We can define the L-function as before, and the completed L-function by adding in various factors of the gamma function to make it satisfy a nice functional equation.
This L-function is known to have analytic continuation to C for curves over Q, and holomorphic continuation to C for totally real fields.
Recall that E[n] denotes the group of n-torsion points on E over the algebraic closure of our field, and for E/Q, E[n] = C_n^2.
Identifying E[2] and E[3] is easy enough, but identifying E[4] involves taking the roots of a sextic.
However noting that the absolute Galois group of our field acts on E[n] we get a map, the mod n representation, from the absolute Galois group to GL2(Z/nZ). The image of this map will be the Galois group of Q(E[n])/Q, and since E[n] is the product of E[p^e_i] the primes dividing n, we can simply study prime power torsion.
To start, we define the (l-adic) Tate module, TlE, the inverse limit over n of E[l^n], which is isomorphic as a group to Z_l^2. Then we define VlE the l-adic representation to be TlE tensor Ql. (sometimes TlE or VlE* are also called this)
Now we can also take the inverse limit of our mod l^n representations to get a map from the absolute Galois group to Aut(TlE)=GL2(Z_l)
Theorem (Serre): if E has no CM, then the above representation is surjective for almost all l. This is false if E has CM, however.
It is conjectured that the above theorem is true for l>163, but it is unproven for any universal bound.
So we now want to study E/Qp and describe the action of the absolute Galois group there on E[l^n], TlE, and VlE.
If we take a finite extension K/Qp with Galois group G, define Gi = {g in G| g(x) is equivalent to x mod pi^i+1 for all x in O_k}, where pi is the prime in K sitting under p. This gives us the ramification filtration, G > G0 > G1 >... >{1} with each containment normal.
We call G0 Inertia I and G1 Wild Inertia.
This also seemed to skip around a fair amount. Hopefully when I write up the lectures from today things will get pulled together a bit.
No comments:
Post a Comment