group law of complex torus is divisible?

I need help with this exercise:

Show that the group law of a complex torus (the definition I have is that of Rick Miranda's book Algebraic curves and Riemann surfaces, the one that he constructs from a lattice) X is divisible: for any point $p\in X$ and any integer $n\geq 1$ here is a point $q\in X$ with $n*q=p$. Indeed, show that there are exactly $n^2$ such points.

Thanks.

• What is your definition of a complex torus? Plane modulo a lattice? Commutative group of semisimple matrices? An elliptic curve? – Jyrki Lahtonen Oct 15 '13 at 19:31
• Edited. The definition I have is that of Rick Miranda's book Algebraic curves and Riemann surfaces, he constructs the complex torus using a lattice yes. – Hernán Ortega Oct 15 '13 at 19:57
• Ok. Can you divide a point $z$ on the complex plane by $n$? What happens if you divide $z+\lambda$ instead of $z$? Here $\lambda$ is a point on the lattice $\Lambda$, and the game is: what may change, if you use $z+\lambda$ to represent $p=z+\Lambda$ instead of $z$? – Jyrki Lahtonen Oct 15 '13 at 20:00

Suppose that $X=\mathbb{C}/L$ where $L=\{m_1\omega_1+m_2 \omega_2: m_1, m_2\in\mathbb{Z}\}$ is a lattice, i.e. $\omega_{1}$ and $\omega_{2}$ are linearly independent over $\mathbb{R}$. Suppose $p\in X$ and $n\geq 1$ is given. We can write $p=\lambda_{1}\omega_{1}+\lambda_{2}\omega_{2}+L$ (note that $p$ is a coset!) for unique $\lambda_{1}, \lambda_{2}\in [0, 1)$. Then for any integers $k_{1}, k_{2}\in [0, n-1]$ $$n\underbrace{\left( \frac{k_{1} + \lambda_1}{n}\omega_{1}+ \frac{k_{2} + \lambda_2}{n}\omega_{2}+L\right)}_{=q} = p$$ This gives $n^2$ solutions in $X$ to $n \cdot q = p$. These solutions are distinct because if $$\frac{k_{1} + \lambda_1}{n}\omega_{1}+ \frac{k_{2} + \lambda_2}{n}\omega_{2}+L = \frac{k'_{1} + \lambda_1}{n}\omega_{1}+ \frac{k'_{2} + \lambda_2}{n}\omega_{2}+L$$ Then $$\frac{k_{1}-k'_{1}}{n}\omega_1 + \frac{k_2-k'_2}{n}\omega_2 = m_1\omega_1+m_2\omega_2$$ for some integers $m_1$ and $m_2$. By linear independence of $\omega_1$ and $\omega_2$, this forces $\dfrac{k_i-k'_i}{n}=m_i$ for $i=1, 2$. Thus, $k_i=k'_i$; here we are using the fact that $k_i, k'_i\in [0, n-1]$.
How do we know we got all the solutions? Well, if $n q = p$, then express $q=\delta_{1} \omega_{1}+\delta_{2}\omega_{2} + L$ for some $\delta_1, \delta_{2}$. We obtain $$n\delta_{1}\omega_{1}+n\delta_{2}\omega_{2}+L = \lambda_{1}\omega_{1}+\lambda_{2}\omega_{2}+L$$ Consequently, there are integers $m_{1}$ and $m_{2}$ such that $$(n\delta_{1}-\lambda_1)\omega_{1}+(n\delta_{2}-\lambda_2)\omega_{2}=m_{1}\omega_{1}+m_{2}\omega_{2}$$ By linear independence of $\omega_1$ and $\omega_{2}$, we get $n\delta_i - \lambda_i= m_{i}$ for $i=1, 2$. So $\delta_{i}=\dfrac{m_{i}}{n}+\dfrac{\lambda_i}{n}$. So $$q=\delta_{1} \omega_{1}+\delta_{2}\omega_{2} + L = \frac{k_{1} + \lambda_1}{n}\omega_{1}+ \frac{k_{2} + \lambda_2}{n}\omega_{2}+L$$ where $k_{i}\equiv m_i \text{ (mod } n)$ satisfying $k_{i}\in [0, n-1]$ for $i=1, 2$.