Title: Daniel's Flashcards / Year 4 / Term 2 / Elliptic Curves

html: Definition (Rational, primitive triangle). A triangle is rational if a , b , c ∈ ℚ.
A triangle is primitive if a , b , c ∈ ℤ and are coprime.
nothing: True

html: Lemma 1.1. Assuming that:
- Δ is a primitive triangle
Then Δ is of the form
for some integers u > v > 0.
proof: Proof. Without loss of generality a odd, b even (work modulo 4). This then forces c odd.
Then
( b 2 ) 2 = c + a 2 ⋅ c − a 2 ,
and note that all the fractions are integers. Also note that the product on the right hand side is a product of positive coprime integers.
Unique prime factorisation in ℤ gives that c + a 2 = u 2, c − a 2 = v 2 for some u , v ∈ ℤ.
Then a = u 2 − v 2, b = 2 u v, c = u 2 + v 2. □
cloze1: True

html: Definition (Congruent number). D ∈ ℚ > 0 is a congruent number if there exists a rational triangle Δ with area ⁡ ( Δ ) = D.
nothing: True

html: Lemma 1.2. Assuming that:
- D ∈ ℚ > 0
Then D is congruent if and only if D y 2 = x 3 − x for some x , y ∈ ℚ, y ≠ 0.
proof: Proof. Lemma 1.1 shows
D congruent ⟺ D w 2 = u v ( u 2 − v 2 )
for some u , v , w ∈ ℚ, with w ≠ 0. Put x = u v and y = w v 2. □
cloze1: True

html: Lemma 1.4. Assuming that:
- u , v ∈ K [ t ] coprime
- α u + β v is a square for 4 distinct ( α : β ) ∈ ℙ 1
Then u , v ∈ K.
proof: Proof. Without loss of generality K = K ¯.
Changing coordinates on ℙ 1, we may assume the ratios ( α : β ) are ( 1 : 0 ) , ( 0 : 1 ) , ( 1 : − 1 ) , ( 1 : − λ ) for some λ ∈ K ∖ { 0 , 1 } (Möbius map).
u = a 2 v = b 2 u − v = ( a + b ) ( a − b ) u − λ v = ( a + μ b ) ( a − μ b )
(where μ is a square root of λ). Unique factorisation in K [ t ] gives that a + b, a − b, a + μ b, a − μ b are squares. But
max ⁡ ( deg ⁡ ⁡ ( a ) , deg ⁡ ⁡ ( b ) ) ≤ 1 2 max ⁡ ( deg ⁡ ⁡ ( u ) , deg ⁡ ⁡ ( v ) ) .
So Fermat’s method of infinite descent, we get a contradiction, unless the degrees of u and v are zero. So u , v ∈ K. □
cloze1: True

html: Definition 1.5 (Elliptic curve (temporary definition)).
- (i) An elliptive curve E ∕ K is the projective closure of the plane affine curve
y 2 = f ( x )
where f ∈ K [ x ] is a monic cubic polynomial with distinct roots in K ¯. We call this equation “a Weierstrass equation”.
- (ii) For L ∕ K any field extension
E ( L ) = { ( x , y ) ∈ L 2 | y 2 = f ( x ) } ∪ { 0 } .
0 is the “point at infinity” that we get because we take the projective closure.
nothing: True

html: Definition 2.1 (Rational plane affine curve). A plane affine curve C = { f ( x , y ) = 0 } ⊂ 𝔸 2 is rational if it has a rational parametrisation, i.e. ∃ ⁡ ϕ ( t ) , ψ ( t ) ∈ K ( t ) such that:
- (i) 𝔸 1 → 𝔸 2, t ↦ ( ϕ ( t ) , ψ ( t ) ) is injective on 𝔸 1 ∖ { finite set }.
- (ii) f ( ϕ ( t ) , ψ ( t ) ) = 0.
nothing: True

html: Proposition 2.4. Assuming that:
- K = K ¯
- C a smooth projective curve
Then
- (i) C is rational (see Definition 2.1) if and only if g ( C ) = 0.
- (ii) C is an elliptic curve (see Definition 1.5) if and only if g ( C ) = 1.
proof: Proof.
- (i) Omitted.
- (ii) ⇒: Check C a smooth plane curve (exercise). Then use Remark 2.3.
⇐: See later.
□
cloze1: True

html: Definition (Divisor). A divisor is a formal sum of points on C, say
D = ∑ P ∈ C n p P
where n p ∈ ℤ and n p = 0 for all but finitely many P ∈ C.
We write deg ⁡ D = ∑ n P.
We say D is effective (written D ≥ 0) if n P ≥ 0 for all P.
nothing: True

html: Proposition 2.7. Assuming that:
- K = K ¯, char ⁡ K ≠ 2
- C ⊂ ℙ 2 a smooth plane cubic
- P ∈ C a point of inflection
Then we may change coordinates such that
C : Y 2 Z = X ( X − Z ) ( X − λ Z )
for some λ ≠ 0 , 1 and P = ( 0 : 1 : 0 ).
proof: Proof. We change coordinates such that P = ( 0 : 1 : 0 ), T p ( C ) = { Z = 0 }, and C : { F ( X , Y , Z ) = 0 } ⊂ ℙ 2.
P ∈ C part of inflection implies F ( t , 1 , 0 ) = const t 3, i.e. F has no terms X 2 Y, X Y 2 or Y 3.
Therefore
F ∈ ⟨ Y 2 Z , X Y Z , Y Z 2 , X 3 , X 2 Z , X Z 2 , Z 3 ⟩ .
The Y 2 Z coefficient must be ≠ 0 otherwise P ∈ C is singular, and the coefficient of X 3 is ≠ 0 otherwize Z | F.
We are free to rescale X, Y, Z and F. Then without loss of generality C is defined by
Y 2 Z + a 1 X Y Z + a 3 Y Z 2 = X 3 + a 2 X z Z + a 4 X Z 2 + a 6 Z 3 .
Weierstrass form.
Substituting Y ← Y − 1 2 a 1 X − 1 2 a 3 Z, we may suppose a 1 = a 3 = 0. Now
Y 2 Z = Z 3 f ( X Z )
for some monic cubic polynomial f.
C is smooth, so f has distinct roots. Without loss of generality say 0 , 1 , λ. Then C is given by
Y 2 Z = X ( X − Z ) ( X − λ Z ) . □
cloze1: True

html: Definition (Degree of a morphism). deg ⁡ ϕ [ K ( C 1 ) : ϕ ∗ K ( C 2 ) ].
nothing: True

html: Definition (Separable morphism). ϕ is separable if K ( C 1 ) ∕ ϕ ∗ K ( C 2 ) is a separable field extension.
nothing: True

html: Definition (Ramification index). Suppose P ∈ C 1, Q ∈ C 2, ϕ : P ↦ Q. Let t ∈ K ( C 2 ) be a uniformiser at Q.
The ramification index of ϕ at P is
e ϕ ( P ) = ord ⁡ P ( ϕ ∗ t )
(always ≥ 1, independent of choice of t).
nothing: True

html: Theorem 2.8. Assuming that:
- ϕ : C 1 → C 2 a non-constant morphism of smooth projective curves
Then
∑ P ∈ ϕ − 1 Q e ϕ ( P ) = deg ⁡ ϕ ∀ ⁡ Q ∈ C 2 .
Moreover, if ϕ is separable then e ϕ ( P ) = 1 for all but finitely many P ∈ C 1. In particular:
- (i) ϕ is surjective (on K ¯-points)
- (ii) # ϕ − 1 ( Q ) ≤ deg ⁡ ϕ
- (iii) If ϕ is separable then equality holds in (ii) for all but finitely many Q ∈ C 2.
proof: Proof of Theorem 3.1 . L ( 2 . 0 E ) ⊂ L ( 3 . 0 E ). Pick basis 1 , x for L ( 2 . 0 E ) and 1 , x , y for L ( 3 . 0 E ).
Note: ord ⁡ 0 E ( x ) = − 2, ord ⁡ 0 E ( y ) = − 3.
The 7 elements 1 , x , y , x 2 , x y , x 3 , y 2 in the 6-dimensional space L ( 6 . 0 E ) must satisfy a dependence relation.
Leaving out x 3 or y 2 gives a basis for L ( 6 . 0 E ) since each term has a different order pole at 0 E.
Therefore the coefficients of x 3 and y 2 are non-zero. Rescaling x and y (if necessary) we get
y 2 + a 1 x y + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6
for some a i ∈ K.
Let E ′ be the curve defined by this equation (or rather its projective closure). There is a morphism
ϕ : E → E ′ ⊂ ℙ 2 P ↦ ( x ( P ) : y ( P ) : 1 ) = ( x y ( P ) : 1 : 1 y ( P ) ) 0 E ↦ ( 0 : 1 : 0 )
Then
[ K ( E ) : K ( x ) ] = deg ⁡ ( E → x ℙ 1 ) = ord ⁡ 0 E ( 1 x ) = 2 [ K ( E ) : K ( y ) ] = deg ⁡ ( E → y ℙ 1 ) = ord ⁡ 0 E ( 1 y ) = 3
This gives us a diagram of field extensions:
By the Tower Law (since 2 , 3 are coprime), we get that [ K ( E ) : K ( x , y ) ] = 1. Hence K ( E ) = K ( x , y ) = ϕ ∗ K ( E ′ ), so deg ⁡ ϕ = 1, so ϕ is birational.
If E ′ is singular then E and E ′ are rational, contradiction.
So E ′ is smooth. Then Remark 2.9 implies that ϕ − 1 is a morphism. So ϕ is an isomorphism. □
cloze1: True

html: Definition (Elliptic curve). An elliptic curve E ∕ K is a smooth projective curve of genus 1, defined over K with a specified K-rational point 0 E ∈ E ( K ).
nothing: True

html: Theorem 3.1. Assuming that:
- E is an elliptic curve
Then E is isomorphic over K to a curve in Weierstrass form via an isomorphism taking 0 E to ( 0 : 1 : 0 ).
proof: Proof of Theorem 3.1 . L ( 2 . 0 E ) ⊂ L ( 3 . 0 E ). Pick basis 1 , x for L ( 2 . 0 E ) and 1 , x , y for L ( 3 . 0 E ).
Note: ord ⁡ 0 E ( x ) = − 2, ord ⁡ 0 E ( y ) = − 3.
The 7 elements 1 , x , y , x 2 , x y , x 3 , y 2 in the 6-dimensional space L ( 6 . 0 E ) must satisfy a dependence relation.
Leaving out x 3 or y 2 gives a basis for L ( 6 . 0 E ) since each term has a different order pole at 0 E.
Therefore the coefficients of x 3 and y 2 are non-zero. Rescaling x and y (if necessary) we get
y 2 + a 1 x y + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6
for some a i ∈ K.
Let E ′ be the curve defined by this equation (or rather its projective closure). There is a morphism
ϕ : E → E ′ ⊂ ℙ 2 P ↦ ( x ( P ) : y ( P ) : 1 ) = ( x y ( P ) : 1 : 1 y ( P ) ) 0 E ↦ ( 0 : 1 : 0 )
Then
[ K ( E ) : K ( x ) ] = deg ⁡ ( E → x ℙ 1 ) = ord ⁡ 0 E ( 1 x ) = 2 [ K ( E ) : K ( y ) ] = deg ⁡ ( E → y ℙ 1 ) = ord ⁡ 0 E ( 1 y ) = 3
This gives us a diagram of field extensions:
By the Tower Law (since 2 , 3 are coprime), we get that [ K ( E ) : K ( x , y ) ] = 1. Hence K ( E ) = K ( x , y ) = ϕ ∗ K ( E ′ ), so deg ⁡ ϕ = 1, so ϕ is birational.
If E ′ is singular then E and E ′ are rational, contradiction.
So E ′ is smooth. Then Remark 2.9 implies that ϕ − 1 is a morphism. So ϕ is an isomorphism. □
cloze1: True

html: Proposition 3.2. Assuming that:
- K is perfect
- E, E ′ are elliptic curves over K
- E , E ′ are in Weierstrass form
Then E ≅ E ′ over K if and only if the equations are related by a change of variables x = u 2 x ′ + r y = u 3 y ′ + u 2 s x ′ + t
where u , r , s , t ∈ K with u ≠ 0.
proof: Proof.
- ⇐ Obvious.
- ⇒ ⟨ 1 , x ⟩ = L ( 2 . 0 E ) = ⟨ 1 , x ′ ⟩ hence x = λ x ′ + r for some λ , r ∈ K with λ ≠ 0.
⟨ 1 , x , y ⟩ = L ( 3 . 0 E ) = ⟨ 1 , x ′ , y ′ ⟩ implies y = μ y ′ + σ x ′ + t for some μ , σ , t ∈ K with μ ≠ 0.
Looking at coefficients of x 3 and y 2, we get λ 3 = μ 2. So λ = u 2, μ = u 3 for some 0 ≠ u ∈ K.
Put s = σ u 2. □
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html: Corollary 3.3. Assuming that:
- char ⁡ K ≠ 2 , 3 and K is perfect
- elliptic curves E : y 2 = x 3 + a x + b, E ′ : y 2 = x 3 + a ′ x + b ′
Then E and E ′ are isomorphic over K if and only if a ′ = u 4 a, b ′ = u 6 b for some u ∈ K ∗.
proof: Proof. E and E ′ are related by a substitution as in Proposition 3.2 with r = s = t = 0. □
cloze1: True

html: Definition (j-invariant). The j-invariant is
j ( E ) = 1 7 2 8 ( 4 a 3 ) 4 a 3 + 2 7 b 2 .
nothing: True

html: Corollary 3.4. Assuming that:
- E ≅ E ′
Then j ( E ) = j ( E ′ ). Moreover, the converse holds if K = K ¯.
proof: Proof.
E ≅ E ′ ⟺ { a ′ = u 4 a b ′ = u 6 b for some u ∈ K ∗ ⟹ ( a 3 : b 2 ) = ( ( a ′ ) 3 : ( b ′ ) 2 ) ⟺ j ( E ) = j ( E ′ )
and the converse holds if K = K ¯ (to go backwards on the ⟹ step, we only need to take some kind of n-th root). □
cloze1: True

html: Theorem 4.1. ( E , ⊕ ) is an abelian group.
proof: Proof.
- (i) ⊕ is clearly commutative.
- (ii) 0 E is the identity:
0 E ⊕ P = P.
- (iii) Inverses:
Let S be the third point of intersection of T 0 E E and E, and let Q be the third point of intersection of P S and E.
Then P ⊕ Q = 0 E.
- (iv) Associativity: much harder! We will develop a bit of theory first. □

html: Definition (Linearly equivalent). D 1 , D 2 ∈ Div ⁡ ( E ) are linearly equivalent if there exists f ∈ K ¯ ( E ) ∗ such that div ⁡ ( f ) = D 1 − D 2.
Write D 1 ∼ D 2 and [ D ] = { D ′ : D ′ ∼ D }.
nothing: True

html: Proposition 4.2. Assuming that:
- we define
ψ : E → Pic ⁡ 0 ( E ) P ↦ [ ( P ) − ( 0 E ) ]
Then
- (i) ψ ( P ⊕ Q ) = ψ ( P ) + ψ ( Q ).
- (ii) ψ is a bijection.
proof: Proof.
- (i) div ⁡ ( l ∕ m ) = ( P ) + ( S ) + ( Q ) − ( 0 E ) − S − ( R ) = ( P ) + ( Q ) − ( 0 E ) − ( P ⊕ Q )
Hence ( P ) + ( Q ) ∼ ( P ⊕ Q ) + ( 0 E ). Therefore ( P ⊕ Q ) − ( 0 E ) ∼ ( P ) − ( 0 E ) + ( Q ) − ( 0 E ). So ψ ( P ⊕ Q ) = ψ ( P ) + ψ ( Q ).
- (ii) Injectivity: Suppose ψ ( P ) = ψ ( Q ) with P ≠ Q.
Then there exists f ∈ K ¯ ( E ) ∗ such that div ⁡ ( f ) = ( P ) − ( Q ), so E → f ℙ 1 has degree 1, hence E ≅ ℙ 1, contradiction.
Note: We now compute
ψ ( ( P ⊕ Q ) ⊕ R ) = ψ ( P ) + ψ ( Q ) + ψ ( R ) = ψ ( P ⊕ ( Q ⊕ R ) )
hence ( P ⊕ Q ) ⊕ R = P ⊕ ( Q ⊕ R ) for all P , Q , R ∈ E (using injectivity). So ( E , ⊕ ) is associative.
- (iii) Surjectivity: Let [ D ] ∈ Pic ⁡ 0 ( E ). Then D + ( 0 E ) has degree 1.
Riemann Roch gives dim ⁡ L ( D + ( 0 E ) ) = 1. So there exists 0 ≠ f ∈ K ¯ ( E ) such that
div ⁡ ( f ) + D + ( 0 E ) ⏟ degree 1 ≥ 0 .
So
div ⁡ ( f ) + D + ( 0 E ) = ( P )
for some P ∈ E. Then ( P ) − ( 0 E ) ∼ D, so ψ ( P ) = [ D ]. □
cloze1: True

html: Corollary 4.3. E ( K ) is an abelian group.
proof: Proof. It is a subgroup of ( E , ⊕ ). Identity 0 E ∈ E ( K ) by definition.
Closure / inverses: see formulae above.
Associative / commutative: inherited. □

html: Theorem 4.4. Elliptic curves are group varieties, i.e. [ − 1 ] : E → E; P ↦ ⊖ P and ⊕ : E × E → E; ( P , Q ) ↦ P ⊕ Q are morphisms of algebraic varieties.
proof: Proof.
- (i) Above formulae imply [ − 1 ] : E → E is a rational map, and hence a morphism (by Remark 2.9).
- (ii) Above formulae imply ⊕ : E × E → E is a rational map regular on
U = { ( P , Q ) ∈ E × E | P , Q , P ⊕ Q , P ⊖ Q ≠ 0 E } .
For P ∈ E, let τ P : E → E; X ↦ X ⊕ P “translation by P”.
τ P is a rational map, and hence a morphism (by Remark 2.9).
Take any A , B ∈ E. We factor ⊕ as
E × E → τ ⊖ A × τ ⊖ B E × E → ⊕ E → τ A ⊕ B E .
This shows ⊕ is regular on ( τ A × τ B ) ( U ) for all A , B ∈ E. Therefore ⊕ is regular on E × E. □
cloze1: True

html: Theorem (Uniformisation Theorem). Every elliptic curve over ℂ arises this way (one proof uses modular forms).
proof: Proof. Let P = ( x , y ) ∈ E. Then
[ 2 ] P = 0 ⟹ P = − P ⟹ ( x , y ) = ( x , − y ) ⟹ y = 0 □
nothing: True

html: Definition. For n ∈ ℤ, let [ n ] : E → E be defined by
P ↦ P ⊕ P ⊕ ⋯ ⊕ P ⏟ n copies
for n ≥ 0, and [ − n ] = [ − 1 ] ∘ [ n ].
cloze1: True

html: Definition (n-torsion subgroup). The n-torsion subgroup of E is
E [ n ] = ker ⁡ ( E → [ n ] E ) .
nothing: True

html: Lemma 4.5. Assuming that:
- char ⁡ K ≠ 2
- E: y 2 = f ( x ) = ( x − e 1 ) ( x − e 2 ) ( x − e 3 ), e 1 , e 2 , e 3 ∈ K ¯
Then E [ 2 ] = { 0 E , ( e 1 , 0 ) , ( e 2 , 0 ) , ( e 3 , 0 ) } ≅ ( ℤ ∕ 2 ℤ ) 2.
proof: Proof. Let P = ( x , y ) ∈ E. Then
[ 2 ] P = 0 ⟹ P = − P ⟹ ( x , y ) = ( x , − y ) ⟹ y = 0 □
cloze1: True

html: Definition (Isogeny). An isogeny ϕ : E 1 → E 2 is a nonconstant morphism with ϕ ( 0 E 1 ) = 0 E 2 (by Theorem 2.8, a morphism is nonconstant if and only if surjective on K ¯-points).
We say E 1 and E 2 are isogenous in this case.
nothing: True

html: Proposition 5.1. Assuming that:
- 0 ≠ n ∈ ℤ
Then [ n ] : E → E is an isogeny.
proof: Proof. [ n ] is a morphism by Theorem 4.4. We must show [ n ] ≠ [ 0 ].
Assume char ⁡ K ≠ 2.
Case n = 2: Lemma 4.5 implies E [ 2 ] ≠ E implies [ 2 ] ≠ [ 0 ].
Case n odd: Lemma 4.5 implies ∃ ⁡ 0 ≠ T ∈ E [ 2 ]. Then n T = T ≠ 0 which gives [ n ] ≠ [ 0 ].
Now use [ m n ] = [ m ] ∘ [ n ].
If char ⁡ K = 2, then could rpelace Lemma 4.5 with an explicit lemma about 3-torsion points. □
cloze1: True

html: Corollary 5.2. Hom ⁡ ( E 1 , E 2 ) is a torsion free ℤ-module.
proof: Proof (sketch). ϕ incudes
ϕ ∗ : Div ⁡ 0 ( E 1 ) → Div ⁡ 0 ( E 2 ) ∑ P ∈ E 1 n P P ↦ ∑ P ∈ E 1 n P ϕ ( P )
Recall ϕ ∗ : K ( E 2 ) ↪ K ( E 1 ).
Fact: If f ∈ K ¯ ( E 1 ) ∗ then
div ⁡ ( N K ( E 1 ) ∕ K ( E 2 ) f ) = ϕ ∗ ( div ⁡ f ) .
So ϕ ∗ sends principal divisors to principal divisors.
Since ϕ ( 0 E 1 ) = 0 E 2, the following diagram commutes:
ϕ ∗ a group homomorphism implies ϕ is a group homomorphism. □
cloze1: True

html: Theorem 5.3. Assuming that:
- ϕ : E 1 → E 2 an isogeny
Then ϕ ( P + Q ) = ϕ ( P ) + ϕ ( Q ) for all P , Q ∈ E 1.
proof: Proof (sketch). ϕ incudes
ϕ ∗ : Div ⁡ 0 ( E 1 ) → Div ⁡ 0 ( E 2 ) ∑ P ∈ E 1 n P P ↦ ∑ P ∈ E 1 n P ϕ ( P )
Recall ϕ ∗ : K ( E 2 ) ↪ K ( E 1 ).
Fact: If f ∈ K ¯ ( E 1 ) ∗ then
div ⁡ ( N K ( E 1 ) ∕ K ( E 2 ) f ) = ϕ ∗ ( div ⁡ f ) .
So ϕ ∗ sends principal divisors to principal divisors.
Since ϕ ( 0 E 1 ) = 0 E 2, the following diagram commutes:
ϕ ∗ a group homomorphism implies ϕ is a group homomorphism. □
cloze1: True

html: Lemma 5.4. Assuming that:
- ϕ : E 1 → E 2 an isogeny
Then there exists a morphism ξ making the following diagram commute:
(x i = x-coordinate on a Weierstrass equation for E i). Moreover if ξ ( t ) = r ( t ) s ( t ), r , s ∈ K [ t ] coprime, then
deg ⁡ ϕ = deg ⁡ ξ = max ⁡ ( deg ⁡ ⁡ ( r ) , deg ⁡ ⁡ ( s ) ) .
proof: Proof. For i = 1 , 2. K ( E i ) ∕ K ( x i ) is a degree 2 Galois extension, with Galois group generated by [ − 1 ] ∗.
Theorem 5.3 implies that ϕ ∘ [ − 1 ] = [ − 1 ] ∘ ϕ.
So if f ∈ K ( x 2 ) then
[ − 1 ] ∗ ϕ ∗ f = ϕ ∗ [ − 1 ] ∗ f = ϕ ∗ f .
Therefore ϕ ∗ f ∈ K ( x 1 ).
In particular, ϕ ∗ = ξ ( x 1 ) for some rational function ξ.
Tower Law implies deg ⁡ ϕ = deg ⁡ ξ.
Now K ( x 2 ) ↪ K ( x 1 ), x 2 ↦ ξ ( x 1 ) = r ( x 1 ) s ( x 1 ), r , s ∈ K [ t ] coprime.
We claim the minimal polynomial of x 1 over K ( x 2 ) is
F ( t ) = r ( t ) − s ( t ) x 2 ∈ K ( x 2 ) [ t ] .
Check:
- F ( x 1 ) = 0: true by the definition of our embedding.
- F is irreducible in K [ x 2 , t ] (since r , s coprime), hence irreducible in K ( x 2 ) [ t ] by Gauss’s lemma.
Therefore
deg ⁡ ξ = [ K ( x 1 ) : K ( x 2 ) ] = deg ⁡ F = max ⁡ ( deg ⁡ ⁡ r , deg ⁡ ⁡ s ) . □
cloze1: True

html: Lemma 5.5. deg ⁡ [ 2 ] = 4.
proof: Proof. Assume char ⁡ K ≠ 2 , 3 (the result is true even in the case of char ⁡ K ∈ { 2 , 3 }, but we will only prove the simpler case).
E: y 2 = x 3 + a x + b = f ( x ).
If P = ( x , y ) ∈ E, then
x ( 2 P ) = ( 3 x 2 + a 2 y ) 2 − 2 x = ( 3 x 2 + a ) 2 − 8 x f ( x ) 4 f ( x ) = x 4 + ⋯ f ( x )
The numerator and denominator are coprime. Indeed, otherwise there exists 𝜃 ∈ K ¯ with f ( 𝜃 ) = f ′ ( 𝜃 ) = 0 and hence f has a multiple root (contradiction).
Now Lemma 5.4 implies deg ⁡ [ 2 ] = max ⁡ ( 4 , 3 ) = 4. □

html: Definition (Quadratic form in an abelian group). Let A be an abelian group. We say q : A → ℤ is a quadratic form if
- (i) q ( n x ) = n 2 q ( x ) for all n ∈ ℤ, x ∈ A.
- (ii) ( x , y ) ↦ q ( x + y ) − q ( x ) − q ( y ) is ℤ-bilinear.
nothing: True

html: Lemma 5.6. q : A → ℤ is a quadratic form if and only if it satisfies the parallelogram law:
q ( x + y ) + q ( x − y ) = 2 q ( x ) = 2 q ( y ) ∀ ⁡ x , y ∈ A .
proof: Proof.
- ⇒ Let ⟨ x , y ⟩ = q ( x + y ) − q ( x ) − q ( y ). Then
⟨ x , x ⟩ = q ( 2 x ) − 2 q ( x ) = 2 q ( x )
(the equality in the last step is using property (i) with n = 2).
But by (ii),
⟨ x + y , x + y ⟩ + ⟨ x − y , x − y ⟩ = 2 ⟨ x , x ⟩ + 2 ⟨ y , y ⟩ .
Hence (upon dividing by 2),
q ( x + y ) + q ( x − y ) = 2 q ( x ) + 2 q ( y ) .
- ⇐ See Example Sheet 2. □
cloze1: True

html: Lemma 5.8. There exists W 0 , W 1 , W 2 ∈ ℤ [ a , b ] [ x 1 , x 2 ] of degree in x 1 and of degree ≤ 2 in x 2 such that
( 1 : x 3 + x 4 : x 3 x 4 ) = ( W 0 : W 1 : W 2 ) .
proof: Proof. Two methods.
- (1) Direct calculation: W 0 = ( x 1 − x 2 ) 2 W 1 = ⋯ W 2 = ⋯
see formula sheet.
- (2) Let y = λ x + ν be the line through P and Q. x 3 + a x + b − ( λ x + ν ) 2 = ( x − x 1 ) ( x − x 2 ) ( x − x 3 ) = x 3 − s 1 x 2 + x 2 x − s 3
where s i is the i-th elementary symmetric polynomial in x 1 , x 2 , x 3.
Comparing coefficients:
λ 2 = s 1 − 2 λ ν = s 2 − a ν 2 = s 3 + b
Eliminating λ and ν gives
( s 2 − a ) 2 − 4 s 1 ( s 3 + b ) = 0 ⏟ = F ( x 1 , x 2 , x 3 )
where F has degree ≤ 2 in x i.
x 3 is a root of the quadratic
W ( t ) = F ( x 1 , x 2 , t ) .
Repeating for the line through P and − Q shows that x 4 is the other root. Therefore
W 0 ( t − x 3 ) ( t − x 4 ) = W 0 t 2 − W 1 t + W 2 ,
hence
( 1 : x 3 + x 4 : x 3 x 4 ) = ( W 0 : W 1 : W 2 ) . □

html: Corollary 5.9. deg ⁡ ( n ϕ ) = n 2 deg ⁡ ( ϕ ) for all n ∈ ℤ, ϕ ∈ Hom ⁡ ( E 1 , E 2 ). In particular, deg ⁡ [ n ] = n 2.
proof: Proof. Let T i = ( e i , 0 ), E [ 2 ] = { 0 , T 1 , T 2 , T 3 }.
div ⁡ ( y ) = ( T 1 ) + ( T 2 ) + ( T 3 ) − 3 ( 0 ) . (1)
For P ∈ E ∖ { 0 },
div ⁡ ( x − x p ) = ( P ) + ( − P ) − 2 ( 0 ) .
If P ∈ E ∖ E [ 2 ], then ord ⁡ p ( x − x p ) = 1 ⟹ ord ⁡ p ( d x ) = 0.
If P ∈ E ∖ E [ 2 ] then ord ⁡ p ( x − x p ) = 1 ⟹ ord ⁡ p ( d x ) = 0.
If P = T i then ord ⁡ p ( x − x p ) = 2 ⟹ ord ⁡ p ( d x ) = 1.
If P = 0 then ord ⁡ p ( x ) = − 2 ⟹ ord ⁡ p ( d x ) = − 3.
Therefore
div ⁡ ( d x ) = ( T 1 ) + ( T 2 ) + ( T 3 ) − 3 ( 0 ) . (2)
(1) and (2) implies div ⁡ ( d x y ) = 0. □

html: Definition (Space of differentials). The space of differentials Ω C is the K ( C )-vector space generated by d f for f ∈ K ( C ), subject to the relations
- (i) d ( f + g ) = d f + d g
- (ii) d ( f g ) = f d g + g d f
- (iii) d a = 0 ∀ ⁡ a ∈ K
nothing: True

html: Definition (Regular differential). A differential ω ∈ Ω C is regular if div ⁡ ( ω ) ≥ 0, i.e. it has no poles.
nothing: True

html: Lemma 6.1. Assuming that:
- char ⁡ K ≠ 2
- E: y 2 = ( x − e 1 ) ( x − e 2 ) ( x − e 3 ), e 1 , e 2 , e 3 ∈ K distinct
Then ω = d x y is a differential on E with no zeroes or poles ⟹ g ( E ) = 1. In particular, the K-vector space of regular differentials on E is 1-dimensional, spanned by ω.
proof: Proof. Let T i = ( e i , 0 ), E [ 2 ] = { 0 , T 1 , T 2 , T 3 }.
div ⁡ ( y ) = ( T 1 ) + ( T 2 ) + ( T 3 ) − 3 ( 0 ) . (1)
For P ∈ E ∖ { 0 },
div ⁡ ( x − x p ) = ( P ) + ( − P ) − 2 ( 0 ) .
If P ∈ E ∖ E [ 2 ], then ord ⁡ p ( x − x p ) = 1 ⟹ ord ⁡ p ( d x ) = 0.
If P ∈ E ∖ E [ 2 ] then ord ⁡ p ( x − x p ) = 1 ⟹ ord ⁡ p ( d x ) = 0.
If P = T i then ord ⁡ p ( x − x p ) = 2 ⟹ ord ⁡ p ( d x ) = 1.
If P = 0 then ord ⁡ p ( x ) = − 2 ⟹ ord ⁡ p ( d x ) = − 3.
Therefore
div ⁡ ( d x ) = ( T 1 ) + ( T 2 ) + ( T 3 ) − 3 ( 0 ) . (2)
(1) and (2) implies div ⁡ ( d x y ) = 0. □
cloze1: True

html: Lemma 6.2. Assuming that:
- P ∈ E,
τ P : E → E X ↦ P + X
- ω = d x y as above
Then τ P ∗ ω = ω (we say ω is the invariant differential).
proof: Proof. τ P ∗ ω is a regular differential on E. So τ P ∗ ω = λ p ω.
The map E → ℙ 1, P ↦ λ P is a morphism of smooth projective curves but not surjective (misses 0 and ∞). Therefore it is constant (by Theorem 2.8), i.e. there exists λ ∈ K ∗ such that τ P ∗ ω = λ ω for all P ∈ E.
Taking P = 0 E shows λ = 1. □
cloze1: True

html: Lemma 6.3. Assuming that:
- ϕ , ψ ∈ Hom ⁡ ( E 1 , E 2 )
- ω an invariant differential on E 2
Then ( ϕ + ψ ) ∗ ω = ϕ ∗ ω + ψ ∗ ω.
proof: Proof. Write E = E 2
E × E → E μ : ( P , Q ) ↦ P + Q p r 1 : ( P , Q ) ↦ P p r 2 : ( P , Q ) ↦ Q
Fact: Ω E × E is a 2-dimensional K ( E × E )-vector space with basis p r 1 ∗ ω and p r 2 ∗ ω.
Therefore
μ ∗ ω = f p r 1 ∗ ω + g p r 2 ∗ ω (1)
for some f , g ∈ K ( E × E ).
For fixed Q ∈ E, let
ι Q : E → E × E P ↦ ( P , Q )
Applying ι Q ∗ to (1) gives
( μ ι Q ) ∗ ω = ( ι Q ∗ f ) ( p r 1 ι Q ) ∗ ω + ( ι Q ∗ g ) ( p r 2 ι Q ) ∗ ω τ Q ∗ ω ⏟ = ω = ( ι Q ∗ f ) ⏟ by Lemma 6.2 ω + 0
Therefore ι Q ∗ f = 1 for all Q ∈ E, so f ( P , Q ) = 1 for all P , Q ∈ E.
Similarly g ( P , Q ) = 1 for all P , Q ∈ E.
( 1 ) ⟹ μ ∗ ω = p r 1 ∗ ω + p r 2 ∗ ω .
Now pull back by
E 1 → E × E P ↦ ( ϕ ( P ) , ψ ( P ) )
to get
( ϕ + ψ ) ∗ ω = ϕ ∗ ω + ψ ∗ ω . □
cloze1: True

html: Lemma 6.4. Assuming that:
- ϕ : C 1 → C 2 a non-constant morphism
Then ϕ is separable if and only if ϕ ∗ : Ω C 1 → Ω C 2 is non-zero
proof: Proof. Omitted. □
cloze1: True

html: Theorem 6.5. Assuming that:
- char ⁡ K ∤ n
Then E [ n ] ≅ ( ℤ ∕ n ℤ ) 2.
proof: Proof. Lemma 6.3 + induction gives [ n ] ∗ ω = n ω.
char ⁡ K ∤ n implies [ n ] is separable. So # [ n ] − 1 ( Q ) = deg ⁡ [ n ] for all but finitely many Q ∈ E. But [ n ] is a group homomorphism, so [ n ] − 1 ( Q ) = # E [ n ] for all Q ∈ E.
Putting these two statements together gives
# E [ n ] = deg ⁡ [ n ] = Corollary 5.9 n 2 .
Group theory (structure theorem) gives that
E [ n ] ≅ ℤ ∕ d 1 ℤ × ⋯ × ℤ ∕ d t ℤ
for some 1 < d 1 | d 2 | ⋯ | d t | n.
Let p be a prime with p | d 1. Then E [ p ] ≅ ( ℤ ∕ p ℤ ) t. But # E [ p ] = p 2, so t = 2. But # E [ p ] = p 2, so t = 2, i.e. E [ n ] ≅ ℤ ∕ d 1 ℤ × ℤ ∕ d 2 ℤ. Since d 1 | d 2 | n and d 1 d 2 = n 2, we get d 1 = d 2 = n.
Thus E [ n ] ≅ ( ℤ ∕ n ℤ ) 2. □
cloze1: True

html: Lemma 7.1. Assuming that:
- A an abelian group
- q : A → ℤ a positive definit quadratic form
Then
| q ( x + y ) − q ( x ) − q ( y ) ⏟ = ⟨ x , y ⟩ | ≤ 2 q ( x ) q ( y ) ∀ ⁡ x , y ∈ A .
proof: Proof. We may assume x ≠ 0, otherwise the result is clear. So q ( x ) ≠ 0.
Let m , n ∈ ℤ. Then
0 ≤ q ( m x + n y ) = 1 2 ⟨ m x + n y , m x + n y ⟩ = m 2 q ( x ) + m n ⟨ x , y ⟩ + n 2 q ( y ) = q ( x ) ( m + ⟨ x , y ⟩ 2 q ( x ) n ) 2 + ( q ( y ) − ⟨ x , y ⟩ 2 4 q ( x ) ) n 2
Take m = − ⟨ x , y ⟩, n = 2 q ( x ) ≠ 0 to deduce
q ( y ) − ⟨ x , y ⟩ 2 4 q ( x ) ≥ 0
hence ⟨ x , y ⟩ 2 ≤ 4 q ( x ) q ( y ) and hence
| ⟨ x , y ⟩ | ≤ 2 q ( x ) q ( y ) .
□
cloze1: True

html: Theorem 7.2 (Hasse’s Theorem). Assuming that:
- E ∕ 𝔽 q an elliptic curve
Then
| # E ( 𝔽 q ) − ( q + 1 ) | ≤ 2 q .
proof: Proof. Recall Gal ⁡ ( 𝔽 q r ∕ 𝔽 q ) is cyclic of order r and generated by Frobenius x ↦ x q.
Let E have Weierstrass equation with coefficients a 1 , … , a 6 ∈ 𝔽 q (so a i q = a i). Define the Frobenius endomorphism
ϕ : E → E ( x , y ) ↦ ( x q , y q )
This is an isogeny of degree q.
Then
E ( 𝔽 q ) = { P ∈ E | ϕ ( P ) = P } = ker ⁡ ( 1 − ϕ ) .
ϕ ∗ ω = ϕ ∗ ( d x y ) = d ( x q ) y q = q x q − 1 d x y q = 0 ( q ≡ 0 ( m o d p ) ) Lemma 6.3 tells us that
( 1 − ϕ ) ∗ ω = ω − ϕ ∗ ( ω ) = ω ≠ 0 .
Hence 1 − ϕ is separable.
By Theorem 2.8 and the fact that 1 − ϕ is a group homomorphism, we argue as in the proof of Theorem 6.5 that
# ker ⁡ ⁡ ( 1 − ϕ ) ⏟ = # E ( 𝔽 q ) = deg ⁡ ⁡ ( 1 − ϕ ) .
deg ⁡ : Hom ⁡ ( E , E ) → ℤ is a positive definite quadratic form (Theorem 5.7, and positive definiteness is obvious since non-constant morphisms have positive degree).
Lemma 7.1 gives
| deg ⁡ ( 1 − ϕ ) − 1 − deg ⁡ ϕ | ≤ 2 deg ⁡ ϕ .
Hence
| # E ( 𝔽 q ) − 1 − q | ≤ 2 q . □
nothing: True

html: Corollary 7.3. Assuming that:
- E ∕ 𝔽 q an elliptic curve
- ϕ ∈ End ⁡ ( E ) the q-power Frobenius
Then # E ( 𝔽 q ) = q + 1 − tr ⁡ ( ϕ ) and | tr ⁡ ( ϕ ) | ≤ 2 q.
proof: Proof. Let ϕ : E → E be the q power Frobenius map. By Corollary 7.3
# E ( 𝔽 q ) = q + 1 − tr ⁡ ( ϕ ) .
Hence tr ⁡ ( ϕ ) = a, deg ⁡ ( ϕ ) = q.
Example Sheet 2, Q6(iii) implies ϕ 2 − a ϕ = q = 0, hence ϕ n = 2 − a ϕ n + 1 + q ϕ n = 0, so
tr ⁡ ( ϕ n + 2 ) − a tr ⁡ ( ϕ n + 1 ) + q tr ⁡ ( ϕ n ) = 0 .
This second order difference equation with initial conditions tr ⁡ ( 1 ) = 2, tr ⁡ ( ϕ ) = a has solution
tr ⁡ ( ϕ n ) = α n + β n ,
where α , β are roots of X 2 − a X + q = 0.
Again by Corollary 7.3,
# E ( 𝔽 q n ) = q n + 1 − tr ⁡ ( ϕ n ) = 1 + q n − α n − β n
Therefore
Z E ( T ) = exp ⁡ ∑ n = 1 ∞ ( T n n + ( q T ) n n − ( α T ) n n − ( β T ) n n ) = ( 1 − α T ) ( 1 − β T ) ( 1 − T ) ( 1 − q T ) = 1 − a T + q T 2 ( 1 − T ) ( 1 − q T ) □
cloze1: True

html: Definition (Zeta function). The zeta function Z C ( T ) of a smooth projective curve C ∕ 𝔽 q is defined by
Z C ( T ) = exp ⁡ ( ∑ n = 1 ∞ # C ( 𝔽 q n ) n T n ) .
nothing: True

html: Theorem 7.4. Assuming that:
- E ∕ 𝔽 q an elliptic curve with # E ( 𝔽 q ) = q + 1 − a
Then
Z E ( T ) = 1 − a T + q T 2 ( 1 − T ) ( 1 − q T ) .
proof: Proof. Let ϕ : E → E be the q power Frobenius map. By Corollary 7.3
# E ( 𝔽 q ) = q + 1 − tr ⁡ ( ϕ ) .
Hence tr ⁡ ( ϕ ) = a, deg ⁡ ( ϕ ) = q.
Example Sheet 2, Q6(iii) implies ϕ 2 − a ϕ = q = 0, hence ϕ n = 2 − a ϕ n + 1 + q ϕ n = 0, so
tr ⁡ ( ϕ n + 2 ) − a tr ⁡ ( ϕ n + 1 ) + q tr ⁡ ( ϕ n ) = 0 .
This second order difference equation with initial conditions tr ⁡ ( 1 ) = 2, tr ⁡ ( ϕ ) = a has solution
tr ⁡ ( ϕ n ) = α n + β n ,
where α , β are roots of X 2 − a X + q = 0.
Again by Corollary 7.3,
# E ( 𝔽 q n ) = q n + 1 − tr ⁡ ( ϕ n ) = 1 + q n − α n − β n
Therefore
Z E ( T ) = exp ⁡ ∑ n = 1 ∞ ( T n n + ( q T ) n n − ( α T ) n n − ( β T ) n n ) = ( 1 − α T ) ( 1 − β T ) ( 1 − T ) ( 1 − q T ) = 1 − a T + q T 2 ( 1 − T ) ( 1 − q T ) □
cloze1: True

html: Definition (I-adic topology). Let R be a ring and I ⊂ R an ideal. The I-adic topology on R has basis { r + I n | r ∈ R , n ≥ 1 }.
nothing: True

html: Definition (Cauchy sequence). A sequence ( x n ) in R is Cauchy if ∀ ⁡ k , ∃ ⁡ N such that ∀ ⁡ m , n ≥ N, x m − x n ∈ I k.
nothing: True

html: Definition (Complete). R is complete if
- (i) ⋂ ⁡ n ≥ 0 I n = { 0 }
- (ii) every Cauchy sequence converges
nothing: True

html: Lemma 8.1 (Hensel’s Lemma). Assuming that:
- R is complete with respect to an ideal I
- F ∈ R [ X ], s ≥ 1
- a ∈ R satisfies F ( a ) ≡ 0 ( m o d I s ), F ′ ( a ) ∈ R ×
Then there exists a unique b ∈ R such that F ( b ) = 0 and b ≡ 0 ( m o d I s ).
proof: Proof. Let u ∈ R × with F ( a ) = u ( m o d I ) (e.g. we could take u = F ′ ( a )). Replacing F ( X ) by F ( X + a ) u, we may assume a = 0, F ′ ( 0 ) ≡ 1 ( m o d I ).
We put x 0 = 0,
x n + 1 = x n − F ( x n ) (1)
Easy induction gives
x n ≡ 0 ( m o d I s ) ∀ ⁡ n (2)
Let
F ( X ) − F ( Y ) = ( X − Y ) ( F ′ ( 0 ) + X G ( X , Y ) + Y H ( X , Y ) ) (3)
for some G , H ∈ R [ X , Y ].
Claim: x n + 1 ≡ x n ( m o d I n + s ) for all n ≥ 0.
Proof: By induction on n. Case n = 0 is true.
Suppose x n ≡ x n = 1 ( m o d I n + s − 1 ). Then
F ( x n ) − F ( x n − 1 ) = ( x n − x n − 1 ) ( 1 + c ) ( m o d I n + s )
for some c ∈ I. Hence
F ( x n ) − F ( x n − 1 ) = x n − x n − 1 ( m o d I n + s )
Hence
x n − F ( x n ) ≡ x n − 1 − F ( x n − 1 ) ( m o d I n + s )
and hence x n + 1 ≡ x n ( m o d I n + s ).
This proves the claim.
Therefore ( x n ) n ≥ 0 is Cauchy. Since R is complete, we have x n → b as n → ∞ for some b ∈ R.
Taking limit n → ∞ in (1) gives b = b − F ( b ), hence F ( b ) = 0.
Taking limit n → ∞ in (2) gives b ≡ 0 ( m o d I s ).
Uniqueness is proved using (3) and the “useful remark” (if x ∈ I then 1 − x ∈ R ×). □
nothing: True

html: Lemma 8.2. Assuming that:
- R an integral domain which is complete with respect to an ideal I
- a 1 , … , a 6 ∈ R
- K = Frac ⁡ ( R )
Then
Ê Ê ( I ) : = { ( t , w ) ∈ E ( K ) | t , w ∈ I }
is a subgroup of E ( K ).
proof: Proof. Taking ( t , w ) = ( 0 , 0 ) show Ê 0 E ∈ Ê ( I ).
So it suffices to show that Ê P 1 , P 2 ∈ Ê ( I ) then
Ê − P 1 − P 2 ⏟ = P 3 ∈ Ê ( I ) .
Ê P 1 , P 2 ∈ Ê ( I ) implies t 1 , t 2 ∈ I, w 1 = w ( t 1 ) ∈ I, w 2 = w ( t 2 ) ∈ I.
w ( t ) = ∑ n = 2 ∞ A n − 2 t n + 1 ( A 0 = 1 ) λ = { w ( t 2 ) − w ( t 1 ) t 2 − t 1 t 1 ≠ t 2 w ′ ( t 1 ) t 1 = t 2 = ∑ n = 2 ∞ A n − 2 ( t 1 n + t 1 n − 1 t 2 + ⋯ + t 2 n ) ∈ I ν = w 1 − λ t 1 ∈ I
Substituting w = λ t + ν into w = f ( t , w ) gives
λ t + ν = t 3 + a 1 t ( λ t + ν ) + a 2 t 2 ( λ t + ν ) + a 3 ( λ t + ν ) 2 + a 4 t ( λ t + ν ) 2 + a 6 ( λ t + ν ) 3 A = ( coefficient of t 3 ) = 1 + a 2 λ + a 4 λ 2 + a 6 λ 3 B = ( coefficient of t 2 ) = a 1 λ + a 2 ν + a 3 λ 2 + 2 a 4 λ ν + 3 a 6 λ 2 ν
We have A ∈ R × and B ∈ I. Hence t 3 = − B A − t 1 − t 2 ∈ I, w 3 = λ t 3 + ν ∈ I. □
cloze1: True

html: Definition (Formal group). Let R be a ring. A formal group over R is a power series F ( X , Y ) ∈ R [ [ X , Y ] ] satisfying
- (i)
F ( X , Y ) = F ( Y , X )
- (ii)
F ( X , 0 ) = X and F ( 0 , Y ) = Y
- (iii)
F ( X , F ( Y , Z ) ) = F ( F ( X , Y ) , Z )
This looks like it would only define a monoid, but in fact inverses are guaranteed to exist in this context.
nothing: True

html: Definition (Morphism / isomorphic (formal groups)). Let F and G be formal groups over R given by power series F and G.
- (i)
A morphism f : F → G is a power series f ∈ R [ [ T ] ] such that f ( 0 ) = 0 satisfying
f ( F ( X , Y ) ) = G ( f ( x ) , f ( Y ) ) .
- (ii)
F ≅ G if there exists F → f G and G → g F morphisms such that f ( g ( X ) ) = g ( f ( X ) ) = X.
nothing: True

html: Theorem 8.3 (All formal groups are isomorphic). Assuming that:
- char ⁡ R = 0
Then any formal group F over R is isomorphic to 𝔾 ^ a over R ⊗ ℚ. More precisely
- (i)
There is a unique power series
log ⁡ ( T ) = T + a 2 2 T 2 + a 3 3 T 3 + ⋯
with a i ∈ R such that
log ⁡ ( F ( X , Y ) ) = log ⁡ ( X ) + log ⁡ ( Y ) . (∗)
- (ii)
There is a unique power series
exp ⁡ ( T ) = T + b 2 2 ! + b 3 3 ! T 3 + ⋯
with b i ∈ R such that
exp ⁡ ( log ⁡ ( T ) ) = log ⁡ ( exp ⁡ ( T ) ) = T .
proof: Proof.
- (i) Notation F 1 ( X , Y ) = ∂ F ∂ x ( X , Y ).
Uniqueness: Let
p ( T ) = d d T log ⁡ ( T ) = 1 + a − 2 T + a 3 T 2 + ⋯ .
Differentiating (∗) with respect to X gives
p ( F ( X , Y ) ) F 1 ( X , Y ) = p ( X ) + 0 .
Putting X = 0 gives p ( Y ) F 1 ( 0 , Y ) = 1, hence p ( Y ) = F 1 ( 0 , Y ) − 1, so p is uniquely determined by F and hence log ⁡ is too.
Existence: Let
p ( T ) = F 1 ( 0 , T ) − 1 = 1 = a 2 T + a 3 T 2 + ⋯
(say). Let
log ⁡ ( T ) = T + a 2 2 T 2 + a 3 3 T 3 + ⋯ .
Calculate
F ( F ( X , Y ) , Z ) = F ( X , F ( Y , Z ) ) F 1 ( F ( X , Y ) , Z ) F 1 ( X , Y ) = F 1 ( X , F ( Y , Z ) ) d d x F 1 ( Y , Z ) F 1 ( 0 , Y ) = F 1 ( 0 , F ( Y , Z ) ) put X = 0 ⟹ F 1 ( Y , Z ) p ( Y ) − 1 = p ( F ( Y , Z ) ) − 1 ⟹ F 1 ( Y , Z ) p ( F ( Y , Z ) ) = p ( Y ) log ⁡ ( F ( Y , Z ) ) = log ⁡ ( Y ) + h ( Z ) integrate w.r.t. Y
for some power series h.
Symmetry Y ↔ Z gives h ( Z ) = log ⁡ ( Z ).
This proves existence.
- (ii) We prove a lemma first. □
nothing: True

html: Lemma 8.4. Assuming that:
- f ( T ) = a T + ⋯ ∈ R [ [ T ] ] with a ∈ R ×
Then there exists a unique g ( T ) = a − 1 T + ⋯ ∈ R [ [ T ] ] such that f ( g ( T ) ) = g ( f ( T ) ) = T.
proof: Proof. We construct polynomials g n ( T ) ∈ R [ T ] such that
f ( g n ( T ) ) ≡ T ( m o d T n + 1 ) g n + 1 ( T ) ≡ g n ( T ) ( m o d T n + 1 )
Then g ( T ) = lim ⁡ n → ∞ g n ( T ) satisfies f ( g ( T ) ) = T.
To start the induction, we set g 1 ( T ) = a − 1 T. Now suppose n ≥ 2 and g n − 1 ( T ) exists. Then
f ( g n − 1 ( T ) ) ≡ T + b T n ( m o d T n + 1 ) .
We put g n ( T ) = g n − 1 ( T ) + λ T n for some λ ∈ R to be chosen later.
Then
f ( g n ( T ) ) = f ( g n − 1 ( T ) + λ T n ) ≡ f ( g n − 1 ( T ) ) + λ a T n ( m o d T n + 1 ) ≡ T + ( b + λ a ) T n ( m o d T n + 1 )
We take λ = − b a (a ∈ R × , b ∈ R ⟹ λ ∈ R).
This completes the induction step.
We get g ( T ) = a − 1 T + ⋯ ∈ R [ [ T ] ] such that f ( g ( T ) ) = T. Applying the same construction to g gives h ( T ) = a T + ⋯ ∈ R [ [ T ] ] such that g ( h ( T ) ) = T.
Now note f ( T ) = f ( g ( h ( T ) ) ) = h ( T ). □
cloze1: True

html: Corollary 8.5. Assuming that:
- F a formal group over R
- n ∈ ℤ such that n ∈ R ×
Then
- (i) [ n ] : F → F is an isomorphism of formal groups
- (ii) If R is complete with respect to ideal I then F ( I ) → × n F ( I ) is an isomorphism of groups. In particular, F ( I ) has no n-torsion.
proof: Proof. We have
[ 1 ] ( T ) = T [ n ] ( T ) = F ( [ n − 1 ] ( T ) , T )
(for n < 0 use [ − 1 ] ( T ) = ι ( T )).
Since
F ( X , Y ) = X + Y + X Y ( ⋯ )
we get
[ 2 ] ( T ) = F ( T , T ) 2 T + ⋯ ∈ R [ [ T ] ]
and by induction we get
[ n ] ( T ) = n T + ⋯ ∈ R [ [ T ] ] .
Lemma 8.4 shows that if n ∈ R × then [ n ] is an isomorphism. This proves (i), and (ii) follows. □
cloze1: True

html: Definition (Integral / minimal Weierstrass equation). A Weierstrass equation for E with coefficients a 1 , … , a 6 ∈ K is integral if a 1 , … , a 6 ∈ O K and minimal if v ( Δ ) is minimal among all integral Weierstrass equations for E.
nothing: True

html: Lemma 9.1. Assuming that:
- E ∕ K have integral Weierstrass equation
y 2 + a 1 x y + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6
- 0 ≠ P = ( x , y ) ∈ E ( K )
Then either x , y ∈ O K or
{ v ( x ) = − 2 s v ( y ) = − 3 s
for some s ≥ 1.
proof: Proof. Throughout this proof, LHS and RHS refer to the Weierstrass equation of the curve.
Case v ( x ) ≥ 0:. If v ( y ) < 0 then v ( LHS ⁡ ) < 0 and v ( RHS ⁡ ) ≥ 0. Therefore x , y ∈ O K.
Case v ( x ) < 0: v ( LHS ⁡ ) ≥ min ⁡ ( 2 v ( y ) , v ( x ) + v ( y ) , v ( y ) ) and v ( RHS ⁡ ) = 3 v ( x ). We get 3 possible inequalities from this, and each of them gives v ( y ) < v ( x ).
Now
v ( LHS ⁡ ) = 2 v ( y ) v ( RHS ⁡ ) = 3 v ( x )
so
{ v ( x ) = − 2 s v ( y ) = − 3 s
for some s ≥ 1. □
cloze1: True

html: Theorem 9.2. Assuming that:
- K a local field with char ⁡ K = 0, char ⁡ k = p > 0
- F a formal group over O K
- e = v ( p )
- r > e p − 1
Then
log ⁡ : F ( π r O K ) → ≅ 𝔾 ^ a ( π r O K )
is an isomorphism of groups with inverse
exp ⁡ : 𝔾 ^ a ( π r O K ) → ≅ F ( π r O K ) .
proof: Proof. For x ∈ π r O K we must show the power series log ⁡ ( x ) and exp ⁡ ( x ) converge to elements in π r O K.
Recall
exp ⁡ ( T ) = T + b 2 2 ! T 2 + b 3 3 ! T 3 + ⋯
for some b i ∈ O K.
Claim: v p ( n ! ) ≤ n − 1 p − 1.
Proof of claim:
v p ( n ! ) = ∑ r = 1 ∞ ⌊ n p r ⌋ < ∑ r = 1 ∞ n p r = n ⋅ 1 p 1 − 1 p = n p − 1 .
Therefore
( p − 1 ) v p ( n ! ) < n ( p − 1 ) v p ( n ! ) ≤ n − 1
(we go from < to ≤ ∙ − 1 by noting that the LHS is in ℤ).
This proves the claim.
Now
v ( b n x n n ! ) ≥ n r − e ( n − 1 p − 1 ) = ( n − 1 ) ( r − e p − 1 ) ⏟ > 0 + r
This is always ≥ r and → ∞ as n → ∞. Therefore exp ⁡ ( x ) converges to an element in π r O K.
Same method works for log ⁡. □
cloze1: True

html: Lemma 9.3. F ( π r O K ) F ( π r + 1 O K ) ≅ ( k , + ) for all r ≥ 1.
proof: Proof. Definition of formal group gives
F ( X , Y ) = X + Y + X Y ( ⋯ ) .
So if x , y ∈ O K,
F ( π r x , π r y ) ≡ π r ( x + y ) ( m o d π r + 1 ) .
Therefore
F ( π r O K ) → ( k , + ) π r x ↦ x mod π
is a surjective group homomorphism with kernel F ( π r + 1 O K ). □
cloze1: True

html: Corollary. Assuming that:
- | k | < ∞
Then F ( π O K ) has a subgroup of finite index isomorphic to ( O K , + ).
proof: Proof. Say Weierstrass equations are related by [ u ; r , s , t ], u ∈ K ∗, r , s , t ∈ K. Then Δ 1 = u 1 2 Δ 2. Both equations minimal gives us that v ( Δ 1 ) = v ( Δ 2 ), hence u ∈ O K ∗.
Transformation formula for the a i and b i + O K is integrally closed, hence r , s , t ∈ O K.
The Weierstrass equations obtained by reducing mod π are now related by ũ [ ũ ; r ~ , s ~ , t ~ ], ũ ũ ∈ k ∗, r ~ , s ~ , t ~ ∈ k. □
cloze1: True

html: Proposition 9.4. Assuming that:
- E ∕ K be an elliptic curve
Then the reduction mod π of any two minimal Weierstrass equations for E define isomorphic curves over k.
proof: Proof. Say Weierstrass equations are related by [ u ; r , s , t ], u ∈ K ∗, r , s , t ∈ K. Then Δ 1 = u 1 2 Δ 2. Both equations minimal gives us that v ( Δ 1 ) = v ( Δ 2 ), hence u ∈ O K ∗.
Transformation formula for the a i and b i + O K is integrally closed, hence r , s , t ∈ O K.
The Weierstrass equations obtained by reducing mod π are now related by ũ [ ũ ; r ~ , s ~ , t ~ ], ũ ũ ∈ k ∗, r ~ , s ~ , t ~ ∈ k. □
cloze1: True

html: Definition (E 0 ( K )). Ẽ E 0 ( K ) = { P ∈ E ( K ) | P ~ ∈ Ẽ ns ( k ) }.
nothing: True

html: Proposition 9.5. E 0 ( K ) is a subgroup of E ( K ) and reduction modulo π is a surjective group homomorphism Ẽ E 0 ( K ) → Ẽ ns ( k ).
proof: Proof. Group homomorphism: A line l in ℙ 2 defined over K has equation
l : a X + b Y + c Z = 0 a , b , c ∈ K .
We may assume min ⁡ ( v ( a ) , v ( b ) , v ( c ) ) = 0. Reduction modulo π gives a line
ã l ~ : ã X + b ~ Y + c ~ Z = 0 .
If P 1 , P 2 , P 3 ∈ E ( K ) with P 1 + P 2 + P 3 = 0 then these points lie on a line l. So P ~ 1 , P ~ 2 , P ~ 3 lie on the line l ~. If P ~ 1 , P ~ 2 ∈ ( ~ ns k ) then P ~ 3 ∈ ( ~ ns k ).
So if P 1 , P 2 ∈ E 0 ( K ) then P 3 ∈ E 0 ( K ) and P ~ 1 + P ~ 2 + P ~ 3 = 0.
[Exercise: check this still works if # { P ~ 1 , P ~ 2 , P ~ 3 } < ∞]
Surjective: Let f ( x , y ) = y 2 + a 1 x y + a 3 y − ( x 3 + ⋯ ). Let Ẽ P ~ ∈ Ẽ ns ( k ) ∖ { 0 }, say ỹ P ~ = ( x ~ 0 , ỹ 0 ) for some x 0 , y 0 ∈ O K.
Since P ~ non-singular, either:
- (i) ∂ f ∂ x ( x 0 , y 0 ) ⁄ ≡ 0 ( m o d π ).
- (ii) ∂ f ∂ y ( x 0 , y 0 ) ⁄ ≡ 0 ( m o d π ).
If (i) then put g ( t ) = f ( t , y 0 ) ∈ O K [ t ]. Then
{ g ( x 0 ) ≡ 0 ( m o d π ) g ′ ( x 0 ) ∈ O K ∗
Hensel’s lemma gives us that there exists b ∈ O K such that
{ g ( b ) = 0 b ≡ x 0 ( m o d π )
Then ( b , y 0 ) ∈ E ( K ) has erduction P ~. asdfadsf □
cloze1: True

html: Lemma 9.6. Assuming that:
- | k | < ∞
Then E 0 ( K ) ⊂ E ( K ) has finite index.
proof: Proof. | k | < ∞ implies that O K π r O K is finite for all r ≥ 1.
Hence
O K ≅ lim ⁡ r [ ] ← O K π r O K
is a profinite group, hence compact.
Then ℙ n ( K ) is the union of sets
{ ( a 0 : ⋯ : a i − 1 : 1 : a i + 1 : ⋯ : a n ) : a j ∈ O K }
and hence compact (for the π-adic topology).
Now note E ( K ) ⊂ ℙ 2 ( K ) is a closed subset, hence compact.
So E ( K ) is a compact topological group.
If Ẽ Ẽ has a singular point ỹ ( x ~ 0 , ỹ 0 ) then
E ( K ) ∖ E 0 ( K ) = { ( x , y ) ∈ E ( K ) | v ( x − x 0 ) ≥ 1 , v ( y − y 0 ) ≥ 1 }
is a closed subset of E ( K ) hence E 0 ( K ) is an open subgroup of E ( K ).
The cosets of E 0 ( K ) are an open cover of E ( K ). Hence [ E ( K ) : E 0 ( K ) ] < ∞. □
cloze1: True

html: Definition (Tamagawa number). c K ( E ) = [ E ( K ) : E 0 ( K ) ] is called the Tamagawa number.
nothing: True

html: Theorem 9.7. Assuming that:
- [ K : ℚ p ] < ∞
Then E ( K ) contains a subgroup of finite index isomorphic to ( O K , + ).
proof: Proof. For each m ≥ 1 there is a short exact sequence
Ẽ 0 → E 1 ( K m ) → E ( K m ) → Ẽ ( K m ) → 0 .
Taking ⋃ ⁡ m ≥ 1 gives a commutative diagram with exact rows:
An isomorphism by Corollary 8.5 applied over each K m (using p ∤ n here).
- surjective by Theorem 2.8
- kernel ≅ ( ℤ ∕ n ℤ ) 2 by Theorem 6.5 (again using p ∤ n).
Snake lemma gives
{ E ( K nr ) [ n ] ≅ ( ℤ ∕ n ℤ ) 2 E ( K nr ) n E ( K nr ) = 0
So if P ∈ E ( K ) then there exists Q ∈ E ( K nr ) such that n Q = P and
[ n ] − 1 P = { Q + T : T ∈ E [ n ] } ⊂ E ( K nr ) .
Hence K ( [ n ] − 1 P ) ⊂ K nr and so K ( [ n ] − 1 P ) ∕ K is unramified. □
cloze1: True

html: Definition (Unramified). L ∕ K is unramified if e = 1.
nothing: True

html: Definition (Maximal unramified extension). K nr = ⋃ ⁡ m ≥ 1 K m (inside K ¯). “maximal unramified extension”
nothing: True

html: Theorem 9.8. Assuming that:
- [ K : ℚ p ] < ∞
- E ∕ K has good reduction
- p ∤ n
- P ∈ E ( K )
Then K ( [ n ] − 1 P ) ∕ K is unramified.
proof: Proof. For each m ≥ 1 there is a short exact sequence
Ẽ 0 → E 1 ( K m ) → E ( K m ) → Ẽ ( K m ) → 0 .
Taking ⋃ ⁡ m ≥ 1 gives a commutative diagram with exact rows:
An isomorphism by Corollary 8.5 applied over each K m (using p ∤ n here).
- surjective by Theorem 2.8
- kernel ≅ ( ℤ ∕ n ℤ ) 2 by Theorem 6.5 (again using p ∤ n).
Snake lemma gives
{ E ( K nr ) [ n ] ≅ ( ℤ ∕ n ℤ ) 2 E ( K nr ) n E ( K nr ) = 0
So if P ∈ E ( K ) then there exists Q ∈ E ( K nr ) such that n Q = P and
[ n ] − 1 P = { Q + T : T ∈ E [ n ] } ⊂ E ( K nr ) .
Hence K ( [ n ] − 1 P ) ⊂ K nr and so K ( [ n ] − 1 P ) ∕ K is unramified. □
cloze1: True

html: Definition (Good reduction (prime)). 𝔭 is a prime of good reduction for E ∕ K if E ∕ K 𝔭 has good reduction.
nothing: True

html: Lemma 10.1. E ∕ K has only finitely many primes of bad reduction.
proof: Proof. Take a Weierstrass equation for E with a 1 , … , a 6 ∈ O K. E non-singular implies that 0 ≠ Δ ∈ O K.
Write ( Δ ) = 𝔭 α 1 ⋯ 𝔭 r α r (factorisation into prime ideals).
Let S = { 𝔭 1 , … , 𝔭 r }. If 𝔭 ∉ S then v 𝔭 ( Δ ) = 0. Hence E ∕ K 𝔭 has good reduction.
Therefore { bad primes for E } ⊂ S, hence is finite. □
cloze1: True

html: Lemma 10.2. E ( K ) tors is finite.
proof: Proof. Take any prime 𝔭. We saw that E ( K 𝔭 ) has a finite index subgroup A (say) with A ≅ ( O 𝔭 , + ).
In particular, A is torsion free
E ( K ) tors ↪ E ( K 𝔭 ) tors ↪ E ( K 𝔭 ) A ⏟ finite . □

html: Lemma 10.3. Assuming that:
- 𝔭 is a prime of good reduction
- 𝔭 ∤ n
Then reduction modulo 𝔭 gives an injective group homomorphism
Ẽ E ( K ) [ n ] ↪ Ẽ ( k 𝔭 ) .
proof: Proof. Proposition 9.5 gives that Ẽ E ( K 𝔭 ) → Ẽ ( k 𝔭 ) is a group homomorphism, with kernel E 1 ( K 𝔭 ).
Corollary 8.5 and 𝔭 ∤ n gives that E 1 ( K 𝔭 ) has no n-torsion. □
cloze1: True

html: Lemma 10.4. Assuming that:
- E ∕ ℚ is given by a Weierstrass equation with a 1 , … , a 6 ∈ ℤ
- 0 ≠ T = ( x , y ) ∈ E ( ℚ ) tors
Then
- (i) 4 x , 8 y ∈ ℤ
- (ii) If 2 | a 1 or 2 T ≠ 0 then x , y ∈ ℤ
proof: Proof.
- (i) The Weierstrass equation defines a formal group Ê Ê over ℤ. For r ≥ 1,
Ê Ê ( p r ℤ p ) = { ( x , y ) ∈ E ( ℚ p ) | v p ( x ) ≤ − 2 r , v p ( y ) ≤ − 3 r } ∪ { 0 } .
Then Theorem 9.2 gives Ê Ê ( p r ℤ p ) ≅ ( ℤ p , + ) if r > 1 p − 1. Hence Ê Ê ( 4 ℤ 2 ) and Ê Ê ( p ℤ p ) for p are odd are torsion free.
So if 0 ≠ T = ( x , y ) ∈ E ( ℚ ) tors then
v 2 ( x ) ≥ − 2 v 2 ( y ) ≥ − 3 v p ( x ) ≥ 0 v p ( y ) ≥ 0
for all odd primes p.
- (ii) Suppose Ê T ∈ Ê ( 2 ℤ 2 ), i.e. v 2 ( x ) = − 2, v 2 ( y ) = − 3.
Since
Ê Ê Ê ( 2 ℤ 2 ) Ê ( 4 ℤ 2 ) ≅ ( 𝔽 2 , + )
and Ê Ê ( 4 ℤ 2 ) is torsion free, we get 2 T = 0.
Also ( x , y ) = T = − T = ( x , − y − a 1 x − a 3 ). So
2 y + a 1 x + a 3 = 0 8 y ⏟ odd + a 1 ( 4 x ) ⏟ odd + 4 a 3 ⏟ even = 0
Hence a 1 is odd.
So if 2 T ≠ 0 or a 1 is even then Ê T ∉ Ê ( 2 ℤ 2 ), so x , y ∈ ℤ. □
cloze1: True

html: Theorem 10.5 (Lutz Nagell). Assuming that:
- E ∕ ℚ: y 2 = x 3 + a x + b, a , b ∈ ℤ
- 0 ≠ T = ( x , y ) ∈ E ( ℚ ) tors
Then x , y ∈ ℤ and either y = 0 or y 2 | ( 4 a 3 + 2 7 b 2 ).
proof: Proof. Lemma 10.4 gives that x , y ∈ ℤ.
If 2 T = 0 then y = 0. Otherwise 0 ≠ 2 T = ( x 2 , y 2 ) ∈ E ( ℚ ) tors.
Lemma 10.4 gives x 2 , y 2 ∈ ℤ. But
x 2 = ( f ′ ( x ) 2 y ) 2 − 2 x
hence y | f ′ ( x ).
E non-singular gives that f ( X ) adn f ′ ( X ) are coprime, so f ( X ) and f ′ ( X ) 2 are coprime. Therefore there exists g , h ∈ ℚ [ X ] satisfying
g ( X ) f ( X ) + h ( X ) f ′ ( X ) 2 = 1 .
Doing this and clearing denominators gives
( 3 X 2 + 4 a ) f ′ ( X ) 2 − 2 7 ( X 3 + a X − b ) f ( X ) = 4 a 3 + 2 7 b 2 .
Since y | f ′ ( x ) and y 2 = f ( x ), we get y 2 | ( 4 a 3 + 2 7 b 2 ). □
nothing: True

html: Lemma 11.1. Assuming that:
- Δ ⊂ K ∗ ∕ ( K ∗ ) n a finite subgroup
- L = K ( Δ n )
Then L ∕ K is Galois and Gal ⁡ ( L ∕ K ) ≅ ⁡ Hom ⁡ ( Δ , μ n ).
proof: Proof. μ n ⊂ K gives L ∕ K normal, and char ⁡ K ∤ n gives that L ∕ K is separable. So L ∕ K is Galois.
Define the Kummer pairing
⟨ , ⟩ : Gal ⁡ ( L ∕ K ) × Δ → μ n ( σ , x ) ↦ σ ( x n ) x n
Well-defined: Suppose α , β ∈ L with α n = β n = x. Then ( α β ) n = 1, so α β ∈ μ n ⊂ K. Then σ ( α β ) = α β for all σ ∈ Gal ⁡ ( L ∕ K ), hence σ ( α ) α = σ ( β ) β for all σ ∈ Gal ⁡ ( L ∕ K ).
Bilinear:
⟨ σ τ , x ⟩ = σ ( τ x n ) τ x n τ x n x n = ⟨ σ , x ⟩ ⟨ τ , x ⟩ ⟨ σ , x y ⟩ = σ x y n x y n = σ x n x n σ y n y n = ⟨ σ , x ⟩ ⟨ σ , y ⟩
Non-degenerate: Let σ ∈ Gal ⁡ ( L ∕ K ). If ⟨ σ , x ⟩ = 1 for all x ∈ Δ, then
σ x n = x n ∀ ⁡ x ∈ Δ
adn hence σ fixes L pointwise, i.e. σ = 1.
Let x ( K ∗ ) n ∈ Δ. If ⟨ σ , x ⟩ = 1 for all σ ∈ Gal ⁡ ( L ∕ K ), then
σ x n = x n ∀ ⁡ σ ∈ Gal ⁡ ( L ∕ K ) .
So x n ∈ K, so x ∈ ( K ∗ ) n, i.e. x ( K ∗ ) n ∈ Δ is the identity element.
We get injective group homomorphisms
- (i) Gal ⁡ ( L ∕ K ) ↪ Hom ⁡ ( Δ , μ n ).
- (ii) Δ ↪ Hom ⁡ ( Gal ⁡ ( L ∕ K ) , μ n ).
(i) implies Gal ⁡ ( L ∕ K ) is abelian and of exponent dividing n.
Fact: If G is a finite abelian group of exponent dividing n then Hom ⁡ ( G , μ n ) ≅ ⁡ G (non canonically).
So
| Gal ⁡ ( L ∕ K ) | ≤ (i) | Δ | ≤ (ii) | Gal ⁡ ( L ∕ K ) | .
Therefore (i) and (ii) are isomorphisms. □
cloze1: True

html: Definition (Abelian extension). We say L ∕ K is abelian if it is Galois, and has abelian Galois group.
Similarly for other group terminology (e.g. we can say that L ∕ K has exponent dividing n to mean that it is Galois, with Galois group having exponent dividing n).
nothing: True

html: Theorem 11.2. There is a bijection
↤ { finite subgroups Δ ⊂ K ∗ ∕ ( K ∗ ) n } ↔ { finite abelian extension L ∕ K of exponent dividing n } Δ ↦ K ( Δ n ) ( L ∗ ) n ∩ K ∗ ( K ∗ ) n ← ↤ L
proof: Proof.
- (i) Let Δ ⊂ K ∗ ∕ ( K ∗ ) n be a finite subgroup. Let L = K ( Δ n ) and Δ ′ = ( L ∗ ) n ∩ K ∗ ( K ∗ ) n.
We must show Δ = Δ ′.
Clearly Δ ⊂ Δ ′. So
L = K ( Δ n ) ⊂ K ( Δ ′ n ) ⊂ L .
So K ( Δ n ) = K ( Δ ′ n ). So Lemma 11.1 gives | Δ | = | Δ ′ |.
Since Δ ⊂ Δ ′, it follows that Δ = Δ ′.
- (ii) Let L ∕ K be a finite abelian extension of exponent dividing n. Let Δ = ( L ∗ ) n ∩ K ∗ ( K ∗ ) n. Then K ( Δ n ) ⊂ L and we aim to prove this inclusion is an equality.
Let G = Gal ⁡ ( L ∕ K ). Thu Kummer pairing gives an injection Δ ↪ Hom ⁡ ( G , μ n ).
Claim: This map is surjective.
Granted the claim,
[ K ( Δ n ) : K ] = Lemma 11.1 | Δ | = by claim | G | = [ L : K ]
Since K ( Δ n ) ⊂ L it follows that L = K ( Δ n ).
Proof of claim: Let χ : G → μ n be a group homomorphism. Distinct automorphisms are linearly independent. So there exists a ∈ L such that
∑ τ ∈ G χ ( τ ) − 1 τ ( a ) ⏟ = y ≠ 0 .
Let σ ∈ G. Then
σ ( y ) = ∑ τ ∈ G χ ( τ ) − 1 σ τ ( a ) = ∑ τ ∈ G χ ( σ − 1 τ ) − 1 τ ( a ) = χ ( σ ) ∑ τ ∈ G χ ( τ ) − 1 τ ( a ) = χ ( σ ) ⋅ y ( ∗ )
Therefore σ ( y n ) = y n for all σ ∈ G. Hence y n ∈ K.
Let x = y n. Then x ∈ K ∗ ∩ ( L ∗ ) n. Then x ( K ∗ ) n ∈ Δ.
Also, by (∗), χ : σ ↦ σ ( y ) y = σ x n x n. So the map Δ ↪ Hom ⁡ ( G , μ n ) sends x ↦ χ. This proves the claim. □
cloze1: True

html: Proposition 11.3. Assuming: - K a number field - μ n ⊂ K - S a finite set of promes of K Then there are only finitely many extensions L ∕ K such that
- (i) L ∕ K is a finite abelian extension of exponent dividing n
- (ii) L ∕ K is unramified at all 𝔭 ∉ S
proof: Proof. Theorem 11.2 gives L = K ( Δ n ) for some Δ ⊂ K ∗ ∕ ( K ∗ ) n a finite subgroup.
Let 𝔭 be a prime of K.
𝔭 O L = P 1 e 1 ⋯ P r e r
for P 1 , … , P r some distinct primes in O L.
If x ∈ K ∗ represents an element of Δ then
n v P i ( x n ) = v P i ( x ) = e i v 𝔭 ( x ) .
If 𝔭 ∉ S then all e i = 1, so v 𝔭 ( x ) ≡ 0 ( m o d n ). Therefore Δ ⊂ K ( S , n ) where
K ( S , n ) = { x ∈ K ∗ ∕ ( K ∗ ) n | v 𝔭 ( x ) ≡ 0 ( m o d n ) ∀ ⁡ 𝔭 ∉ S } .
The proof is completed by the next lemma. □

html: Lemma 11.4. K ( S , n ) is finite.
proof: Proof. The map
K ( S , n ) → ( ℤ ∕ n ℤ ) | S | x ↦ ( v 𝔭 ( x ) ( m o d n ) ) 𝔭 ∈ S
is a group homomorphism with kernel K ( ∅ , n ).
Since | S | < ∞, it suffices to prove the lemma with S = ∅.
If x ∈ K ∗ represents an element of K ( ∅ , n ) then ( x ) = 𝔞 n for some fractional ideal 𝔞. There is a short exact sequence
0 → O K ∗ ∕ ( O K ∗ ) n → K ( ∅ , n ) → Cl ⁡ K → 0 x ( K ∗ ) n ↦ [ 𝔞 ]
| Cl ⁡ K | < ∞ and O K ∗ being a finitely generated abelian group (Dirichlet’s unit theorem) gives us that K ( ϕ , n ) is finite. □
cloze1: True

html: Theorem 12.1. Assuming that:
- E ∕ K an elliptic curve
- L ∕ K a finite Galois extension
Then the natural map E ( K ) n E ( K ) → E ( L ) n E ( L ) has finite kernel.
proof: Proof. For each element in the kernel we pick a coset representative P ∈ E ( K ) and then Q ∈ E ( L ) such that n Q = P.
For any σ ∈ Gal ⁡ ( L ∕ K ) we have
n ( σ Q − Q ) = σ P − P = 0 .
So σ Q − Q ∈ E [ n ].
Since Gal ⁡ ( L ∕ K ) and E [ n ] are finite, there are only finitely many possibilities for the map
Gal ⁡ ( L ∕ K ) → E [ n ] σ ↦ σ Q − Q
(even without requiring it to be a group homomorphism!).
So we have a map
ker ⁡ ( E ( K ) n E ( K ) → E ( L ) n E ( L ) ) → ( ∗ ) Maps ⁡ ( Gal ⁡ ( L ∕ L ) , E [ n ] ) P + n E ( K ) ↦ ( σ ↦ σ Q − Q ) where n Q = P
It remains to show (∗) is injective.
So suppose P 1 , P 2 ∈ E ( K ), P i = n Q i for i = 1 , 2, and suppose σ Q 1 − Q 1 = σ Q 2 − Q 2 for all σ ∈ Gal ⁡ ( L ∕ K ). Then σ ( Q 1 − Q 2 ) = Q 1 − Q 2 for all σ ∈ Gal ⁡ ( L ∕ K ), hence Q 1 − Q 2 ∈ E ( K ), so P 1 − P 2 ∈ n E ( K ). Hence P 1 + n E ( K ) = P 2 + n E ( K ) as desired. □
cloze1: True

html: Theorem (Weak Mordell-Weil Theorem). Assuming that:
- K a number field
- E ∕ K a elliptic curve
- n ≥ 2 an integer
Then E ( K ) n E ( K ) is finite.
proof: Proof. Theorem 12.1 tells us that we may replace K by a finite Galois extension.
So without loss of generality μ n ⊂ K and E [ n ] ⊂ E ( K ). Let
S = { 𝔭 | n } ∪ { primes of bad reduction for E ∕ K } .
For each P ∈ E ( K ), the extension K ( [ n ] − 1 P ) ∕ K is unramified outisde S by Theorem 9.8. Since Gal ⁡ ( K ¯ ∕ K ) acts on [ n ] − 1 P, it follows that Gal ⁡ ( K ¯ ∕ K ( [ n ] − 1 P ) ) is a normal subgroup of Gal ⁡ ( K ¯ ∕ K ) and hence K ( [ n ] − 1 P ) ∕ K is a Galois extension.
Let Q ∈ [ n ] − 1 P. Since E [ n ] ⊂ E ( K ), we have K ( Q ) = K ( [ n ] − 1 P ). Consider
Gal ⁡ ( K ( Q ) ∕ K ) → E [ n ] ≅ ( ℤ ∕ n ℤ ) 2 σ ↦ σ Q − Q
Group homomorphism: σ τ Q − Q = σ ( τ Q − Q ) + ( σ Q − Q ).
Injective: If σ Q = Q then σ fixes K ( Q ) pointwise, i.e. σ = 1.
Therefore K ( Q ) ∕ K is an abelian extension of exponent n, unramified outside S.
Proposition 11.3 shows that as we vary P ∈ E ( K ) there are only finitely many possibilities for K ( Q ).
Let L be the composite of all such extensions of K. Then L ∕ K is finite and Galois, and E ( K ) n E ( K ) → E ( L ) n E ( L ) is the zero map. Theorem 12.1 gives | E ( K ) n E ( K ) | < ∞. □
nothing: True

html: Theorem (Mordell-Weil Theorem). Assuming that:
- K a number field
- E ∕ K an elliptic curve
Then E ( K ) is a finitely generated abelian group.
proof: Proof. Fix an integer n ≥ 2.
Weak Mordell-Weil Theorem implies that | E ( K ) n E ( K ) | ∞. Pick coset representatives P 1 , P 2 , … , P m. Let
ĥ Σ = { P ∈ E ( K ) | ( P ) ^ ≤ max ⁡ 1 ≤ i ≤ m ĥ ( P i ) } .
Claim: Σ generates E ( K ).
If not, then there exists P ∈ E ( K ) ∖ ( subgroup generated by Σ ) of minimal height (exists by (∗)). Then P = P i + n Q for some i and Q ∈ E ( K ). Note that Q ∈ E ( K ) ∖ ( subgroup generated by Σ ).
Minimal choice of P gives
ĥ ĥ ĥ ĥ ĥ ĥ ĥ ĥ ĥ 4 ĥ ( P ) ≤ 4 ĥ ( Q ) ≤ n 2 ĥ ( Q ) = ĥ ( n Q ) = ĥ ( P − P i ) ≤ ĥ ( P − P i ) + ĥ ( P + P i ) = 2 ĥ ( P ) + 2 ĥ ( P i ) parallelogram law
Therefore ĥ ĥ ĥ ( P ) ≤ ĥ ( P i ). Hence P ∈ Σ (by definition of Σ), which contradicts the choice of P.
This proves the claim.
By (∗), Σ is finite. □
nothing: True

html: Definition (Height of a point). H ( P ) = max ⁡ 0 ≤ i ≤ n | a i |.
nothing: True

html: Lemma 13.1. Assuming that:
- f 1 , f 2 ∈ ℚ [ X 1 , X 2 ] coprime homogeneous polynomials of degree d
- let
F : ℙ 1 → ℙ 1 ( x 1 : x 2 ) ↦ ( f 1 ( x 1 , x 2 ) : f 2 ( x 1 , x 2 ) )
Then there exist c 1 , c 2 > 0 such that for all P ∈ ℙ 1 ( ℚ ),
c 1 H ( P ) d ≤ H ( F ( P ) ) ≤ c 2 H ( P ) d .
proof: Proof. Without loss of generality f 1 , f 2 ∈ ℤ [ X 1 , X 2 ].
Upper bound: Write P = ( a 1 : a 2 ), a 1 , a 2 ∈ ℤ coprime.
H ( F ( P ) ) ≤ max ⁡ ( | f 1 ( a 1 , a 2 ) | , | f 2 ( a 1 , a 2 ) | ) ≤ c 2 max ⁡ ( | a 1 | d , | a 2 | d )
where
c 2 = max ⁡ i = 1 , 2 (sum of absolute values of coefficients of f i ) .
Therefore H ( F ( P ) ) ≤ c 2 H ( P ) d.
Lower bound: We claim that there exists g i j ∈ ℤ [ X 1 , X 2 ] homogeneous of degree d − 1 and κ ∈ ℤ > 0 such that
∑ j = 1 2 g i j f j = κ X i 2 d − 1 , i = 1 , 2 . (†)
Indeed, running Euclid’s algorithm on f 1 ( X , 1 ) and f 2 ( X , 1 ) gives r , s ∈ ℚ [ X ] of degree < d such that
r ( X ) f 1 ( X , 1 ) + s ( X ) f 2 ( X , 1 ) = 1 .
Homogenising and clearing denominators gives (†) for i = 2. Likewise for i = 1.
Write P = ( a 1 : a 2 ) with a 1 , a 2 ∈ ℤ coprime. (†) gives
∑ j = 1 2 g i j ( a 1 , a 2 ) f j ( a 1 , a 2 ) = κ a i 2 d − 1 , i = 1 , 2 .
Therefore gcd ⁡ ( f 1 ( a 1 , a 2 ) , f 2 ( a 1 , a 2 ) ) divides gcd ⁡ ( κ a 1 2 d − 1 , κ a 2 2 d − 1 ) = κ.
| κ a i 2 d − 1 | ≤ max ⁡ j = 1 , 2 | f j ( a 1 , a 2 ) | ⏟ ≤ κ H ( F ( P ) ) ∑ j = 1 2 | g i j ( a 1 , a 2 ) | ⏟ ≤ γ i H ( D ) d − 1
where
γ i = ∑ j = 1 2 (sum of absolute values of coefficients of g i j ) .
Therefore
κ | a i | 2 d − 1 ≤ κ H ( F ( P ) ) γ i H ( P ) d − 1
so
H ( P ) 2 d − 1 ≤ max ⁡ ( γ 1 , γ 2 ) H ( F ( P ) ) H ( P ) d − 1
so
1 max ⁡ ( γ 1 , γ 2 ) ⏟ = c 1 H ( P ) d ≤ H ( F ( P ) ) . □
cloze1: True

html: Definition (Height of a point). Let E ∕ ℚ be an elliptic curve, y 2 = x 3 + a x + b.
Define the height
H : E ( ℚ ) → ℝ ≥ 1 P ↦ { H ( x ) if P = ( x , y ) 1 if P = 0 E
Alsdefine logarithmic height
h : E ( ℚ ) → ℝ ≥ 0 P ↦ log ⁡ H ( P )
nothing: True

html: Lemma 13.2. Assuming that:
- E, E ′ are elliptic curves over ℚ
- ϕ : E → E ′ an isogeny defined over ℚ
Then there exists c > 0 such that
| h ( ϕ ( P ) ) − ( deg ⁡ ϕ ) h ( P ) | ≤ c ∀ ⁡ P ∈ E ( ℚ ) .
proof: Proof. Recall (Lemma 5.4)
deg ⁡ ϕ = deg ⁡ ξ (= d say). Lemma 13.1 tells us that there exists c 1 , c 2 > 0 such that
c 1 H ( P ) d ≤ H ( ϕ ( P ) ) ≤ c 2 H ( P ) d ∀ ⁡ P ∈ E ( ℚ ) .
Taking logs gives
| h ( ϕ ( P ) ) − d h ( P ) | ≤ max ⁡ ( log ⁡ ⁡ c 2 − log ⁡ ⁡ c 1 ) ⏟ = c □
cloze1: True

html: Definition (Canonical height of a point). The canonical height is
ĥ ĥ ( P ) = lim ⁡ n → ∞ 1 4 n h ( 2 n P ) .
nothing: True

html: Lemma 13.3. ĥ | h ( P ) − ĥ ( P ) | is bounded for P ∈ E ( ℚ ).
proof: Proof. Put n = 0 in above calculation to get
| 1 4 m h ( 2 m P ) − h ( P ) | < c 3 .
Take limit m → ∞. □
cloze1: True

html: Lemma 13.4. Assuming that:
- B > 0
Then
ĥ # { P ∈ E ( ℚ ) | ĥ ( P ) ≤ B } < ∞ .
proof: Proof. ĥ ĥ ( P ) bounded means we have a bound on h ( P ) (by Lemma 13.3). So only finitely many possibilities for x. Each x gives ≤ 2 choices for y. □
cloze1: True

html: Lemma 13.5. Assuming that:
- ϕ : E → E ′ an isogeny defined over ℚ
Then
ĥ ĥ ĥ ( ϕ P ) = ( deg ⁡ ϕ ) ĥ ( P ) ∀ ⁡ P ∈ E ( ℚ ) .
proof: Proof. By ?? 66, there exists c > 0 such that
| h ( ϕ P ) − ( deg ⁡ ϕ ) h ( P ) | < c ∀ ⁡ P ∈ E ( ℚ ) .
Replace P by 2 n P, divide by 4 n and take limit n → ∞. □
cloze1: True

html: Lemma 13.6. Assuming that:
- E ∕ ℚ an elliptic curve
Then there exists c > 0 such that for all P , Q ∈ E ( ℚ ) with P , Q , P + Q , P − Q ≠ 0, we have
H ( P + Q ) H ( P − Q ) ≤ c H ( P ) 2 H ( Q ) 2 .
proof: Proof. Let E have Weierstrass equation y 2 = x 3 + a x + b, a , b ∈ ℤ. Let P , Q , P + Q , P − Q have x coordinates x 1 , … , x 4. By Lemma 5.8, there exists W 0 , W 1 , W 2 ∈ ℤ [ x 1 , x 2 ] of degree ≤ 2 in x 2 such that
( 1 : x 3 + x 4 : x 3 x 4 ) = ( W 0 ⏟ = ( x 1 − x 2 ) 2 : W 1 : W 2 ) .
Write x i = r i s i with r i , s i ∈ ℤ coprime.
( s 3 s 4 : r 3 s 4 + r 4 s 3 : r 3 r 4 ) ⏟ gcd ⁡ = 1 = ( ( r 1 s 2 − r 2 s 1 ) 2 : ⋯ ) H ( P + Q ) H ( P − Q ) = max ⁡ ( | r 3 | , | s 3 | ) max ⁡ ( | r 4 | , | s 4 | ) ≤ 2 max ⁡ ( | s 3 s 4 | , | r 3 s 4 + r 4 s 3 | , | r 3 r 4 | ) ≤ 2 max ⁡ ( | r 1 s 2 − r 2 s 1 | 2 , … ⁡ ) ≤ c max ⁡ ( | r 1 | 2 , | s 1 | ) 2 max ⁡ ( | r 2 | , | s 2 | ) 2 = c H ( P ) 2 H ( Q ) 2
where c depends on E, but not on P and Q. □
cloze1: True

html: Theorem 13.7. ĥ ĥ : E ( ℚ ) → ℝ ≥ 0 is a quadratic form.
proof: Proof. Lemma 13.6 and | h ( 2 P ) − 4 h ( P ) | bounded gives that there exists c ∈ ℝ such that
h ( P + Q ) + h ( P − Q ) ≤ 2 h ( P ) + 2 h ( Q ) + c ∀ ⁡ P , Q ∈ E ( ℚ ) .
Replacing P, Q by 2 n P, 2 n Q, dividing by 4 n and taking the limit n → ∞ gives
ĥ ĥ ĥ ĥ ĥ ( P + Q ) + ĥ ( P − Q ) ≤ 2 ĥ ( P ) + 2 ĥ ( Q ) .
Replacing P , Q by P + Q, P − Q and ĥ ĥ ĥ ( 2 P ) = 4 ĥ ( P ) gives the reverse inequality. Therefore ĥ ĥ satisfies the parallelogram law, and hence ĥ ĥ is a quadratic form. □

html: Proposition 14.1. Assuming that:
- Φ ⊂ E ( K ¯ ) be a finite Gal ⁡ ( K ¯ ∕ K )-stable subgroup
Then there exists an elliptic curve E ′ ∕ K and a separable isogeny ϕ : E → E ′ defined over K, with kernel Φ, such that every isogeny ψ : E → E ″ with Φ ⊂ ψ factors uniquely via ϕ:
proof: Proof. Omitted (see Silverman, Chapter III, Proposition 4.12). □
cloze1: True

html: Proposition 14.2. Assuming that:
- ϕ : E → E ′ an isogeny of degree n
Then there exists a unique isogeny ϕ ^ : E ′ → E such that ϕ ^ ϕ = [ n ].
proof: Proof. Case ϕ is separable: We have | ker ⁡ ϕ | = n, hence ker ⁡ ϕ ⊂ E [ n ]. Apply Proposition 14.1 with ψ = [ n ].
Case ϕ is inseparable: omitted.
Uniqueness: Suppose ψ 1 ϕ = ψ 2 ϕ = [ n ]. Then ( ψ 1 − ψ 2 ) ϕ = 0, so deg ⁡ ( ψ 1 − ψ 2 ) deg ⁡ ϕ = 0. Then deg ⁡ ( ψ 1 − ψ 2 ) = 0, hence ψ 1 = ψ 2. □
cloze1: True

html: Lemma 14.3. Assuming that:
- ϕ , ψ ∈ Hom ⁡ ( E , E ′ )
Then ϕ + ψ ^ = ϕ ^ + ψ ^.
proof: Proof.
- (i) If E = E ′, then this follows from Tr ⁡ ( ϕ + ψ ) = Tr ⁡ ( ϕ ) + Tr ⁡ ( ψ ).
- (ii) In general, let α : E ′ → E be any isogeny (e.g. ϕ ^). Then: ( α ϕ + α ψ ) ^ = α ϕ ^ + α ψ ^ ( ϕ + ψ ) ^ α ^ = ( ϕ ^ + ψ ^ ) α ^
Hence the result follows. □
cloze1: True

html: Lemma 14.4. Assuming that:
- D ∈ Div ⁡ ( E )
Then D ∼ 0 if and only if deg ⁡ D = 0 and sum ⁡ D = 0 E.
proof: Proof.
- (i) Linearity in first argument: e ( S 1 + S 2 , T ) = g ( X + S 1 + S 2 ) g ( X + S 2 ) g ( X + S 2 ) g ( X ) = e ( S , T ) e ( S 2 , T )
- (ii) Linearity in second argument: Let T 1 , T 2 ∈ E ′ [ ϕ ^ ]. div ⁡ ( f 1 ) = n ( T 1 ) − n ( 0 ) ϕ ∗ f = g 1 n div ⁡ ( f 2 ) = n ( T 2 ) − n ( 0 ) ϕ ∗ f 2 = g 2 n
There exists h ∈ K ¯ ( E ′ ) such that
div ⁡ ( h ) = ( T 1 ) + ( T 2 ) − ( T 1 + T 2 ) − ( 0 ) .
Then put f = f 1 f 2 h n and g = g 1 g 2 ϕ ∗ h.
Check: div ⁡ ( f ) = n ( T 1 + T 2 ) − n ( 0 ). Yes.
ϕ ∗ f = ϕ ∗ f 1 ϕ ∗ f 2 ( ϕ ∗ h ) n = ( g 1 g 2 ϕ ∗ ( h ) ) n = g n .
Therefore
e ( S , T 1 + T 2 ) = g ( X + S ) g ( X ) = g 1 ( X + S ) g 1 ( X ) g 2 ( X + S ) g 2 ( X ) h ( ϕ ( X ) ) h ( ϕ ( X + S ) ) ⏟ = 1 since S ∈ E [ ϕ ] = e ( S , T 1 ) e ( S , T 2 )
- (iii) e is non-degenerate. Fix T ∈ E ′ [ ϕ ^ ]. Suppose e ( S , T ) = 1 for all S ∈ E [ ϕ ]. Then τ S ∗ g = g for all S ∈ E [ ϕ ].
Have K ¯ ( E ) ∕ ϕ ∗ K ¯ ( E ′ ) is a Galois extension with Galois group E [ ϕ ] (S ∈ E [ ϕ ] acts on K ¯ ( E ) via τ S ∗).
Therefore g = ϕ ∗ h for some h ∈ K ¯ ( E ′ ). So ϕ ∗ f = g n = ϕ ∗ ( h n ). So f = h n, and hence div ⁡ ( h ) = ( T ) − ( 0 ). So T = 0. □
cloze1: True

html: Proposition 14.5. e is bilinear and non-degenerate.
proof: Proof.
- (i) Linearity in first argument: e ( S 1 + S 2 , T ) = g ( X + S 1 + S 2 ) g ( X + S 2 ) g ( X + S 2 ) g ( X ) = e ( S , T ) e ( S 2 , T )
- (ii) Linearity in second argument: Let T 1 , T 2 ∈ E ′ [ ϕ ^ ]. div ⁡ ( f 1 ) = n ( T 1 ) − n ( 0 ) ϕ ∗ f = g 1 n div ⁡ ( f 2 ) = n ( T 2 ) − n ( 0 ) ϕ ∗ f 2 = g 2 n
There exists h ∈ K ¯ ( E ′ ) such that
div ⁡ ( h ) = ( T 1 ) + ( T 2 ) − ( T 1 + T 2 ) − ( 0 ) .
Then put f = f 1 f 2 h n and g = g 1 g 2 ϕ ∗ h.
Check: div ⁡ ( f ) = n ( T 1 + T 2 ) − n ( 0 ). Yes.
ϕ ∗ f = ϕ ∗ f 1 ϕ ∗ f 2 ( ϕ ∗ h ) n = ( g 1 g 2 ϕ ∗ ( h ) ) n = g n .
Therefore
e ( S , T 1 + T 2 ) = g ( X + S ) g ( X ) = g 1 ( X + S ) g 1 ( X ) g 2 ( X + S ) g 2 ( X ) h ( ϕ ( X ) ) h ( ϕ ( X + S ) ) ⏟ = 1 since S ∈ E [ ϕ ] = e ( S , T 1 ) e ( S , T 2 )
- (iii) e is non-degenerate. Fix T ∈ E ′ [ ϕ ^ ]. Suppose e ( S , T ) = 1 for all S ∈ E [ ϕ ]. Then τ S ∗ g = g for all S ∈ E [ ϕ ].
Have K ¯ ( E ) ∕ ϕ ∗ K ¯ ( E ′ ) is a Galois extension with Galois group E [ ϕ ] (S ∈ E [ ϕ ] acts on K ¯ ( E ) via τ S ∗).
Therefore g = ϕ ∗ h for some h ∈ K ¯ ( E ′ ). So ϕ ∗ f = g n = ϕ ∗ ( h n ). So f = h n, and hence div ⁡ ( h ) = ( T ) − ( 0 ). So T = 0. □
cloze1: True

html: Corollary 14.6. Assuming that:
- E [ n ] ⊂ E ( K )
Then μ n ⊂ K.
proof: Proof. Let T ∈ E [ n ] have order n. Non degeneracy of e n implies that there exists S ∈ E [ n ] such that e n ( S , T ) is a primitive n-th root of unity, say ζ n.
Then
σ ( ζ n ) = σ ( e n ( S , T ) ) = e n ( σ S , σ T ) = ζ n
for all σ ∈ Gal ⁡ ( K ¯ ∕ K ). So ζ n ∈ K. □
cloze1: True

html: Definition (H 0). Define
H 0 ( G , A ) = A G = { a ∈ A | σ ( a ) = a ∀ ⁡ σ ∈ G } .
nothing: True

html: Definition (Cochains). Define
C 1 ( G , A ) = { maps G → A }
(called “cochains”).
nothing: True

html: Definition (Cocycles). Define
Z 1 ( G , A ) = { ( a σ ) σ ∈ G | a σ τ = σ ( a τ ) + a σ ∀ ⁡ σ , τ ∈ G }
(called “cocycles”).
nothing: True

html: Definition (Coboundaries). Define
B 1 ( G , A ) = { ( σ b − b ) σ ∈ G | b ∈ A }
(called “coboundaries”).
nothing: True

html: Definition (H 1). Define
H 1 ( G , A ) = Z 1 ( G , A ) B 1 ( G , A ) .
nothing: True

html: Theorem 15.1. Assuming that:
- we have a short exact sequence of G-modules
0 → A → ϕ B → ψ C → 0 .
Then it gives rise to a long exact sequence of abelian groups:
0 → A G → ϕ B G → ψ C G → δ H 1 ( G , A ) → ϕ ∗ H 1 ( G , B ) → ψ ∗ H 1 ( G , C ) .
proof: Proof. Omitted. □
cloze1: True

html: Definition (d e l t a). Let c ∈ C G. Then b ∈ B such that ψ ( b ) = c. Then
ψ ( σ b − b ) = σ c − c = 0 ∀ ⁡ σ ∈ G
so σ b − b = ϕ ( a σ ) for some a σ ∈ A.
Can check ( a σ ) σ ∈ G ∈ Z 1 ( G , A ).
Then δ ( c ) = class of ( a σ ) σ ∈ G in H 1 ( G , A ).
nothing: True

html: Theorem 15.2. Assuming that:
- A is a G-module
- H ⊴ G a normal subgroup
Then there is an inflation restriction exact sequence
0 → H 1 ( G ∕ H , A H ) → inf ⁡ H 1 ( G , A ) → res ⁡ H 1 ( H , A )
proof: Proof. Omitted. □
cloze1: True

html: Theorem (Hilbert’s Theorem 90). Assuming that:
- L ∕ K a finite Galois extension
Then
H 1 ( Gal ⁡ ( L ∕ K ) , L ∗ ) = 0 .
proof: Proof. Let G = Gal ⁡ ( L ∕ K ). Let ( a σ ) σ ∈ G ∈ Z 1 ( G , L ∗ ). Distinct automorphisms are linearly independent, so there exists y ∈ L such that
∑ τ ∈ G a τ − 1 τ ( y ) ⏟ = x ≠ 0 .
Then
σ ( x ) = ∑ τ ∈ G σ ( a τ ) − 1 σ τ ( y ) = a σ ∑ τ ∈ G a σ τ − 1 σ τ ( y ) ⏟ = x
Then a σ = σ ( x ) x for all σ ∈ G. (a σ τ = σ ( a τ ) a σ ⟹ σ ( a τ ) − 1 = a σ a σ τ − 1).
So ( a σ ) σ ∈ G ∈ B 1 ( G , L ∗ ). Therefore H 1 ( G , L ∗ ) = 0. □
nothing: True

html: Definition (Selmer-group). We define the ϕ-Selmer group
S ( ϕ ) ( E ∕ K ) = ker ⁡ ( H 1 ( K , E [ ϕ ] ) → ∏ v H 1 ( K v , E ) ) = { α ∈ H 1 ( K , E [ ϕ ] ) | res ⁡ v ( α ) ∈ Im ⁡ ( δ v ) ∀ ⁡ v }
(the map H 1 ( K , E [ ϕ ] ) → ∏ ⁡ v H 1 ( K v , E ) is as in the commutative diagram above).
nothing: True

html: Definition (Tate-Shafarevich group). The Tate-Shafarevich group is
T S ( E ∕ K ) = ker ⁡ ( H 1 ( K , E ) → ∏ v H 1 ( K v , E ) ) .
nothing: True

html: Theorem 15.3. S ( n ) ( E ∕ K ) is finite.
proof: Proof. For L ∕ K a finite Galois extension,
Therefore by extending our field, we may assume E [ n ] ⊂ E ( K ) and hence by the Weil pairing μ n ⊂ K.
Therefore E [ n ] ≅ μ n × μ n as a Gal ⁡ ( K ¯ ∕ K )-module. Then
H 1 ( K , E [ n ] ) ≅ H 1 ( K , μ n ) × H 1 ( K , μ n ) ≅ K ∗ ∕ ( K ∗ ) n × K ∗ ∕ ( K ∗ ) n
Let
S = { primes of bad reduction for E } ∪ { v | n ∞ }
(a finite set of places).
Define the subgroup of H 1 ( K , A ) unramified outside of S as
H 1 ( K , A ; S ) = ker ⁡ ( H 1 ( K , A ) → ∏ v ∉ S H 1 ( K v nr , A ) ) .
There is a commutative diagram with exact rows:
The bottom map × n is surjective ∀ ⁡ v ∉ S (see Theorem 9.8).
Therefore
S ( n ) ( E ∕ K ) ⊂ H 1 ( K , E [ n ] ; S ) ≅ H 1 ( K , μ n ; S ) × H 1 ( K , μ n ; S )
But
H 1 ( K , μ n ; S ) = ker ⁡ ( K ∗ ( K ∗ ) n → ∏ v ∉ S ( K v nr ) ∗ ( K v nr ) ∗ n ) ⊂ K ( S , n )
which is finite by Lemma 11.4. □
cloze1: True

html: Theorem 16.1. Assuming that:
- f ∈ K ( E ′ ) and g ∈ K ( E )
- div ⁡ ( f ) = n ( T ) − n ( 0 )
- ϕ ∗ f = g n
Then α ( P ) = f ( P ) ( m o d ( K ∗ ) n ) for all P ∈ E ; ( K ) ∖ { 0 , T }.
proof: Proof. Let Q ∈ ϕ − 1 P. Then δ ( P ) ∈ H 1 ( K , μ n ) is represented by σ ↦ σ Q − Q ∈ E [ ϕ ] ≅ μ n.
e ( σ Q − Q , T ) = g ( σ Q − Q + X ) g ( X ) for any X ∈ E avoiding zeroes and poles of g = g ( σ Q ) g ( Q ) pick X = Q = σ ( g ( Q ) ) g ( Q ) = σ f ( P ) n f n ( P )
(ϕ ∗ f = g n, f ( P ) = g ( Q ) n).
But
↤ H 1 ( K , μ n ) ≅ K ∗ ∕ ( K ∗ ) n ( σ ↦ x n x n ) ← ↤ x
Hence α ( P ) = f ( P ) ( m o d ( K ∗ ) n ). □
cloze1: True

html: Proposition 16.2. There is a group homomorphism
E ′ ( K ) → K ∗ ∕ ( K ∗ ) 2 ( x , y ) ↦ { x mod ( K ∗ ) 2 if x ≠ 0 b ′ mod ( K ∗ ) 2 if x = 0
with kernel ϕ E ( K ).
proof: Proof. Either: Apply Theorem 16.1 with f = x ∈ K ( E ′ ), g = y x ∈ K ( E ).
Or: direct calculation – see Example Sheet 4. □
cloze1: True

html: Lemma 16.3. 2 rank ⁡ E ( K ) = | Im ⁡ α E | ⋅ | Im ⁡ α E ′ | 4.
proof: Proof. If A → f B → g C are homomorphisms of abelian groups, then there is an exact sequence
Since ϕ ^ ϕ = [ 2 ] E, we get an exact sequence
0 → E ( K ) [ ϕ ] ⏟ ≅ ⁡ ℤ ∕ 2 ℤ → E ( K ) [ 2 ] → ϕ E ′ ( K ) [ ϕ ′ ] ⏟ ≅ ⁡ ℤ ∕ 2 ℤ → E ′ ( K ) ϕ E ( K ) ⏟ ≅ ⁡ Im ⁡ ( α E ′ ) → ϕ ^ E ( K ) 2 E ( K ) → E ( K ) ϕ ^ E ′ ( K ) ⏟ ≅ ⁡ Im ⁡ ( α E ) → 0 .
Therefore
| E ( L ) ∕ 2 E ( K ) | | E ( K ) [ 2 ] | = | Im ⁡ α E | ⋅ | Im ⁡ α E ′ | 4 . (1)
Mordell-Weil: E ( K ) ≅ Δ × ℤ r, where Δ is a finite group and r = rank ⁡ E ( K ).
E ( K ) 2 E ( K ) ≅ Δ 2 Δ × ( ℤ 2 ℤ ) r.
E ( K ) [ 2 ] ≅ Δ [ 2 ].
Since Δ is finite, we have that Δ 2 Δ and Δ [ 2 ] have the same order, and therefore
| E ( K ) ∕ 2 E ( K ) | | E ( K ) [ 2 ] | = 2 r . (2)
So we are done, by using (1) and (2). □
cloze1: True

html: Lemma 16.4. Assuming that:
- K is a number field
- a , b ∈ O K
Then Im ⁡ ( α E ) ⊂ K ( S , 2 ), where S = { 𝔭 | b }.
proof: Proof. We must show that x , y ∈ K, y 2 = x ( x 2 + a x + b ) and v 𝔭 ( b ) = 0 then v 𝔭 ( x ) ≡ 0 ( m o d 2 ).
Case v 𝔭 ( x ) < 0: Lemma 9.1 gives that for some r ≥ 1, v 𝔭 ( x ) = − 2 r and v 𝔭 ( y ) = − 3 r, so done.
Case v 𝔭 ( x ) > 0: Then v 𝔭 ( x 2 + a x + b ) = 0. So v 𝔭 ( x ) = v 𝔭 ( y 2 ) = 2 v 𝔭 ( y ). □
cloze1: True

html: Lemma 16.5. Assuming that:
- b 1 b 2 = b
Then b 1 ( K ∗ ) ∈ Im ⁡ ( α E ) if and only if
w 2 = b 1 u 4 + a u 2 v 2 + b 2 v 4 (∗)
is soluble for u , v , w ∈ K not all zero.
proof: Proof. If b 1 ∈ ( K ∗ ) 2 or b 2 ∈ ( K ∗ ) 2 then both conditions are satisfied. So we may assume b 1 , b 2 ∉ ( K ∗ ) 2.
Now note b 1 ( K ∗ ) 2 ∈ Im ⁡ ( α E ) if and only if there exists ( x , y ) ∈ E ( K ) such that x = b 1 t 2 for some t ∈ K ∗. This implies
y 2 = ( b 1 t 2 ) ( b 1 2 t 2 + a b 1 t 2 + b )
hence
( y b 1 t ) 2 = b 1 t 4 + a t 2 + b b 1 ⏟ = b 2 .
So (∗) has solution ( u , v , w ) = ( t , 1 , y b 1 t ).
Conversely, if ( u , v , w ) is a solution to (∗) then u v ≠ 0 and ( b 1 ( u v ) 2 , b 1 u w v 3 ) ∈ E ( K ). □
cloze1: True

html: Definition (L ( E , s )). Define
L ( E , s ) = ∏ p L p ( E , s )
where
L p ( E , s ) = { ( 1 − a p p − s + p 1 − 2 s ) − 1 if good reduction ( 1 − p − s ) − 1 if split multiplicative reduction ( 1 + p − s ) − 1 if nonsplit multiplicative reduction 1 if additive reduction
where Ẽ # Ẽ ( 𝔽 p ) = 1 + p − a p.
nothing: True