# Image of the cuspidal subgroup of J_0(N) in J_1(N)

Let N be a prime integer. We know that the element $c=(0)-(\infty)$ generates the torsion subgroup of $J_0(N)$ and it has order Num( (N-1)/12). Now, there is a natural map $\pi^*:J_0(N) \rightarrow J_1(N)$, coming from the covering map $\pi:X_1(N) \rightarrow X_0(N)$. My question is what is the image of c under this map? Specifically, is it possible for $\pi^*(c)=0$?

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This can be made "explicit" in the special case $N=11$, for which both curves have genus 1. By moduli interpretation of cusps, $X_1(p)$ has $p-1$ geometric cusps: $(p-1)/2$ are rational points (the $p$-gon cusps) and $(p-1)/2$ are a single Galois orbit (the $1$-gon cusps, residue field $\mathbf{Q}(\zeta_p)^+$). Making $X_1(11)$ an elliptic curve using a rational cusp and $X_0(11)$ an elliptic curve using its image (the cusp $0$, not $\infty$!!), the deg-5 map $\pi$ has kernel given by the rat'l cusps: $\ker \pi = \mathbf{Z}/(5)$. Hence, dual map has kernel $\mu_5$, and $\mu_5(\mathbf{Q})=1$. –  BCnrd Sep 27 '10 at 5:53
Thanks! This example was actually confusing me, and that clarifies the situation. –  Soroosh Sep 27 '10 at 17:41

The fact you mention about $(0) - (\infty)$ generating the torsion subgroup of $J_0(N)$ is Theorem 1 (Ogg's conjecture) on the first page of Mazur's paper "The Eisenstein Ideal". I recommend you actually read this paper. If you get as far as page 2, you will find a "Theorem 2 (twisted Ogg's conjecture)" which concerns the Shimura subgroup $\Sigma$. The construction of this subgroup essentially identifies it with the kernel of $J_0(N) \rightarrow J_1(N)$, and Proposition 11.6 of ibid. shows that $\Sigma$ is of multiplicative type, so BCnrd's remarks apply to the general case.
It's zero when $N=2$, right? So maybe you mean "...can be zero for $N$ suff large" or something. –  Kevin Buzzard Sep 27 '10 at 12:32