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I am trying to compute the divisor of $\Delta(z)/\Delta(pz)$ on the modular curve $X_0(p)$ where $p$ is a prime. I know that as a function on the full modular group, the $\Delta$ function has only a simple zero at infinity, but I can't get much further than this. I know that $X_0(p)$ has two cusps, $0$ and $\infty$, so I'm thinking I can compute this by considering the map from $X_0(p)$ to to the half plane modulo the full modular group, but I can't quite figure this out either.

If I'm to believe Gross in his paper on Heegner points, I expect the answer to be $$(p-1)\{(0)-(\infty)\}$$

Thanks for any insight.

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up vote 5 down vote accepted

Just compute the $q$-expansion! We have $\Delta(z) / \Delta(pz) = (q + \dots) / (q^p + \dots) = q^{1-p} + \dots$, so there is a pole of order $(p-1)$ at the cusp $\infty$. Since the function is obviously non-vanishing away from the cusps, and there are only two cusps, then (since the divisor of a function has degree 0) we see that the divisor is $(p-1)\{ (0) - (\infty) \}$ as expected.

(You can also see directly that there is a zero of order $(p-1)$ at $0$, by calculating the $q$-expansion of $F(-1/z)$ where $F(z)$ is your function, but you need to remember to correct for the fact that the cusp $0$ has width $p$.)

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