Let $E$ be an elliptic curve over the rational number field $\mathbb{Q}$, and I have seen $\text{End}_{\mathbb{Q}}(E)=\mathbb{Z}$ in many places ,but no proof is given. So I want to ask for a proof or a refernce!
Thanks
Let $E$ be an elliptic curve over the rational number field $\mathbb{Q}$, and I have seen $\text{End}_{\mathbb{Q}}(E)=\mathbb{Z}$ in many places ,but no proof is given. So I want to ask for a proof or a refernce!
Thanks
The point is that $E$ may have complex multiplication, so it has endomorphisms over $\Bbb C$ which are outside $\Bbb Z$. But these are not defined over $\Bbb Q$.
To see this note that the endomorphism ring acts on the vector space of invariant differentials on $E$ (spanned by $\omega=dx/(2y)$ for $y^2=x^3+ax+b$). An endomorphism defined over $\Bbb Q$ takes $\omega$ to a rational multiple of $\omega$, and so we get a ring map $\text{End}_{\Bbb Q}(E)\to \Bbb Q$. No element of $\text{End}_{\Bbb C}(E)$ other that those in $\Bbb Z$ can map to and element of $\Bbb Q$ as in the CM case the endomorphism ring in an order in an imaginary quadratic field.