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See Advanced Topics in elliptic curves for the full question(see also errata:

2.30 (pg 184) Given $E/L$ an elliptic curve with complex multiplication by the ring of integers $R_K$ of $K$. Assume that $L$ does not contain $K$. Let $L'=LK$, so $L'$ is a quadratic extension of $L$, and let $\mathfrak{P}$ be a prime of $L$. Assume that $E$ has good reduction at $\mathfrak{P}$. Prove that $\mathfrak{P}$ is unramified in $L'$.

I think I should be using the theorem that $\mathfrak{P}$ ramifies in $L'$ $\iff$ $\mathfrak{P} \mid \text{disc}(L'/L)$, but how do I show that $\mathfrak{P}\nmid 2$? I think I would need to use good reduction, but I'm not sure how. Thanks for your help in advance!

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Help us help you... please cite the page where you find this question. – Álvaro Lozano-Robledo Jun 29 '13 at 17:41
It seems to be asking to show that, if $E$ has a good reduction at $\mathfrak P$, then $\mathfrak P$ cannot divide the discriminant of $L'$. But some doubts remain: what are $K$ and $L$? And I think there should be some relation between the discriminant of $L'/L$ and the discriminant of $E$? As a suggestion, type all your question here for the best answer, just as one should show full work for the best credit in a test. – awllower Jun 30 '13 at 3:43
@ÁlvaroLozano-Robledo I've just edited the question to show the source, terribly sorry for the oversight! – BlackAdder Jun 30 '13 at 9:04
@awllower This is the full question. – BlackAdder Jun 30 '13 at 9:06
Sorry, but you still have not explained what $K$ and $L$ are? That was what I meant: background references. Thanks still for responding. – awllower Jun 30 '13 at 9:15
up vote 1 down vote accepted

Answering my own question:
First, I'll answer something from the comments: $L'$ is a quad extension of $L$ (see Silverman's Arithmetic of Elliptic Curves III.9.4)

Now let $\rho\in\text{Aut}(L'/L)$.

Claim 1: $K\subseteq L$ if and only if every element of $\text{End}(E)$ is defined over $L$.
$\blacktriangleleft$Let $\sigma$ be an automorphism of the algebraic closure $\overline{L}$ over $L$. Choosing $\omega$ the invariant differential defined over $L$, we have $$\omega[\alpha]^{\sigma}=\alpha^{\sigma}\omega$$ where $[\cdot]$ is defined in ATAECII.1.1 as an isomorphism such that $(E,[\cdot])$ is normalised. So $[\alpha]^{\sigma}=[\alpha]$ if and only if $\alpha^{\sigma}=\alpha$ which gives us the result. $\blacktriangleright$

Claim 2: If $\mathfrak{P}$ ramifies in $L'$, then the effect of $\rho$ is trivial on the the residue class field.
$\blacktriangleleft$ See Prove automorphism is trivial $\blacktriangleright$

Claim 3: $\text{End}(E)\rightarrow\text{End}(\tilde{E})$ is injective.
$\blacktriangleleft$ See Silverman's AECVII.2.1 $\blacktriangleright$

Now we know from the claim 1 that there exist an endomorphism $f\in\text{End}(E)$ defined over $L'$ but not $L$. So we have $f^{\rho}\neq f$. If $\mathfrak{P}$ ramifies in $L'$, then the effect of $\rho$ is trivial on the the residue class field (claim 2), so reducing mod $\mathfrak{P}$ gives $\widetilde{f^{\rho}}=\widetilde{f}$. But this contradicts with the injectivity of the reduction map on $\text{End}(E)$ (claim 3). Hence $\mathfrak{P}$ is unramified.

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