Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $p$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.