The isotopy classes of oriented simple closed curves on the torus, are classified by primitive vectors in $\mathbb{Z}^2$, namely, $\{(p, q)| gcd(p,q) = 1 \}$. Prove that for two simple closed curves, $a_{1}$ represented by ($p_{1}$,$q_{1}$), and $a_{2}$ by ($p_{2}$,$q_{2}$), we have:
\begin{equation} i(a_{1},a_{2})= |p_{1}q_{2}-p_{2}q_{1}| \end{equation}
where $i(a,b)$ is the geometric intersection number of two isotopy classes a,b defined by:
\begin{equation} i(a,b)= min\{|\alpha\cap \beta| | \alpha\in a, \beta\in b,\alpha,\beta -simple \} \end{equation}
My best idea was to actually use the universal cover $\mathbb{R}^2$ to count all of the intersections of the all of the vectors in the unit square, but I can't find a way to write a general formula for this.
Any idea\clue how to prove this? Maybe it is related to the fact that the intersection number is the absolute value of the determinant?
Thank you in advance!