Why complex multiplication between torus is holomorphic?

Question

I am reading Rational Points on Elliptic Curves by Silverman and Tate. At page233, I'm having trouble. So I need your help.

Let $L＝｛aw1+bw2:a,b∈\Bbb Z｝$ Then complex multiplication $φ：C（\Bbb C）→C（\Bbb C）$　induces a map $f:\Bbb C /L→\Bbb C /L$.

The book states f is holomorphic without no explanation.

Why can we say f is holomorpchic? Thank you in advance.

Answer 1

Consider the diagram

$\require{AMScd}$ \begin{CD} C(\Bbb C) @>\varphi>> C(\Bbb C)\\ @V\cong VV @VV\cong V\\ \Bbb C/L @>>f> \Bbb C/L \end{CD}

as a definition of $f$. By theorem 6.16, same page 233 in loc. cit. the map $\phi$ is a homomorphism, thus holomorphic. The vertical arrows are automorphism, holomorphic "translations" of the structure. Thus $f$, defined by the commutative square, is holomorphic.

Answer 2

We can use the following facts

1. $\mathbb{C}/L$ is torus $\mathbf{T}^{2}\cong \mathbf{S}^{1}\times \mathbf{S}^{1}$, which is compact.
2. Homomorphism: $$ \varphi(z_1+z_2)=\varphi(z_1)+\varphi(z_2),\;\text{for all}\;z_1,z_2\in \mathbf{T}^{2} $$

to show

1. The map $\varphi$ is continuous.

2. The map $\varphi$ has the form $\varphi(z)=c z$, where $c$ is a complex number.

First, it's sufficient to verify the continuity of $\varphi$ at $z=0$. Choose sequence $\{z_{i}\}$ approximating $0$ and there is a convergent subsequence $\{w_{j}=\varphi(z_{j})\}$ by compactness. Then apply $\varphi(z_1-z_2)=\varphi(z_1)-\varphi(z_2)$ we can deduce $$ \varphi(z_{k}-z_{l})\to 0, \;\text{as}\;\{z_{k}-z_{l}\} \to 0. $$ by triangle inequality. At every argument $\theta$, $\varphi(re^{i\theta})$ tends to $0$ when $r\to 0$, which yields the continuity of $\varphi$.

Second, write (we do NOT say it converges to a holomorphic function) $$ \varphi(z)=c_{0}+c_{1}z+c_{2} z^{2}+\cdots. $$ By convention, we set $\varphi(0)=c_{0}=0$. Then use the homomorphism again to give $c_{i}=0,i\geq 2$.