# Showing there are $m^2$ points in $\mathbb{C}/L$ with order dividing $m$

Let $E(\mathbb{C})$ be an elliptic curve. Let $L=\{a\omega_1+b\omega_2\mid a,b\in\mathbb{Z}\}$ be a lattice. Consider the map $\phi:\mathbb{C}\rightarrow E(\mathbb{C})$ given by $\phi(z)=(\wp_L(z),\wp_L'(z))$, where $\wp_L(z)=\frac{1}{z^2}+\sum_{w\in L}\frac{1}{(z-w)^2}-\frac{1}{w^2}$ is the Weierstrass $\wp$-function. (Note that this series converges for every $z\not\in L$). It follows from addition identities of $\wp$ that $\phi$ is a homomorphism from $\mathbb{C}$ to $E(\mathbb{C})$, with $\ker\phi=L$, where $E(\mathbb{C})$ is the elliptic curve $y^2=4x^2-g_2x-g_3$ for some constants $g_2,g_3$. By the first isomorphism theorem, we have $\mathbb{C}/L\cong E(\mathbb{C})$. Further, we may write $\mathbb{C}/L=\{a\omega_1+b\omega_2\mid 0\leq a<1,0\leq b<1\}$.

I am trying to understand the proof that if $m\geq 1$, then there are precisely $m^2$ points in $\mathbb{C}/L$ with order dividing $m$. The proof I have been given is as follows: Suppose $z\in \mathbb{C}$ is such that $z=a\omega_1+b\omega_2$ where $0\leq a<1$ and $0\leq b<1$ with $|\phi(z)|$ dividing $m$. Then $m\phi(z)=(0:1:0)$ and since $\phi$ is a homomorphism, $\phi(mz)=(0:1:0)$, which implies that $mz\in L$ (the kernel of $\phi$). So then we have that $mz=ma\omega_1+mb\omega_2=k_1\omega_1+k_2\omega_2$, where $k_1,k_2\in\mathbb{Z}$ and $0\leq k_1,k_2<m$. So we now have that $z=(k_1/m)\omega_1+(k_2/m)\omega_2$. It follows that the $m^2$ points of order dividing $m$ are $\{\phi((k_1/m)\omega_1+(k_2/m)\omega_2)\mid 0\leq k_1,k_2<m\}$.

What I don't understand is why we are allowed to assume from the start that $z$ can be written in the form $a\omega_1+b\omega_2$ where $0\leq a <1$ and $0\leq b <1$. This would mean that $z$ is an element of the lattice. But shouldn't $z$ be an arbitrary element of $\mathbb{z}$? Why are we allowed to pick our $z$ so that it fits this condition?

What I don't understand is why we are allowed to assume from the start that $z$ can be written in the form $a\omega_1+b\omega_2$ where $0 \le a < 1$ and $0 \le b < 1$. This would mean that $z$ is an element of the lattice.
No, the lattice consists of the points $a\omega_1+b\omega_2$ where $a$ and $b$ are integers, not values between 0 and 1.
If $z \in \mathbb{C}$ then it can be written as $a \omega_1 + b \omega_2$ where $a, b \in \mathbb{R}$. For points of $\mathbb{C}/L$, we can freely add and subtract elements of $L$, i.e., integer multiples of $\omega_1$ and $\omega_2$, so we may assume $0 \le a < 1$ and $0 \le b < 1$.
• But why are we allowed to assume that $z\in C/L$ and so can be written in that form? I understand that $C/L$ absorbs elements of $L$, but I still don't see why $z$, as an arbitrary complex number, must be able to be written in the above form. – ponchan Oct 15 '17 at 1:51
• An arbitrary complex number cannot be written in that form, but for any $z \in \mathbb{C}$, we may find a $z' \in \mathbb{C}$ such that $\phi(z) = \phi(z')$, and $z'$ can be written in that form. The rest of the argument is unaffected if we replace $z$ with $z'$. – Ted Oct 15 '17 at 2:19
• And why are we guaranteed that we may find a $z'\in\mathbb{C}$ such that $\phi(z)=\phi(z')$? – ponchan Oct 15 '17 at 19:55