[ Pobierz całość w formacie PDF ]

3 than any of the form
2
Y Z = X3 + aXZ2 + bZ3.
For example, it may be possible to find one of the first form that gives a nonsingular curve
over F2, whereas all equations of the second form become singular over F2 (see 1.3).
Other fields. Throughout this section, we can replace Q and Z with Qp and Zp, or in fact
with any local field and its ring of integers. Also, we can replace Q and Z with a number field
K and its ring of integers, with the caution that, for a number field K with class number
= 1, it may not be possible to find an equation for the elliptic curve that is minimal for all
primes simultaneously.
Exercise 6.2. (a) Find examples of elliptic curves E over Q such that
(i)  has a cusp S which lifts to a point in E(Qp);
(ii)  has a node S which lifts to a point in E(Qp);
(iii)  has a node S which does not lift to a point in E(Qp).
Here  is the reduction of the curve modulo a prime p = 2, 3. The equation you give for
2
E should be a minimal equation of the standard form Y Z = X3 + aXZ2 + bZ3.
(b) For the example you gave in (a)(i), decide whether it acquires good or nodal reduction
in a finite extension of Q.
ELLIPTIC CURVES 29
7. Elliptic Curves over Qp
Notation: A nonzero rational number a can be written a = pm r with r and s not divisible
s
by p. We then set ordp(a) = m. The following rule is obvious:
ordp(a + b) e" min{ordp(a), ordp(b)}, with equality unless ordp(a) = ordp(b).
Similarly, for an a " Qp, we set ordp(a) = m if a " pmZp \ pm+1Zp. The same rule holds,
and the two definitions of ordp agree on Q. In both cases, we set ordp(0) = ". Note that
ordp is a homomorphism Q× ’!Z.
p
Consider a curve
2
E : Y Z = X3 + aXZ2 + bZ3, a, b " Qp, 4a3 + 27b2 = 0.
After a change of variables X ’! X/c2, Y ’! Y/c3, Z ’! Z, we may suppose that a, b " Zp.
As in the last section, we obtain from E a curve  over Fp and a reduction map
¯
P ’! P : E(Qp) ’! (Fp).
We shall define a filtration
E(Qp) ƒ" E0(Qp) ƒ" E1(Qp) ƒ" · · · ƒ" En(Qp) ƒ" · · ·
and identify the quotients. First, define
¯
E0(Qp) = {P | P is nonsingular}.
It is a subgroup because, as we observed on p26, a line through two nonsingular points on a
cubic (or tangent to a nonsingular point), will meet the cubic again at a nonsingular point.
Write ns for  \ {any singular point}. The reduction map
¯
P ’! P : E0(Qp) ’! ns(Fp)
is a homomorphism, and we define E1(Qp) be its kernel. Thus E1(Qp) consists of the points
P that can be represented as (x : y : z) with x and z divisible by p but y not divisible by p.
In particular, P " E1(Qp) =Ò! y(P ) = 0.
Define
x(P )
En(Qp) = {P " E1(Qp) | " pnZp}.
y(P )
Theorem 7.1. The filtration E(Qp) ƒ" E0(Qp) ƒ" E1(Qp) ƒ" · · · ƒ" En(Qp) ƒ" · · · has the
following properties:
(a) the quotient E(Qp)/E0(Qp) is finite;
¯
(b) the map P ’! P defines an isomorphism E0(Qp)/E1(Qp) ’! (Fp);
(c) for n e" 1, En(Qp) is a subgroup of E(Qp), and the map P ’! p-n x(P ) mod p is an
y(P )
isomorphism En(Qp)/En+1(Qp) ’! Fp;
(d) the filtration is exhaustive, i.e., )"nEn(Qp) = {0}.
Proof. (a) We prove that E(Qp) has a natural topology with respect to which it is compact
and E0(Qp) is an open subgroup. Since E(Qp) is a union of the cosets of E0(Qp), it will
follow that there can only be finitely many of them.
Endow Qp×Qp×Qp with the product topology, Q3\{(0, 0, 0)} with the subspace topology,
p
and P2(Qp) with the quotient topology via Q3 \ {(0, 0, 0)} ’! P2(Qp). Then P2(Qp) is the
p
union of the images of the sets Z× × Zp × Zp, Zp × Z× × Zp, Zp × Zp × Z×, each of which is
p p p
30 J.S. MILNE
compact and open. Therefore P2(Qp) is compact. Its subset E(Qp) is closed, because it is
the zero set of a polynomial. Relative to this topology on P2(Qp) two points that are close
will have the same reduction modulo p. Therefore E0(Qp) is the intersection of E(Qp) with
an open subset of P2(Qp).
(b) Hensel s lemma implies that the reduction map E0(Qp) ’! (Fp) is surjective, and we
defined E1(Qp) to be its kernel.
(c) We assume (inductively) that En(Qp) is a subgroup of E(Qp). If P = (x : y : 1) lies in
E1(Qp), then y " Zp. Set x = p-mx0 and y = p-m y0 with x0 and y0 units in Zp. Then
/
2
p-2m y0 = p-3mx3 + ap-mx0 + b.
On taking ordp of the two sides, we find that -2m = -3m. Since m and m are integers,
this implies that there is an integer n such m = 2n and m = 3n; in fact, n = ordp(x).
y
The above discussion shows that if P = (x : y : z) " En(Qp) \ En+1(Qp), n e" 1, then
ordp(x) = ordp(z) - 2n
ordp(y) = ordp(z) - 3n.
Hence P can be expressed P = (pnx0 : y0 : p3nz0) with ordp(y0) = 0 and x0, z0 " Zp. In fact,
this is true for all P " En(Qp). Since P lies on E,
2 2 3
p3ny0z0 = p3nx3 + ap7nx0z0 + bp9nz0,
and so P0 =df (x0 : y0 : z0) lies on the curve
¯ ¯ ¯
2
E0 : Y Z = X3.
As y0 = 0, P0 is not the singular point of E0. From the description of the group laws in
¯
terms of chords and tangents, we see that the map
P ’! P0 : En(Qp) ’! E0(Fp)
is a homomorphism. Its kernel is En+1(Qp), which is therefore a subgroup, and it follows
from Hensel s lemma that its image is the set of nonsingular points of E0(Fp). We know
x(Q)
(see p27) that Q ’! is an isomorphism E0(Fp) \ {singularity} ’! Fp. The composite
y(Q)
x(P0) p-nx(P )
P ’! P0 ’! is P ’! mod p.
y(P0) y(P )
(d) If P " )"En(Qp), then x(P ) = 0, y(P ) = 0. This implies that either z(P ) = 0 or
2
Y = bZ2, but the second equation would contradict P " E1(Qp). Hence z(P ) = 0 and
P = (0 : 1 : 0).
Remark 7.2. In the above, Qp can be replaced with any local field.
Remark 7.3. It is possible to say much more about the structure of E(Qp). A one-
parameter commutative formal group over a (commutative) ring R is a power series
F (X, Y ) " R[[X, Y ]] satisfying the following conditions:
(a) F (X, Y ) = X + Y + terms of degree e" 2;
(b) F (X, F (Y, Z)) = F (F (X, Y ), Z);
(c) F (X, Y ) = F (Y, X);
(d) there is a unique power series i(T ) " R[[T ]] such that F (T, i(T )) = 0.
(e) F (X, 0) = X and F (0, Y ) = Y .
ELLIPTIC CURVES 31
In fact, (a) and (b) imply (d) and (e). If F is such a formal group over Zp, then the series
F (a, b) converges for a, b " pZp, and so F makes pZp into a group. One can show ([S1]
Chapter IV) that an elliptic curve E over Qp defines a formal group F over Zp, and that
there are power series x(T ) and y(T ) such that t ’! (x(t) : y(t) : 1) is an isomorphism of
pZp (endowed with the group structure provided by F ) onto E1(Qp). This is useful because
it allows us to derive results about elliptic curves from results about formal groups, which
are generally easier to prove.
An algorithm to compute intersection numbers. For f(X, Y ), g(X, Y ) " k[X, Y ], set I(f, g) =
I(origin, {f = 0} )" {g = 0}). We explain how to compute I(f, g) using only the following
properties of the symbol: I(X, Y ) = 1; I(f, g) = I(g, f); I(f, gh) = I(f, g) + I(f, h);
I(f, g + hf) = I(f, g) for all h; I(f, g) = 0 if g(0, 0) = 0. Regard f(X, Y ) and g(X, Y ) as
elements of k[X][Y ]. The theory of resultants allows us to construct polynomials a(X, Y )
and b(X, Y ) such that af + bg = r(X) with r(X) " k[X] and degY (b)
degY (g). Now
I(f, g) = I(f, bg) - I(f, b) = I(f, r) - I(f, b). [ Pobierz całość w formacie PDF ]
  • zanotowane.pl
  • doc.pisz.pl
  • pdf.pisz.pl
  • arachnea.htw.pl