Endomorphism of a CM elliptic curve E can always be defined over its CM field

Question

Let $E$ be an elliptic curve which has CM in an imaginary quadratic field $K$ and let $\tau$ denote the element of $K$ for which $End(E)=\mathbb{Z}[\tau]$. How can one prove that the action of $\tau$ on $E(\overline{\mathbb{Q}})$ is given by $(x,y) \to (\phi(x,y),\psi(x,y))$, where $\phi(x,y), \psi(x,y) \in K(X,Y)$ are rational functions defined over $K$? I am interested in the proof of the fact that the rational functions are defined over $K$, rather than that the isogeny is given by rational functions. Also, can one say something more precise than that about $\phi(x,y), \psi(x,y)$?

Answer 1

COMMENT.- What made me doubt a bit was your writing "....that the action of $\tau$ on $E(\overline{\mathbb{Q}})$ is given by $(x,y) \to (\phi(x,y),\psi(x,y))$".

Now I know that your $(x,y)$ is an element of $E$ and therefore your question is almost trivial since $E$ is an abelian group (whose addition is rational) and the operation involved is that of the group itself so it can not give results outside the group. If you put the equation of $E$ in its homogeneous form, all operation in $E$ gives an element of $E$ whose three coordinates are polynomials so that returning to the afinne form you have rational functions in $K(X,Y)$ (when dividing by $Z$; see the given example with $x^3+y^3=Az^3$ ) to coefficients in your field $K$. I suggest you very cordially delete your post by not relevant (I will not be the one to put a vote to "close" because I do not like to object to the posters). Regards. $$*******$$

Hello Holz. I try to answer your comment before deleting mine. The ring $ \ End (E)$ is either isomorphe to $\mathbb Z$ (when $E$ is not with CM) or a subring of the ring of integers of an imaginary quadratic field (when $E$ is with CM).

Very briefly, the endomorphisms of the torus $\mathbb C/L$ (the plane modulo the lattice L giving the elliptic curve $E$) are determined by the numbers $\alpha$ such that $\alpha L\subseteq L$ and it happens that sometimes there are non-rational $\alpha$ (necessarily imaginary quadratic) fulfilling this condition.

In the case of $x^3+y^3=Az^3$, the ring of integers of $\mathbb Q(\sqrt{-3}$), because of $-3\equiv1\pmod4$, is $$A=\dfrac 12(a+b\sqrt{-3})\text { with } a,b\in\mathbb Z$$ so $\ End (E)$ shoul be a subring of $A$. In particular, for $(a,b)=(0,2)$, you have that $[\sqrt{-3}]$ is an endomorfism of $E$ and every allowed operation with it or in conjunction with himself or others give necessarily rational function over $K=\mathbb Q(\sqrt{-3})$ by the rational definition of the group law in $E$.