aa r X i v : . [ m a t h . HO ] D ec Introduction to Mordell Weil Theorem
Shenghao LiAugust 9, 2019
Abstract
This is an article about Mordell Weil Theorem. Mordell-Weil Theoremis one of the greatest theorems about ellpitic curve. In this article, I willintroduce the proof of Mordell-Weil theorem and some simple ways tocompute the torsion part of the group.
In this part, we will introduce some basic notations and properties about ellipticcurves. Some classic results won’t be proved, but I will list some books whereyou can look them up.
Definition 1.1.
An elliptic curve is a pair (E,O), where E is a nonsingularcurve of genus 1 and O ∈ E . The elliptic curve E is defined over K, writtenE/K, if E is defined over K(a field) as a curve and O ∈ E ( K ) . The definition above is not very clear and it’s hard for us to study them. So,by Riemann-Roch theorem, we have the following equivalent definition:
Definition 1.2.
An elliptic curve over K can be defined as a nonsingular pro-jective plane curve over K of the form Y Z + a XY Z + a Y Z = X + a X Z + a XZ + a Z Here O=[0,1,0] is ths basepoint.This is called the Weierstrass equation of an elliptic curve.
If we let Z = 0 , then we find that the point must be O. So we can assumethat Z = 0 , and we can get the following equation E : y + a xy + a y = x + a x + a x + a If char ( ¯ K ) = 0 , then we can simplify the equation by completing the square.Thus replacing y by y − a x − a gives an equation of the form E : y = 4 x + b x + 2 b x + b where b = a + 4 a b = 2 a + a a b = a + 4 a
1e also define quantities b = a a + 4 a a − a a a + a a − a c = b − b c = b + 36 b b − b ∆ = − b b − b − b + 9 b b b If further char( ¯ K ) = (( x − b ) / , y/ we canget a simpler equation E : y = x − c x − c Thus we often use the equation E : y = x + Ax + B to denote an elliptic curve,and such curves are nonsingular if and only if ∆ = 0 (i.e. − A + 27 B ) = 0 ).Next we will intoduce one of the most important structures on elliptic curves,the group law. Definition 1.3.
Let
P, Q ∈ E , L the line connecting P and Q(tangent line if P = Q ). According to the Bezout theorem, the line L and E must intersect onthe third point R(may be the same as P, Q).Let L’ be the line connecting R andO. Then P ⊕ Q is the point that such that L’ intersects E at R, O , and P ⊕ Q . Now we justify the use of the symbol ⊕ . Proposition 1.1.
The composition law( ⊕ ) satifies the following properties.(a) If a line L intersects E at the points P,Q and R, then ( P ⊕ Q ) ⊕ R = O (b) P ⊕ O = P for all P ∈ E (c) P ⊕ Q = Q ⊕ P for all P, Q ∈ E (d) Let P ∈ E . There is a point of E, denoted -P, so that P ⊕ ( − P ) = O (e) Let P, Q, R ∈ E . Then ( P ⊕ Q ) ⊕ R = P ⊕ ( Q ⊕ R ) In other words, the composition law makes E into an abelian group with identityelement O. We further have:(f ) Suppose E is defined over K. Then E ( K ) = (cid:8) ( x, y ) ∈ K : y + a xy + a y = x + a x + a x + a (cid:9) ∪ { O } is a subgroup of E.Proof. Only (e) is not trivial. One can laboriously verify the associative lawcase by case by checking the equations. However, we will use the RiemannRoch theorem to prove that, and also use a bit of divisors.(For definition ofdivisors, one can read [Sil], p.31) 2 roposition 1.2.
Let (E,O) be an elliptic curve.(a) For every divisor D ∈ Div ( E ) there exists a unique point P ∈ E so that D ∼ ( P ) − ( O ) Let σ : Div ( E ) → E be the map given by this association. (b) The map σ is surjective. (c) Let D , D ∈ Div ( E ) . Then σ ( D ) = σ ( D ) if f D ∼ D Thus σ induces a bijection of sets(which we also denote by σ ) σ : P ic ( E ) → E (d) The inverse to σ is the map κ : E → P ic ( E ) P → class of ( P ) − ( O ) (e) If E is given by a Weierstrass equation, then the composition law we men-tioned above and the group law from P ic ( E ) by using σ are the same. Thus,the composition law satisfied the associative law.Proof. (a) Since E has genus 1, the Riemann-Roch theorem says that dim L ( D + ( O )) = 1 Let f ∈ ¯ K ( E ) be a generator for L ( D + ( O )) . Since div ( f ) ≥ − D − ( O ) and deg ( div ( f )) = 0 it follows that div ( f ) = − D − ( O ) + ( P ) for some P ∈ E . Hence D ∼ ( P ) − ( O ) To prove that P is unique, we assume that there are two points P and P’ bothsatisfy the condition. Then we get that P ∼ P ′ . So there exists f ∈ ¯ K ( E ) sothat div ( f ) = ( P ) − ( P ′ ) Then f ∈ L (( P ′ )) , and by the Riemann-Roch theorem we have dim L (( P ′ )) = 1 .However we already know that the constant function is in L (( P ′ )) , so we canget f is a constant function. Thus P = P ′ . Hence P is unique.(b) For any P ∈ E σ (( P ) − ( O )) = P (c) Suppose σ ( D ) = P , σ ( D ) = P . Then we can get ( P ) − ( P ) ∼ D − D .Thus σ ( D ) = σ ( D ) we can imply that D ∼ D . Also if D ∼ D , we have P ∼ P , so P = P .(d) Directly from (b) and (c). 3e) Let E be given by a Weierstrass equation, and let P, Q ∈ E . It clearly sufficesto show that κ ( P + Q ) = κ ( P ) + κ ( Q ) Let f ( X, Y, Z ) = aX + bY + cZ = 0 give the line L in P going through P and Q, let R be the third point ofintersection of L with E, and let f ′ ( X, Y, Z ) = a ′ X + b ′ Y + c ′ Z = 0 be the line L in P through R and O. Then from the definition of addition onE and the fact that Z = 0 intersects E at O with multiplicity 3, we have div ( f /Z ) = ( P ) + ( Q ) + ( R ) − O ) and div ( f ′ /Z ) = ( P + Q ) + ( R ) − O ) Thus ( P + Q ) − ( P ) − ( Q ) + ( O ) = div ( f ′ /f ) Hence κ ( P + Q ) = κ ( P ) + κ ( Q ) Remark 1.
Here we will directly write out the equation of the composition law.Let E be an elliptic curve given by a Weierstrass equation E : y + a xy + a y = x + a x + a x + a (a) Let P = ( x , y ) ∈ E . Then − P = ( x , − y − a x − a ) Now let P + P = P with P i = ( x i , y i ) ∈ E (b) If x = x and y + y + a x + a = 0 , then P + P = O Otherwise, let λ = y − y x − x , ν = y x − y x x − x if x = x λ = 3 x + 2 a x + a − a y y + a x + a , ν = − x + a x + 2 a − a y y + a x + a if x = x (c) P = P + P is given by x = λ + a λ − a − x − x y = − ( λ + a ) x − ν − a (d) As special cases of (c), we have for P = ± P x ( P + P ) = y − y x − x + a y − y x − x − a − x − x and the duplication formula for (x,y) ∈ E x ([2] P ) = x − b x − b x − b x + b x + 2 b x + b Definition 1.4.
Let V and V ⊂ P n be two projective varieties. A rationalmap from V to V is a map of the form φ : V → V φ = [ f , . . . , f n ] where f , . . . , f n ∈ ¯ K ( V ) have the property that for every point P ∈ V at which f , . . . , f n are all defined, φ ( P ) ∈ V . Definition 1.5.
A rational map φ is regular(or defined) at P if there is a func-tion g ∈ ¯ K ( V ) such that (a) each gf i is regular at P (b) for some i, ( gf i )( P ) = 0 A rational map which is regular at every point in V is called a morphism. Next we will state two very important results for morphisms on curves. Wewon’t prove them and , and for those who want to see the proofs, you can check[Har, Chapter 2 Thm 6.8].
Theorem 1.1.
For every Q ∈ C , we have the following equationLet φ : C → C be a morphism of curves. Then φ is either constant or surjective. Theorem 1.2.
Let φ : C → C be a non-constant map of smooth curves. Forall but finitely many Q ∈ C φ − ( Q ) = deg s ( φ ) Now we go back to the ellpitic curves. Because an elliptic curve contains apoint O, so the map between elliptic curves should contains more imformation.Therefore we have the following definition.
Definition 1.6.
Let E and E be elliptic curves. An isogeny between E and E is a morphism φ : E → E satisfying φ ( O ) = O . E and E are isogenous if there is an isogeny φ betweenthem with φ ( E ) = { O } From the definition, we can clearly see that the map defined by multiplyingm is an isogeny, and we use [m] to denote it.
Proposition 1.3.
Let E/K be an elliptic curve and let m ∈ Z , m = 0 . Thenthe multiplication by m map [ m ] : E → E is non-constant.Proof. We start by showing that [2] = [0] . From the duplication formula, if apoint P=(x,y) ∈ E has order 2, then it must satisfy x + b x + 2 b x + b = 0 which only has finitely many solutions. Therefore [2] = [0] . Now, using the factthat [ mn ] = [ m ][ n ] , we are reduced to considering the case of odd m.5sing the long division, one can easily find out that the polynomials x + b x + 2 b x + b does not divide x − b x − b x − b (If it does, then ∆ =0, contradiction). Hence we can find an x ∈ ¯ K so that theformer vanishes to a higher order at x = x than the latter. Choosing y ∈ ¯ K so that P = ( x , y ) ∈ E , the doubling formula implies that [2] P =O.In otherwords, we have shown that E has a non-trivial point of order 2. But then for modd [ m ] P = P = O so clearly [ m ] = [0] . Theorem 1.3.
Consider the isogeny [ m ] : E → E . For every Q ∈ E m ] − ( Q ) = deg s [ m ] Proof.
From Theorem 1.2 we know that m ] − ( Q ) = deg s [ m ] for all but finitely many Q ∈ E . But for any P, P ′ ∈ E , if [ m ] P = [ m ] P ′ , then P − P ′ ∈ [ m ] − ( O ) . Thus for every Q ∈ E , [ m ] − ( Q ) is a coset of [ m ] − ( O ) .So for all Q, we have m ] − ( Q ) = deg s [ m ] By now, we have introduced some important properties of elliptic curves,and next we will introduce the Mordell Weil theorem.
Theorem 2.1. (Mordell-Weil)
Let E be an elliptic curve defined over a numberfield K. The group E(K) is a finitely generated Abelian group
The proof is given in two parts: The first part is called the Weak Mordell-Weil Theorem, which proves that E ( K ) /nE ( K ) is finite, and the second partuses height function to prove E ( K ) is finitely generated. In this section, we will give two proofs of the Weak Mordell-Weil Theorem.
Theorem 2.1.1. (Weak Mordell-Weil)
Let E be an elliptic curve defined overa number field K. Then E(K)/mE(K) is finite for any n ≥ The first proof, given by Silverman, is based on theories about field extension.
Lemma 2.1.1.
Let L/K be a finite Galois extension. If E(L)/mE(L) is finite,then E(K)/mE(K) is finite. roof. Let Φ be the kernel of the natural map E ( K ) /nE ( K ) → E ( L ) /nE ( L ) .Therefore, Φ = ( E ( K ) ∩ mE ( L )) /mE ( K ) and for each P (mod mE ( K ) ) in Φ , we can choose a point Q p ∈ E ( L ) with [ m ] Q p = P . Having done this, we define a map of sets λ p : G L/K → E [ m ] , λ p ( σ ) = Q σp − Q p Here Q p is fixed for each P .We notice that λ p ( σ ) = [ m ]( Q σp − Q p ) = [ m ] Q σp − [ m ] Q p = 0 So λ p ( σ ) is in E [ m ] .Suppose that λ p = λ p ′ for two points P, P ′ ∈ E ( K ) ∩ mE ( L ) . Then we have ( Q p − Q p ′ ) σ = Q p − Q p ′ f or all σ ∈ G L/K so Q p − Q p ′ ∈ E ( K ) . Therefore P − P ′ = [ m ]( Q p − Q p ′ ) ∈ mE ( K ) ⇔ P ≡ P ′ (mod mE ( K )) So the map Φ → M ap ( G L/K , E [ m ]) , P → λ p is an injection. But G L/K and E [ m ] are finite sets, so Φ is a finite set.Finally, the exact sequence → Φ → E ( K ) /mE ( K ) → E ( L ) /mE ( L ) implies that E ( K ) /mE ( K ) is finite(because it is between two finite sets).In view of the lemma above, we can enlarge the number field K and supposethat E [ m ] ⊂ E ( K ) (because E [ m ] is finite). We will assume this is true for theremainder of this section.The next step we will do is to translate the question into a question about acertain field extension of K. Definiton 2.1.1.
The Kummer pairing κ : E ( K ) × G K/K → E [ m ] is defined as follows. Let P ∈ E ( K ) , and choose any Q ∈ E ( K ) satisfying [ m ] Q = P . Then κ ( P, σ ) = Q σ − Q Actually, from the definition we can see that it is similar to the definition of λ p . It is well-defined because E [ m ] ⊂ E ( K ) . Theorem 2.1.2. (a) The Kummer pairing is bilinear.(b) The kernel of the Kummer pairing on the left is mE ( K ) .(c) The kernel of the Kummer pairing on the right is G K/L , where L = K ([ m ] − E ( K )) s the compositum of all fields K ( Q ) as Q ranges over the points of E ( K ) sat-isfying [ m ] Q ∈ E ( K ) .Hence the Kummer pairing induces a perfect bilinear pairing E ( K ) /mE ( K ) × G L/K → E [ m ] Proof. (a) The linearity of P is trivial. For σ , let σ, τ ∈ G K/K . Then κ ( P, στ ) = Q στ − Q = ( Q σ − Q ) τ + Q τ − Q = κ ( P, σ ) τ + κ ( P, τ ) However, κ ( P, σ ) ∈ E [ m ] ⊂ E ( K ) , so it is fixed by τ . Therefore, κ ( P, στ ) = κ ( P, σ ) + κ ( P, τ ) (b) Suppose κ ( P, σ ) = 0 for all σ ∈ G K/K . Then we have Q σ = Q for all σ ∈ G K/K .Therefore, Q ∈ E ( K ) and P = [ m ] Q ∈ mE ( K ) And if P ∈ mE ( k ) ,it is obvious that κ ( P, σ ) = 0 for all σ ∈ G K/K . Therefore, the kernel on theleft is mE ( K ) .(c) Suppose κ ( P, σ ) = 0 for all P ∈ E ( K ) , then Q σ − Q = 0 for all Q satisfying [ m ] Q ∈ E ( K ) . But L is the compositum of K ( Q ) over all such Q , so σ fixes L .Hence σ ∈ G K/L . Conversely, if σ ∈ G K/L , then we have κ ( P, σ ) = Q σ − Q = 0 since Q ∈ E ( L ) from the definition. Thus the kernel on the right is G K/L .Finally, for the last statement of the theorem, we firstly claim that
L/K isGalois because it is normal from the definition( [ m ] Q ′ = [ m ] Q ∈ E ( K ) if Q ′ is aconjugate of Q ). Since L/K is Galois, we have G K/K /G K/L = G L/K
Thus it is a perfect bilinear pairing.From Theorem 2.1.2 we can see that if we can prove L is a finite extension,or in other words, G L/K is finite, then the group E ( K ) /mE ( K ) is finite. So thenext step is to analyze this extension. Theorem 2.1.3.
Let L be the field defined in Theorem 2.1.2.(a) L/K is an abelian extension of exponent m.(I.e. G L/K is abelian and everyelement has order dividing m.)(b) Let S = (cid:8) v ∈ M K : E has bad reduction at v (cid:9) ∪ (cid:8) v ∈ M K : v ( m ) = 0 (cid:9) ∪ M ∞ K Then
L/K is unramified outside S.Proof. (a) This follows immediately from the last statement of Theorem 2.2,which implies that there is an injection G L/K → Hom ( E ( K ) , E [ m ]) σ → κ ( · , σ ) v ∈ M K with v / ∈ S . Choose an arbitrary element Q in m − E ( K ) ,and the only thing we have to show is that K ′ = K ( Q ) is unramified at v,because L is the compositum of all such K ′ . Let v ′ ∈ M K ′ be a place of K ′ such that v | v ′ , and let k v ′ /k v be the corresponding extension of residue fields.Since E has good reduction at v , E also has good reduction at v ′ (because thediscriminants are the same). Thus we have the usual reduction map E ( K ′ ) → ˜ E v ′ ( k ′ v ′ ) Now let I v ′ /v ⊂ G K ′ /K be the inertia group for v ′ /v , and let σ ∈ I v ′ /v . Bydefinition of inertia, σ acts trivially on ˜ E v ′ ( k ′ v ′ ) , so ˜ Q σ − Q = ˜ Q σ − ˜ Q = ˜0 On the other hand, Q σ − Q ∈ E ( K )[ m ] , so Q σ = Q . Thus Q is fixed by allelements of I v ′ /v , which implies that the action of inertia group on K ′ is trivial.Hence K ′ is unramified over K at v ′ .Next we will prove that all field extensions L/K satifying the condition inTheorem 2.1.3 must be a finite field extension.
Theorem 2.1.4.
Let K be a number field, S ⊂ M K a finite set of places con-taining M ∞ K , and m ≥ an integer. Let L/K be the maximal abelian extensionof K having exponent m which is unramified outside of S. Then L/K is a finiteextension.Proof.
First, we can assume that K contains the m t h − roots of unity µ m . Thatis because if K doesn’t contain it, we can choose K ′ = K ( µ m ) and LK ′ /K ′ isalso an abelian extension of exponent m unramified at S ′ , where S ′ is the setof places of K ′ lying over S. And if LK ′ /K ′ is finite, L/K is also finite. So wecan assume that K contains the m t h − roots of unity µ m .Furthermore, we may increase the set S, because this can only make the fieldextension larger. Using the fact that the class number of K is finite, we can thusadd a finite number of elements to S so that the ring of S-integers R s = { a ∈ K : v ( a ) ≥ f or all v ∈ M K , v / ∈ S } is a principle ideal domain. We may also enlarge S so that v ( m ) = 0 for all v / ∈ S .Next, according to Kummer theory, we know that L is the largest subfieldof K ( m √ a : a ∈ K ) which is unramified outside S.Let v ∈ M K , v / ∈ S . Looking at the equation X m − a = 0 over local field K v , and remembering that v ( m ) = 0 , it is clear that K v ( m √ a ) /K v is unramified iff ord v ( a ) ≡ m ) Therefore, L = K ( m √ a : a ∈ T S ) , where T S = { a ∈ K ∗ / ( K ∗ ) m : ord v ( a ) ≡ m ) } T S is a finite group, we can see that L is a finite extension over K. To prove T S isfinite, we first consider the natural map R ∗ S → T S We claim that the map is surjective. To see this, suppose a ∈ K ∗ represents anelement of T S . Then the ideal aR S is the m th -power of an ideal in R S , since theprime ideals of R S correspond to the valuations v / ∈ S . Since R S is a principleideal domain, we can find b ∈ K ∗ s.t. aR S = b m R S , which means that a = ub m for u ∈ R ∗ S . Then u and a give the same element of T S , showing that the mapis surjective. Now the kernel of the map certainly contains ( R ∗ S ) m , so we havea surjection R ∗ S / ( R ∗ S ) m → T S According to Dirichlet’s unit theorem, which shows that the group of units isfinitely generated, we know that R ∗ S / ( R ∗ S ) m is a finite group. Thus T S is a finitegroup, and the proof is completed.From Theorem 2.1.4 we can see that L/K is a finite galois extension, so G ( L/K ) is finite. Therefore, the group E ( k ) /mE ( K ) is finite, and we have theWeak Mordell Weil theorem correct.Next we will use cohomology to prove the Weak Mordell Weil theorem. Firstwe will introduce group cohomology. Definiton 2.1.2.
Let G be a finite group acting on an abelian group M. Wedefine H ( G, M ) = M G = { m ∈ M | σm = m, all σ ∈ G } A crossed homomorphism is a map f : G → M such that f ( στ ) = f ( σ ) + σf ( τ ) all σ, τ ∈ G and a crossed homomorphism is said to be principal if given an m ∈ Mf ( σ ) = σm − m, all σ ∈ G Next we define H ( G, M ) = { crossed homomorphisms }{ principle crossedhomomorphisms } We then state the most important and basic properties of cohomology.
Proposition 2.1.1.
For any exact sequence of G-modules → M → N → P → there is a canonical exact sequence → H ( G, M ) → H ( G, N ) → H ( G, P ) δ −→ H ( G, M ) → H ( G, N ) → H ( G, P ) Definiton 2.1.3.
Let K be a perfect field, ¯ K its algebraic closure, and let G := Gal ( ¯
K/K ) = G K be its Galois group. Then we can dress G in the Krull topology: a subgroup isopen if it fixes a finite extension of K. Thus all these subgroups form a base of G . And thus they form a base for every point g ∈ G . So we can a topology onG.Next a G-module M is said to be discrete if the map G × M → M is continuousrelative to the discrete topology on M and the Krull topology on G. This isequivalent to requiring that every element of M is fixed by the subgroup of Gfixing some finite extension of K. For a dicrete G − module M , every principle crossed homomorphism f : G → M is continuous. That is because every element of M is fixed by an opennormal subgroup of G . Definiton 2.1.4. H ( G, M ) = { continuous crossed homomorphisms }{ principle crossedhomomorphisms } And still, we have Theorem correct.Also, we have the short exact sequence → E ( ¯ Q )[ m ] → E ( ¯ Q ) m −→ E ( ¯ Q ) → Therefore, we can get the long exact sequence → E ( Q )[ m ] → E ( Q ) m −→ E ( Q ) δ −→ H ( Q , E [ m ]) → H ( Q , E ) m −→ H ( Q , E ) From this, we can get another short exact sequence → E ( Q ) /mE ( Q ) δ −→ H ( Q , E [ m ]) → H ( Q , E )[ m ] → Since δ is an injection here, if we can prove the group H ( Q , E ( Q )[ m ]) is finite,then we can prove the Weak Mordell-Weil Theorem. However, this might notbe true. So we will use the local field Q p to solve the problem.First, we choose the algebraic closure ¯ Q for Q , and ¯ Q p for Q p . The em-bedding Q ֒ → Q p extends to an embedding ¯ Q ֒ → ¯ Q p . Moreover, the action of Gal ( ¯ Q p / Q p ) on ¯ Q ⊂ ¯ Q p defines a homomorphism ψ : G Q p → G Q by restrictionof the Galois action.Therefore, a crossed homomorphism f : G Q → E ( ¯ Q ) defines a crossedhomomorphism ˜ f : G ¯ Q p → E ( ¯ Q p ) by composition ˜ f = f ◦ ψ . To check this iswell defined, for any σ, τ ∈ G Q p , ˜ f ( στ ) = f ( ψ ( στ )) = f ( ψ ( σ ))+ ψ ( σ ) f ( ψ ( τ )) = f ( ψ ( σ ))+ σf ( ψ ( τ )) = ˜ f ( σ )+ σ ˜ f ( τ ) f is a principle crossed homomorphism, then ˜ f is also principle.Thus we can get a map φ : H ( Q , E ) → H ( Q p , E ) by taking f to ˜ f .We can get the following commutative diagram → E ( Q ) /mE ( Q ) δ −→ H ( Q , E [ m ]) → H ( Q , E )[ m ] → ↓ ↓ ↓ → E ( Q p ) /mE ( Q p ) δ −→ H ( Q p , E [ m ]) → H ( Q p , E )[ m ] → where the top and bottom lines are exact and the vertical maps are embedding.Next we reach a crucial argument. If some γ ∈ H ( Q , E [ m ]) comes from theclass of an element of E ( Q ) , then its image γ p ∈ H ( Q p , E [ m ]) arises from anelement of E ( Q p ) . We want to quantify those γ whose local versions γ p comesfrom E ( Q p ) and all those γ which vanish locally.Here comes two definitions that we will mainly talk about. Definiton 2.1.5.
The n − Selmergroup is defined by S ( n ) ( E/ Q ) : = (cid:8) γ ∈ H ( Q , E [ n ]) | ∀ p, γ p comes f rom E ( Q p ) (cid:9) = ker ( H ( Q , E [ n ]) → Y p prime H ( Q p , E )) Definiton 2.1.6.
The
T ate − Shaf arevich group is defined by X ( E/ Q ) = ker ( H ( Q , E ) → Y p prime H ( Q p , E )) And we need the following lemma, which is easy to prove.
Lemma 2.1.2.
For any chain of modules A α −→ B β −→ C , we can get a longexact sequence → ker ( α ) → ker ( βα ) → ker ( β ) → coker ( α ) → coker ( βα ) → coker ( β ) → We won’t prove this because all the maps are natural.If we apply the lemma to the maps H ( Q , E [ n ]) → H ( Q , E )[ n ] → Y p prime H ( Q p , E )[ n ]) , we obtain the fundamental exact sequence → E ( Q ) /nE ( Q ) → S ( n ) E/ Q → X ( E/ Q )[ n ] → We shall prove E ( Q ) /nE ( Q ) to be finite by showing that S ( n ) E/ Q is finite.First we will prove the Selmer group is finite in a special case. Lemma 2.1.3.
If all the pointos of order 2 on an elliptic curve given by theWeierstrass equation Y Z + a XY Z + a Y Z = X + a X Z + a XZ + a Z have coordinates in Q , then the Selmer group S (2) ( E/ Q ) is finite. roof. Since they all have coordinates in Q , we can imply that E ( ¯ Q )[2] = E ( Q )[2] ∼ = ( Z / Z ) × ( Z / Z ) (One can check [Sil, Cor 6.4(b)] for the proof)And the group Gal ( ¯ Q / Q ) acts trivially on E[2]. Thus we have H ( Q , E [2]) ∼ = ( Q × / Q × ) (One can get this by using the long exact sequence of cohomology on the shortexact sequence → Z / Z → Q × −→ Q × → ) Let γ ∈ S (2) ( E/ Q ) ⊂ H ( Q , E [2]) . For each prime p not dividing 2 ∆ , thereexists a finite unramified extension K of Q p such that γ maps to zero underthe vertical arrows: H ( Q , E [2]) ∼ = −→ ( Q × / Q × ) ↓ ↓ H ( K, E [2]) ∼ = −→ ( K × /K × ) We choose a representative element (( − ε ( ∞ ) Q p p ε ( p ) , ( − ε ′ ( ∞ ) Q p p ε ′ ( p ) ) ∈ ( Q × / Q × ) for γ . Here each ε or ε ′ is either 0 or -1. Therefore we can see that ord p (( − ε ( ∞ ) Y p p ε ( p ) ) = ε ( p ) and so if ( − ε ( ∞ ) Q p p ε ( p ) is a square in K, then ε ( p ) = 0 . Therefore the onlyp that can occur in the factorizations are those dividing 2 ∆ , which allows onlyfinitely many possibilities for γ .After proving the special case, we will now turn to prove the general case. Theorem 2.1.5.
The Selmer group S ( n ) ( E/ Q ) is finite.Proof. Actually, instead of proving S ( n ) ( E/ Q ) is finite, we want to prove that S ( n ) ( E/L ) := ker ( H ( L, E [ n ]) → Y v ∈ M K H ( Q p , E )) is finite for any suitably large L. And according to the next lemma, we will showthat if it is correct for L, then it is correct for Q . Lemma 2.1.4.
For any finite Galois extension L of Q and integer n ≥ , thekernel of S ( n ) ( E/ Q ) → S ( n ) ( E/L ) is finiteProof. Since S ( n ) ( E/ Q ) and S ( n ) ( E/L ) are subgroups of H ( Q , E [ n ]) and H ( L, E [ n ]) respectively, it suffices to prove that the kernel of H ( Q , E [ n ]) → H ( L, E [ n ]) is finite. However, we can easily verify that the kernel of the map is H ( Gal ( L/ Q ) , E ( L )[ n ]) ,which is finite because both Gal(L/ Q ) and E(L)[n] are finite.13ere we still need some preparations from algebraic number theory. Lemma 2.1.5.
When T is a finite set of prime ideals in L, the groups U T and C T defined by the exactness of the sequence → U T → L × a → ( ord p ( a )) −−−−−−−−→ M p / ∈ T Z → C T → are, respectively, finitely generated and finite.Proof. First, let’s consider the kernel of the map f : L × → M p Z An element a is kerf iff ord p ( a ) = 0 for all p , thus a is in the kernel iff it is aunit of O L . And the cokernel of f is obviously finite due to the finiteness of theclass number. Hence we get an exact sequence → U → L × a → ( ord p ( a )) −−−−−−−−→ M p Z → C → where U is the unit group of O L , and C is the ideal class group. So U is finitelygenerated due to the Dedekind Unit theorem and C is a finite group.Next, use the kernel-cokernel exact sequence of L × → M p Z → M p / ∈ T Z is an exact sequence → U → U T → M p ∈ T Z → C → C T → o Thus we can see that U T and C T are finitely generated and finite recpectively.Now we come back to the proof of the theorem.Let’s review the proof of the special case. Actually, we can see that the proofused the following facts:(a) Q contains a primitive square root of 1(b) The points of order 2 all have coordinates in Q (c) For any finite set T of prime numbers, the kernel of r → ( ord p ( r ) (mod 2)) : Q × / Q × → M p ∈ T Z / Z is finite.Therefore, according to the above discussion, what we have to do is to provethe following lemma and the proof will be completed.14 emma 2.1.6. Assume that L contains the n th -unity root. For any finite subsetT of M L containing M ∞ K , let N be the kernel of a → ( ord p ( a ) (mod n )) : L × /L × n → M p ∈ T Z /n Z Then there is an exact sequence → U T /U nT → N → C T [ n ] Therefore N is a finite group.Proof.
This can be proved by a diagram chase in → U T → L × → M p / ∈ T Z → C T → ↓ n ↓ n ↓ n ↓ n → U T → L × → M p / ∈ T Z → C T → ↓ ↓ L × /L × n → M p ∈ T Z /n Z Since we have the lemma correct, we have the theorem correct, and the proofis cmpleted.Actually, we can see that the proof above can prove that the Selmer group S ( n ) ( E/K ) is finite for any number field K. Therefore we prove the Weak MordellWeil theorem by using cohomology. Q In this section, we will prove the Mordell Weil theorem on Q . Proposition 2.2.1. (Descent theorem) Let A be an abelian group. Supposethere is a ’height’ funtion h : A → R with the following three properties:( i ) Let Q ∈ A . There is a constant C depending on A and Q, so that for all P ∈ A , h ( P + Q ) ≤ h ( P ) + C ( ii ) There is an integer m ≥ and a constant C , depending on A, so that forall P ∈ A , h ( mP ) ≥ m h ( P ) − C ( iii ) For every constant C , { P ∈ A : h ( P ) ≤ C } is a finite set.Suppose further that for the integer m in ( ii ), the quotient group A/mA is finite.Then A is finitely generated. roof. Choose elements Q , . . . , Q r ∈ A to represent the finitely many cosets inA/mA. The idea is to show that by substracting an appropriate linear combi-nation of Q , . . . , Q r from P, we will be able to make the height of the resultingpoint less than a constant which is independent of P. Then the Q , . . . , Q r andthe finitely many points with height less than this constant will generate A.Write P = mP + Q i f or some ≤ i ≤ r Continuing in this fashion, P = mP + Q i ...P n − = mP n + Q i n Now for any j, we have h ( P j ) ≤ m [ h ( mP j ) + C ] f rom ( iii )= 1 m [ h ( P j − − Q i j ) + C ] ≤ m [2 h ( mP j − ) + C ′ + C ] f rom ( i ) where we take C ′ to be the maximum of the constants from ( i ) for Q = − Q i , ≤ i ≤ r . Note that C ′ and C do not depend on P. Now use the aboveinequality repeatedly, starting from P n and working back to P. This yields h ( P n ) ≤ ( 2 m ) n h ( P ) + [ 1 m + 2 m + 4 m + · · · + 2 n − m n ]( C ′ + C ) (1) < ( 2 m ) n h ( P ) + C ′ + C m − (2) ≤ − n h ( P ) + ( C ′ + C ) / (3)It follows that by taking n sufficiently large, we will have h ( P n ) ≤ C ′ + C ) / Since P = m n P n + Σ nj =1 m j − Q [ i j ] it follows that every P ∈ A is a linear combination of the points in the set { Q , . . . , Q r } ∪ { Q ∈ A : h ( Q ) ≤ C ′ + C ) / } And from the third property, this is a finite set, which proves that A is finitelygenerated.Therefore, to solve the problem, all we have to do is to find a height functionon E ( K ) satisfying the three properties. First let’s talk about how to define aheight function on E ( Q ) .Fix a Weierstrass equation for E/ Q of the form E : y = x + Ax + B with A, B ∈ Z . 16 efiniton 2.2.1. Let t ∈ Q and write t = p/q as a fraction in lowest terms.The height of t, denoted H(t), is defined by H ( t ) = max {| p | , | q |} Definiton 2.2.2.
The height on E ( Q ) (relative to the given Weierstrass equa-tion) is the function h x : E ( Q ) → R h x ( P ) = (cid:26) logH ( x ( P )) if P = O if P = O We want to prove that the height function defined above has the three prop-erties. Therefore we should prove the following lemma
Lemma 2.2.1. (a) Let P ∈ E ( Q ) . There is a constant C , depending on P ,A, B, so that for all P ∈ E ( Q ) , h x ( P + P ) ≤ h x ( P ) + C (b) There is a constant C , depending on A, B, so that for all P ∈ E ( Q ) , h x ([2] P ) ≥ h x ( P ) − C (c) For every constant C , the set { P ∈ E ( Q ) : h x ( P ) ≥ C } is finite.Proof. Taking C > max { h x ( P ) , h x ([2] P ) } , we may assume P = O and P = O, ± P . Then writing P = ( x, y ) = ( ad , bd ) P = ( x , y ) = ( a d , b d ) (we can write the coordinates in this form because of the form of the WeierstrassEquation) where the indicated fractions are in lowest terms. Thus we have x ( P + P ) = ( y − y x − x ) − x − x . Now multiplying this out and using that P and P satisfy the Weierstrassequation yields x ( P + P ) = ( xx + A )( x + x ) + 2 B − yy ( x − x ) = ( aa + Ad d )( ad + a d ) + 2 Bd d − bdb d ( ad − a d ) In computing the height of a rational number, cancellation between numeratorand denominator can only decrease the height, so we find by an easy estimationthat H ( x ( P + P )) ≤ C ′ max (cid:8) | a | , | d | , | bd | (cid:9) H ( x ( P )) = max (cid:8) | a | , | d | (cid:9) , and from the equation below b = a + Aad + Bd we can get that | b | ≤ C ′′ max n | a | / , | d | o which implies that H ( x ( P + P )) ≤ C max (cid:8) | a | , | d | (cid:9) = C H ( x ( P )) Now taking logarithms gives the desired result.(b) By choosing C ≥ h x ( T ) for each of the points T ∈ E ( Q [2]) , we mayassume that [2] P = O . Then writing P = ( x, y ) , the duplication formula reads x ([2] P ) = x − Ax − Bx + A x + 4 Ax + 4 B It is convenient to define homogeneous polynomials F ( X, Z ) = X − AX Z − BXZ + A Z G ( X, Z ) = 4 X Z + 4 AXZ + 4 BZ Then if we write x = x ( P ) = a/b as a fraction in lowest terms, x([2]P) can bewritten as a quotient of integers x ([2] P ) = F ( a, b ) /G ( a, b ) Unlike what we’ve done in (a), we have to find a lower bound for H(x([2]P)),so it will be important to bound how much cancellation can occur betweennumerator and denominator. The idea is to use the fact F(X,1) and G(X,1) arerelative prime polynomials, so they generate the unit ideal in Q . Sublemma 2.2.1.
Let ∆ = 4 A + 27 B f ( X, Z ) = 12 X Z + 16 AZ g ( X, Z ) = 3 X − AXZ − BZ f ( X, Z ) = 4(4 A + 27 B ) X − A BX Z + 4 A (3 A + 22 B ) XZ + 12 B ( A + 8 B ) Z g ( X, Z ) = A BX + A (5 A + 32 B ) X Z + 2 B (13 A + 96 B ) XZ − A ( A + 8 B ) Z Then the following identities hold in Q [ X, Z ] : f ( X, Z ) F ( X, Z ) − g ( X, Z ) G ( X, Z ) = 4∆ Z f ( X, Z ) F ( X, Z ) − g ( X, Z ) G ( X, Z ) = 4∆ X Let δ = gcd ( F ( a, b ) , G ( a, b )) be the cancellation in our fraction for x([2]P). From equations f ( a, b ) F ( a, b ) − g ( a, b ) G ( a, b ) = 4∆ b f ( a, b ) F ( a, b ) − g ( a, b ) G ( a, b ) = 4∆ a
18e see that δ divides 4 ∆ . Hence we obtain the bound δ ≤ | | and so H ( x ([2] P )) ≥ max { F ( a, b ) , G ( a, b ) } / | | On the other hand, the same identites give the estimates | b | ≤ max { f ( a, b ) , g ( a, b ) } max { F ( a, b ) , G ( a, b ) }| a | ≤ max { f ( a, b ) , g ( a, b ) } max { F ( a, b ) , G ( a, b ) } Now looking at the expressions for f , f , g , g , we have max { f ( a, b ) , g ( a, b ) , f ( a, b ) , g ( a, b ) } ≥ Cmax (cid:8) | a | , | b | (cid:9) where C is a constant relying on A and B. Combining the last three inequlitiesyields max (cid:8) | a | , | b | (cid:9) ≤ Cmax (cid:8) | a | , | b | (cid:9) max { F ( a, b ) , G ( a, b ) } And so cancelling max (cid:8) | a | , | b | (cid:9) gives max { F ( a, b ) , G ( a, b ) } / | | ≥ (2 C ) − max (cid:8) | a | , | b | (cid:9) Since max (cid:8) | a | , | b | (cid:9) = H ( x ( P )) , this gives the desired estimate H ( x ([2] P )) ≥ (2 C ) − H ( x ( P )) and now taking logarithms gives the desired result.(c) For any constant C, the set { t ∈ Q : H ( t ) ≤ C } is obviously finite. And given any x, there will be at most two values of ysatisfying the Weierstrass equation. Thus we have { P ∈ Q : h x ( P ) ≤ C } is finite.Using the Decent theorem, the Weak Mordell-Weil theorem for m = 2 andthe lemma above, we can see that E ( Q ) is finite generated. We want to prove the Weak Mordell-Weil theorem for any number field K, sowe have to find a height function satisfying the three properties, and then byapplying the Desecent theorem we can finish the proof. However, unlike Q , it’snot easy to define a height function on other number fields. So we have to provea lot of things to develop a height function in general cases.19 efiniton 2.3.1. The set of standard absolute value on Q , which we againdenote by M Q , consists of the following:( i ) M Q contains one archimedean absolute value, given by | x | ∞ = usual absolute value ( ii ) For each prime p ∈ Z , M + Q contains one non-archimedean (p-adic)absolute value, given by | p n ab | = p − n for a, b ∈ Z , gcd ( p, ab ) = 1 The set of standard absolute values on K, denoted M K , consists of all absolutevalues on K whose restriction to Q is one of the absolute values in M Q . Definiton 2.3.2.
For v ∈ M K , the local degree at v, denoted n v , is given by n v = [ K v : Q v ] Here K v and Q v denote the completion of the field with respect to the absolutevalue v. With these definitions, we can state two basic facts from algebraic numbertheory which will be needed.
Proposition 2.3.1.
Let
L/K/ Q be a tower of number fields, and v ∈ M K .Then X w ∈ MLw | v n w = [ L : K ] n v (4) Proposition 2.3.2.
Let x ∈ K ∗ . Then Y v ∈ M K | x | n v = 1 Next we will define the height of a point in projective space.
Definiton 2.3.3.
Let P ∈ P N ( K ) be a point with homogeneous coordinates P = [ x , . . . , x N ] , x i ∈ K The height of P (relative to K) is defined by H K ( P ) = Y v ∈ M K max {| x | v , . . . | x N | v } n v As we can see, when K = Q , this definition is the same as H ( P ) = max {| x | , . . . , | x N |} where x , . . . , x N ∈ Z and gcd ( x , . . . , x N ) = 1 . We will state some important proerties of the given height function.20 roposition 2.3.3.
Let P ∈ P N ( K ) (a) The height H K ( P ) does not depend on the choice of homogeneous coordinatesfor P.(b) H K ( P ) ≥ (c) Let L/K be a finite extension. Then H L ( P ) = H K ( P ) [ L : K ] Proof. (a) It is directly from Proposition 2.3.2.(b) For any point in projective space, one can find homogeneous coordinates bymultiplying a number so that one of the coordinates is 1. Then every factor inthe product defining H K ( P ) is at least 1.(c) We compute H L ( P ) = Y w ∈ M L max {| x | w , . . . | x N | w } n w = Y v ∈ M K Y w ∈ MLw | v max {| x | v , . . . | x N | v } n w since x i ∈ K = Y v ∈ M K max {| x | v , . . . | x N | v } [ L : K ] n v = H K ( P ) [ L : K ] Sometimes, when a field is not given, it’s easier to use a height function notrelative to a field.
Definiton 2.3.4.
Let P ∈ P N ( ¯ Q ) . The absolute height of P, denoted H(P), isdefined as follows. Choose any field K such that P ∈ P N ( K ) . Then H ( P ) = H K ( P ) / [ K : Q ] In view of Proposition 2.3.3, it’s easy to see that this is well defined.
We now investigate how the height changes under mappings between pro-jective spaces.
Definiton 2.3.5.
A morphism of degree d between projective spaces is a map F : P N → P M F ( P ) = [ f ( P ) , . . . , f M ( P )] where f , . . . , f M ∈ ¯ Q [ X , . . . , X N ] are homogeneous polynomials of degree dwith no commone zero in ¯ Q other than X = · · · = X N = 0 . To prove the height function has the three properties, we have to find thelower bound and upper bound of the height function. Therefore we have thefollowing theorem:
Theorem 2.3.1.
Let F : P N → P M be a morphism of degree d. Then there are constants C and C , depending onF, so that for all points P ∈ P N ( ¯ Q ) , C H ( P ) d ≤ H ( F ( P )) ≤ C H ( P ) d roof. Write F = [ f , . . . , f M ] with homogeneous polynomials f i , and let P =[ x , . . . , x N ] ∈ P N ( ¯ Q ) . Choose some number field K containing x , . . . , x N andall of the coefficients of all of the f ′ i s . Then for each v ∈ M K , let | P | v = max ≤ i ≤ N {| x i | v } , | F ( P ) | v = max ≤ j ≤ M {| f j ( P ) | v } and | F | v = max {| a | v : a is a coef f icient of some f i } Then from the definition of height, H K ( P ) = Y v ∈ M K | P | n v v and H K ( F ( P )) = Y v ∈ M K | F ( P ) | n v v so it makes sense to define H K ( F ) = Y v ∈ M K | F | n v v Finally, we let C , /dots, denote constants which depend only on M,N and d,and set ε ( v ) = (cid:26) if v ∈ M ∞ K if v ∈ M K Having set notation, we turn to the proof of the theorem. The upper bound isrelatively easy. Let v ∈ M K . The triangle inequality yields | f i ( P ) | v ≤ C ε ( v )1 | F | v | P | dv Now raise to the n v − power , multiply over all v i nM K , and take the [ K : Q ] th − root . This yields the desired upper bound H ( F ( P )) ≤ C Σ v ∈ MK ε ( v ) n v / [ K : Q ]1 H ( F ) H ( P ) d = C Σ v ∈ M ∞ K n v / [ K : Q ]1 H ( F ) H ( P ) d = C H ( F ) H ( P ) d Notice that we don’t use the fact that the f ′ i s have no common non-trivial zero.But for the lower bound, we have to use this condition.From the Nullstellensatz theorem that the ideal generated by f , . . . , f M in ¯ mathbf Q [ X , . . . , X N ] contains some power of each X , . . . , X N , since each (0 , . . . , . Thus for an approriate integer e ≥ , there are polynomials g ij ∈ ¯ mathbf Q [ X , . . . , X N ] such that X ei = M X j =0 g ij f j for each ≤ i ≤ N (5)Replacing K by a finite extension, we may assume that each g ij ∈ K [ X , . . . , X N ] .Further, by discarding all terms except those which are homogeneous of degreee, we may assume that each g ij is homogeneous of degree e-d. Let us set thefurther reasonable notation | G | v = max {| b | v : b is a coef f icient of some g ij } K ( G ) = Y v ∈ M K | G | n v v Recalling that P = [ x , . . . , x N ] , the equation described above imply that foreach i, | x i | ev = | M X j =0 g ij ( P ) f j ( P ) | v ≤ C ε ( v )2 max ≤ j ≤ M {| g ij ( P ) f j ( P ) | v } Now taking the maximum over i gives | P | ev ≤ C ε ( v )2 max ≤ j ≤ M ≤ i ≤ N {| g ij ( P ) | v } | F ( P ) | v But since each g ij has degree e-d, the usual application of the triangle inequalityyields | g ij ( P ) | v ≤ C ε ( v )3 | G | v | P | e − dv Substituting this in above and multiplying through by | P | d − ev gives | P | dv ≤ C ε ( v )4 | G | v | F ( P ) | v and now the usual raising to the n v -power, multiplying over v ∈ M K and takingthe [ K : Q ] th -root yields the desired lower bound. Definiton 2.3.6.
For x ∈ ¯ Q , let H ( x ) = H ([ x, Similarly, if x ∈ K , then H K ( x ) = H K ([ x, Theorem 2.3.2.
Let f ( T ) = a T d + a T d − + · · · + a d = a ( T − α ) . . . ( T − α d ) ∈ ¯ Q [ T ] be a polynomial of degree d. Then − d d Y j =1 H ( α j ) ≤ H ([ a , . . . , a d ]) ≤ d − d Y j =1 H ( α j ) Proof.
First note that the inequality remains unchanged if f ( T ) is replaced by (1 /a ) f ( T ) . Thus we can prove the inequality under the condition of a = 1 .Let Q ( α , . . . , α d ) , and for v ∈ M K , set ε ( v ) = (cid:26) if v ∈ M ∞ K if v ∈ M K (Note that this function is different from the ε ( v ) we defined before, that isbecause we can see that | x + y | v ≤ ε ( v ) max {| x | v , | y | v } ε ( v ) − d d Y j =1 max {| α j | v , } ≤ max ≤ i ≤ d {| a i | v } ≤ ε ( v ) d − d Y j =1 max ≤ i ≤ d {| a i | v , } Once this is done, raising to the n v -power, multiplying over v ∈ M K , and taking [ K : Q ] th -roots gives the desired result.The proof is by induction of d = deg ( f ) . For d = 1 , the inequality is clear.Assume for polynomials of degree d-1, the result is true. Choose an index k sothat | α k | v ≥ | α j | v for all ≤ j ≤ d And we define a polynomial g ( T ) = ( T − α ) . . . ( T − α k − )( T − α k +1 ) . . . ( T − α d )= b T d − + · · · + b d − Therefore we can see that f ( T ) = ( T − α k ) g ( T ) . By comparing coefficients wecan get a i = b i − α k b i − We now prove the upper above bound. max ≤ i ≤ d {| a i | v } = max ≤ i ≤ d {| b i − α k b i − |}≤ ε ( v ) max ≤ i ≤ d {| b i | , | α k b i − |}≤ ε ( v ) max ≤ i ≤ d {| b i |} max {| α k | , }≤ ε ( v ) d − d Y j =1 max ≤ i ≤ d {| a i | v , } (The last step is by the induction hypothesis applied to g).Next, to prove the lower bound, we consider two cases. First, if | α k | v ≤ ε ( v ) ,then by the choice of the index k, d Y j =1 max {| α j | v , } ≤ max {| α k | v } ≤ ε ( v ) d And remember that a = 1 , so we have max ≤ i ≤ d {| a i | v } ≥ Therefore ε ( v ) − d d Y j =1 max {| α j | v , } ≤ max ≤ i ≤ d {| a i | v } Second, if | α k | v ≥ ε ( v ) , then max ≤ i ≤ d {| a i | v } = max ≤ i ≤ d {| b i − α k b i − | v } = max ≤ i ≤ d − {| b i | v } | α k | v v ∈ M K . And for v ∈ M ∞ K , max ≤ i ≤ d {| b i − α k b i − } ≥ ( | α k | v − max ≤ i ≤ d − {| b i | v } > ε ( v ) − | α k | v max ≤ i ≤ d − {| b i | v } Combining the two situations we can get that max ≤ i ≤ d {| a i | v } ≥ ε ( v ) − | α k | v max ≤ i ≤ d − {| b i | v } And now applying the induction hypothesis to g gives the desired lower bound,which completes the proof.The reason we prove this theorem is that we want to prove that this heightfunction ’satisfies’ the third property. But before we do that, we have to proveanother lemma.
Lemma 2.3.1.
Let P ∈ P N ( ¯ Q ) and σ ∈ G ¯ Q / Q . Then H ( P σ ) = H ( P ) Proof.
Let K/ Q be a field with P ∈ P N ( K ) and σ gives an isomorphism σ : K → K σ . It likewise identifies the sets of absolute values, σ : M K → M K σ v → v σ Clearly σ also gives an isomorphism K v → K σv σ , so n v = n v σ . We now compute H K σ ( P σ ) = Y w ∈ M Kσ max {| x σi | w } n w = Y v ∈ M K max {| x σi | v σ } n vσ = Y v ∈ M K max {| x i | v } n v = H K ( P ) . Since [ K : Q ] = [ K σ : Q ] , this is the desired result. Theorem 2.3.3.
Let C and d be constants. Then the set (cid:8) P ∈ P N ( ¯ Q ) : H ( P ) ≤ C and [ Q ( P ) : Q ] ≤ d (cid:9) contains only finitely many points. In particular, for any number field K, (cid:8) P ∈ P N ( K ) : H K ( P ) ≤ C (cid:9) is a finite set.Proof. Let P ∈ P N ( ¯ Q ) . Take homogeneous coordinates for P, say P = [ x , . . . , x N ] x j = 1 . Then Q ( P ) = Q ( x , . . . , x N ) , and we have the easy estimate H Q ( P ) ( P ) = Y v ∈ M Q ( P ) max ≤ i ≤ N {| x i | v } n v ≥ max ≤ i ≤ N ( Y v ∈ M Q ( P ) max {| x i | v , } n v )= max ≤ i ≤ N H Q ( P ) ( x i ) Thus if H ( P ) ≤ C and [ Q ( P ) : Q ] ≤ d , then max ≤ i ≤ N H ( x i ) ≤ C max ≤ i ≤ N [ Q ( x i ) : Q ] ≤ d It thus suffices to prove that the set (cid:8) x ∈ ¯ Q : H ( x ) ≤ C and [ Q ( x i ) : Q ] ≤ d (cid:9) is finite, which means that we only have to prove the case N = 1 .Suppose x ∈ ¯ Q is in this set, and let e = [ Q ( x ) : Q ] ≤ d . Further let x = x , . . . , x e be the conjugates of x, so the minimal polynomial of x over Q is f x ( T ) = ( T − x ) . . . ( T − x e ) = T e + · · · + a e ∈ Q ( T ) Now H ([1 , a , . . . , a e ]) ≤ e − e Y j =1 H ( x j )= 2 e − H ( x ) e ≤ (2 C ) d Thus we can see that there are only finitely many choices for a i , so the set isfinite, which completes the proof. Now we have already developed enough theorems about height functions onprojective space, so we will focus on elliptic curves, and finish the proof of theWeak-Mordell theorem.
Definiton 2.4.1.
Let f,g be two real-valued functions on a set Φ . Then we write f = g + O (1) if there’s constants C and C so that C ≤ f ( P ) − g ( P ) ≤ C for all P ∈ Φ In only the lower(respectively upper) inequality is satisfied, then we naturallywrite f ≥ g + O (1) (respecively f ≤ g + O (1) ). Definiton 2.4.2.
The height on projective space is the function h : P N ( ¯ Q ) → R ( P ) = logH ( P ) Note that h ( P ) ≥ for all P since H ( P ) ≥ .And let E/K be an elliptic curve and f ∈ ¯ K ( E ) a function. The height onE(relative to f ) is the function h f : E ( ¯ K ) → R h f ( P ) = h ( f ( P )) Proposition 2.4.1.
Let E/K be an elliptic curve and f ∈ K ( E ) a non-constantfunction. The for any constant C, { P ∈ E ( K ) : h f ( P ) ≤ C } is a finite set.Proof. The function f gives a finite-to-one map of the set in question to the set (cid:8) Q ∈ P ( K ) : H ( Q ) ≤ e C (cid:9) Now apply Theorem to this last set and we can get the desired result.The next step helps us find the relation between the additive law on anelliptic curve and the height function.
Theorem 2.4.1.
Let E/K be an elliptic curve and let f ∈ K ( E ) be an evenfunction(i.e. f ◦ [ −
1] = f ). Then for all P, Q ∈ E ( ¯ K ) , h f ( P + Q ) + h f ( P − Q ) = 2 h f ( P ) + 2 h f ( Q ) + O (1) Proof.
Choose a Weierstrass equation for E/K of the form E : y = x + Ax + B We start by proving the theorem for the particular function f = x (Note that itis an even function). The general case will be an easy corollary.Since h x ( O ) = 0 and h x ( − P ) = h x ( P ) , the result clearly holds if P = O or Q = O . We now assume that P, Q = O , and write x ( P ) = [ x , x ( Q ) = [ x , x ( P + Q ) = [ x , x ( P − Q ) = [ x , Thus we have x + x = 2( x + x )( A + x x ) + 4 B ( x + x ) − x x x x = ( x x − A ) − B ( x + x )( x + x ) − x x Define a map g : P → P by g ([ t, u, v ]) = [ u − tv, u ( At + v ) , ( v − At ) − Btu ] x and x shows that there is a commutative diagram E × E G −→ E × E ↓ ↓ σ P × P P × P σ ↓ ↓ P g −→ P where G ( P, Q ) = ( P + Q, P − Q ) and the vertical map σ is the composition of the two maps E × E → P × P and P × P → P ( P, Q ) → ( x ( P ) , x ( Q )) ([ α , β ] , [ α , β ]) → [ β β , α β + α β , α α ] The next step is to show that g is morphism, so as to be able to applyTheorem. By definition, this is equivalent to prove that the three polynomialshave no common non-trivial zeros. Suppose that g ([ t, u, v ]) = [0 , , . If t = 0 ,then from u − tv = 0 and ( v − At ) − Btu = 0 we see that u = v = 0 . Thus we may assume that t = 0 , and so it makes senseto define a new quantity x = u/ t . Notice that the equation u − tv = 0 canbe written as x = v/t . Now dividing the equalities u ( At + v ) + 4 Bt = 0 and ( v − At ) − Btu = 0 by t and rewriting them interms of x yields the two equations ψ ( x ) = 4 x + 4 Ax + 4 B = 0 φ ( x ) = x − Ax − Bx + A = 0 And we can see that (12 X + 16 A ) φ ( X ) − (3 X − AX − B ) ψ ( X ) = 4(4 A + 27 B ) = 0 This completes the proof that g is a morphism.We return to our commutative diagram, and compute h ( σ ( P + Q, P − Q )) = h ( σ ◦ G ( P, Q ))= h ( g ◦ σ ( P, Q ))= 2 h ( σ ( P, Q )) + O (1) f rom T heorem since g is a morphism of degree 2. Now to complete the proof for f = x , we willshow that for all R , R ∈ E ( ¯ K ) , there is a relation h ( σ ( R , R )) = h x ( R ) + h x ( R ) + O (1) h ( σ ( P + Q, P − Q )) = 2 h ( σ ( P, Q )) + O (1) will give the desired result.One verifies that if either R = O or R = O , then h ( σ ( R , R )) = h x ( R ) + h x ( R ) . Otherwise, we may write x ( R ) = [ α , and x ( R ) = [ α , and so h ( σ ( R , R )) = h ([1 , α + α , α α ]) and h x ( R ) + h x ( R ) = h ( α ) + h ( α ) Then from Theorem applied to the polynomial ( T + α )( T + α ) , we obtain thedesired estimate h ( α ) + h ( α ) − log ≤ h ([1 , α + α , α α ]) ≤ h ( α ) + h ( α ) + log Finally, to deal with the general case, we prove that h f = deg ( f ) h x O (1) Once this is proved, the theorem follows immediately from multiplying theknown relation for h x by deg ( f )2 Lemma 2.4.1.
Let f, g ∈ K ( E ) be even functions. Then ( deg g ) h f = ( deg f ) h g + O (1) Proof.
Let x, y ∈ K ( E ) be Weierstrass coordinates for E/K. The subfield con-sisting of all even functions is exactly K(x), so we can find a rational function ρ ( X ) ∈ K ( X ) so that f = ρ ◦ x Hence using theorem and the fact that ρ is a morphism, we can get that h f = ( deg ρ ) h x + O (1) But from the equation above, we have degf = degxdegρ = 2 degρ So we find h f = ( deg f ) h x + O (1) The same reasoning for g yields h g = ( deg g ) h x + O (1) and combining the last two equations gives the desired result.29 orollary 1. Let E/K be an elliptic curve and f ∈ E ( K ) an even function.(a) Let Q ∈ E ( ¯ K ) . Then for all P ∈ E ( ¯ K ) , h f ( P + Q ) ≤ h f ( P ) + O (1) (b) Let m ∈ Z . Then for all P ∈ E ( ¯ K ) , h f ([ m ] P ) = m h f ( P ) + O (1) Proof. (a) This is directly from Theorem 2.4.2, because h f ( P − Q ) ≥ .(b) Since f is even, it suffices to consider m ≥ . Further, this result is trivialfor m = 0 , . We will finish the proof by induction. Assume it is known for m-1and m. Replacing P,Q in Theorem 2.4.3 by [m]P and P, we find h f ([ m + 1] P ) = − h f ([ m − P ) + 2 h f ([ m ] P ) + 2 h f ( P ) + O (1)= ( − ( m − + 2 m + 2) h f ( P ) + O (1)= ( m + 1) h f ( P ) + O (1) Theorem 2.4.2. (Mordell-Weil)
Let E be an elliptic curve defined over a num-ber field K. The group E(K) is a finitely generated Abelian groupProof.