The Kummer Pairing. Alexander J. Barrios Purdue University. 12 September 2013

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1 The Kummer Pairing Alexander J. Barrios Purdue University 12 September 2013 Preliminaries Theorem 1 (Artin. Let ψ 1, ψ 2,..., ψ 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 = σ. Then (a for β E, N(β = 1 if and only if there exist E such that β = /σ( (b for β E, Tr(β = 0 if and only if there exist a E such that β = σ(. Proof. (a ( = Suppose β = σ(. Then N(β = N(a N(σ( = 1 since N( = N(σ(. (= Suppose that N(β = 1. Since Id K, βσ, βσ(β σ 2,..., σj (β σ n 1 are all distinct group homomorphisms from K to K, we have that χ = Id K +βσ + βσ(β σ σ j (β σ n 1 is a map which is not identically 0 by Artin s Theorem. So there exist a nonzero θ K such that = χ(θ 0. Now consider, σ( = σ(χ(θ = σ(θ + σ(β σ 2 (θ + σ(β σ 2 (β σ σ j+1 (β σ n (θ n 1 = σ(θ + σ(β σ 2 (θ + σ(β σ 2 (β σ σ j (β σ n 1 (θ + σ j (β θ. 1

2 PRELIMINARIES Now consider, n 1 βσ( = βσ(θ + βσ(β σ 2 (θ + βσ(β σ 2 (β σ β σ j (β σ n 1 (θ + β σ j (β θ σ(. = βσ(θ + βσ(β σ 2 (θ + βσ(β σ 2 (β σ σ j (β σ n 1 (θ + N(β θ = θ + βσ(θ + βσ(β σ 2 (θ + βσ(β σ 2 (β σ σ j (β σ n 1 (θ = and so β = (b ( = Suppose β = σ(. Then Tr(β = Tr( σ( = Tr( p Tr(σ( = 0 since Tr( p = Tr(σ(. (= Now suppose Tr(β = 0. By Artin s Theorem we have that χ = βσ + (β + σ(β σ σ j (β σ n 1 is not identically zero on K. So there exist θ K with 0 and χ(θ 0. Set = 1 χ(θ. Then 1 σ( = σ(β σ(θ + ( σ(β + σ 2 (β σ 2 (θ + + σ j+1 (β σ n (θ σ( = 1 σ(β σ(θ + ( σ(β + σ 2 (β n 1 σ 2 (θ + + σ j (β θ. So we get that σ( = 1 βσ(θ + (β + σ(β σ 2 (θ + + σ j (β σ n 1 (θ 1 σ(β σ(θ + ( σ(β + σ 2 (β n 1 n 1 σ 2 (θ + + σ j (β σ n 1 (θ + σ j (β θ = 1 σ(θ (β σ(β + σ 2 (θ ( β σ 2 (β + + σ n 1 (θ ( β σ n 1 (β n 1 σ j (β θ = 1 (β and so β = σ(. 2

3 PRELIMINARIES Remark Part (a and (b above are called the multiplicative form of Hilbert s 90 th and the additive form of Hilbert s 90 th, respectively. Proposition 3. Let F be a field and n be a natural number not dividing char F = p if p > 0. Suppose ζ n is a primitive n th root of unity lying in F. (a If E/F is cyclic of degree n, then there exist E such that E = F ( and satisfies X n a = 0 for some a F. (b If is a root of X n a where a F, then F ( is cyclic over F of order d, where d n and d F. Proof. (a Let ζ be a primitive n-th root of unity in F. Let G = Gal (E/F = σ, since E/F is cyclic. Since N ( ζ 1 = ( ζ 1 n = 1, we have by Hilbert s 90 th Theorem that there exist E such that ζ 1 = σ( σ( = ζ. Since ζ F, we have that σ(ζ = ζ and so σ2 ( = σ(ζ σ( = ζ 2 and this in turn implies that for j {1,..., n}, we have σ j ( = σ j. In particular, each ζ j is a conjugate of over F, and so [F ( : F ] n. Since [E : F ] = n and F ( K, we conclude that E = F (. Moreover, σ( n = σ( n = (ζ n = n and therefore n is fixed by σ, i.e., n F. Let a = n, then X a is a minimal polynomial for over F. (b Conversely, let a F and be a root of X n a. Then each ζ j for j {1,..., n} is also a root X n a. Therefore all roots lie in F ( and hence F ( /F is Galois. Let G = Gal (F ( /F. If σ G, then σ( is also a root of X n a. Thus σ( = ω σ where ω σ is an n-th root of unity. In particular, the map σ ω σ is an injective group homomorphism of G into µ n. Since µ n is cyclic, we have that G must by cyclic of order d where d n. If σ = G, then ω σ is a primitive d-th root of unity and we get which implies that d F, as claimed. σ ( d = σ( d = (ω σ d = d, 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 E such that E = F ( and satisfies X p X a = 0 for some a F. (b Conversely, given a F, the polynomial f(x = X p 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 is a root then F ( is cyclic of degree p over F. Proof. (a Suppose that E/F is cyclic of order p and let G = Gal (E/F = σ. Since Tr( 1 = p ( 1, we have by Hilbert s 90 th Theorem that there exist E so that σ( = + 1. In particular, σ 2 ( = σ(+1 = +2 and in general we have σ j ( = +j for j {1,... p}. Therefore has p distinct conjugates and so [F ( : F ] p. But by assumption [E : F ] = p which forces E = F ( since F ( E. Note that σ( p = σ( p σ( = σ( p σ( = ( + 1 p ( + 1 = p = p, thus p is fixed by G and hence p F. Now take a = p and therefore satisfies X p X a = 0. (b Now let a F and consider the polynomial f(x = X p X a. Suppose is a root of f(x. Then + j for j {1,..., p} are also roots of f(x since ( + j p ( + j a = p + j p j a = p a = 0. In particular, f(x has p distinct roots. If some root lies in F, it follows that every root is in F. So suppose that no root 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 F [X] with their degrees being strictly less than p. If is a root of f, then p 1 f(x = (X j. 3

4 ABELIAN KUMMER THEORY It follows that both f and g are products of certain distinct integers j. {0,..., p 1} such that I K = and I J = {0,..., p 1} and That is, there exist I, J g(x = i I (X i and h(x = k K (X k. Let d = deg g and write g(x = d l=0 g lx l. Then g d 1 = i I ( + i by the theory on symmetric polynomials. Since I = d, we have that g d 1 = d + m for some m F p. Since d 0 and g d 1 F, it follows that F, which is a contradiction and therefore f is irreducible. Moreover, each root of f then lies in F ( and so F ( /F is Galois. Let G = Gal (F ( /F. Then there exist σ G such that σ( = + 1. But this implies that σ j ( = + j for each j {0,..., p 1}. Since each + j is a distinct root of f(x, we conclude that G is cyclic and it is generated by σ, as desired. Abelian Kummer Theory Definition A group G is said to be exponent m > 0 if σ m = 1 for each σ G. Let µ m denote the group of m-th roots of unity. Throughout this section we will assume that m char F = p if p > 0. We will also assume that µ m F and we denote by F, a fixed algebraic closure of F. Set F m = {a m F a F }. Let a F and consider F ( a 1/m. This is well-defined since µ m F implies that for any, β F satisfying m = β m = a, F ( = F (β. Now let B be a subgroup of F containing F m. We denote by K B = F ( B 1/m the compositum of all F ( a 1/m as a ranges over B. In particular, if B = F m, then K B = F. Notation 5. By A G B we mean that A is a subgroup of B. Lemma 6. Let F be a field, m a natural number prime to char F = p if p > 0, and suppose µ m F. Let B G F such that B contains F m and let K B = F ( B 1/m. Then the extension K B /F is Galois and G = Gal (K B /F is abelian and of exponent m. Proof. Let a B and let be a m-th root of a. Then X m a F [X] splits into linear factors in K B, and thus K B is Galois over F since this holds for each a B. Definition If B is a subgroup of F containing F m, we call its associated field extension K B /F a Kummer m-extension. Definition Let B G F such that F m B. The Kummer pairing is defined as κ : Gal (K B /F B µ m where κ(σ, a = ω σ = σ( where m = a. Theorem 7. Let F be a field, m a natural number prime to char F = p if p > 0, and suppose µ m F. Let B G F such that B contains F m and let K B = F ( B 1/m. Then (a The Kummer pairing κ is a well-defined bilinear map; (b The kernel on the left is 1; (c The kernel on the right is F m. Moreover, the Kummer pairing induces a perfect bilinear pairing κ : G B/F m µ m. 4

5 ABELIAN KUMMER THEORY Proof. (a Let m = β m = a. Then there is some ζ µ m such that β = ζ. It follows that σ(β β = σ(ζ ζ = ζσ( ζ = σ( and so κ is independent of the m-th root of a. Moreover, the σ κ(σ, a is a homomorphism for each a B, since for σ, τ G, we have that κ(στ, a = στ( = σ(ζ τ = ζ σζ τ = ζ σ ζ τ = σ( τ( = κ(σ, a κ(τ, a and a κ(σ, a is a homomorphism since a, b B with m = a and β m = b, we have κ(σ, ab = σ(β β = σ( σ(β = κ(σ, a κ(σ, b, β thus κ is bilinear. (b Let σ G and suppose κ(σ, a = 1 for each a B. Then for every generator of K B with m = a we have that σ( = 1, i.e., σ( =. Hence σ induces the identity on K B and we conclude that the kernel on the left is 1. (c Let a B and suppose κ(σ, a = 1 for each σ G. Consider the subfield F ( a 1/m of K B. If a 1/m is not in F, then there exist an automorphism of F ( a 1/m /F which is not the identity. Extending this automorphism to K B, we have that its extension by construction is not 1, and therefore κ(a, σ 1 if a 1/m F. If a 1/m F, then σ ( a 1/m = a 1/m and so we conclude that the kernel on the right consists of F m. Theorem 8. Let F be a field, m a natural number prime to char F = p if p > 0, and suppose µ m F. Let B G F such that B contains F m and let K B = F ( B 1/m. Then (a The map B K B gives a bijection of the set of subgroups of F containing F m and the abelian extensions of k of order m. (b The extension is K B /F is finite if and only if (B : F m is finite. Moreover, if this is the case we have that B/F m = Hom (G, µ m and [K B : F ] = (B : F m. ( ( Proof. (a Let B 1, B 2 be subgroups of F that contain F m. If B 1 B 2, then F B 1/m 1 F B 1/m 2. ( ( Conversely, assume that F B 1/m 1 F B 1/m 2. We claim that B 1 B 2. Let b B 1. Then F ( b 1/m ( ( F and it is a finite generated subextension of F. WLOG, suppose B 2 /F m is finitely B 1/m 2 generated and( therefore finite. ( Let B 3 = B 2, b. Then B 3 is a finitely generated subgroup of F and in particular, K = K.Moreover, (B 2 : F m = (B 3 : F m and so B 2 = B 3 which gives us that B 1/m 2 B 1/m 3 B 1 B 2. We conclude that we have an injection of our set of groups B into the set of abelian extensions of F of exponent m. Now suppose that E is an abelian extension of F of exponent m. Any finite subextension is a composite of cyclic extensions of exponent m because any finite abelian group is a product of cyclic groups. In particular, it has only a finite number of intermediate fields. But we have seen that every cyclic extension can be obtained B 1/m 2 by adjoining a family of m-th roots of unity, say {b j } j J with each b j F. Let B = b = ba m with a, b F, then F ( b 1/m = F ( b 1/m and so F ( B 1/m = E. {b j } j J, F m. If Example Let E be the splitting field of X 3 a with a Q 3. Then ζ 3, a 1/3 E. Let F = Q (ζ 3 where ζ 3 is a primitive 3-rd root of unity. We have that F F E. Note that L/F is Galois with group Z 3, and therefore it is abelian of exponent 3, therefore it is a Kummer 3-extension. 5

6 ELLIPTIC CURVES Example As an example consider E = Q(ζ 7 where ζ 7 is a primitive 7-th root of unity. Let F = Q(2 Re ζ 7. One can show 2 Re ζ 7 has minimal polynomial m F (x = x 3 + x 2 2x 1 and using Cardano s formula one can attain an exact value for 2 Re ζ 7. Moreover, [F : Q] = 3 and F/Q is Galois since m F (x splits into linear factors over F. That is, F is abelian of exponent 3. However Q(2 Re ζ 7 Q ( a 1/3 for any a Q. This follows since if Q(2 Re ζ 7 = Q ( a 1/3 for some a Q, then X 3 a must split into linear factors over Q(2 Re ζ 7. This in turn implies that ζ 3 Q(2 Re ζ 7, which implies that Q(2 Re ζ 7 R, a contradiction. This shows why it is essential that we make the assumption that the field contain the m-th roots of unity. We now extend the theory to abelian extensions of exponent p equal to the characteristic of F. We will only prove the results for extensions of exponent p. However, the case of exponent p n for n > 1 is due to Ernst Witt Let F be a field of characteristic p. Let P : F F by P(a = a p a for a F. Note that P is an additive homomorphism. In what follows, P (F := F p will be the analogue of F m above. For a F, we set P 1 (a to be a root of the polynomial X p X a. Let B be an additive subgroup of F containing F p. We define K B = F ( P 1 B to be the field obtained by adjoining P 1 (a to F for each a B. Then we have results analogous to the above: Let F be a field of characteristic p Lemma 9. Let F be a field of characteristic p. Let B G F such that B contains F p and let K B = F ( P 1 B. Then the extension K B /F is Galois and G = Gal (K B /F is abelian and of exponent p. Definition If B is a subgroup of F containing F p, we call its associated field extension K B /F a Kummer p-extension. Definition Let B G F such that F p B. The (additive Kummer pairing is defined as κ : G B Z/pZ where κ + (σ, a = σ( where P ( = a. Theorem 10. Let F be a field of characteristic p. Let B G F such that B contains F p and let K B = F ( P 1 B. Then (a The Kummer pairing κ + is a well-defined bilinear map; (b The kernel on the left is 1; (c The kernel on the right is F p. Moreover, the Kummer pairing induces a perfect bilinear pairing Elliptic Curves κ + : G B/F p Z/pZ. We shall now construct the Kummer pairing in the context of elliptic curves. Let E/K be an elliptic curve over K. Let E(K denote the group of K-rational points on the elliptic curve E. Let m 2. In this section we will assume that the m-torsion subgroup E [m] = {P E [m] P = O} E(K. By me(k = {mp P E(K}. Definition The Kummer pairing is defined as κ e : E(K Gal ( K/K E [m] where κ(p, σ = σ(q Q where [m] Q = P. 6

7 ELLIPTIC CURVES Theorem 11. Let E/K be an elliptic curve with group E(K and suppose that E [m] E(K. (a The Kummer pairing κ e is a well-defined bilinear map; (b The kernel of κ e on the left is me(k; (c The kernel of κ e on the right is Gal ( K/L where L = K ([m] 1 E(K is the compositum of all fields K(Q as Q ranges over the points E ( K satisfying [m] Q E(K. Hence the Kummer pairing induces a perfect bilinear pairing where L is the field given in (d. κ e : E(K /me(k Gal ( K/K E [m] Proof. (a We first show that κ(p, σ E [m]. Note that [m] κ(p, σ = [m] σ(q [m] Q = σ(p P = O since P E (K. If P = [m] Q and P = [m] R, then R = Q + T for some T E [m] and therefore since E [m] E (K and so σ fixes T. to σ(q + T (Q + T = σ(q + σ(t Q T = σ(q Q If we consider the isogeny [m] : E E where P [m] P, then we have that the Weil pairing reduces e m : E [m] E [m] µ m. We have seen that the Weil pairing is bilinear, non-degenerate, and Galois invariant. With this in mind we show: Corollary 12. There exist points S, T E [m] such that e m (S, T is a primitive m-th root of unity. In particular, if E [m] E(K, then µ m K. Proof. We have that e m (E [m] E [m] µ m is a subgroup. Let s say it is equal to µ d where d n. Then 1 = e m (S, T d = e m ([d] S, T for each S, T E [m]. Since e m is non-degenerate, we have that [d] S = O. But then d = m since E [m] = Z/mZ Z/mZ. If E [m] E (K, then the Galois invariance of the Weil-pairing implies that e m (S, T K for each S, T E [m]. Hence µ m K. Since µ m K, we can consider the Kummer pairing κ. We have that is a perfect pairing and so we have an isomorphism κ : Gal (L/K K /K m µ m δ κ : K /K m Hom ( Gal ( K/K, µm where δκ (a (σ = κ(σ, a. Similarly, we have that the following isomorphism from the Kummer pairing on elliptic curves: δ E : E(K /me(k Hom ( Gal ( K/K, E [m] where δe (P (σ = κ e (σ, P. Theorem 13. There is a bilinear pairing satisfying The pairing is nondegenerate on the left. b : E (K /me (K E [m] K /K m e m (δ E (P, T = δ κ (b(p, T. 7

8 REFERENCES REFERENCES References [Lan02] Serge Lang. Algebra, volume 211 of Graduate Texts in Mathematics. Springer-Verlag, New York, third edition, [Sil09] Joseph H. Silverman. The arithmetic of elliptic curves, volume 106 of Graduate Texts in Mathematics. Springer, Dordrecht, second edition,