MA 254 notes: Diophantine Geometry

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1 MA 254 notes: Diophantine Geometry (Distilled from [Hindry-Silverman], [Manin], [Serre]) Dan Abramovich Brown University Janoary 27, 2016 Abramovich MA 254 notes: Diophantine Geometry 1 / 16

2 Mordell Weil and Weak Mordell-Weil Theorem (The Mordell-Weil theorem) Let A/k be an abelian variety over a number field. Then A(k) is finitely generated. We deduce this from Theorem (The weak Mordell-Weil theorem) Let A/k be as above and m 2. Then A(k)/mA(k) is finite. Abramovich MA 254 notes: Diophantine Geometry 2 / 16

3 Mordell-Weil follows from Weak Mordell-Weil The reduction comes from the following. Lemma (Infinite Descent Lemma) Let G be an abelian group, q : G R a quadratic form. Assume (i) for all C R the set {x G q(x) C} is finite, and (ii) there is m 2 such that G/mG is finite. Then G is finitely generated. In fact if {g i } are representatives for G/mG and C 0 = max{q(g i )} then the finite set S := {x G q(x) C 0 } generates G. indeed the canonical Néron-Tate height ĥ D on A(k) associated to an ample symmetric D gives a quadratic form satisfying (i), and (ii) follows from Weak Mordell-Weil. The Mordell-Weil theorem follows. Abramovich MA 254 notes: Diophantine Geometry 3 / 16

4 Proof of the Infinite Descent Lemma for m 3 q(x) 0 otherwise the elements {nx} would violate (i). Write x = q(x), c 0 = max{ g i }, S m k c0 = {x G : x m k c 0 }. Clearly G = S m k c0, so enough to show S m k c0 S for all k. We use induction on k 0. Clearly S c0 = S S. Assume S m k c0 S and let x S m k+1 c0 S m k c0. x g i + mg for some i so there is x 1 such that m x 1 = x g i. So x 1 = x g i m since m 3. x + g i m m k c 0 m k c 0 m Abramovich MA 254 notes: Diophantine Geometry 4 / 16

5 Proof of Weak Mordell-Weil - Finite Kernel Lemma Lemma (Finite Kernel Lemma) Fix a finite k /k. Then Ker ( A(k)/mA(k) A(k )/ma(k ) ) is finite. Proof of lemma Note this kernel is B k /k := (A(k) ma(k ))/ma(k). Since for a further extension k /k we have B k /k B k /k, we may replace k by a further extension. In particular we may assume k /k Galois, with Galois group G. The lemma now follows from the following, since G and A m ( k) are finite: Sublemma There is an injective function B k /k Maps(G, A m ( k)). Abramovich MA 254 notes: Diophantine Geometry 5 / 16

6 Proof of Weak Mordell-Weil - reduction - sublemma Proof of sublemma For x B k /k fix y A(k ) such that [m]y represents x. Consider the function f x : G A(k ) where f (σ) = y σ y. Note [m]f x (σ) = [m](y σ y) = [my] σ [m]y = x σ x = 0. This defines B k /k Maps(G, A m ( k)). We claim it is injective. Assume f x = f x, and let y, y be the chosen points in A(k ). So y σ y = y σ y, hence (y y) σ = y y for all σ. So y y A(k) and [m](y y) ma(k). Hence x x = 0 in B k /k, as needed. Abramovich MA 254 notes: Diophantine Geometry 6 / 16

7 Theorem (Chevalley-Weil+Hermite for [m]) Fix an abelian variety A/k and an integer m. There is a finite extension k /k such that if x A(k) then there is y A(k ) such that [m]y = x. Proof of Weak Mordell-Weil assuming Chevalley-Weil+Hermite Let k be as in Chevalley-Weil+Hermite for [m]. Then the map A(k)/mA(k) A(k )/ma(k ) is zero, hence by the Finite Kernel Lemma A(k)/mA(k) is finite. We will present two proofs of Chevalley-Weil+Hermite: one quick and dirty using Scheme Theory. One going through with rings and differentials explicitly. Abramovich MA 254 notes: Diophantine Geometry 7 / 16

8 Hermite s theorem(s) The Mordell-Weil theorem The following classical theorem appears in a first course: Theorem (Hermite 1) For any real B there are only finitely many number fields k with d k B. This one might not be as visible: Theorem (Hermite 2) For any integer n and finite set of primes S there are only finitely many number fields k unramified outside S with [k : Q] n. Hermite 2 follows from Hermite 1 given the following: Proposition (Discriminant bound (Serre), not proven here) Write [k : Q] = n and N = p d k p. Then d k (N n) n. Abramovich MA 254 notes: Diophantine Geometry 8 / 16

The Mordell-Weil theorem Why number-theorists want to understand schemes What I stated as Chevalley-Weil+Hermite is a consequence of Proposition (Spreading out) Let φ : Y X be a finite unramified

9 The Mordell-Weil theorem Why number-theorists want to understand schemes What I stated as Chevalley-Weil+Hermite is a consequence of Proposition (Spreading out) Let φ : Y X be a finite unramified morphism of projective varieties, all over k. Then there is a finite set of primes S, projective schemes X Spec Ok,S, Y Spec Ok,S and a finite unramified morphism ϕ : Y X, whose restriction to Spec k is φ : Y X. Abramovich MA 254 notes: Diophantine Geometry 9 / 16

10 The Mordell-Weil theorem Spreading out: Picture with point Abramovich MA 254 notes: Diophantine Geometry 10 / 16

11 Spreading out implies Chevalley-Weil The reduction works as follows: In our case take X = Y = A and φ = [m]. Since X Spec O k,s pojective, a point x X (k) extends to a morphism x : Spec O k,s X. Write Y x = Spec O k,s X Y. Since Y X is unramified we have Ω Y/X = 0. Ω Yx / Spec O k,s is the pullback of this to Y x, so also 0. If [m]y = x then y Y x and its closure ỹ is unramified over x. This means that O(ỹ) is finite unramified over O k,s, so k(y)/k unramified away from S. By Hermite 2 there are only finitely many such k(s). Let k be the Galois closure of their compositum. so y A(k ). Abramovich MA 254 notes: Diophantine Geometry 11 / 16

12 Exorcising schemes: Spreading Out explained Step 1: coordinates Say Y P m and X P n. We may replace Y P m by the graph of φ. We now have Y P m P n, and φ is the restriction of the natural projection P m P n P n. For each coordinate X i of P n we have U i = X Z(X i ) affine, with coordinate ring A i = k[x 0,..., x i,..., x n ]/(f 1i,..., f ri i). φ finite means: preimage of affine is affine, corresponding to a finite ring extension. So V i = φ 1 U i affine, with ring B i finite over A i. Let {y 1i,..., y ki i} be module generators and g ji the module relations and Y ji Y j i = h jj i the ring relations, with g ji, h jj i A i -linear in Y 1i,..., Y ki i : B i = A i [Y 1i,..., Y ki i]/({g ji }, {Y ji Y j i h jj i}). Abramovich MA 254 notes: Diophantine Geometry 12 / 16

13 Exorcising schemes: Spreading Out explained Step 2: shrinking to define rings and maintain finiteness There are finitely many nonzero coefficients {a α } in f j,i, g ji, h jj i, giving a subring O k,s0 := O k [{a α, aα 1 }] of k in which a α are units. Here S 0 is the set of places appearing in the factorizations of the a α. Let A i = O k,s [x 0,..., x i,..., x n ]/(f 1i,..., f ri i) and B i = A i [Y 1i,..., Y ki i]/({g ji }, {Y ji Y j i h jj i}). We clearly have A i = A i Ok,S k and B i = B i Ok,S k. By construction A i Y ji B i is a surjective module homomorphism. So B i is a finite A i -algebra. Abramovich MA 254 notes: Diophantine Geometry 13 / 16

14 Exorcising schemes: Spreading Out explained Step 3: shrinking to evade ramification The statement Y X unramified is equivalent to V i U i unramified. This is equivalent to Ω Bi /A i = 0. Consider the finitely generated B i -module Ω Bi /A i. We have Ω Bi /A i Ok,S k = Ω Bi /A i = 0. So Ann Ok,S (Ω Bi /A i ) 0. In other words there is nonzero c O k,s such that cω Bi /A i = 0. Replacing O k,s by O k,s [c 1 ] k we may assume Ω Bi /A i = 0, hence B i is an unramified A i -algebra. In the language of spreading out, Y := Spec B i and X := Spec A i. (Spreading Out) Abramovich MA 254 notes: Diophantine Geometry 14 / 16

15 Exorcising schemes: Chevalley-Weil explained Proposition Let x X (k) and φ(y) = x. Then k(y)/k is unramified outside S. Fix a nonzero prime p O k,s. We show k(y)/k is unramified at p. Let x = (α 0,..., α n ) X. Since O k,p is principal we can take α i O k,p relatively prime. Without loss of generality the uniformizer π p α 0. Replacing α i by α i /α 0 we may assume α 0 = 1. Consider the epimorphism A 0 k given by the maximal ideal (x 1 α 1,..., x n α n ). It gives (A 0 ) p x Ok,p (the image is no bigger). Abramovich MA 254 notes: Diophantine Geometry 15 / 16

16 The Mordell-Weil theorem Exorcising schemes: Chevalley-Weil explained Proof of the proposition Consider C := B 0 A0 O k,p. Since A 0 B 0 is finite and unramified, also O k,s C is finite and unramified. Consider the commutative diagram B 0 C A 0 O k,p B 0 k(y) A 0 k The universal property of tensor gives an arrow C k(y). Its image is a subring R y,p k(y) finite unramified over O k,p, which must be O k(y),p. Abramovich MA 254 notes: Diophantine Geometry 16 / 16

Proof of the Shafarevich conjecture

Proof of the Shafarevich conjecture Rebecca Bellovin We have an isogeny of degree l h φ : B 1 B 2 of abelian varieties over K isogenous to A. We wish to show that h(b 1 ) = h(b 2 ). By filtering the kernel