Elliptic Curves over p-adic Fields

Transcription

1 Elliptic Curves over p-adic Fields Robert L. Benedetto Amherst College University of Connecticut Saturday, May 17, 2014

2 Quick review of p-adic numbers The p-adic absolute value p on Q has 0 p = 0 and r p s pn = p n for r, s Z not divisible by p. p is non-archimedean: x p 0, with equality iff x = 0, xy p = x p y p, x + y p max{ x p, y p }. Q p is the completion of Q w.r.t. p. (All p -Cauchy sequences converge in Q p ). Fun Fact: Let {a n } n 0 be a sequence in Q p. Then n 0 a n converges if and only if lim n a n = 0.

3 The Residue Field and Value Group The ring of integers and (unique) maximal ideal of Q p are O p = Z p = {x Q p : x p 1} and M p = pz p = {x Q p : x p < 1}. The residue field of Q p is O p /M p = Z p /pz p = Fp. The value group of K is Q p p = p Z (0, ).

4 A Sketch of Q Z Z 3 Z 3 1+3Z 3 3Z 3 2+3Z Z Z 3

5 Extension Fields Let K be a finite (or more generally, algebraic) extension of Q p. Then p extends uniquely to K. The new value group K contains Q p as a subgroup. The ramification degree is e = [ K : Q p ]. If e <, a uniformizer π M K = {x K : x p < 1} is an element of maximum absolute value less than 1. M K = πo K, where O K = {x K : x p 1}. The new residue field k is a finite (respectively, algebraic) extension of F p. The residue field extension degree is f = [k : F p ]. Fact: [K : Q p ] = ef.

6 Examples K = Q p ( n p) has e(k/q p ) = n, f (K/Q p ) = 1. K = Q p (ζ p n 1) has e(k/q p ) = 1, f (K/Q p ) = n. K = Q ur p, the unramified closure of Q p, has e = 1 and f =. In fact, k = F p. K = Q p, the algebraic closure of Q p, has e = and f =. In fact, Q p p = p Q and k = F p. K = C p, the completion of Q p, has e = and f =. In fact, Q p p = p Q and k = F p.

7 Elliptic Curves over p-adic fields Let K/Q p be a p-adic field with e(k/q p ) finite, so K has a uniformizer π. For this talk, an elliptic curve over K is given by a Weierstrass equation E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6. After a change of coordinates, we may assume this is an integral model, i.e., a i p 1. Mod out by πo K = M K, (giving coefficients a i k), E may or may not still be an elliptic curve. (May be singular.) If it s still an elliptic curve: good reduction. Otherwise: bad reduction

8 Examples: Good Reduction p 2, and E : y 2 = x 3 + Ax + B, where p = 16(4A B 2 ) p = 1. Then E : y 2 = x 3 + Ax + B has 0 and hence is nonsingular. So good reduction. p anything, and E : y 2 + a 1 xy = x 3 + a 6, has = a 6 (a a 6 ). Again, p = 1 implies good reduction.

9 An Example of Multiplicative Reduction p anything and E : y 2 + xy = x 3 + π n has = π n ( π n ), so p = π n p < 1. E : y 2 + xy = x 3 is singular at (0, 0). Note: Node, not cusp, because near (0, 0) it looks like y 2 + xy = 0, i.e. y(x + y) = 0: two crossing lines.

10 Multiplicative Reduction Example: Part 2 E : y 2 + xy = x 3 + π n reduces to E : y 2 + xy = x 3. Blow up E at (0, 0) via y = tx, giving E : t 2 x 2 + tx 2 = x 3, i.e. x = t 2 + t, y = tx = t 3 + t 2. [Note: bad at t = 0, 1.] So E ns is a copy of P 1 with two points missing. But that was blowing up E at the point x = 0, y = 0 in Spec k[x, y]. Instead, let s blow up E at the point x = 0, y = 0, π = 0 in Spec O K [x, y].

11 Multiplicative Reduction Example: Part 3 Blowing up E : y 2 + xy = x 3 + π n at x = 0, y = 0, π = 0: We ve already seen one component on the special fiber is E : y 2 + xy = x 3, i.e. x = t(t + 1). [Via y = tx.] If n 2, blowing up via x = πx 1 and y = πy 1 [and cancelling π 2 ] gives E 1 : y1 2 + x 1y 1 = πx1 3 + πn 2 which reduces (mod π) to y 1 (x 1 + y 1 ) = 0, i.e., two lines. And each of those points really should have two points removed: (0, 0), where they cross, and each one s point at, where it meets the original component.

12 Multiplicative Reduction Example: Part 4 Blowing up E : y 2 + xy = x 3 + π n at x = 0, y = 0, π = 0: Component 0: E : y 2 + xy = x 3, i.e. x = t(t + 1). Components 1 and n 1: E 1 : y x 1y 1 = πx πn 2, with E 1 : y 1 (x 1 + y 1 ) = 0, where x = πx 1, y = πy 1. If n 4, blowing up via x 1 = πx 2 and y 1 = πy 2 [and cancelling π 2 ] gives E 2 : y2 2 + x 2y 2 = π 2 x2 3 + πn 4 And so on, until we stop at E m : y 2 m + x m y m = π m x 3 m + π (if n = 2m + 1 is odd) or E m : y 2 m + x m y m = π m x 3 m + 1 (if n = 2m is even)

13 Néron Models over a p-adic field K The set of all the equations E, E 1,..., E m, with the understanding that the singular points on the special fiber have been removed, is the Néron Model E for E/K. Key properties: Each equation E i /K is simply a change of coordinates of E/K. That is, the generic fiber of E is E. E is smooth: any singular points, even on the special fiber (i.e., mod π), have been removed. Every K-rational point P E(K) has a reduction P on one of the E i. If we replace K by an unramified extension L/K, like L = K ur, the Néron model doesn t change. (Well, technically it base-changes to E Spec O L.) But if we replace K by a ramified extension L/K, E is usually not a Néron model for E/L.

14 E : y 2 + xy = x 3 + π n Revisited Special fiber of E has n components (called type I n ): but if we work over L = K( π), then points (x, y) with, say, x p = π p, will want to reduce to those missing singular points. Instead, E L should have 2n components on the special fiber, since equation is E : y 2 + xy = x 3 + π 2n.

15 Another Example: Additive Reduction Consider p 2, and E : y 2 = x 3 π 2 x. = 64π 6, so p < 1. E : y 2 = x 3 is singular at (0, 0). (Cusp, not node, because near (0, 0) it looks like y 2 = 0, i.e., doubled line.) Blowing up y = tx gives t 2 = x. So again reduced curve is P 1, but this time with only one bad point (t = 0) removed.

16 E : y 2 = x 3 π 2 x, Part 2 Let s blow up in Spec O K [x, y] more at x = 0, y = 0, π = 0: x = πx 1 and y = πy 1 gives E 1 : y 2 1 = πx 3 1 πx 1 = πx 1 (x 1 + 1)(x 1 1). Reduction: E 1 : y 2 1 = 0; double copy of P1 But there are K-rational points with y 1 p < 1 and x 3 1 x 1 p < 1. We need to blow up more: (0, 0): x 1 = πx 2,0, y 1 = πy 2 gives E 2,0 : y 2 2 = π2 x 3 2,0 x 2,0. Reduction: E 2,0 : y 2 2 = x 2,0; single P 1. (1, 0): x 1 = 1 + πx 2,1, y 1 = πy 2 gives E 2,1 : y 2 2 = x 2,1(1 + πx 2 2,1 )(2 + πx 2 2,1 ) Reduction: E 2,1 : y 2 2 = 2x 2,1; single P 1. ( 1, 0): x 1 = 1 + πx 2, 1, y 1 = πy 2 gives E 2, 1 : y 2 2 = x 2, 1( 1 + πx 2 2, 1 )( 2 + πx 2 2, 1 ) Reduction: E 2, 1 : y 2 2 = 2x 2, 1; single P 1.

17 E : y 2 = x 3 π 2 x, Part 3 The Néron model for E consists of the four copies of P 1 : E, E 2,0, E 2,1, and E 2, 1. The double copy of P 1 (from E 1 ) is entirely singular, so remove it. [E 1 is part of the minimal proper regular model, but not part of the Néron model.] E : y 2 = x 3 π 2 x is said to have type I 0 reduction. (Recall p 2.) Note: over L = K( π), E becomes y 2 = x 3 π 4 L x. The change of coordinates x = π 2 L x, y = π3 Lỹ gives E : ỹ 2 = x 3 x, which has good reduction (type I 0 ).

18 The Tate Curve Let s now view Q p K C p. D(a, r) denotes the open disk D(a, r) = {x C p : x a p < r}. Theorem (Tate) There are power series a 4 (q) = q + O(q 2 ) and a 6 (q) = q + O(q 2 ) in Z[[q]] converging for q D(0, 1), so that for q 0, E q : y 2 + xy = x 3 + a 4 (q)x + a 6 (q) has multiplicative reduction, with = q + O(q 2 ). Moreover, for fixed q, there is a map φ q : C p /q Z E q (C p ). If q K, then φ q : K /q Z E q (K).

19 The Tate Curve, Continued Idea: For fixed q K with 0 < q p < 1, consider the annulus A q = {x C p : q p x p 1}, and glue the two ends of the annulus to each other: for y p = 1, glue y to qy. We get a p-adic analog of a torus. In fact, any elliptic curve E/K of (split) multiplicative reduction is isomorphic to the Tate curve for a unique q. If E has reduction type I n, then q p = π n p, and the n components of the Néron model E correspond to the sets: C i = {x C p : x p = π i p}, for i = 0, 1,..., n 1. Each would be a copy of all of P 1 (C p ), except it is missing two residue classes: at 0 and. Hence the two missing points on each component of E.

20 Berkovich Disks Idea: Make a bigger space containing the disk D(a, r) = {x C p : x a p r} C p. The new space D Ber (a, r) will include all the points of D(a, r), plus one point ζ(b, s) for each closed disk D(b, s) D(a, r). In particular, the Berkovich version A q,ber of the annulus is A q = {x C p : q p x p 1} A q,ber = D Ber (0, 1) D Ber (0, q ). The Tate curve glueing will glue ζ(0, q ) to ζ(0, 1).

21 Berkovich Disks Are Connected D(0,1) ζ(0,1) D(x, x-y )=D(y, x-y ) ζ(x, x-y ) D(x,r) x y D(y,r) ζ(x,r) ζ(y,r) x y

22 The Berkovich Unit Disk ζ(0,1)

23 Berkovich Elliptic Curves: Good Reduction Any algebraic variety over K can be Berkovichized; the construction is functorial. If E has good reduction, then E Ber has one special point ζ 0 : The branches emanating from ζ 0 are in natural one-to-one correspondence with the points of E(F p ) Each branch is a copy of the open Berkovich disk D Ber (0, 1).

24 Berkovich Elliptic Curves: Multiplicative Reduction If E has multiplicative reduction, then the Tate curve for E glues the two endpoints of the line segment producing a circle. ζ(0, 1) to ζ(0, q ), For each point ζ(0, r) on this circle, with r C p : There are two branches, towards zero and, pointing around the circle, The other branches emanating from ζ(0, r) are in natural one-to-one correspondence with the points of F p, Each such branch is a copy of the open Berkovich disk D Ber (0, 1). Moreover, the K-rational points on this circle, i.e., the ones of the form ζ(0, π i ), correspond to the components C i of the Néron model.

25 Moral: A Berkovich point ζ is a choice of coordinates for the defining equation of E. and The topology of E Ber determines the reduction type of E (over a large enough extension of K).