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 .