The Kummer Pairing Alexander J. Barrios Purdue University 12 September 2013 Preliminaries ∗ Theorem 1 (Artin). Let 1; 2; : : : ; n be distinct group homomorphisms from a group G into K , where K is a field. Then the group homomorphisms are linearly independent over K. Theorem 2 (Hilbert's Theorem 90). Let E=F be a cyclic extension of degree n with G = Gal (E=F ) = hσi. Then (a) for β 2 E, N(β) = 1 if and only if there exist α 2 E∗ such that β = α/σ(α) (b) for β 2 E; Tr(β) = 0 if and only if there exist a 2 E such that β = α − σ(α). α Proof. (a)((=) Suppose β = σ(α) . Then N(a) N(β) = = 1 N(σ(α)) since N(α) = N(σ(α)). 2 Qn−2 j n−1 (=)) Suppose that N(β) = 1. Since IdK∗ , βσ; βσ(β) σ ;:::; j=1 σ (β) σ are all distinct group homomorphisms from K∗ to K∗, we have that n−2 2 Y j n−1 χ = IdK∗ +βσ + βσ(β) σ + ··· + σ (β) σ j=0 is a map which is not identically 0 by Artin's Theorem. So there exist a nonzero θ 2 K∗ such that α = χ(θ) 6= 0. Now consider, n−2 Y σ(α) = σ(χ(θ)) = σ(θ) + σ(β) σ2(θ) + σ(β) σ2(β) σ3 + ··· + σj+1(β) σn(θ) j=0 n−2 n−1 Y Y = σ(θ) + σ(β) σ2(θ) + σ(β) σ2(β) σ3 + ··· + σj(β) σn−1(θ) + σj(β) θ: j=1 j=1 1 PRELIMINARIES Now consider, n−2 n−1 Y Y βσ(α) = βσ(θ) + βσ(β) σ2(θ) + βσ(β) σ2(β) σ3 + ··· + β σj(β) σn−1(θ) + β σj(β) θ j=1 j=1 n−2 Y = βσ(θ) + βσ(β) σ2(θ) + βσ(β) σ2(β) σ3 + ··· + σj(β) σn−1(θ) + N(β) θ j=0 n−2 Y = θ + βσ(θ) + βσ(β) σ2(θ) + βσ(β) σ2(β) σ3 + ··· + σj(β) σn−1(θ) j=0 = α α and so β = σ(α) . (b)((=) Suppose β = α − σ(α). Then Tr(β) = Tr(α − σ(α)) = Tr(α) p − Tr(σ(α)) = 0 since Tr(α) p = Tr(σ(α)). (=)) Now suppose Tr(β) = 0. By Artin's Theorem we have that n−2 X χ = βσ + (β + σ(β)) σ2 + ··· + σj(β) σn−1 j=0 ∗ ∗ 1 is not identically zero on K . So there exist θ 2 K with Tr(θ) 6= 0 and χ(θ) 6= 0. Set α = Tr(θ) χ(θ). Then 0 n−2 1 1 X σ(α) = σ(β) σ(θ) + σ(β) + σ2(β) σ2(θ) + ··· + σj+1(β) σn(θ) σ(Tr(θ)) @ A j=0 0 n−1 1 1 X = σ(β) σ(θ) + σ(β) + σ2(β) σ2(θ) + ··· + σj(β) θ : Tr(θ) @ A j=1 So we get that 0 n−2 1 1 X α − σ(α) = βσ(θ) + (β + σ(β)) σ2(θ) + ··· + σj(β) σn−1(θ) Tr(θ) @ A j=0 0 n−1 n−1 1 1 X X − σ(β) σ(θ) + σ(β) + σ2(β) σ2(θ) + ··· + σj(β) σn−1(θ) + σj(β) θ Tr(θ) @ A j=1 j=1 0 n−1 1 1 X = σ(θ)(β − σ(β)) + σ2(θ) β − σ2(β) + ··· + σn−1(θ) β − σn−1(β) − σj(β) θ Tr(θ) @ A j=1 1 = (β Tr(θ)) Tr(θ) and so β = α − σ(α). 2 PRELIMINARIES Remark Part (a) and (b) above are called the multiplicative form of Hilbert's 90th and the additive form of Hilbert's 90th, respectively. Proposition 3. Let F be a field and n be a natural number not dividing char F = p if p > 0. Suppose ζn is a primitive nth root of unity lying in F . (a) If E=F is cyclic of degree n, then there exist α 2 E such that E = F (α) and α satisfies Xn − a = 0 for some a 2 F . (b) If α is a root of Xn − a where a 2 F , then F (α) is cyclic over F of order d, where djn and αd 2 F . Proof. (a) Let ζ be a primitive n-th root of unity in F: Let G = Gal (E=F ) = hσi, since E=F is cyclic. n Since N ζ−1 = ζ−1 = 1, we have by Hilbert's 90th Theorem that there exist α 2 E∗ such that −1 α 2 2 ζ = σ(α) () σ(α) = ζα. Since ζ 2 F , we have that σ(ζ) = ζ and so σ (α) = σ(ζ) σ(α) = ζ α and this in turn implies that for j 2 f1; : : : ; ng, we have σj(α) = σjα. In particular, each ζjα is a conjugate of α over F , and so [F (α): F ] ≥ n. Since [E : F ] = n and F (α) ⊂ K, we conclude that E = F (α). Moreover, σ(αn) = σ(α)n = (ζα)n = αn and therefore αn is fixed by σ, i.e., αn 2 F . Let a = αn, then X − a is a minimal polynomial for α over F . (b) Conversely, let a 2 F and α be a root of Xn − a. Then each αζj for j 2 f1; : : : ; ng is also a root Xn − a. Therefore all roots lie in F (α) and hence F (α) =F is Galois. Let G = Gal (F (α) =F ) . If σ 2 G, n then σ(α) is also a root of X − a. Thus σ(α) = !σα where !σ is an n-th root of unity. In particular, the map σ 7−! !σ is an injective group homomorphism of G into µn. Since µn is cyclic, we have that G must by cyclic of order d where djn. If hσi = G, then !σ is a primitive d-th root of unity and we get d d d d σ α = σ(α) = (!σα) = α ; which implies that αd 2 F , as claimed. Theorem 4 (Artin-Schreier). Let F be a field of characteristic p > 0. (a) Let E=F be cyclic of order p. Then there exists α 2 E such that E = F (α) and α satisfies Xp−X−a = 0 for some a 2 F . (b) Conversely, given a 2 F , the polynomial f(X) = Xp − X − a either has one root in F , in which case all its roots are in F , or it is irreducible. In this latter case, if α is a root then F (α) is cyclic of degree p over F . Proof. (a) Suppose that E=F is cyclic of order p and let G = Gal (E=F ) = hσi. Since Tr(−1) = p (−1), we have by Hilbert's 90th Theorem that there exist α 2 E so that σ(α) = α + 1. In particular, σ2(α) = σ(α)+1 = α+2 and in general we have σj(α) = α+j for j 2 f1; : : : pg. Therefore α has p distinct conjugates and so [F (α): F ] ≥ p. But by assumption [E : F ] = p which forces E = F (α) since F (α) ⊂ E. Note that σ(αp − α) = σ(αp) − σ(α) = σ(α)p − σ(α) = (α + 1)p − (α + 1) = αp + 1 − α − 1 = αp − α; thus αp −α is fixed by G and hence αp −α 2 F . Now take a = αp −α and therefore α satisfies Xp −X −a = 0. (b) Now let a 2 F and consider the polynomial f(X) = Xp − X − a. Suppose α is a root of f(X). Then α + j for j 2 f1; : : : ; pg are also roots of f(X) since (α + j)p − (α + j) − a = αp + jp − α − j − a = αp − α − a = 0: In particular, f(X) has p distinct roots. If some root α lies in F , it follows that every root is in F . So suppose that no root α lies in F . We claim that f(X) is an irreducible polynomial. Suppose on the contrary that it is reducible over F , then f(X) = g(X) h(X) for some g(X) ; h(X) 2 F [X] with their degrees being strictly less than p. If α is a root of f, then p−1 Y f(X) = (X − α − j) : j=0 3 ABELIAN KUMMER THEORY It follows that both f and g are products of certain distinct integers j. That is, there exist I;J ⊂ f0; : : : ; p − 1g such that I \ K = ? and I [ J = f0; : : : ; p − 1g and Y Y g(X) = (X − α − i) and h(X) = (X − α − k) : i2I k2K Pd l P Let d = deg g and write g(X) = l=0 glX . Then gd−1 = i2I − (α + i) by the theory on symmetric polynomials. Since jIj = d, we have that gd−1 = −dα + m for some m 2 Fp. Since d 6= 0 and gd−1 2 F , it follows that α 2 F , which is a contradiction and therefore f is irreducible. Moreover, each root of f then lies in F (α) and so F (α) =F is Galois. Let G = Gal (F (α) =F ). Then there exist σ 2 G such that σ(α) = α + 1. But this implies that σj(α) = α + j for each j 2 f0; : : : ; p − 1g. Since each α + j is a distinct root of f(X), we conclude that G is cyclic and it is generated by σ, as desired.