Why function $j(\tau)$ has degree 1?

Question

We have

$$ j(\tau)=\frac{1}{q}+\sum_{n=0}^{\infty}a_nq^n, a_n\in\mathbb{Z},q=e^{2\pi i\tau} $$

Then it is said that because $j$'s only pole is simple, $j$ has degree 1 as a map $j:X(\text{SL}_2(\mathbb{Z}))\rightarrow\mathbb{C}\cup\{\infty\}$.

My question is, why can we deduce this from that "$j$'s only pole is simple"?

Answer 1

The $j$-functions is a holomorphic map from one compact Riemman surface (the completion of $\mathcal H/\mathrm{SL}_2(\mathbb Z)$) to another (the Riemann sphere). The only place it has a pole is at the "cusp" $\infty$ of its domain, and this pole is simple. Thus the preimage of $\infty$ in the Riemann sphere (i.e. the polar part of the divisor of $j$) consists of a single point. But the number of points (counted with correct multiplicties) in the preimage of any point of a degree $d$ map between compact Riemann surfaces is equal to $d$. (If the points lying above a given point have ram. degrees $e_1,\ldots,e_n$, then this is the formula $\sum_i e_i = d$.) Applying this to $j$ we find that $d = 1$.

Answer 2

$j$ is periodic of period 1, so $j$ is actually defined on $\mathcal{H}/\mathbb{Z}$ which is biholomorphic to the unit disc under the map $\tau \to e^{2 \pi i \tau}$. Thus when we regard $j$ as a meromophic function on the unit disk, by the series you have written down, you can see that $j$ only has one pole (at 0 of the disk, corresponding to $\infty$ of the upper half plane). The leading term being $1/q$ means that the pole is simple.