Why so few complex multiplication Let's take I lattice $\Lambda$ and $\alpha\in\Lambda$. Then we have $\alpha\Lambda\subseteq\Lambda$ so $z\mapsto \alpha z$ induces an isogenie $\mathbb{C}/\Lambda\to\mathbb{C}/\Lambda$ wich has no reason to be one $[n]:\overline{z}\mapsto \overline{nz}$.
What's wrong? Normally They are very few lattice $\Lambda$ for which there exist a complex multiplication on $\mathbb{C}/\Lambda$.
 A: Let $\Lambda$ be a lattice, generated by $w_1$ and $w_2$. Then,
$$\Lambda = \{aw_1+bw_2 : a,b\in\mathbb{Z}\}.$$
If $\alpha\in\Lambda$ there is no reason, in general, to expect that $\alpha\Lambda \subseteq \Lambda$. In other words, $\alpha w_1$ and/or $\alpha w_2$ may not be elements of $\Lambda$.
Since $\mathbb{C}/\Lambda \cong \mathbb{C}/\lambda\Lambda$, we may assume $\Lambda = \langle 1,\tau\rangle$ for some $\tau\in\mathbb{C}$ (note $\tau\not\in \mathbb{R}$, otherwise $\Lambda$ would not be a lattice in $\mathbb{C}$). Now, let $\alpha\in \Lambda$ such that $\alpha \Lambda \subseteq \Lambda$. Since $\alpha\in\Lambda$, we have $\alpha=a+b\tau$ for some $a,b\in\mathbb{Z}$. And since $\alpha \Lambda \subseteq \Lambda$, we must have $\alpha\tau = c+d\tau$ for some $c,d\in\mathbb{Z}$. Putting this together we obtain
$$(a+b\tau)\tau = c+d\tau,$$
or,  $b\tau^2+(a-d)\tau -c = 0$. In other words, $\tau$ satisfies a quadratic equation with integer coefficients, and therefore $\mathbb{Q}(\tau)$ is an imaginary quadratic extension of $\mathbb{Q}$. But most lattices $\Lambda=\langle 1,\tau\rangle$ do not satisfy this hypothesis, as $\tau$ would be arbitrary (for instance, the elements $\tau\in\mathbb{C}$ such that $\mathbb{Q}(\tau)/\mathbb{Q}$ is quadratic is countable, while its complement in  $\mathbb{C}$ is not).
