DIOPHANTINE GEOMETRY { TAUGHT BY JOE SILVERMAN NOTES BY SHAMIL ASGARLI 1. Diophantine Equations Objective. Solve polynomial equations using integers (or rationals). Example. Linear equation: Solve ax + by = c where a; b; c 2 Z. It is a classical fact from elementary number theory that this equation has a solution (for x and y) if and only if gcd(a; b) j c. Example. Quadratic equation: x2 + y2 = z2. We are interested in non-zero solutions (x; y; z) 2 Z, i.e. (x; y; z) 6= (0; 0; 0). Since the equation is homogeneous, it is enough to understand the solutions of X2 + Y 2 = 1 where X; Y 2 Q (points on the unit circle). Anyways, the complete solution is known for this problem. WLOG gcd(x; y; z) = 1, x is odd and y is even. Then the solutions are given by x = s2 − t2, y = 2st, and z = s2 + t2. s2 − t2 Analogously, the solutions (X; Y ) 2 2 of X2 + Y 2 = 1 are parametrized by: X = Q s2 + t2 2st and Y = . s2 + t2 Example. Consider the equation y2 = x3 − 2. It turns out that (3; ±5) are the only solutions in Z2, while there are infinitely many solutions (x; y) 2 Q. Goal. Given f1; f2; :::; fk 2 Z[X1; :::; Xn]. For R = Z; Q, or any other ring. Let n V (R) = f(x1; x2; :::; xn) 2 R : fi(x1; :::; xn) = 0 for all ig Describe V (R). Two questions naturally arise. 1) Is V (R) = ;? This is undecidable for R = Z. 2) Is V (R) finite? 2 variables, 1 equation. Let C be a plane curve given by f(x; y) = 0 where f(x; y) 2 Z[x; y]. The goal would be to describe the solutions (x; y) 2 Z2; Q2; R2 or C2. As the ring gets bigger and bigger, the task progressively becomes easier. In other words, we are concerned with the solution set C(R) = f(x; y) 2 R2 : f(x; y) = 0g. Example. C : x3 + y2 = 1. THM C(Q) = C(Z) = f(1; 0); (0; 1)g Example. C : xn + yn = 1. FLT C(Q) ⊆ f(±1; 0); (0; ±1)g where FLT stands for Fermat's Last Theorem (proved by Wiles). Idea. If degf(x; y) is big, does that necessarily mean fewer solutions? Not necessarily, e.g. y = xd still has plenty of solutions. Guiding Principle. Geometry (solutions to polynomial equations over an algebraically closed field) determines the arithmetic (number theory, i.e. solutions over integers or non- closed fields). 1 Consider the plane curve C : f(x; y) = 0. There are some extra points \at infinity". Let C = C [ fpoints at infinityg. Sometimes, C is a nice curve (smooth). Not so nice curves are the ones with singularities (think of a cuspidal or nodal cubic curve). We can blow-up these curves at their singular points to make them smooth. Assuming C is nice, the set C(C) is a nonsingular compact 1-dimensional complex mani- fold, i.e. a Riemann surface of genus g. Intuitively, g counts the number of holes. So g = 0 corresponds to the 2-sphere, while g = 1 corresponds to the usual torus, etc. So here, the genus g is the \geometry" side (see the Guiding Principle above). Theorem. Consider the plane curve C : f(x; y) = 0 for f 2 Q[x; y]. Suppose there are no singularities, so C(C) = g-holed torus, where g is the genus of C: There are three cases to consider. Case 1. g = 0. Then C(Q) = ; or C(Q) = Q [ f;g. There exists an algorithm to determine the conclusion. Case 2. g = 1. Then C(Q) = ; or C(Q) = finitely-generated abelian group: This is the Mordell-Weil Theorem. In the latter case, we know that any finitely-generated abelian group is of the form r finite abelian group × Z | {z } =torsion part . The non-negative integer r is called the rank. It is a theorem of Mazur that the torsion part has order at most 16. Furthermore, there exists an algorithm to determine the torsion part. It is not known if the rank r can be unbounded or not. Current record for an example with high rank is r = 28 due to Elkies. There is no known algorithm to determine the rank in general. Case 3. g ≥ 2. Then C(Q) is finite. This is a theorem of Faltings (this result was previously known as Mordell's Conjecture). There is no algorithm in general to find the solution set. Goals for the class. We will prove Mordell-Weil Theorem and Faltings' Theorem (but not Faltings' original proof). Key Tools. (1) Diophantine Approximation: how closely can a rational quantity approximate an irrational quantity? We will learn about results of Roth, Baker and others. (2) Height Functions: measuring complexity of objects. 2. Diophantine Approximation Let us say few words about Diophantine Approximation. First, since is dense in , it is a p Q p R true that inf j − 2j = 0. However, we are interested in approximating 2 with rationals a=b2Q b whose denominators are not so large (relatively speaking). For example, here are two facts that are easy to prove: a a p 1 (1) There are infinitely many 2 with gcd(a; b) = 1 satisfying − 2 < . b Q b b2 a a p 1 (2) There are only finitely many 2 with gcd(a; b) = 1 satisfying − 2 < . b Q b b3 In fact, let's prove a more general result that implies the second statement. 2 Theorem. (Liouville) Let > 0. If x 2 R satisfies a degree n polynomial with coefficients in Q, then a 1 − x < b bn+ a has only finitely many solutions for with gcd(a; b) = 1. b Remark. The following proof was communicated to me by Ming Hao Quek. Proof. Consider the set S defined by a a 1 S := 2 : − x < b Q b bn+ Assume, to the contrary, that S is infinite. Say x satisfies a monic polynomial P (X) 2 Q[X] Qn a a of degree n. Let P (X) = i=1(X − xi) where xi 2 C and x1 = x. Given b 2 S, P b is a rational number with denominator at most Dbn for some fixed D > 0. Since P (X) has only finitely many roots, and S is infinite, the subset na a o S0 := 2 S : P 6= 0 b b a 0 must be infinite as well. For all b 2 S , we have a 1 P ≥ b Dbn a 0 On the other hand, for all b 2 S , a a a a P = − x − x ··· − x b b b 2 b n 0 1 n a Y B a C a n−1 a ∆ ≤ − x B − x + jx − xijC ≤ − x (1 + δ) = ∆ − x ≤ b b b b bn+ i=2 @ | {z }A | {z } <δ ≤1 where δ is any upper bound for the difference of the roots of P (X), and ∆ := (1 + δ)n−1 only depends on x. Combining the upper and lower bounds, we get 1 ∆ ≤ ) bn+ ≤ D∆bn Dbn bn+ a 0 0 a 0 for all b 2 S . Since S is infinite, we can choose b 2 S where b is arbitrarily large, and this leads to a contradiction. a Using the above theorem, we easily see that there are only finitely many 2 with b Q a p 1 gcd(a; b) = 1 satisfying − 2 < . Similarly, for any > 0, there are only finitely many b b3 a a p 1 2 with gcd(a; b) = 1 satisfying − 3 2 < . Q 3+ b b b a Fact. It is also true, but much harder to prove, that there are only finitely many b 2 Q a p 1 with gcd(a; b) = 1 satisfying − 3 2 < . This would instantly follow from Roth's b b2:5 celebrated theorem. 3 Theorem. (Roth) Let > 0. If x is an irrational number, then there are only finitely a many 2 with gcd(a; b) = 1 satisfying b Q a 1 − x < b b2+ 3. Algebraic Geometry Background We have the standard definitions of affine and projective spaces. n A (K) = f(x1; :::; xn): xi 2 Kg n n+1 ∗ ∗ P (K) = A (K) n f0g=K = f[x0 : ··· : xn] = [λx0 : ··· : λxn] for λ 2 K g Let K be a number field, and fix an algebraic closure K. The Galois group GK = Gal(K=K) n acts on A (K) by σ(P ) = (σ(x1); :::; σ(xn)) for P = (x1; :::; xn). n GK n n n Then A (K) = fixed points of A (K) = A (K). Likewise, GK acts on P (K) by σ(P ) = [σ(x0): ··· : σ(xn)] for P = [x0 : ··· : xn]. Proposition. Pn(K)GK = Pn(K). Proof. This is an application of Hilbert's Theorem 90. Definition. We say that f(x0; x1; :::; xn) is a homogeneous of degree d if X i0 in f = aI x0 ··· xn I=(i0;:::;in) i0+···+in=d n or equivalently, f(λx0; :::; λxn) = λ f(x0; :::; xn) in the ring K[λ, x0; :::; xn]. For P 2 Pn(K), f(P ) is not well-defined but fP : f(P ) = 0g is well-defined. n m Definition. A rational map f : P ! P is f = [f0; :::; fm] with f0; :::; fm 2 K[x0; :::; xn] homogeneous of degree d. To be more pedantic, we could have written f : Pn(K) ! Pm(K). Then f(P ) is (almost) well-defined: If P = [a0; :::; an], then f(P ) = [f0(a0; :::; an); :::; fm(a0; :::; an)] is well-defined when fi(a0; :::; an) 6= 0 for some i.