Let $E$ be an elliptic curve over $\Bbb Q$. What is the relation between $End(E)$ and $End(E_{\Bbb C})$ ? We clearly have an inclusion $End(E) \subset End(E_{\Bbb C})$ : given $f :E\to E$, we can base change it to get $f_{\Bbb C} : E_{\Bbb C} \to E_{\Bbb C}$. But is it possible to have a strict inclusion? Apparently I've seen that Chow proved $End(E_{\overline{\Bbb Q}}) = End(E_{\Bbb C})$.
Of course, if $End(E_{\Bbb C}) = \Bbb Z$, then we have equality. But what if $E_{\Bbb C}$ has larger endomorphism ring?
The elliptic curves with larger endomorphism rings than $\Bbb Z$ are said to have “complex multiplication”. Here are the most familiar two examples: \begin{align} y^2&=x^3+1\,,&\varphi(x,y)&=(\zeta_3x,y)\,,&(\zeta_3^2+\zeta_3+1&=0)\\ y^2&=x^3-x\,,&\varphi(x,y)&=(-x,iy)\,. \end{align}
The elliptic curves with complex multiplication play a surprisingly large role in number theory. Others will point you to references, where you may read all about it.
Edit (addition)
In a word, the only endomorphisms not in $\Bbb Z$ are truly complex, in this sense: such an endomorphism $\varphi$ will satisfy an equation of type $\varphi^2+b\varphi + c=0$, with $b,c\in\Bbb Z$ and $b^2-4c<0$. And in particular, locally at the neutral point, the action on the tangent space is just that, $z\mapsto\lambda z$ with $\lambda^2+b\lambda+c=0$, same $b$ and $c$. In other words, the endomorphism itself can not have rational definition. Thus $\text{End}_{\Bbb Q}(E)=\Bbb Z$ always.