Selmer Groups and Galois Representations

Transcription

1 Selmer Groups and Galois Representations Edray Herber Goins July 31, 2016

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3 Contents 1 Galois Cohomology Notation Galois Groups as Topological Spaces Abstract Algebra Review Commutative Groups Fields Galois Theory Profinite Groups as Topological Spaces Absolute Galois Groups Krull Topology Profinite Groups Example: l-adic Numbers G-Modules and X-Torsors Galois Modules Abelian Groups with Continuous Galois Action Example: Additive Group G a and Multiplicative Group G m Example: Elliptic Curves Tate Modules Isogenies of Galois Modules Sheaves of Galois Modules Principal Homogeneous Spaces Torsors via Translation Maps Example: Quadratic Curves Example: Cubic Curves Selmer s Cubic Example: Quartic Curves Elliptic Curves: 2-Isogeny Elliptic Curves: 2-Torsion Galois Cohomology Continuous Maps Sections Cochain Complexes

4 4.1.3 Where are these maps coming from? Example: Free G-Modules Cohomology Groups Weil-Châtalet Groups Example: Hilbert s Theorem Cup Product Long Exact Sequence for Cohomology Example: Kummer Theory Example: Elliptic Curve Example: Weil Pairing Pushforward on Cohomology Example: Decomposition and Inertia Groups I Selmer Groups History Lind and Reichardt s Quartic Selmer and Shafarevich-Tate Groups Classical Definitions Elliptic Curves

5 List of Figures 3.1 A Geometric Proof of the Associativity of the Group Law

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8 Chapter 1 Galois Cohomology 1.1 Notation k, F, K, L Fields k I G, Γ, U A separable, algebraic closure of k A partially ordered set Absolute Galois group, namely Galk/k, with composition : G G G A profinite group Open subgroup of G U, V, W Open sets in G σ, τ, ν, ω Elements of G X,, G-module with maps : X X X and : G X X P, Q, R O Y,, Elements of X Identity element of X Principal homogeneous space with maps : X Y Y and : G Y Y Q, R Elements of Y f g {Y/X} Bijection X Y from G-module to a principal homogeneous space Bijection Y Z between principal homogeneous spaces Equivalence class of principal homogeneous spaces Z Y E, E Elliptic curves φ, φ Dual m-isogenies φ : E E and φ : E E C n G, X All continuous maps G G X ξ, α, β Elements of C n G, X n Boundary map C n G, X C n+1 G, X Z n G, X B n G, X Kernel of n, i.e., the n-cocycles Image of n 1, i.e., the n-coboundaries H n G, X nth cohomology group, as Z n G, X/B n G, X 8 W CE/k Weil-Châtalet group of E, as H 1 Gal k/k, Ek f Pushforward H n G, X H n G, Y of the map f : X Y δ n Connecting homomorphism H n G, X H n+1 G, X ϕ Pullback H n G, X H n Γ, X of the map ϕ : Γ G

9 Chapter 2 Galois Groups as Topological Spaces 2.1 Abstract Algebra Review Commutative Groups Let X be a set. We say that X is a commutative group or an abelian group if there exists a well-defined binary operation : X X X such that Associativity P Q R = P Q R for all P, Q, R X. Commutativity P Q = Q P for all P, Q X. Identity There exists a unique O X such that O P = P for all P X. Inverses For each P X there exists a unique P X such that P P = O. Say that X,, Y,, and Z, are commutative groups. A map f : X Y is a homomorphism if fp Q = fp fq for all P, Q X. Define the kernel, image, and cokernel of a homomorphism as, respectively, ker f = { P X fp = O } = X[f] im f = { Q Y Q = fp for some P X } = fx coker f = { Q = Q im f Q Y } = Y/im f If g : Y Z is another homomorphism, we say that the composition of maps X f Y g Z is an exact sequence if im f = ker g as subsets of Y. The following is a well-known result about homomorphisms between commutative groups which we state without proof. Proposition Say that f : X Y is a homomorphism of commutative groups X and Y. 1. fo = O and f P = fp for all P X. 2. ker f,, im f,, and coker f, are commutative groups. 9

10 3. The following is an exact sequence: {O} ker f X f Y coker f {O} For example, Z, + is a commutative group under addition with identity 0. For any integer n, the map f : Z Z defined by a n a is a homomorphism. The kernel and image are ker f = {0} and im f = n Z, respectively. In particular, coker f = Z/n Z is a commutative group under addition Fields Let k be a set. We say that k is a field if there are well-defined binary operations + : k k k and : k k k such that Addition G a k = k is a commutative group under addition + with identity 0. Multiplication G m k = k {0} is a commutative group under multiplication with identity 1. Distributivity a b + c = a b + a c for all a, b, c k. The sets of rational numbers Q, real numbers R, complex numbers C, and residue classes modulo a prime F p = Z/p Z are each examples of fields. For any field k, exists an integer n such that the ideal n Z is the kernel of the morphism 1 } + {{ + 1 } if m > 0; Z k defined by m m times }{{} m times 0 if m = 0. if m < 0; and We call n the characteristic of k. For example, the sets of rational numbers Q, real numbers R, complex numbers C each have characteristic 0, whereas the set of residue classes modulo a prime F p has characteristic p Galois Theory Let k be a field. The polynomial ring R = k[x 1,..., x n ] is a unique factorization domain. We say that a field K is a normal, separable extension of k if 1 k K and K is the splitting field of a family of polynomials in R not having repeated roots. Define Aut and Emb! Let L be a normal, separable extension of k. Denote G = GalL/k = AutL/k = EmbL/k as its Galois group. Say that K is a subfield of L containing k; then K is a separable extension of k. Let H denote all σ G such that σa = a for all a K. Then H is a subgroup of G, and H = GalL/K. Conversely, say that H is a subgroup of G. Let K denote all a L such that σa = a for all σ H. Then K is a subfield of L, and K = L H. In either case, K is a normal extension of k if and only if H is a normal subgroup of G. 10

11 2.2 Profinite Groups as Topological Spaces Absolute Galois Groups Let k be a field, and denote k as a separable, algebraic closure of k. In practice, we will choose k as either the rational numbers Q, a completion Q v, or a finite field F p. We define G = Gal k/k as follows. Let I k denote the category of finite, normal, separable extensions K of k with the morphisms being set inclusion: F K L. We have a directed family of groups given by restriction of the action to subfields: proj L/F : GalL/k proj L/K GalK/k proj K/F GalF/k Define the projective limit as Gal k/k = lim GalK/k = K..., σ K,... GalK/k K I k proj L/Kσ L = σ K It is easy to check that this is a group under composition. Since G = Gal k/k is the projective limit of finite groups GalK/k we say that G is a profinite group. It may be useful to think of this using the following commutative diagram: Gal k/k GalL/k proj L proj L/K proj K GalK/k proj F proj K/F GalF/k Krull Topology We may turn G into a topological space using the Krull topology: We say that a subgroup U G is open if it has finite index in G. One may think of this as an open ball centered at the origin. In general, we say that a subset V G is open if for each σ V we can find a subgroup U σ of finite index in G such that the coset σ U σ is contained in V. One may think of this coset as an open ball centered at σ. Here is a standard result: Theorem For any field k, denote its absolute Galois group as G = Gal k/k. 1. Both and G are open. The union α V α of arbitrarily many open sets is open. The intersection α V α of finitely many open sets is open. 2. Let U G be a subgroup of finite index. Then each coset σ U is both open and closed. 3. The maps G G G and G G defined as σ, τ σ τ and σ σ 1, respectively, are continuous. 4. G is a topological group which is Hausdorff, totally disconnected, and compact. Proof. Clearly and G are open. Fix σ in an arbitrary union V = α V α. Then σ V α for some α, so that σ U σ V α for some open subgroup U σ. As σ U σ V, we see that that V must be open. Similarly, fix σ in a finite intersection V = α V α. Then σ V α for each α, so that σ U σ,α V α 11

12 for some open subgroups U σ,α. Denote U α = α U σ,α. As the index G : U σ α G : U σ,α is finite and σ U σ V, we see that the intersection of finitely many open sets must be open as well. Let U G is an open subgroup, and denote V = σ U. As τ U = V for each τ V, we see that V is open. Since U has finite index in G, we have a finite disjoint union G = n i=1 σi U. As V = σ j U for some j, the compliment G V = i j σi U is the union of open sets, we see that V is closed. We show that multiplication and inverses are continuous maps. Fix an open set W G. Fix σ, τ in the inverse image V = { σ, τ G G σ τ W }. We can find an open subgroup U σ τ in W such that σ τ U σ τ W. Denote the open subgroup U σ,τ = τ U σ τ τ 1 U σ τ in G G. It is easy to see that σ, τ U σ,τ V, so that V is indeed open. Similarly, fix σ in the inverse image V = { σ G σ 1 W }. We can find an open subgroup U σ 1 in W such that σ 1 U σ 1 W. Denote the open subgroup U σ = σ 1 U σ 1 σ in G. It is easy to see that σ U σ V, so that V is indeed open. Clearly G is a topological group. We show that G is both Hausdorff and totally disconnected. Choose distinct σ 1 and σ 2 in a subset S G. As σ1 1 σ 2 1, there exists a finite, normal, separable extension K of k such that σ1 1 σ 2 does not lie in the kernel U = ker projk of the canonical projection proj K : G GalK/k. As σ 1 U σ 2 U = we see that σ 1 and σ 2 can be housed ofh into disjoint open sets. Similarly, as S = σ S α G/U α is a nontrivial disjoint union of subsets S α = σ α U S, we see that S is disconnected in the subspace topology. Finally, we show that G is compact. As each finite group GalK/k is compact in the finite topology, Tychonoff s Compactness Theorem states that the product K I k GalK/k is also compact in the product topology. It suffices then to show that G K I k GalK/k is closed, which we do by showing that every convergent sequence in G has a limit in G. To this end, let I be a totally ordered set. We say that a subset {σ α α I} G is a Cauchy sequence if, given an open subgroup U G, we can find N U I such that σ 1 β σ α U whenever α, β N U. In particular, σ α β N U σβ U is contained in the intersection of closed sets, so that the subsequence {σ α α N U } is contained in a closed subset of G. Hence must have an accumulation point σ = lim α σ α in G. Any subgroup U of finite index in G contains a normal subgroup U = τ G τ 1 U τ also of finite index. Hence we may as well define an open subgroup as one which is a normal subgroup of finite index. In this case, the quotient group G/U GalK/k for some finite, normal, separable extension K of k, which shows that by definition G lim G/U. Often we abuse notation and U say U = Galk/K Profinite Groups In general, let I be some partially ordered set, and say that we have a collection of finite groups { Γα α I }. This is a directed family of groups whenever we have a collection of compatible group homomorphisms proj γ,β proj γ,α : Γ γ Γ β whenever α β γ. Define the projective limit as { Γ = lim α Γ α =..., g α,... α I proj β,α Γ α Γ α proj β,α g β = g α }

13 Using the ideas from above, it is easy to check that this is a group under composition. As Γ is the projective limit of finite groups, we say that Γ is a profinite group. In fact, Γ is a topological group which is Hausdorff, totally disconnected, and compact. Serre, in his Galois cohomology, shows the converse: If Γ is a topological group which is Hausdorff, totally disconnected, and compact, then G is the projective limit of of finite groups Example: l-adic Numbers Fix a prime l. Denote I = {1, 2,... } = Z >0 as the collection of positive integers. For each positive integer α I, the subset l α Z is an additive subgroup of the integers Z which is also closed under multiplication. The quotients Γ α = Z/l α Z are then finite abelian groups. We have projection maps proj β,α : Z/l β Z Z/l α Z which sends a mod l β a mod l α whenever α β. We define the set { Z l = lim Z/l α Z =..., a α α,... } Z/l α Z a β a α mod l α α I It is easy to check that Γ α = Z/l α Z Z l /l α Z l for any α I It is easy to show that Z l is a ring; we call it the collection of l-adic integers. The quotient field Q l is the collection of l-adic numbers. 13

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15 Chapter 3 G-Modules and X-Torsors 3.1 Galois Modules Abelian Groups with Continuous Galois Action Continue to denote G = Gal k/k as above. Let X be a set. In practice, X is a subset of the multiplicative group k or a subset of Ek for some elliptic curve defined over k. We say that X is a G-module or a Galois module over k if there exist binary operations : X X X and : G X X such that Associativity P Q R = P Q R and σ τ P = σ τ P for all σ, τ G and P, Q, R X. Commutativity P Q = Q P for all P, Q X. Distributivity σ P Q = σ P σ Q for all σ G and P, Q X. Identity There exists a unique O X such that O P = 1 P = P for all P X. Inverses For each P X there exists a unique P X such that P P = O. Continuity For each P X, the stabilizer V = { σ G σ P = P } of P is open. To be more precise, we may also say X is a Z[G]-module on which G acts continuously, where we view X as a topological space via the discrete topology. Condition Continuity needs further explanation: The set U V G is the kernel of the permutation representation G AutX defined as that map which sends σ to the homomorphism P σ P. Hence U is a normal subgroup of G; the condition Continuity assumes its index is finite. As each P X is contained in some subset X U = { P X σ P = P for all σ U } corresponding to some normal, open subgroup U G, we have X = U XU Example: Additive Group G a and Multiplicative Group G m Say that X = G a k = k under = + or X = G m k = k {0} under =. The absolute Galois group G = Gal k/k acts in the canonical way; denote this action by the binary operation : G X X. We show that X is a G-module. 15

16 It is easy to see that axioms Associativity, Commutativity, Distributivity, Identity, and Inverses are satisfied. It suffices to show that axiom Continuity is safisfied: Choose P X, and let V G denote its stabilizer. As this is an element of k, it lies in some finite, normal separable extension K of k. Denote the open subgroup U = ker proj K as the kernel of the canonical projection proj K : G GalK/k. It is easy to see that the coset σ U acts trivially on P for any σ V. Hence σ U V, so we see that V is indeed open. Note that X U = G a K = K or X U = G m K = K {0}, respectively, so that X = U XU Example: Elliptic Curves Fix coefficients a 1, a 2, a 3, a 4, a 6 k, and consider the homogeneous cubic polynomial fx 1, x 2, x 0 = x 2 2 x 0 + a 1 x 1 x 2 x 0 + a 3 x 2 x 2 0 x a 2 x 2 1 x 0 + a 4 x 1 x a 6 x For each normal, separable extension K of k we denote the set { } EK = x 1 : x 2 : x 0 P 2 K fx 1, x 2, x 0 = 0 { } x, y A 2 K y2 + a 1 x y + a 3 = x 3 + a 2 x 2 + a 4 x + a 6 {O} in terms of O = 0 : 1 : 0. The only solution over k to the simultaneous polynomial equations f = f = f = f = 0 is the trivial solution x 1 = x 2 = x 0 if and only if the discriminant x 1 x 2 x 0 = a 4 1 a 2 a a 2 1 a 2 2 a a 3 2 a a 3 1 a a 1 a 2 a a a 5 1 a 3 a a 3 1 a 2 a 3 a a 1 a 2 2 a 3 a 4 30 a 2 1 a 2 3 a a 2 a 2 3 a 4 + a 4 1 a a 2 1 a 2 a a 2 2 a a 1 a 3 a a 3 4 a 6 1 a 6 12 a 4 1 a 2 a 6 48 a 2 1 a 2 2 a 6 64 a 3 2 a a 3 1 a 3 a a 1 a 2 a 3 a a 2 3 a a 2 1 a 4 a a 2 a 4 a a 2 6 is nonzero. In this case, we say that E : y 2 + a 1 x y + a 3 = x 3 + a 2 x 2 + a 4 x + a 6 is an elliptic curve defined over k. Proposition For any field k, denote its absolute Galois group as G = Gal k/k. Let E be an elliptic curve defined over k. Then X = Ek is a G-module. Sketch of Proof. The absolute Galois group G = Gal k/k acts in the canonical way; denote this action by the binary operation : G X X. We construct a binary operation : X X X with certain properties: Consider two points P = p 1 : p 2 : p 0 and Q = q 1 : q 2 : q 0 in X. Draw a projective line through them, say a x 1 + b x 2 + c x 0 = 0 in terms of the projective point p 2 p 0 q 2 q 0 : p 0 p 1 q 0 q 1 : p 1 p 2 q 1 q 2 a : b : c = f f f P : P : P x 1 x 2 x 0 16 if P Q, or if P = Q

17 Denote P Q as the point of intersection between the line a x 1 + b x 2 + c x 0 = 0 and the curve fx 1, x 2, x 0 = 0. We define : X X X by P Q = P Q O. This is the so-called Group Law for elliptic curves. We must show that six properties hold for the binary operations : X X X and : G X X. Commutativity holds because P Q = Q P. Distributivity holds because σ P Q = σ P σ Q and σ O = O. Identity holds for O because the line through P O and O goes through one other point, namely P, so that P O = P O O = P. Similarly, Inverses holds for P = P O because the line through P and P O goes through one other point, namely O, so that P P = P P O = O O = O. As for Continuity, choose P X and let V G denote its stabilizer. Then the coordinates of P = p 1 : p 2 : p 0 lie in some finite, normal separable extension K of k. Denote the open subgroup U = ker proj K as the kernel of the canonical projection proj K : G GalK/k. It is easy to see that the coset σ U acts trivially on P for any σ V. Hence σ U V, so we see that V is indeed open. Note that X U = EK, so that X = U XU. It suffices then to show that Associativity holds for : X X X. For this, it suffices to show that P Q R = P Q R for all P, Q, R X. We omit a rigorous proof, and instead give a geometric one for k R. See Figure Tate Modules Let X and Y be G-modules, and say that we have a G-module homomorphism f : X Y. We say that f is an isogeny if either fp = O for all P X or f has the following properties: Homomorphism fp Q = ϕp ϕq for all P, Q X. G-Map σ fp = fσ P for all σ G and P X. Kernel The kernel X[f] = { P X fp = O } is a finite group. Image The image fx = Y. That is, for each Q Y, there exists at least one P X such that fp = Q. If f : X Y is an isogeny, define the separable degree as the integer deg f = #X[f]. For each integer m, define the multiplication-by-m map [m] : X X as that map which sends P } {{ P } if m > 0; [m] P = m times P P }{{} m times O if m = 0. if m < 0; and This is a G-module homomorphism. Let m = l α be a power of a prime l; then X[l α ] is a G- submodule of X. We have projection maps proj β,α : X[l β ] X[l α ] which sends P [l β α ] P

18 3 P*Q 2.5 R Q 2 Q*R 1.5 P Figure 3.1: A Geometric Proof of the Associativity of the Group Law Q+R P+Q*R = P*Q+R P+Q -3

19 whenever α β. We define the l-adic Tate module for X as the projective limit { T l X = lim X[l α ] =..., P α α,... } X[l α ] [lβ α ] P β = P α α I This is a Z l [G]-module. To see why, define : Z l [G] T l X T l X by [ ] mσ σ P =..., σ [m α σ] P α, σ G σ G Each m α Z/l α Z. If m α = n α + a α l α, then [m α ] P α = [n α ]P α [a α ] [l α ]P α = [nα ]P α O = [n α ]P α Define V l X = T l X Zl Q l as a Q l -module. Since G acts continuously on V l X, we have a continuous representation ρ X,l : Gal k/k GL V l X We define this as the l-adic representation associated to X Isogenies of Galois Modules Let X, Y, and Z be G-modules. Say that we have G-module homomorphisms f : X Y and h : X Z. Whenever ker f ker h and im f = Y, there exists a G-module homomorphism g : Y Z which makes the following diagram commute: X h Z f Y g Let f : X Y be an isogeny, and denote m = deg f as its separable degree. There exists a G-module homomorphism f : Y X such that f f = [m] is the multiplication-by-m map. Proof. Write ϕ 1 Q = R f 1 O for some R f 1 Q; then #f 1 Q = #f 1 O = m. For the composition f f : X X we have f f P = R = fp = R f fp 1 R f 1 O For the composition f f : Y Y we have f f Q = ϕ R = R f 1 Q R f 1 Q ϕr = Hence f f = [m] is indeed the multiplication-by-m map. 19 R f 1 Q Q = [m] Q

20 3.1.6 Sheaves of Galois Modules Continue to denote G = Gal k/k and X as a G-module, both as above. Given any element ν G, define the subset X ν = { P X σ P = P for all σ V } ν V as the union over those open subsets V G containing ν. We call X ν the stalk over ν, and any element P ν X ν a germ of the stalk. Given any nonempty open set V G, define U V = τ G τ 1 V τ as the largest normal, open subgroup contained in the group generated by V ; for V being empty, define U = G. We wish to consider the subset X V =..., Pν,... for each ν V, there exists a normal, X ν open subgroup U ν G such that ν U ν V ν V, and σ P ν ω = P ν ω for all σ U V and ω U ν where we set X = {O}. We may identify each point P =..., P ν,... in X V as a morphism P : V X ν defined by ν P ν ν V Proposition For any field k, denote its absolute Galois group as G = Gal k/k. Let X be a G-module, X ν X be a stalk for each ν G, and X V as above for each open set V G. 1. X ν is a G-module. 2. X is a sheaf of G-modules. Proof. We begin by showing that each stalk X ν is a G-module. Since X ν = ν U XU is the intersection over normal, open subgroups U containing ν, it suffices to show that each X U is a G-module. First, we show that there are restrictions : X U X U X U and : G X U X U. Each subset X U is closed under addition because σ P Q = σ P σ Q = P Q for any σ U and P, Q X U. As U is a normal subgroup in G, the conjugate σ = τ 1 σ τ is also in U for any τ G. Hence σ τ P = τ σ P = τ P, showing that τ P is also in X U for any τ G. Hence properties Associativity, Commutativity, Distributivity, and Continuity hold. Property Identity holds because σ O σ O = σ O O = σ O, so that σ O = O for any σ G. Property Inverses holds because σ P σ P = σ P P = σ O = O, so that σ P = σ P for any σ G. In particular, σ P = P for any σ U and P X U. Proposition shows that G is a topological space. In order to show that X is a sheaf, we must show that the following three properties hold: G-Modules X V is a G-module for every open set V G, where X = {O}. We define binary operations : X V X V X V and : G X V X V componentwise: P =..., P ν,... } { P Q =..., Pν Q ν,... Q =..., Q ν,... = τ P =..., τ P ν,

21 We explain why these are well-defined. Fix P, Q X V. For each ν V, we can find normal, open subgroups U ν,p, U ν,q G such that ν U ν,p, ν U ν,q V and σ P ν ω = P ν ω, σ Q ν ω = Q ν ω for all σ U V and ω U ν,p, ω U ν,q. Recall that for V nonempty we set U V = τ G τ 1 V τ, and U = G otherwise. Define U ν = U ν,p U ν,q ; this is a normal, open subgroup such that ν U ν V. Property Distributivity for the stalks X ν ω implies that σ P ν ω Q ν ω = σ Pν ω σ Qν ω = Pν ω Q ν ω for all σ U V and ω U ν, showing that P Q is in X V. As U V is a normal subgroup, the conjugate σ = τ 1 σ τ is also in U V for any τ G. Then σ τ P ν ω = τ σ P ν ω = τ Pν ω for all σ U V and ω U ν, showing that τ P is in X V. It is easy to see that properties Associativity, Commutativity, and Distributivity hold. As σ O = O and σ P ν = σ P ν for all σ G, the elements O =..., O,... and P =..., P ν,... are in X V ; hence properties Identity and Inverses hold. It remains to show that property Continuity holds. Fix a point P =..., P ν,... in X V, and consider an element σ of the stabilizer { σ G σ P = P } = { } ν V σ G σ P ν = P ν. This group contains the open set σ U V, so that it is indeed open. Hence X V is a G-module for every open set V G. Restriction Morphisms Whenever we have open sets U V W, there exist G-module homomorphisms such that res V/V = 1. res W/U : X W res W/V X V res V /U X U Define the map res W/V : X W X V as that which takes a tuple P =..., P ν,... for ν W to that tuple res W/V P =..., P ν,... for ν V which is formed by removing those coordinates corresponding to ν W V. We explain why this is well-defined for V nonempty. For each ν W, there exists a normal, open subgroup U ν G such that ν U ν W and σ P ν ω = P ν ω for all σ U W and ω U ν. As V is open, we can find a normal, open subgroup U ν G such that ν U ν V. Denote U ν = U ν U ν ; this is a normal, open subgroup such that ν U ν V. As V W, we have U V U W recall that V is assumed nonempty so that σ P ν ω = P ν ω for all σ U V and ω U ν. This shows that res W/V P is indeed in X V. As σ O = O, it is clear that σ [ res W/V P ] [ ] = res W/V σ P and resw/v P Q = res W/V P res W/V Q for all σ G and P, Q X W ; hence res W/V is a G-module homomorphism. Similarly, it is clear that res W/U = res V/U res W/V and res V/V = 1 whenever U V W. Gluing Property Say that V = α V α is a covering of open sets. Whenever we have a collection of points P α X V α such that res Vα/Vα V β P α = res Vβ /V α V β P β for all α and β, then there exists a unique P X V such that res V/Vα P = P α. V res V /Vα P res V /Vβ V α V β P α P β V α V β res V/Vα V β P 21

22 Say that V = α V α is an open cover, and that we have a collection of points P α =..., P α,ν,... in X V α such that res Vα/Vα V β P α = res Vβ /V α V β P β for all α and β. We must show that there is a unique P X V such that res V/Vα P = P α. If V is empty, we must have P = O because X = {O}, so assume V is nonempty. Each ν V lies in some V αν, so denote P =..., Pαν,ν,... as that tuple constructed by choosing the ν-coordinate of P αν X V αν. We call X the sheafification of the G-module X. We view X as a contravariant functor from the category of open sets V in G to the category of G-submodules X V of X. Note that when V is nonempty, U = τ G τ 1 V τ is an open, normal subgroup. In fact, since G/U GalK/k for some finite, normal separable extension K of k, we may identify X V as the K-rational points of X Principal Homogeneous Spaces Continue to denote G = Gal k/k and X as a G-module, both as above. Let A be a set. In practice, A is a subset of C d k for some quartic curve associated to an elliptic curve defined over k. We say that A is an X-torsor or a principal homogeneous space for X if there exist binary operations : X A A and : G A A such that Associativity P Q R = P Q R and σ τ R = σ τ R for all σ, τ G, P, Q X and R A. Distributivity σ P Q = σ P σ Q for all σ G, P X, and Q A. Identity O Q = 1 Q = Q for all Q A. Inverses For each Q, R A there exists a unique P X such that P Q = R. Continuity For each Q A, the stabilizer V = { σ G σ Q = Q } of Q is open. Note that X is a principal homogeneous space of itself. In a sense, A looks like X, yet we do not fix an identity O. Indeed, we even have A = U AU. Proposition For any field k, denote its absolute Galois group as G = Gal k/k. Let A be a principal homogeneous space for a G-module X. 1. Both X and G act continuously on A in the discrete topology. Moreover, X acts transitively, and the stabilizer of any element via this action is trivial. 2. There exists a bijection Ξ : X A defined over a finite, normal, separable extension K of k such that ΞP = P ΞO for all P X. 3. Define : A A X as that map which sends R, Q to the unique P such that P Q = R. For σ, τ G and Q A, we have the following identity in X: [ σ τ Q Q ] = [ σ Q Q ] [ σ τ Q Q ]

23 4. Define the relation A B when B is a principal homogeneous space for X and there is a bijection f : A B, commuting with the action by G, satisfying fp Q = P fq for all P X and Q A. Then is an equivalence relation. Moreover, for σ G, Q A, and R B, we have the following identity in X: [ σ Q Q ] = [ σ R R ] [ σ P P ] where P = fq R is in X. Three principal homogeneous spaces A, B, and C for a G-module X are said to be equivalent when we can find maps f : A B and g : B C which make the following diagram commute: X X X Ξ A Ξ B Ξ C A f B g C In fact, it is easy to verify that top row is a translation map P P O B,A P O C,A for all P X, in terms of O B,A = Ξ B 1 f Ξ A O, etc. Milne, in his Étale Cohomology, denotes P SHX/k as the set of equivalence classes {A/X} of such principal homogeneous spaces for X. Proof. The action of X on A is Q P Q for any fixed P X. As X, is an abelian group with identity O, properties Associativity and Identity show that both X and G do indeed act on A. Property Inverses shows that the only P = O if the only element such that P Q = Q; hence the stabilizer of any Q A is the trivial group {O} X. As one point sets are open in the discrete topology, this shows that the group action is continuous. Similarly, property Inverses shows that any R A is in the orbit of a given Q A, so this action is transitive. Fix Q A. Define Ξ : X A by ΞP = P Q. Clearly this is well-defined and bijective. Let V be the stabilizer of Q. As G acts on A continuously, we can find an open normal subgroup U V G, so that G/U GalK/k for some finite, normal, separable extension K of k. Clearly, Ξ is defined over K. Continue to fix Q A. For each σ G, denote ξσ = σ Q Q as that unique element such that ξσ Q = σ Q. It suffices to show that ξσ τ = ξσ σ ξτ. We have the identity ξσ τ Q = σ τ Q = σ τ Q = σ ξτ Q = σ ξτ σ Q = [ σ ξτ ξσ ] Q, where we have used properties Associativity and Distributivity. As ξσ is unique, we must have ξσ τ = ξσ σ ξτ as desired. We show that is an equivalence relation. Clearly A A since we may choose f = 1. Assume that A B. Denote f 1 : B A as the inverse of the map f : B A. Choose P A and R B, and denote Q = f 1 R A. Then we have f 1 P R = f 1 P fq = f 1 fp Q = P Q = P f 1 R Hence B A. Now assume that A B and B C. Denote f : A B and g : B C as bijections such that for all P X, Q A, and R B we have fp Q = P fq and gp R = P gr. The composition h = g f : A C is also a bijection. We have the identity hp Q = g fp Q = g P fq = P g fp = P hq

24 Hence A C. This shows that is indeed an equivalence relation. Fix σ G, Q A, and R B. Define ξ A σ, ξ B σ, P X via the relations ξ A σ Q = σ Q, ξ B σ R = σ R, and P R = fq. We have the identity ξa σ P fq = P [ ξ A σ fq ] = P [ f ξ A σ Q ] = P f σ Q = P σ fq = P σ [ P R ] = P σ P σ R = P σ P ξ B σ R = ξ B σ σ P P R = ξ B σ σ P fq This shows that ξ A σ P = ξ B σ σ P, so that ξ A σ = ξ B σ [ σ P P ] Torsors via Translation Maps If A is a principal homogeneous space for X, then there exists a bijection Ξ : X A. The following result gives a partial converse: if there exists a certain type of bijection Ξ : X A, then there is a binary operation : X A A such that A is a principal homogeneous space for X. Proposition For any field k, denote its absolute Galois group as G = Gal k/k. Let X and Y be G-modules, f : X Y be a G-module homomorphism, and A and B be sets on which G acts continuously. Assume that we have bijections Ξ A : X A and Ξ B : Y B such that for each σ G there exists ξ B σ = f ξ A σ Y for ξ A σ X satisfying σ Ξ A P = Ξ A σ P ξa σ σ Ξ B Q = Ξ B σ Q ξb σ for all P X, Q Y The map : X A A defined by P Q = Ξ A P Ξ 1 A Q makes A a principal homogeneous space for X. Moreover, Ξ A P = P Ξ A O for all P X. 2. Then there is a map f : A B such that f P Q = fp f Q and σ f Q = f σ Q for all σ G, P X, and Q A. Moreover, the following diagram commutes: ξ A G ξ B X f Y Ξ A A f Ξ B B This proposition explains how the existence of a bijection Ξ A : X A which depends on a map ξ A : G X constructs homogeneous spaces. We say that A is an f-descendant of B whenever the conditions above hold. Note that if ξ B = O, then B Y as G-modules and im ξ A ker f. In this case, we say A an f-cover for Y, with f : A Y being the covering map. Proof. We show that : X A A satisfies the following properties: Associativity P Q R = P Q R for all P, Q X and R A. 24

25 Distributivity σ P Q = σ P σ Q for all σ G, P X, and Q A. Identity O Q = Q for all Q A. Inverses For each Q, R A there exists a unique P X such that P Q = R. Translation Ξ A P = P Ξ A O for all P X. For Associativity, we have P Q R = ΞA P Q Ξ 1 A R [ = Ξ A P Ξ 1 A ΞA Q Ξ 1 A R] = P Q R For Distributivity, σ P Q = σ Ξ A P Ξ 1 A Q = f σ P [ σ Ξ 1 A Q] ξ A σ [ = Ξ A σ P σ Ξ 1 A Q ξσ] = Ξ A σ P Ξ 1 A = σ P σ Q. [ σ Q ] For Identity, we have O Q = Ξ A O Ξ 1 A Q = Ξ A Ξ 1 A Q = Q. For Inverses, let Q, R A be given, and define P = Ξ 1 A R Ξ 1 A Q. It is easy to see that P X is that unique element such that P Q = R. For Translation, let Q = Ξ A O. Then Ξ A P = Ξ A P O = ΞA P Ξ 1 A Q = P Q = P Ξ A O. Define f : A B by the composition f = Ξ B f Ξ 1 A. For any P X and Q A, we have f P Q = ΞB f P Ξ 1 A Q = Ξ B [fp 1 f Ξ ] A Q = fp f Q Now choose σ G and Q A, and set P = Ξ 1 A Q. Since σ Q = Ξ A σ P ξa σ, we have f σ Q = Ξ B f σ P ξ A σ = Ξ B [f σ P f ξ A σ ] [ ] σ = Ξ B fp ξb σ = σ Ξ B fp = σ f Q. By construction, the diagram commutes Example: Quadratic Curves The following proposition explains how to determine conic sections as principal homogeneous spaces for Pell s equation x 2 D y 2 = 1. Proposition Let k be a field of characteristic different from 2, and denote its absolute Galois group as G = Gal k/k. Fix elements a i k such that the determinants a 11, D = a 11 a 12 a 11 a 12 a 13 a 12 a 22 and a 12 a 22 a a 13 a 23 a 33 are nonzero. 25

26 1. Denote X as the collection of k-rational points P = x, y satisfying x 2 D y 2 = 1. Then X is a G-module via the binary operation : X X X defined by x 1, y 1 x 2, y 2 = x 1 x 2 + D y 1 y 2, x 1 y 2 + x 2 y Denote A as the collection of k-rational points Q = z, w on the conic section a 11 z a 12 z w + a 22 w a 13 z + 2 a 23 w + a 33 = Then A is a principal homogeneous space for X. Proof. Consider the injective group homomorphism [ ] x D y X SL 2 k defined by x, y y x The obvious map X k defined by P x + D y is a homomorphism of G-modules, but it is not injective! As SL 2 k is G-module, the induced structure turns X into a G-module as well. Note that x, y 1, 0 = x, y, while O = 1, 0 is the identity and P = x, y is the inverse. Consider the bijection Ξ : X A defined by Ξx, y = z 0 + x a 12 y, w 0 + a 11 y z d d 0 = a 12 a 23 a 13 a 22 a 11 a 22 a 2, 12 where ΞO = z in terms of, w 0 w 0 = a 12 a 13 a 11 a 23 d a 11 a 22 a This is defined over the finite, normal, separable extension K = k d of k, where a d = a 11 a 12 a 11 a 12 a a 12 a 22 a 12 a 22 a 23 a 13 a 23 a 33 1 a 11 = a 13 z 0 + a 23 w 0 + a 33 Geometrically, the point Q 0 = z 0, w 0 is the center of the conic section; it is not actually on the curve. It is easy to check that σ [ ΞP ] = Ξ σ P ξσ { 1, 0 if σ d = d; where ξσ = O if σ d = d. Proposition states that A is a principal homogeneous space for X. For example, the binary operation : X A A defined by x, y z, w = f x, y f 1 z, w = z 0 + x z z 0 y a 12 z + a 22 w + a 23, w 0 + x w w 0 + y a 11 z + a 12 w + a

27 which may also be realized via the injective group homomorphism Y SL 2 k defined by a 12 z + a 22 w + a 23 a 11 z + a 12 w + a 13 a z, w 13 z 0 + a 23 w 0 + a 33 z z. 0 w w 0 a 13 z 0 + a 23 w 0 + a The proposition follows Example: Cubic Curves 3.2 Selmer s Cubic Proposition Let k be a field of characteristic different from 2 and 3. Fix D k, and consider the cubic curve C : a u 3 + b v 3 + c w 3 = 0 for a, b, c k satisfying D = a b c. 1. C is a principal homogeneous space for the elliptic curve E : y 2 = x D 2. In particular, E acts continuously and transitively on C via the map : E C C given by x, y u : v : w [ 2 w y + x 2 17 z = x x z 2, w x ] 34 z [ x w z x y 2 x + 17 z 2 + x 2 z 2 ] [ x x z 2] Denoting the elliptic curve E : y 2 = x D 2 x, there are rational maps ϕ : E E and g : C E defined over k which make the following diagram commute: E ϕ E Moreover, gp Q = ϕ P gq for all P on E and Q on C. f C 3. If C has a k-rational point Q 0 = u 0 : v 0 : w 0, then C and E are birationally equivalent over k. Ernst Selmer considered this family of cubic curves for k = Q. In particular, he showed that C : 3 u v w 3 = 0 has a Q v -rational point for every place v of Q, yet it has no Q-rational point. We will see later that we can use properties of the Selmer group to better understand this phenomenon. Proof. Denote X = Ek as the collection of k-rational points P = x, y on the cubic curve E : y 2 = x D 2. As its discriminant is a nonzero = D 4, we see that E is an elliptic curve defined over k. Recall that X is a G-module by Proposition Denote Y = Ck as the collection of k-rational points Q = u, v, w the quartic curve C : a u 3 + b v 3 + c w 3 = 0. The map f : E C defined by fx, y = 6 b x 3 d : 36 a b c y 3 d 2 : 36 a b c + y d 27 g where fo = 0 : 3 d :

28 is a bijection from X to Y which is defined over the finite, normal, separable extension K = k 3, 3 d of k for d = c/b. Using the identity x, y 0, 0 = 17/x, 17 y/x 2 for the group law on E, one checks that for any σ in G = Gal k/k we have the relation σ [ fp ] = f σ P ξσ where ξσ = { 0, 0 if σ 2 = 2; O if σ 2 = Proposition states that Y = Ck is a principal homogeneous space for X = Ek. The map : X Y Y above is defined by x, y u : v : w = f x, y f 1 u : v : w. We have seen before that there is a 2-isogeny ϕ : E E defined by x ϕ x, y =, y x2 17 x x 2 = ϕ f 1 2 z, w = z 2, 4 w z Since im ξ = { 0, 0, O } = E[ϕ ], the second statement in the proposition above follows from Proposition Finally, say that Q 0 = u 0 : v 0 : w 0 is a k-rational point on C. Since x = 1 17 z2 z w w 0 z z 0 2 y = 34 z z 0 z 2 w 0 + z 2 0 w 2 w + w 0 z z if and only if z = z w 0 y + 17 z0 3 x + 17 z 0 x z0 2 x + 17 w = w 0 34 z 0 z 2 0 x x + 17 z0 2 y + w0 3 z 2 0 x z0 4 x2 + x z0 2 x + 17 x z0 2 x we see that the quartic curve 2 w 2 = 1 17 z 4 is birationally equivalent over k to the cubic curve y 2 = x x. Proposition. Let k be a field of characteristic different from 3 containing a cube root ζ 3 of unity. In practice, we choose k = Q 3 since we want a number field. Choose k-rational numbers A, B, C, and D such that = 27 A B C D 3 A B C 3 is nonzero. We want to consider the projective curves C : A z B z C z 3 0 = 3 D z 1 z 2 z 0. These are called Desboves Curves. An example would be 3 z z z3 0 = 0. This family of curves has complex multiplication, i.e., for a primitive cube root ζ 3 of unity, we have an automorphism of order 3 defined by [ζ 3 ] : Ck Ck which sends z1 : z 2 : z 0 ζ3 z 1 : ζ 2 3 z 2 : z

29 First, we show why this is a principal homogeneous space for an elliptic curve. Consider the map f : E C defined by 3 B x 3 D x + 1 ζ 3 y + 1 ζ3 2 fx, y = 3 : D3 A B C 3 d d 2 where fo = 0 : 3 D x + 1 ζ3 2 : y + 1 ζ 3 D 3 A B C d 3 d : for the elliptic curve E : y D x y + D 3 A B C y = x 3 and d = C/B. This is defined over the finite, normal, separable extension K = k 3 d of k, and one checks that σ [ fp ] = f σ P ξσ 0, 0 if σ 3 d = ζ 3 3 d; where ξσ = 0, b if σ 3 d = ζ d; O if σ 3 d = 3 d. Hence C is indeed a principal homogeneous space for E. Second, we show that if C has a rational point P 0 = a 1 : a 2 : a 0, then C is birationally equivalent to a different elliptic curve Example: Quartic Curves The following proposition explains how to determine quartic curves as principal homogeneous spaces for elliptic curves. Proposition Let k be a field of characteristic different from 2, and denote its absolute Galois group as G = Gal k/k. Consider a quartic Qz = c 4 z 4 + c 3 z 3 + c 2 z 2 + c 1 z + c 0 with coefficients in k and a nonzero discriminant DiscQ = c 2 1 c 2 2 c c 0 c 3 2 c c 3 1 c c 0 c 1 c 2 c c 2 0 c c 2 1 c 3 2 c c 0 c 4 2 c c 3 1 c 2 c 3 c 4 80 c 0 c 1 c 2 2 c 3 c 4 6 c 0 c 2 1 c 2 3 c c 2 0 c 2 c 2 3 c Define the coefficients 27 c 4 1 c c 0 c 2 1 c 2 c c 2 0 c 2 2 c c 2 0 c 1 c 3 c c 3 0 c 3 4. a 1 = c3 3 4 c 2 c 3 c c 1 c c 2 4 a 2 = 3 c2 3 8 c 2 c 4 4 c 4 a 4 = 3 c c 2 c 2 3 c c 2 2 c c 1 c 3 c c 0 c c 2 4 c 3 a 6 = 3 4 c 2 c 3 c c 1 c c The polynomial P x = x 3 + a 2 x 2 + a 4 x + a 6 in terms of is a resolvent cubic for Qz such that DiscP = DiscQ. 2. The quartic curve C : w 2 = Qz is a principal homogeneous space for the elliptic curve E : y 2 = P x. If C has a k-rational point Q 0 = z 0, w 0, then C and E are birationally equivalent over k. 29

30 3. Assume that a 1 = 0. In particular, the quartic curves w 2 = 1 z 4 and 2 w 2 = 1 17 z 4 are principal homogeneous spaces for the cubic curves y 2 = x x and y 2 = x x, respectively, as well as g-covers for the cubic curves y 2 = x 3 x and y 2 = x 3 68 x, respectively. We will see later that the former quartic curve has Q-rational points, whereas the latter quartic curve does not. Proof. Using the factorization Qz = c 4 z e 1 z e 2 z e 3 z e 4 over k, we find the factorization e 1 + e 2 e 3 e 4 P x = [x 2 ] e 1 e 2 + e 3 e 4 + c 4 [x 2 ] e 1 e 2 e 3 + e 4 + c 4 [x 2 ] + c also over k. One readily verifies that DiscP = DiscQ. Denote X = Ek as the collection of k-rational points P = x, y satisfying y 2 = P x, and A = Ck as the collection of k-rational points Q = z, w satisfying w 2 = Qz. The discriminant of the cubic curve is = 2 4 DiscQ. As this is nonzero, we see that E is indeed an elliptic curve over k; Proposition states that X is indeed a G-module. The birational transformation Ξ : X A defined by Ξx, y = c 3 x + 2 c 4 a d y d 2 a 6 + a 4 x x 3 +2 c 4 a 1 y, 4 c 4 x 4 c 4 x is defined over the finite, normal, separable extension K = k d of k, where d = c 4. It is easy to check that σ [ ΞP ] = Ξ σ P ξσ 0, a 1 d if σ d = d; where ξσ = O if σ d = d. Proposition states that A is a principal homogeneous space for X. Say that Q 0 = z 0, w 0 is a k-rational point on C. Then the invertible substitution x = b 1 z z 0 + b 2 b 3 + z z 2 w w 0 0 z z0 y = b 4 z z b 5 z z 0 + b 6 z z0 2 + b 7 z z 0 + b 8 z z0 3 w w in terms of the coefficients b 1 = 2 c 4 z c 3 z 0 b 2 = 4 c 4 z c 3 z c 2 z 0 + c 1 b 3 = 2 w 0 b 4 = w 0 4 c4 z 0 + c 3 b 5 = 2 w 0 6 c4 z c 3 z 0 + c 2 b 6 = 2 w 0 4 c4 z c 3 z c 2 z 0 + c 1 b 7 = 4 c 4 z c 3 z c 2 z 0 + c 1 b 8 = 4 w shows that the relation y 2 = P x holds if and only if w 2 = Qz holds. Hence the curves C and E are birationally equivalent over k. 30

31 Proposition Let k be a field of characteristic different from 2, and denote its absolute Galois group as G = Gal k/k. Fix c i k such that the discriminants c 4 0, c c 2 c 3 c c 1 c 2 4 = 0, c 2 1 c 2 2 c c 0 c 3 2 c c 3 1 c c 0 c 1 c 2 c c 2 0 c c 2 1 c 3 2 c c 0 c 4 2 c c 3 1 c 2 c 3 c 4 80 c 0 c 1 c 2 2 c 3 c 4 6 c 0 c 2 1 c 2 3 c c 2 0 c 2 c 2 3 c 4 27 c 4 1 c c 0 c 2 1 c 2 c c 2 0 c 2 2 c c 2 0 c 1 c 3 c c 3 0 c C : w 2 = c 4 z 4 + c 3 z 3 + c 2 z 2 + c 1 z + c E : y 2 = x 3 + c 2 x 2 + c 1 c 3 4 c 0 c 4 x + c0 c c 2 1 c 4 4 c 0 c 2 c 4 E : y 2 = x 3 + c 2 x c 2 c c2 2 c c 1 c 3 c c 0 c c 4 x g z, w = + 3 c2 2 c c3 2 c 4 6 c 1 c 2 c 3 c c 0 c 2 3 c 4 24 c 2 1 c c 0 c 2 c c 4 16 c 2 4 z c 3 c 4 z + 4 c 2 c 4 c 2 3 Moreover, the following diagram commutes: 4c 4, 2Y c 3 + 4Xc E ξ A f G E ξ B O g E Ξ A Ξ B C f C g Proof. Denote X = E k, Y = Ek, and Z = E k. We have an isogeny g : Y Z defined by x 2 e x + e 2 4 c 0 c 4 gx, y =, y x2 2 e x + 4 c 0 c 4 x e x e 2 where e = c2 3 4 c 2 c 4 4 c Similarly, denote A = C k and B = Ck. In the proof of Proposition 3.2.2, we exhibited a birational transformation Ξ B : Y A such that σ [ Ξ B P ] = Ξ B σ P ξb σ { }, where ξ B σ e, 0, O = ker g for all σ G. Proposition states that the map g = g Ξ 1 B makes the diagram above commute. 31

32 3.2.2 Elliptic Curves: 2-Isogeny Let E : y 2 = x 3 + a x 2 + b x be an elliptic curve over a field k having characteristic different from 2, and denote X = Ek as the collection of k-rational points P = x, y. Recall that X, is an abelian group with identity O = 0 : 1 : 0: x 1, y 1 x 2, y 2 y1 y 2 2 = x 1 x 2 a, x 1 x 2 x x 2 + a y 1 x x 1 + a y 2 x 1 x 2 y1 y 3 2. x 1 x For example, x, y 0, 0 = b/x, b y/x 2. Note that 0, 0 is a point of order 2, i.e., [2] 0, 0 = O. For each d k, consider the quartic curve C d : w 2 = d 2 a z 2 + a 2 4 b/d z 4, and denote Y = C d k as the collection of k-rational points Q = z, w. The map f : E C d defined by fx, y = d y x 2 + a x + b, d x 2 b x 2 + a x + b where fo = 0, d is a bijection from X to Y which is defined over the finite, normal, separable extension K = k d of k. You can find these formulas on page 294 of Silverman s The Arithmetic of Elliptic Curves. In fact, one checks that σ [ fp ] = f σ P ξσ { 0, 0 if σ d = d; where ξσ = O if σ d = d. Proposition states that Y = C d k is a principal homogeneous space for X = Ek. We abuse notation and say C d is a principal homogeneous space for E. Rather explicitly, the map : E C d C d is given by x, y z, w = f x, y f 1 z, w d 2 x 2 + a x + b [ w x 2 b 2 a y z ] d w y + d x 2 b z = d x 2 + a x + b a 2 4 b x z 2, + d a 2 4 b z [ x z w x 2 b + 2 y d x + b z 2 + x 2 z 2 ] [ d x 2 + a x + b a 2 4 b x z 2] Elliptic Curves: 2-Torsion Here s a slightly more advanced example using the same elliptic curve. Assume that E[2] Ek, so that we can write E : y 2 = x e 1 x e 2 x e 3 in terms of the k-rational roots e 1 = a + a 2 4 b 2, e 2 = a a 2 4 b, and e 3 = For each d 1, d 2 k, consider the quadric intersection H d1,d 2 : x A x = x B x = 0 defined in terms of the 4 4 matrices d d A = 0 d and B = d 1 d e e 2 32

33 When x = u : v : w : 1 is an affine point, we may express this in terms of the affine equations d 1 u 2 d 2 v 2 = e 1 and d 1 u 2 d 1 d 2 w 2 = e 2. Denote Y = H d1,d 2 k as the collection of k-rational points x. The map f : E H d1,d 2 defined by x 2 e 1 e 2 fx, y = 2 d 1 y : x2 2 e 1 x + e 1 e 2 2 : x2 2 e 2 x + e 1 e 2 d 2 y 2 : 1 d 1 d 2 y where fo = : : : 0 d1 d2 d1 d 2 is a bijection from X to Y which is defined over the finite, normal, separable extension K = k d 1, d 2 of k. It is a bit tedious, but one checks that σ [ fp ] = f σ P ξσ where e 1, 0 if σ d 1 = d 1, σ d 2 = + d 2 ; e 2, 0 if σ d 1 = d 1, σ d 2 = d 2 ; ξσ = e 3, 0 if σ d 1 = + d 1, σ d 2 = d 2 ; O if σ d 1 = + d 1, σ d 2 = + d Proposition states that H d1,d 2 is also a principal homogeneous space for E. Rather explicitly, the map : E H d1,d 2 H d1,d 2 is given by x, y u : v : w : 1 = f x, y f 1 u : v : w : 1 =? :? :? :? I find it unsettling that no one has ever written down these expressions before when it seems obvious to do so... Elliptic Curves: 3-Isogeny Let E : y 2 + a x y + b y = x 3 be an elliptic curve over a field k having characteristic different from 3 as well as a primitive cube root of unity ζ 3, and denote X = Ek as the collection of k-rational points P = x, y. It is easy to check that x, y 0, 0 = b y/x 2, b 2 y/x 3 and x, y [2] 0, 0 = b x/y, b x 3 /y 2. Note that 0, 0 is a point of order 3, i.e., [3] 0, 0 = O. For each d k, consider the cubic curve C d : w 3 = d + 3 a z w + a 3 27 b/d z 3, and denote Y = C d k as the collection of k-rational points Q = z, w. The map f : E C d defined by 3 x fx, y = d 2 a x + 1 ζ3 2 y + 1 ζ 3 b, 3 d a x + 1 ζ 3 y + 1 ζ3 2 b a x + 1 ζ3 2 y + 1 ζ 3 b where fo = 3 0, d is a bijection from X to Y which is defined over the finite, normal, separable extension K = k 3 d of k. Recall that a primitive cube root ζ 3 of unity is assumed to be an element of k. One checks that σ [ fp ] = f σ P ξσ 0, 0 if σ 3 d = ζ 3 3 d; where ξσ = 0, b if σ 3 d = ζ d; O if σ 3 d = 3 d. 33

34 Proposition states that C d is a principal homogeneous space for E. Rather explicitly, the map : E C d C d is given by x, y z, w = f x, y f 1 z, w =?,?

35 Chapter 4 Galois Cohomology 4.1 Continuous Maps Sections Let G = Gal k/k, and X be a G-module as above. Let C 0 G, X = X; and for each positive integer n, let C n G, X denote the collection of continuous maps ξ : G G X. That is, for each normal, open subgroup U G we have maps ξ U : G/U G/U s s G G ξ X U in terms of sections which are continuous maps s : G/U G of profinite groups. We explain why such sections exist. To this end, we will show the following: Proposition Let G = Gal k/k. Given closed subgroups U and W satisfying W U G, there exists a continuous map s such that the composition is the identity map. G/U s G/W G/U Proposition shows that every open subgroup U is also a closed set. Similarly, W = {1} is a closed set: Choose σ G W. Then σ U G W for any nontrivial open subgroup U G. Proof. We follow the ideas in Serre s Galois Cohomology. Consider the set { } V is a closed subgroup of U containing W I = V, s and s : G/U G/V is a continuous section First we show that I contains a maximal element V, s. We may place a partial ordering on I by saying V α, s α V β, s β whenever V β V α and the following diagram is commutative: G/V G/U s s β G/V β G/U s α G/V α 35