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FINITE FIELDS BRIAN OSSERMAN Although the result statements are largely the same as in Artin (§15.7), the proofs are quite different, and follow a different order. Proposition. Let F be a field. (i) If F has characteristic 0, then there is a unique field inclusion of Q into F . (ii) If F has characteristic p > 0, then there is a unique field inclusion of Fp into F . Proof. Both of these are determined by sending 0 to 0 and 1 to 1, and letting additivity and multiplicativity determine the rest. Now, if a field F is finite, it can’t contain Q, so it immediately follows that F has characteristic p for some prime p, and that furthermore, F is an extension of Fp in a unique way. If [F : Fp ] = r, then a basis of F has r elements, so we see that F has pr elements. The main classification theorem for finite fields is the following: Theorem. (i) Every finite field F has characteristic p for some prime p, and then it has pr elements for some positive integer r. (ii) If a finite field F has q = pr elements, then F is a splitting field for the polynomial xq − x over Fp . In particular, any two fields with q elements are isomorphic to one another. (iii) For every prime p and positive integer r, there is a finite field with pr elements. Proof. We have already proved (i). For (ii), the multiplicative group F × has order q − 1, so for all nonzero α ∈ F , we have αq−1 = 1, i.e. α is a root of xq−1 − 1. Of course, 0 is a root of xq − x, so we conclude that every element of F is a root of xq − x. But since F has q elements, we must have the factorization Y xq − x = (x − α), α∈F so F is a splitting field of xq − x. It then follows from our theorem on splitting fields that any two fields of order q are isomorphic. (iii) Let L be a field in which xq − x splits. We observe that the derivative of xq − x is qxq−1 − 1 = −1, since q = pr and we are in characteristic p. Thus, by our proposition on multiple roots of polynomials, we have that xq − x cannot have multiple roots in any extension field of Fp . In particular, xq − x does not have multiple roots in L, so L contains exactly q roots of xq − x. If we let F be the set of roots of xq − x in L, we will show that F is a subfield of L, so that F is the desired field with q elements. Note that F can also be described as the set of α ∈ L such that αq = α. It is clear that F contains 0 and 1, and also that it is closed under multiplication, and also under division. We next check that F is closed under addition. By Exercise 11.3.8 of Artin, we have the Frobenius homomorphism F → F sending α to αp . Composing the Frobenius homomorphism with itself r r times, we find that the map α 7→ αp is also a homomorphism. This means that (α + β)q = αq + β q , so if α, β ∈ F , so is α + β. Finally, we see that in characteristic p, we have −α = (p − 1)α, so it follows from F being closed under addition that it is also closed under taking additive inverses. Thus, F is a subfield, as desired. The theorem justifies the following notation: 1 Notation. Given q ∈ Z a prime power, let Fq be the finite field with q elements. We next consider which finite fields are contained in one another. Note that if Fpr is an extension of Fps , then by the theorem, there is no ambiguity about how it is realized as an extension: Fps s must be the subfield consisting of the roots of xp − x. The basic result is then the following: Proposition. Fix a prime p, and r ∈ Z>0 . (i) Given also s ∈ Z>0 , the field Fpr can be realized as an extension of Fps if and only if s|r. (ii) Given also s ∈ Z>0 , with s|r, and α ∈ Fpr , then α ∈ Fps if and only if the degree of α over Fp divides s. r (iii) The irreducible factors in Fp [x] of the polynomial xp − x consist precisely of the irreducible polynomials in Fp [x] having degree dividing r. Proof. (i) if Fpr can be realized as an extension of Fps , then it is a finite-dimensional vector space over Fps , so if the dimension is equal to d, we have pr = (ps )d = psd , so s divides r. Conversely, If s r r = sd for some d ∈ Z, and if αp = α, then we note that αp is obtained by raising α to the ps r s power d times, so we conclude that αp = α. Phrased differently, any root of xp − x is a root of r s xp − x also, so if we let K/Fpr be a splitting field of xp − x, we see that all the roots already lie in r Fpr , since Fpr is a splitting field for xp − x. We conclude that Fps can be realized as the subfield s of Fpr consisting of roots of xp − x. (ii) First note that α ∈ Fps if and only if Fp (α) ⊆ Fps . If the degree of α over Fp is d, then Fp (α) = Fpd , so the desired statement then follows from (i). r (iii) Fpr consists precisely of roots of xp − x, so if f (x) is an irreducible factor, then Fpr contains some root α of f (x). Then by (ii) with s = r, the degree of α, which is by definition the degree of f (x), must divide r. Conversely, if f (x) is an irreducible polynomial of degree d|r, and we let F = Fp (α) where α is a root of f (x), then F ∼ = Fpd , so by (i) we can realize F as a subfield of Fpr , r and thus f (x) has a root in Fpr . But since all the elements of Fpr are roots of xp − x, this means r that f (x) and xp − x have a nonconstant common factor. Since f (x) is irreducible, this implies r that f (x) divides xp − x. We next consider the structure of the multiplicative group of a finite field. We need the following. Lemma. Let G be a finite abelian group, and let m be the maximal order among all elements of G. Then every element of G has order dividing m. Note that this lemma is false if G is not abelian – for instance, it fails already for S3 . Proof. Suppose that we have g, g 0 ∈ G with order d, d0 respectively. We first show that if d and d0 0 are relatively prime, then ord(gg 0 ) = dd0 . Certainly, (gg 0 )dd = 1. On the other hand, if (gg 0 )n = 1, then (g 0 )n = (g n )−1 , so both (g 0 )n and g n have the same order. But the order of (g 0 )n divides d0 and the order of g n divides d, so by relative primality we conclude that both (g 0 )n and g n have order 1, i.e. (g 0 )n = g n = 1. But then d and d0 both divide n, and again using relative primality we conclude that n is a multiple of dd0 . Thus, gg 0 has order dd0 , as claimed. Now, suppose that d, d0 are not necessarily relatively prime. We claim there exists g 00 ∈ G having order equal to the least common multiple of d and d0 . To see this, choose e, e0 so that e|d, e0 |d0 , e and e0 have no common factors, and the least common multiple of d and d0 is equal to ee0 . (Exercise: 0 0 check this is always possible) Then g d/e has order e, and g 0d /e has order e0 , so by the relatively 0 0 prime case above, we conclude that g 00 = g d/e g 0d /e has order equal to ee0 , as desired. Finally, given any h ∈ G, let n be the order of h. If n doesn’t divide m, then the least common multiple of m and n is strictly greater than m, and by the above, there would be an element of this order, contradicting the hypothesis that m is the maximal order of an element in G. Thus, n divides m, as desired. 2 Corollary. Let F be a field, and G a finite subgroup of F × . Then G is cyclic. In particular, if F is a finite field, then F × is cyclic. Proof. Let m be the maximal order among all elements of G. This is certainly at most |G|. However, according to the previous lemma, every element of G has order dividing m, so it follows that if we have α ∈ G, then αm = 1. Thus, every element of G is a root of the polynomial xm − 1, so we conclude that |G| 6 m, and hence that |G| = m, and G is cyclic. 3