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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.

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  • $\begingroup$ Wait – you want to show that $\mathfrak{P}$ is unramified in $L’$, no? $\endgroup$
    – Aphelli
    Jun 18, 2021 at 18:23
  • $\begingroup$ Yes, I have changed it $\endgroup$
    – Rdrr
    Jun 18, 2021 at 18:25

3 Answers 3

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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).

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  • $\begingroup$ Why would the pullback (inverse image?) of $C_3$ under the mod 2 representation, be equal to $L(\sqrt{\Delta(E)})$? $\endgroup$
    – Rdrr
    Jun 18, 2021 at 19:28
  • $\begingroup$ @Rdrr I've edited to make this clearer $\endgroup$ Jun 18, 2021 at 19:47
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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.

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  • $\begingroup$ I was curious to know if there is a proof that doesn't appeal to Galois representations or Tate module. Perhaps something a little more class field theoretic. $\endgroup$
    – Rdrr
    Jun 21, 2021 at 12:54
  • $\begingroup$ Also, what is $T_qK$? $\endgroup$
    – Rdrr
    Jun 21, 2021 at 13:25
  • $\begingroup$ That one is $T_qE$, sorry. $\endgroup$
    – Aphelli
    Jun 21, 2021 at 13:28
  • $\begingroup$ I have a lot of questions about this proof. I don't understand why $T_qE$ has an endomorphism $u$ just because $q$ is not a square in $\mathbb{F}_q$. Now if $\mathfrak{P}$ ramifies in $L$, why does the existence of such a $\sigma$ follow? Why is the action on $T_qE$ trivial? Why does $u$ commute with $\Gamma_L$? $\endgroup$
    – Rdrr
    Jun 21, 2021 at 13:43
  • $\begingroup$ $u$ is an endomorphism of $E_{L'}$ in the order given by the complex multiplication. I can choose it so that its square is scalar. So the endomorphism of $T_qE$ induced by $u$ commutes to $G_{L'}$ and its square $d$ is a scalar of $\mathbb{Z}_q$. I can choose $q$ so that $d$ is not a square in $\mathbb{Z}_q$. As $L'/L$ has degree $2$, if it is ramified at $\mathfrak{P}$, then its Galois group is the full inertia group at $\mathfrak{P}$. So you can define $\sigma$ on $L'$ at least. And then I think (not sure but it looks true) that you can extend it so that it stays in the inertia group. $\endgroup$
    – Aphelli
    Jun 21, 2021 at 14:06
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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.

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