# Eta quotient and order of a cuspidal divisor

Let $$X_0(N)$$ be the modular curve associated to a congruence subgroup $$\Gamma_0(N)$$. If $$N=p$$ is a prime, then there are two cusps $$0$$ and $$\infty$$ on $$X_0(N)$$. Suppose that $$p>7$$ so that the genus of $$X_0(p)$$ is positive.

I'd like to compute the order of the image of the divisor $$0-\infty$$ in $$J_0(p)$$, the Jacobian variety of $$X_0(p)$$. It is well-known that it is the numerator of $$\frac{p-1}{12}$$.

The computation is as follows: First, consider the eta quotient $$g(a, b)=\eta(z)^a \eta(pz)^b$$ for some $$a, b \in \mathbb{Q}$$, where $$\eta(z)$$ is the Dedekind's eta function. Then, $$g(a, b)$$ is a modular function on $$X_0(p)$$ if and only if all of the following are satisfied. (cf. Proposition 1 in the paper by Ling, "On the Q_rational cuspidal subgroup and the component group of J0(pr)" published in Israel Journal of Mathematics 99 (1997), 29--54)

1. $$a, b \in \mathbb{Z}$$.
2. $$a+pb \in 24\mathbb{Z}$$.
3. $$pa+b \in 24\mathbb{Z}$$.
4. $$a+b=0$$.
5. $$b \in 2\mathbb{Z}$$.

Thus, $$a=-b=\frac{24k}{p-1} \in 2\mathbb{Z}$$ for some $$k\in \mathbb{Z}$$. The smallest possible $$k$$ is $$n=\frac{p-1}{\gcd(12, p-1)}$$, the numerator of $$\frac{p-1}{12}$$.

Next, let $$a=\frac{24n}{p-1}$$ and consider $$\mathrm{div}~ g(a, -a)$$, the principal divisor on $$X_0(N)$$ associated to $$g(a, -a)$$. By direct computation, we have $$\mathrm{div}~ g(a, -a)= \frac{(p-1)a}{24}(0-\infty)=n(0-\infty).$$ Thus, by definition the order of the image of $$0-\infty$$ is a divisor of $$n$$.

Last, suppose that the image of $$m(0-\infty)$$ in $$J_0(p)$$ is zero. By definition, there is a modular function $$f$$ on $$X_0(p)$$ such that $$\mathrm{div}~f=m(0-\infty)$$. Since $$m$$ can be taken as a divisor of $$n$$, let $$n=mk$$ for some $$k\in \mathbb{Z}$$. In order to prove the order of the image of $$0-\infty$$ is $$n$$, we have to show that $$k=1$$.

In this situation, we have $$\mathrm{div}~ f^k=mk(0-\infty)=n(0-\infty)=\mathrm{div} ~g(a, -a).$$ Thus, the function $$\frac{f^k}{g(a, -a)}$$ is a constant because it has no zeros or poles on $$X_0(p)$$. So, we have $$f^k=A \cdot g(a, -a)$$ for some $$A\in \mathbb{C}^*$$.

My question is here:

Q. How can we deduce $$k=1$$ in this situation?

If we could take $$k$$-th roots on both sides, we would have $$f=A^{1/k}\cdot g(a/k, -a/k)$$. If so, then $$g(a/k, -a/k)$$ is also a modular function on $$X_0(p)$$ because $$f$$ is a modular function. By the discussion above $$a$$ is the smallest such one, so $$a/k=a$$, i.e., $$k=1$$, as desired. A generalization of my question is as follows:

"Let $$f$$ be a (meromorphic) function on a compact Riemann surface $$X$$ and let $$g^k$$ be a function on $$X$$ for some $$k\in \mathbb{Z}$$. If $$f^k=g^k$$, then is it true that $$g$$ is also a function on $$X$$?"