I have a problem to see how the galois-theoretic definition of the L-function of an elliptic curve gives the right answer and also the connection between the L-function of a weight 2 Hecke eigenform and the corresponding elliptic curve.
- Let E over $\mathbb{Q}$ be an elliptic curve. At the good primes $p$ the local L-function is defined by the inverse of $L_p(E,s) = 1-a_pp^{-s}+p\cdot p^{-2s}$. On the other hand, the L-function may also be defined via the inverse of the characteristic polynomial of the Frobenius endomorphism at $p$ acting on the (first) etale cohomology of $E$. But the char polynomial of the Frobenius is $f_p(X) = X^2-a_pX+p$ and evaluated at $X=p^{-s}$ we get $f_p(s) = p^{-2s}-a_pp^{-s}+p$ which obviously differs from $L_p(E,s)$ even though they are supposed to be the same? Where is the mistake?
- Let $f=\sum_{n\geq 1} a_n q^n$ be a Hecke eigennewform (is this the right terminology?) of weight 2 of level $N$ which corresponds to the elliptic curve $E$. The $L$-function of this modular form is by definition $L(f,s)= \sum_{n\geq 1} a_n/n^s$. By some computation we can write this as an Euler product $\prod_{\text{good p}} L_p(f,s) \prod_{\text{bad p}} L_p(f,s)$. The modularity theorem then says $L(f,s)=L(E,s)$, the $L$-functions of $E$ and $f$ coincide. But it does not say that $L_p(f,s)=L_p(E,s)$, right? Just the global $L$-functions are the same, but not necessarily the local factors? For a normalized Hecke eigenform $f$ of weight $k$, level $N$ and character $\xi$ the local factors at good primes for the $L$-function $L(f,s)$ are $L_p(f,s) = 1-a_pp^{-s}+p^{k-1}\xi(p)p^{-2s}$. And if this corresponds to an elliptic curve, then $k=2$ and $\xi$ is trivial, which would give $L_p(f,s)=1-a_pp^{-s}+p\cdot p^{-2s}$ which coincides with the local factors of the elliptic curve in (1) above. Is this right?
- To the first question. Is the galois-representational definition of the $L$-function just the Artin $L$-function of the etale cohomology representations of the elliptic curve? It should be, if I am not mistaken, right?