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?
 A: 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 require 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 $\mathbb Q$.
A: 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 $L(E^D/\mathbb{Q},1)$.
