Fix an integer . The group structure on
is defined by
algebraic formulas with coefficients that are elements of
, so the subgroup
We continue to assume that is an elliptic curve over a number field
.
For any positive integer
, the group
is isomorphic as an
abstract abelian group to
. There are various
related ways to see why this is true. One is to use the Weierstrass
-theory to parametrize
by the the complex numbers, i.e.,
to find an isomorphism
, where
is a
lattice in
and the isomorphism is given by
with respect to an appropriate choice of coordinates
on
. It is then an easy exercise to verify that
.
Another way to understand is to use that
is isomorphic
to the quotient
If is a prime, then upon chosing a basis for the two-dimensional
-vector space
, we obtain an isomorphism
. We thus obtain a mod
Galois representation
In order to attach an -function to
, one could try to embed
into
and use the construction of Artin
-functions from Section 9.5.
Unfortunately, this approach is doomed in general, since
frequently does not embed in
.
The following Sage session shows that for
, there are
no 2-dimensional irreducible representations of
,
so
does not embed in
.
(The notation in the output below is [degree of rep, number of times it occurs].)
Instead of using the complex numbers, we use the -adic numbers, as
follows. For each power
of
, we have defined a homomorphism
Let be the fixed field of
. The image of
is infinite, so
is an infinite extension of
.
Fortunately, one can prove that
is ramified at only finitely many
primes (the primes of bad reduction for
and
--see [ST68]). If
is a
prime of
, let
be a choice of decomposition group for
some prime
of
lying over
, and let
be the
inertia group. We haven't defined inertia and decomposition groups
for infinite Galois extensions, but the definitions are almost the
same: choose a prime of
over
, and let
be the
subgroup of
that leaves
invariant. Then the
submodule
of inertia invariants is a module for
and the characteristic polynomial
of
on
is well defined (since inertia
acts trivially). Let
be the polynomial obtained by
reversing the coefficients of
. One can prove that
and that
, for
does
not depend on the choice of
. Define
for
using a different prime
, so the definition of
does not depend on the choice of
.
A prime
of
is a prime of good reduction for
if
there is an equation for
such that
is an elliptic
curve over
.
If
and
is a prime of good reduction for
, then
one can show that that
where
and
is the reduction of a local minimal
model for
modulo
. (There is a similar statement
for
.)
One can prove using fairly general techniques that the product
expression for defines a holomorphic function in some right
half plane of
, i.e., the product converges for all
with
Re
, for some real number
.