Endomorphism of the elliptic curve $y^2 = x^3 - ax$ For the elliptic curve $$E:y^2 = x^3 - ax$$I know that $E\left(\mathbb{C}\right)\cong\mathbb{C}/\left(\mathbb{Z}i + \mathbb{Z}\right)$ and hence $\text{End}_{\mathbb{C}}\left(E\right)\cong\mathbb{Z}\left[i\right]$, so $1+i\in\text{End}_{\mathbb{C}}\left(E\right)$ and this map is surjective. Is it also true that $1+i$ is an element of the groups $\text{End}_{\overline{\mathbb{Q}}}\left(E\right)$ and/or $\text{End}_{\mathbb{Q}\left(i\right)}\left(E\right)$? Does the first one also surjective?
Thanks!
 A: Yes, we can see this explicitly as follows: let $[i]$ be the automorphism of the curve taking $(x, y) \mapsto (-x, -iy)$. This is clearly an automorphism, and induces multiplication by i on the invariant differential $dx/y$, so gives the standard map from $\mathbb{Z}[i] \to \text{End}(E)$; the endomorphism is obviously defined over $\mathbb{Q}(i)$, so lives in both those subrings you mentioned. (I see you're using notation as in Silverman's books where the endomorphism ring is indexed by a field, which is a bit confusing since it conflicts with the more normal notation for varieties, but I will try to stay with it.)
As for the surjectivity of $1+i$, I assume you are asking if it is surjective on the set of $\mathbb{Q}(i)$ rational points of $E$. 
This is false. In general, that set of points is a finitely generated abelian group, and $[2] = [(1+i)(1-i)] = [-i(1+i)^2]$. Since $-i$ is an automorphism, multiplication by $(1+i)$ will be surjective if and only if multiplication by $2$ is, i.e. if the group contains no free part and no $2$-torsion. But this curve always contains the $2$-torsion point $(0, 0)$. 
