Chris' Third Talk

Consider the following elliptic curve:

C: y^2 = x^3 + 5^2

Which has additive reduction over Q5 but good reduction over Q5(5^1/3)

Now I_Q5 (the Inertia group, see previous posts) acts on VlE through C3 for all l other than 5. In our example, I_Q5 injects into VlE tensor Ql bar, and acts by a two by two diagonal matrix with entries phi and phi inverse, where phi is some character of order 3.

Define Xl:G_Qp -> Zl* as the action of G_Qp on the inverse limit of l^n roots of unity; then we call Xl the l-adic cyclotomic character.

Recall that TlE wedge TlE is isomorphic to Xl by the Weil paring.

I missed a bunch in the talk at this point.

Theorem("Tate Curve"): say E/Qp has split multiplicative reduction.

Then there is some q in Zp so that v(q)=v(disc(E))=-v(j)=n>0 (not sure what this v function is)

such that

E(Qp bar) ~ Qp bar cross modulo q^Z

here q^z represents a sort of spiral lattice; think of them as integer powers of a complex number; this allows us to "rotate" the p-adic space to within a certain sort of angle/absolute value pie chunk.

so E[l]= = (Z/lZ)^2, and E[l^n]=, and TlE=Zl^2.

Now G_Qp acts on TlE as an upper diagonal two by two matrix where the upper left entry is the l-adic cyclotomic character and the lower right entry is 1.

And I_Qp acts as an upper triangular matrix with 1s on the diagonal and v(q)*phi is in the upper right.

Then we define this phi to be the tame character; i.e. it is the part of Inertia that acts in a way we can observe like this.

So here I_Qp/G1 = the product of Zl for l not p, and we are projecting onto specific Zl.

Corollary: E/Qp has split multiplicative reduction -> Fp(T)=1-T. Yay local factors!

Now the case of potentially multiplicative reduction:

say E/K: y^2=x^3+ax+b

then the quadratic twist of E by d is

Ed/K: dy^2=x^3+ax+b

which is isogenous to E over K(d^1/2)

Let phi from the Galois group of that field extension be the nontrivial character.

Then Vl(Ed)=VlE tensor phi (this was an exercise to prove. I don't entirely know what it means)

Now say our E has pot. multiplicative reduction. Now consider E twisted by -6B, which has split multiplicative reduction (seeing why was maybe an exercise?)

now phi:Gal -> +-1 nontrivially means in the non-split multiplicative case, inertia acts trivially so Frobp |-> -1, in the additive inertia can act as -1 but this isn't too big a deal I guess? in the first case we can see that Fp(T)=1+T; in the additive case Fp(T)=1.

What this finally allows us to do is characterize Fp(T) for all cases. Yay!