# Quadratic twist of Elliptic curves with complex multiplication

Suppose $E/\mathbb{Q}$ is an elliptic curve that has complex multiplication by $\mathcal{O}_K$, where $K=\mathbb{Q}(\sqrt{D})$, for $D<0$ and squarefree.

In "The main conjectures of Iwasawa theory for imaginary quadratic fields" by Karl Rubin, it is mentioned that the $L$-function of $E$ over $K$ satisfies

\begin{equation*} L(E/K,s) = L(E/\mathbb{Q},s)^2 \end{equation*}

I understand that in general, $L(E/K,s) = L(E/\mathbb{Q},s)\cdot L(E^D/\mathbb{Q},s)$, where $E^D/\mathbb{Q}$ is the quadratic twist of $E$. I'm guessing that the above equation holds because $E^D$ is $\mathbb{Q}$-isogenous to $E$, however I am unable to see why this is so, is there a way of showing this?

## 2 Answers

Are you sure it is written on Rubin's paper? Can you give a more precise reference? I fastly checked the paper but I did not find you assertion.

The formula (1) $$L(E/K,s)=L(E/ \mathbb{Q})^2$$

in general is false, unless you have some extra hypotesis you did not mention.

Indeed, assume that the analytic rank of $E/\mathbb{Q}$ is one.

By a result of Murty-Murty and Waldspurger there exists a quadratic imaginary field $K$ of negative discriminant (and some other properties), such that $\operatorname{ord}_{s=1}L(E/K,1)=1$.

Note that this contradicts the formula (1) , while it is compatible with the equality

$$L(E/K,s) = L(E/\mathbb{Q},s)\cdot L(E^D/\mathbb{Q},s).$$ and in particular implies the non-vanishing of $L(E^D/\mathbb{Q},1)$.

• I was looking at remark 1 on page 26 of the paper. The reason why your above example fails is because K satisfies the Heegner hypothesis, such that the sign of the functional equation of L(E/K,s) is -1. I think the key condition here is complex multiplication. – user132590 Aug 22 '14 at 0:38
• You are right! I checked in literature. If you look at the last page of Wiles-Coates "on the Birch and Swinnerton-Dyer conjecture" there is an analogue of your statement. In that paper there is a reference to Shimura's book "Shimura, G.: Introduction to the Arithmetic Theory of Automorphic Functions" in particular page 219 formula 7.8.11. – And85 Aug 22 '14 at 10:38
• Finally a precise reference! journals.cambridge.org/… – And85 Aug 25 '14 at 16:00

Since $E$ has $CM$ by $K$, the $G_{\mathbb Q}$-reps. on its Tate module $V_{\ell}(E)$ are induced from (one-dimensional $\ell$-adic) characters of $G_K$. (This is one interpretation of what it means for $E$ to have CM.)

These Tate modules are thus invariant under twisting by the quad. char $\chi: Gal(K/\mathbb Q) \cong \{\pm 1\}$. (A general property of inductions of characters.)

Thus $E^D$ has the same $V_{\ell}$ as $E$, and hence is isogenous to $E$, by the Tate conjecture.

You don't the general case of the Tate conj. proved by Faltings here.

For example, CM curves are well-known to be modular (a paper of Shimura from the 70's, but it basically follows from the Weil converse theorem applied to the $L$-function, which for a CM ell. curve is the $L$-function of a Grossenchar. of $K$, which by Hecke is known to have the correct analytic continuation and functional equation), and Ribet proved the Tate conjecture all modular elliptic curves (in a paper in the 80's, if I remember correctly).

Actually, here is how you might make a more direct proof: the curves $E$ and $E^D$ are isomorphic over $K$, but (as noted above) already have isomorphic $V_{\ell}$'s over $\mathbb Q$. I think that some consideration of Galois actions should be enough to conclude that you can find an isogeny (i.e. an isomorphism in the isogeny category) over $\mathbb Q$.