Section 33 Finite fields

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1 Section 33 Finite fields Instructor: Yifan Yang Spring 2007

2 Review Corollary (23.6) Let G be a finite subgroup of the multiplicative group of nonzero elements in a field F, then G is cyclic. Theorem (27.19) A field is either of prime characteristic p containing a subfield isomorphic to Z p, or of characteristic 0 containing a subfield isomorphic to Q. In essence, if a field is of characteristic p, then it is an extension field of Z p. If a field is of characteristic 0, then it is an extension field of Q.

3 Review Theorem If E is an extension field of a field F, then E is a vector space over F. Definition If the dimension of E over F is finite, then E is a finite extension of F, and the dimension of E over F is called the degree of E over F and denoted by [E : F ]. Theorem Every finite-dimensional vector space over a field contains a basis.

4 Finite fields Theorem (33.1) Let E be a finite extension of degree n over a finite field F. If F has q elements, then E has q n elements. Proof. The vector space E contains a basis {α 1,..., α n } and each element β E is uniquely expressed as b 1 α b n α n. Since each b i has q possible different choices, E has q n elements. Corollary (33.2) If E is a finite field of characteristic p, then the number of elements in E is p n for some positive integer n.

5 Explicit construction of finite fields Idea. Theorem 33.1 asserts that if E is a finite extension of degree n over Z p, then E has p n elements. Thus, to construct a field of p n elements, we look for an irreducible polynomial f (x) over Z p of degree n. Let α be a zero of f (x). Then Z [ α] is a field of p n elements. (See the slides for Section 29.)

6 Example Problem. Construct a field of 16 elements. Solution. We look for an irreducible polynomial of degree 4 over Z 2. Such a polynomial takes the form x 4 + a 1 x 3 + a 2 x 2 + a 3 x + a 4, where a i Z 2. 0 cannot be a zero, so a 4 = 1. 1 cannot be a zero, so a 1 + a 2 + a 3 + a 4 = 0 (in Z 2 ). Thus, there are only 4 possibilities remained x 4 + x 3 + 1, x 4 + x 2 + 1, x 4 + x + 1, and x 4 + x 3 + x 2 + x + 1. Among them, we have x 4 + x = (x 2 + x + 1) 2. The others are irreducible. Pick any of x 4 + x + 1, x 4 + x 3 + 1, and x 4 + x 3 + x 2 + x + 1. Let α be a zero of the polynomial. Then Z 2 [α] is a field of 16 elements.

7 Example, continued Problem. Assume that we pick x 4 + x + 1 Z 2 [x], and let α be a zero of x 4 + x + 1. Find an element of Z 2 [α] that generates the cyclic group Z 2 [α]. Solution. Note that the cyclic group of nonzero elements in Z 2 [α] has order 15. Thus, if an element β satisfies β, β 3, β 5 1, then β has order 15. Let α be a zero of x 4 + x + 1 Z 2 [x]. We have α, α 3 1. Also, α 5 = α(α 4 + α + 1) + α 2 + α = α 2 + α 1. Thus, α generates the cyclic group of nonzero elements in Z 2 [α].

8 In-class exercises Find an irreducible polynomial of the given degree over the given field. 1. degree 3 over Z degree 3 over Z 5.

9 Structure of finite fields Theorem (33.3) Let E be a field of p n elements contained in an algebraic closure Z p of Z p. Then the elements of E are precisely the zeros of x pn x in Z p [x]. Proof. By Lagrange s theorem, every nonzero element of E is a zero of x pn 1 1. Thus, every element of E is a zero of x pn x = x(x pn 1 1). On the other hand, since Z p is a field, a polynomial of degree p n has at most p n zeros. Therefore, the elements of E are precisely the zeros of x pn x.

10 Structure of finite fields Corollary (23.6) The multiplicative group F of nonzero elements in a finite field F is cyclic. Definition An element α in a field F is an nth root of unity if α n = 1. It is a primitive nth root of unity if α m 1 for 0 < m < n. Corollary (33.6) A finite extension E of a finite field F is a simple extension of F. Proof. Let α be a generator of the multiplicative group E. Then clearly, E = F(α).

11 Existence of finite fields of p n elements Theorem (33.10) A finite field of p n elements exists for every prime power p n. Main idea of proof. In Theorem 33.3, we have shown that if E is a field of p k elements contained in Z p, then the elements of E are precisely the zeros of x pk x in Z p. Here we will show the converse is true. That is, the set S of zeros of x pn x in Z p forms a field of p n elements. Note that since Z p is algebraically closed, x pn x factors completely into linear factors over Z p. We will show The zeros of x pn x are all distinct so that S has p n elements. S is a subdomain in Z p. Since S has finitely many elements, this implies that S is a field.

12 Derivation Definition Let F be a field. The map D : F[x] F[x] defined by D(a 0 + a 1 x + + a n x n ) = a 1 + 2a 2 x + + na n x n 1 is called the derivation, and D(f (x)) is called the derivative of f (x). Lemma Let F be a field. The derivation D on F[x] satisfies D(f + g) = D(f ) + D(g). D(fg) = gd(f ) + fd(g). Proof. By direct verification.

13 Proof of S = p n Suppose that α is a repeated zero of x pn x, say, x pn x = (x α) 2 g(x) for some g(x) Z p [x]. We have D((x α) 2 g(x)) = 2(x α)g(x) + (x α) 2 D(g(x)). Thus, α is also a zero of D((x α) 2 g(x)) = D(x pn x) = p n x pn 1 1. Since the characteristic is p, the last polynomial is just 1, which has no zeros at all. Therefore, x pn x has no repeated root.

14 Proof that S is an integral domain We need to show that 0, 1 S. If α S, then so is α. If α, β S, then so are α + β and αβ. 0 and 1 clearly satisfy x pn x = 0. Assume that α pn α = 0. If p = 2, then α = α S. If p is odd, then ( α) pn ( α) = (α pn α) = 0, and α S. If α and β are zeros of x pn x, then (αβ) pn = α pn β pn = αβ and αβ is in S. Recall that p ( p k) for 1 k p 1. Thus (u + v) p = u p + v p for all u, v Z p. Then, (α + β) pn = ((α + β) p ) pn 1 = (α p + β p ) pn 1 = = α pn + β pn = α + β. Therefore, α + β S.

15 Galois fields GF (p n ) Corollary (33.11) Let F be a field. Then for every positive integer n, there exists an irreducible polynomial of degree n over F. Theorem (33.12) Let F and F be two fields of order p n. Then F and F are isomorphic. Definition The unique field of p n (up to isomorphism) is called the Galois field of order p n.

16 Proof of Theorem Assume that F and F are fields of p n elements. Since F and F are both algebraic extensions of Z p, F and F are isomorphic to some subfields of order p n in Z p. However, in Z p, there is only one subfield of order p n, namely, the set of all zeros of x pn x in Z p. In other words, F and F are both isomorphic to this subfield of p n elements in Z p.

17 Example Let α be a zero of x 3 + x + 1 in Z 2 [x]. Since x 3 + x + 1 is irreducible over Z 2, Z 2 [α] is a field of 8 elements. Theorem 33.3 says that α is a zero of x 8 x. This implies that x 3 + x + 1 is a factor of x 8 x. Indeed, we have x 8 x = x(x 1)(x 3 + x 2 + 1)(x 3 + x + 1). Note that x 3 + x is also irreducible over Z 2, giving rise to the field Z 2 [x]/ x 3 + x of 8 elements.

18 Example Question. Observe that over Z 2 we have x 16 x = x(x 1)(x 2 +x +1)(x 4 +x +1)(x 4 +x 3 +1)(x 4 +x 3 +x 2 +x +1), where the factors are all irreducible. Why isn t there any polynomial of degree 3 in the factorization? Answer. Suppose that an irreducible factor f (x) of x 16 x has degree 3. Let α be a zero of f (x). Then [Z 2 [α] : Z 2 ] = deg f = 3. However, by Theorem 31.4, we must have 4 = [GF(16) : Z 2 ] = [GF(16) : Z 2 [α]][z 2 [α] : Z 2 ] = 3[GF(16) : Z 2 [α]], which is absurd.

19 Galois fields Theorem In Z p, GF (p m ) is contained in GF(p n ) if and only if m n. Proof. The argument in the above example shows that GF (p m ) GF(p n ) implies m n. Conversely, assume that m n. It suffices to prove that if α Z p is a zero of x pm x, then it is also a zero of x pn x. Now α pn = (α pm ) pn m = α pn m = α pn 2m =. Since m n, eventually we arrive at α pn = α. Corollary The polynomial x pn x is equal to the product of all monic irreducible polynomial over Z p with degree d dividing n.

20 Example Over Z 2, we have x 64 x = x(x + 1)(x 2 + x + 1)(x 3 + x + 1)(x 3 + x 2 + 1) (x 6 + x + 1)(x 6 + x 3 + 1)(x 6 + x 5 + 1) (x 6 + x 4 + x 2 + x + 1)(x 6 + x 4 + x 3 + x + 1) (x 6 + x 5 + x 2 + x + 1)(x 6 + x 5 + x 3 + x 2 + 1) (x 6 + x 5 + x 4 + x + 1)(x 6 + x 5 + x 4 + x 2 + 1).

21 Homework Problems 4, 6, 9, 10, 13, 14 of Section 33.

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