This has been posted on SE, but I haven't gotten a reply, so I thought I'll try my luck here.


I would like to refer you to 2.30 & 2.32 in Silverman's book Advanced Topics in the Arithmetic of Elliptic Curves.


2.30(b)[(c) in errata]: Suppose $\mathfrak{P}$ remains inert in $L'$, say $\mathfrak{P}R\_{L'}=\mathfrak{P}'$. Prove that 
$$q\_{\mathfrak{P}}^2=q\_{\mathfrak{P}'},\quad a\_{\mathfrak{P}}=0,\quad\psi\_{E/L'}(\mathfrak{P}')=-q\_{\mathfrak{P}}.$$
2.32(a) Prove that the local $L$-series of $E$ at $\mathfrak{P}$ is given by
$$ L\_{\mathfrak{P}}(T,E/L) = \left\{ 
 \begin{array}{l l}
 \cdots & \quad \mathfrak{P}\text{ splits}\\
 1-\psi\_{E/L'}(\mathfrak{P}')T & \quad \text{$\mathfrak{P}$ inert}\\
 \cdots & \quad \mathfrak{P}\text{ ramifies}\\
 \end{array} \right.$$


In the beginning of the chapter, we have that if $E$ has good reduction (in this case, we have good reduction because of 2.30(c))$$L\_{\mathfrak{P}}(T,E/L) = 1-a\_{\mathfrak{P}}T+q\_{\mathfrak{P}}T^2$$
So is the question missing a power of 2 in $T$? Should it be $1+\psi\_{E/L'}(\mathfrak{P}')T^2\text{when is $\mathfrak{P}$ inert}$? Thanks for the help! (If it helps, I can print the question; say so in the comments and I'll do it.)


Edit: In Deuring's paper, he has $L\_0(s,Kk\_1,\mathbf{P})=1+N\mathbf{p}^{1-2s}$. So the presence of $2s$ seems to me to tally with $T^2$.



Hecke $L$-series exercise in Silverman's Advanced Topics in Arithmetic of EC
There is nothing wrong with the question. The subtlety is that $T$ is not constant with respect to the fields $L'$ and $L$. My mistake was that I took $T$ to be the same in both fields when in reality it is not.


\begin{align\*}
L(s,E/L)&=\prod\_{\text{$\mathfrak{P}$ in $L$}}\left(1-a\_{\mathfrak{P}}q\_{\mathfrak{P}}^{-2}+q\_{\mathfrak{P}}^{1-2s}\right)^{-1}\\
&=\prod\_{\text{$\mathfrak{P}$ in $L$}}\left(1+q\_{\mathfrak{P}}^{1-2s}\right)^{-1}\\
&=\prod\_{\text{$\mathfrak{P}$ in $L$}}\left(1+q\_{\mathfrak{P}}q\_{\mathfrak{P}}^{-2s}\right)^{-1}\\
&=\prod\_{\text{$\mathfrak{P}$ in $L$}}\left(1+q\_{\mathfrak{P}}q\_{\mathfrak{P}'}^{-s}\right)^{-1}\\
&=\prod\_{\text{$\mathfrak{P}'$ in $L'$}}\left(1-\psi\_{E/L'}(\mathfrak{P}')q\_{\mathfrak{P}'}^{-s}\right)^{-1}\\
&=L(s,\psi\_{E/L'})
\end{align\*}


