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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 $T^2$.

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up vote 2 down vote accepted

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*}

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