Milne. .Elliptic.Curves.And.Algebraic.Geometry.(1996).[sharethefiles.com]

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that the connected component G0 of G containing zero is Gm, and such that G/G0
is a cyclic group of order n.
[[Diagram omitted]]
(c) For an elliptic curve E which has cuspidal reduction and ordp(") = 5, the special
fibres for the three models are: (a) a cubic curve with a cusp; (b) five curves of genus
0, one with multiplicity 2, intersecting as below; (b) an algebraic group G whose zero
component is Ga and such that G/G0 is a group of order 4 killed by 2.
[[Diagram omitted.]]
Finally, the mysterious quotient E(Qp)/E0(Qp) is equal to G(Fp)/G0(Fp) where G is the
special fibre of the smooth N�ron model and G0 is its zero component. In the above three
examples, it is (a) the trivial group; (b) a subgroup of a cyclic group of order n (and equal
to a cyclic group of order n if E has split nodal reduction); (c) a subgroup of (Z/2Z)2.
Summary. [[The top three E s are s]]
Minimal Model Weierstrass complete N�ron smooth N�ron
Plane curve Yes Not always Not always
Regular? Not always Yes Yes
E complete? Yes Yes Not always
E nonsingular? Not always Not always Yes
E a group? Not always Not always Yes
E(Zp) = E(Qp)? Yes Not always Yes
Tate has given an algorithm for determining the N�ron model of an elliptic curve.
10. Elliptic Curves over the Complex Numbers
In this section, we review some of the theory of elliptic curves over C.
Lattices and bases. A lattice in C is the subgroup generated by two complex numbers
linearly independent over R: thus
� = Z�1 + Z�2.
Since neither �1 nor �2 is a real multiple of the other, we can order them so that (�1/�2) >
0. If {�1, �2} is a second pair of elements of �, then
�1 = a�1 + b�2, �2 = c�1 + d�2, a, b, c, d " Z,
that is,
�1 �1
�2 = A �2 ,
with A a 2 � 2 matrix with integer coefficients. The pair (�1, �2) will be a basis for � if and
only if A has determinant �1, and (�1/�2) > 0 if and only if det A > 0. Therefore, if we
let SL2(Z) be the group of matrices with integer coefficients and determinant 1, then SL2(Z)
acts transitively on the set of bases (�1, �2) for � for which (�1/�2) > 0. We have proved
the following statement:
42 J.S. MILNE
Proposition 10.1. Let M be the set of pairs of complex numbers (�1, �2) such that
(�1/�2) > 0, and let L be the set of lattices in C. Then the map (�1, �2) �! Z�1 + Z�2
induces a bijection
SL2(Z)\M �! L.
Here SL2(Z)\M means the set of orbits in M for the action
a b �1 a�1 + b�2
c d �2 = c�1 + d�2
Let H be the complex upper half-plane:
H = {z " C | (z) > 0}.
Let z " C� act on M by the rule z(�1, �2) = (z�1, z�2) and on L by the rule z� = {z� |
� " �}. The map (�1, �2) �! �1/�2 induces a bijection M/C� �! H. The action of SL2(Z)
on M corresponds to the action
a� + b
a b
� =
c d
c� + d
on H. We have bijections
L/C� �!�! SL2(Z)\M/C� �!�! SL2(Z)\H.
Z� + Z (�, 1) �
For a lattice �, the interior of any parallelogram with vertices z0, z0+�1, z0+�2, z0+�1+�2,
where {�1, �2} is a basis for �, is called a fundamental domain or period parallelogram D
for �. We usually choose D to contain 0.
Quotients of C by lattices. . Let � be a lattice in C. Topologically the quotient C/� H"
R2/Z2, which is a one-holed torus (the surface of a donut).
Write � : C �! C/� for the quotient map. Then C/� can be given the structure of a
Riemann surface (i.e., complex manifold of dimension 1) such that a function � : U �! C on
an open subset U of C/� is holomorphic (resp. meromorphic) if and only if the composite
� �% � : �-1(U) �! C is holomorphic (resp. meromorphic) in the usual sense. It is the unique
structure for which � is a local isomorphism of Riemann surfaces.
We shall see that, although any two quotients C/�, C/� are homeomorphic, they will be
isomorphic as Riemann surfaces only if � = z� for some z " C.
Doubly periodic functions. Let � be a lattice in C. According to the above discussion,
a meromorphic function on C/� is simply a meromorphic function f(z) on C such that
f(z + �) = f(z) for all � " �.
This condition is equivalent to
f(z + �1) = f(z), f(z + �2) = f(z)
for {�1, �2} a basis for �. Such a meromorphic function on C is said to be doubly periodic
for �.
Proposition 10.2. Let f(z) be a doubly periodic function for �, not identically zero, and
let D be a fundamental domain for � such that f has no zeros or poles on the boundary of
D. Then
ELLIPTIC CURVES 43
(a) ResP (f) = 0;
P "D
(b) ordP (f) = 0;
P "D
(c) ordP (f) � P a" 0 mod �.
P "D
The first sum is over the points in D where f has a pole, and the other sums are over the
points where it has a zero or pole (and ordP (f) is the order of the zero or the negative of the
order of the pole). Each sum is finite.
Proof. According to the residue theorem,
f(z)dz = 2�i( ResP (f)),
�
P "D
where � is the boundary of D. Because f is periodic, the integrals of it over opposite sides of
D cancel, and so the integral is zero. This gives (a). For (b) one applies the residue theorem
to f /f, noting that this is again doubly periodic and that ResP (f /f) = ordP (f). For (c)
one applies the residue theorem to z � f (z)/f(z). This is no longer doubly periodic, but the
integral of it around � lies in �.
Corollary 10.3. A nonconstant doubly periodic function has at least two poles.
Proof. A holomorphic doubly periodic function is bounded on the closure of any fundamental
domain (by compactness), and hence on the entire plane (by periodicity). It is constant by
Liouville s theorem. It is impossible for a doubly periodic function to have a single simple
pole in a period parallelogram, because by (a) of proposition the residue at the pole would
have to be zero there, which contradicts the fact that it has a simple pole there.
The holomorphic maps C/� �! C/� . Let � and � be lattices in C. The map � :
C �! C/� realizes C as the universal covering space of C/�. Since the same is true of
� : C �! C/� , a continous map � : C/� �! C/� such that �(0) = 0 will lift uniquely to a
continuous map � : C �! C such that �(0) = 0:
�
C - C
�!
�! �!
�
C/� - C/�
�!
(see, for example, Greenberg, Lectures on Algebraic Topology, 5.1, 6.4). The map � will
be holomorphic (i.e., a morphism of Riemann surfaces) if and only if � is holomorphic.
[Nonexperts can take this as a definition of a holomorphic map � : C/� �! C/� .]
Proposition 10.4. Let � and � be lattices in C. A complex number � such that �� " �
defines a holomorphic map
[z] �! [�z] : C/� �! C/�
sending 0 to 0, and every holomorphic map C/� �! C/� is of this form (for a unique �).
Proof. It is obvious that � defines a holomorphic map C/� �! C/� . Conversely, let � :
C/� �! C/� be a holomorphic map such that �(0) = 0, and let � be its unique lifting to a
holomorphic map C �! C sending 0 to 0. For any � " �, z �! �(z + �) - �(z) takes values
in � �" C. But a continuous map from a connected set to a set with the discrete topology
is constant, and so the derivative of this function is zero:
� (z + �) = � (z).
44 J.S. MILNE
Therefore � (z) is doubly periodic. As it is holomorphic, it must be constant, say � (z) = � [ Pobierz całość w formacie PDF ]

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