# 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.

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

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).

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

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.

• 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.
– Rdrr
Jun 21, 2021 at 12:54
• Also, what is $T_qK$?
– Rdrr
Jun 21, 2021 at 13:25
• That one is $T_qE$, sorry. Jun 21, 2021 at 13:28
• 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 $\Gamma_L$?
– Rdrr
Jun 21, 2021 at 13:43
• $u$ 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. Jun 21, 2021 at 14:06

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.