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I'm working on the following problem from Silverman's advanced topic in the arithmetic of elliptic curves: Let $E$ be an elliptic curve defined over a number field $L$ with complex multiplication by $K$, and let $\mathfrak{P}$ be a prime of $L$ lying over $p$ at which $E$ has good ordinary reduction. Prove that $\mathbf{Q}_p$ contains a subfield isomorphic to $K$.

Now, I have the following partial solution which I can't finish. Let $K=\mathbf{Q}(\sqrt{-D})$ for some positive square-free integer. Since $E$ has good reduction at $\mathfrak{P}$, we know that $End(E)$ injects into $End(\tilde{E})$ (this is prop 4.4 in the CM chapter) where $\tilde{E}$ denote reduction at $\mathfrak{P}$. It follows that the ring of integers in $K$, $\mathcal{O}$ injects into $End(\tilde{E})$. Restriction then gives a map to $End(\tilde{E}[p])$. But $\tilde{E}[p]\cong\mathbf{Z}/p\mathbf{Z}$ since $E$ has ordinary reduction here, so this endomorphism ring is just $\mathbf{Z}/p\mathbf{Z}$ itself.

Summing up, I now have a map from $\mathcal{O}\to\mathbf{Z}/p\mathbf{Z}$. Now, under this map, $-D$ just goes to $-D\bmod p$. But then $\sqrt{-D}$ maps to a square root of $-D$, so that $-D$ must be a square $\bmod p$. Now, if $p$ does not divide $D$, this tells us something useful: $-D$ is invertible $\bmod p$ and a square, so by Hensel's lemma, $\sqrt{-D}\in\mathbf{Q}_p$, whence $K$ is a subfield of $\mathbf{Q}_p$. But if $p$ does divide $D$, this tells us nothing.

So, assuming what I have done above is coherent, I need to somehow rule out the case that $p$ divides $D$. Of course, this has to be the case because if $p$ divides $D$ then $K$ cannot be contained in $\mathbf{Q}_p$.

Is there an easy way to show that if $p$ divides $D$ and if $\mathfrak{P}$ lies over $p$, then the reduction must be bad or super singular? I can't think of a reason why this would be the case or how else to proceed. Any insight on this problem would be greatly appreciated.

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