then an elementary calculation shows that a 2 is also a zero of this polynomial. In fact, 1 + X 2 + X 4 + X 6 + X 10 =rem(0, 1 + X + X 2 + X 3 + X 5 ). The five zeros of the polynomial are therefore a, a 2, a 4, a 8, a 16. This result could also have been derived by applying a previous result [] with x x 2. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 17 / 20

The following theorem is the principal result on finite fields. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 18 / 20

Theorem Classification theorem of finite fields For every prime number p and positive integer a there exists a field of order p a. It is unique up to isomorphism. A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 19 / 20

Example To construct a field of order 81=3 4, we look for an irreducible polynomial f of degree 4 in (Z/3Z)[X ]. According to the theory, f is a divisor of the polynomial X 81 X. We first divide out the roots belonging to the subfield of order 9: X 81 X X 9 X =X 72 + X 64 + X 56 + X 48 + X 40 + X 32 + X 24 + X 16 + X 8 + 1. This polynomial will factor into 18 irreducible polynomials of degree 4. We find one by trial and error: Creating a degree 4 polynomial and checking that it is relatively prime with X 9 X. The 18 choices for f that may arise are: X 4 X 2 1 X 4 + X 2 X + 1 X 4 X 3 + X 2 + 1 X 4 + X 3 X + 1 X 4 + X 3 + X 2 X 1 X 4 + X 2 1 X 4 X 3 1 X 4 + X 1 X 4 + X 3 1 X 4 X 3 + X + 1 X 4 X 3 + X 2 + X 1 X 4 + X 2 + X + 1 X 4 X 3 X 2 + X 1 X 4 X 3 + X 2 X + 1 X 4 + X 3 X 2 X 1 X 4 + X 3 + X 2 + X + 1 X 4 X 1 X 4 + X 3 + X 2 + 1 A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 20 / 20