Reduction of CM Elliptic Curves I'm working on Exercise 2.30 of Silverman's Advanced Topics of Elliptic Curves:
Suppose that $E/L$ is an elliptic curve with CM by an imaginary quadratic field $K$. Suppose that $L$ does not contain $K$ and let $L'=LK$ denote the compositum. Let $\mathfrak{P}$ be a prime of $L$ such that $E$ has good reduction. I want to show that $\mathfrak{P}$ is unramified in $L'$.
I believe a natural argument is to suppose that $\mathfrak{P}$ ramifies in $L'$ and to contradict the fact that $E$ has good reduction at $\mathfrak{P}$. Since $[L:L']=2$ it follows readily that $\mathfrak{P} = \mathfrak{Q}^2$ for some prime $\mathfrak{Q}$. However, I am having difficulty getting a handle on the fact that $E$ has good reduction at $\mathfrak{P}$. Any help would be appreciated.
 A: Here is a fun argument which follows some ideas of Serre in his "Lectures on the Mordell-Weil Theorem". For simplicity I'll take $L = \mathbb{Q}$ but this should generalise. Let $K = \mathbb{Q}(\sqrt{-d})$ where $d$ is squarefree and let $\mathcal{O}_K$ be the ring of integers. I claim that each $p$ ramifying in $K$ divides the discriminant of $E$.
First suppose $2$ does not ramify in $K$. Note that the action of $\mathcal{O}_K$ on $E[2]$ factors through $A = \mathcal{O}_K/2\mathcal{O}_K$
, so we consider the mod $2$ galois representation
$$\bar{\rho}_{E, 2} : G_{L} \to GL_2(\mathbb{F}_2).$$
If $2$ is inert then $A^\times$ is isomorphic to $C_3$, and if $2$ splits, then $A^\times$ is trivial. In either case an homomorphism from $A ^\times$ to $GL_2(\mathbb{F}_2)$ is contained in the unique $C_3$ subgroup.
In particular $\bar{\rho}_{E, 2} (G_{K})$ is contained in $C_3$. Now if $E$ is in Weierstrass form $y^2 = f(x)$ then the action of galois on $E[2]$ is just the $S_3$ action on the roots of $f$ which is a $C_3$ action if and only if the discriminant of $f(x)$ is a square (this discriminant is equal to $\Delta(E)$ up to an even power of $2$). Thus the inverse image of $C_3$ under $\bar{\rho}_{E, 2}$ is the absolute galois group of $L(\sqrt{\Delta(E)})$.
Thus $d$ divides $\Delta(E)$ (this is where I'm using $L = \mathbb{Q}$ because I don't want to think) - in fact we've proved the stronger fact that $-d$ and $\Delta(E)$ are equal up to a square factor.
Recall by binary quadratic form considerations that $-d \equiv 0,1 \pmod{4}$ and that $Cl(K)$ has elements of order $2$ if and only if $d$ is divisible by $2$ distinct prime factors. Thus $2$ is ramified if and only if $K = \mathbb{Q}(i), \mathbb{Q}(\sqrt{-2})$. In these cases, just writing down the specific curve shows that $2$ divides $\Delta(E)$ (note you're only allowed to change $\Delta(E)$ by certain powers by twisting).
A: I think (hope?) this argument works in full generality.
Let $q$ be a large prime inert in $K$. So $T_qE$ has a $G_{L’}$-endomorphism $u$ of square $d$ ($d$ being some $t^2\Delta$ – where $\Delta$ is the discriminant of $K$ – and coprime to $q$, so it is not a square in $\mathbb{F}_q$ thus in $\mathbb{Z}_q$).
Let $\Gamma_L$ and $\Gamma_{L’}$ be the images of the Galois actions of $G_L,G_{L’}$ respectively on $T_qE$. As $[L’:L]=2$, $[\Gamma_L:\Gamma_{L’}] \leq 2$.
Assume $\mathfrak{P}$ ramifies in $L’$. Then there is some $\sigma \in G_L$ nontrivial on $L’$ that is in the inertia group of $\mathfrak{P}$. So its action on $T_qE$ is trivial. But as $G_L=G_{L’} \cup G_{L’}\sigma$, $\Gamma_L=\Gamma_{L’}$. In particular, $u$ commutes to $\Gamma_L$ on $T_qE$, so $u$ commutes to $G_L$, so $u \in End(E)$. In particular, the action of $u$ on $\Omega^1(E)$ is a scalar $\lambda \in L$ such that $d=\lambda^2$, and thus $\Delta \in L^2$, so that $K \subset L$. Contradiction.
A: I found another proof.
Suppose $\mathfrak{P}$ ramifies in $L' = LK$.
Since $\mathfrak{P}$ is a prime of good reduction we have by the criterion of Neron-Ogg-Shafarevich that $L(E[N])/L$ is unramified at $\mathfrak{P}$, for any $N$ prime to the norm of $\mathfrak{P}$.
An elementary result in CM theory is that for $N$ odd, $K\subseteq L(E[N])$ (cf. Torsion points on CM elliptic curves over real number fields by Bourdon, Clark, Stankewicz). Thus $L'(E[N])=L(E[N])$ and so $\mathfrak{P}$ is unramified in $L'(E[N])$
However, since $\mathfrak{P}$ ramifies in $LK$ and $L \subseteq LK \subseteq L'(E[N])$, we must have that $\mathfrak{P}$ ramifies in $L'(E[N])$; which is a contradiction.
