Let $N$ be a positive integer and consider the modular curve $X_0(N)$ over $\mathbb{Q}$. Also, consider the Jacobian variety $J_0(N)$ of $X_0(N)$, which is an abelian variety defined over $\mathbb{Q}$.

Let $\mathsf{Cusp}$ denote the group of cuspidal divisors, namely, the group of divisors supported only on cusps and let $\mathsf{Cusp}^0$ denote the group of degree-0 cuspidal divisors. Let $\mathcal{C}(N)$ denote the image of $\mathsf{Cusp}^0$ in $J_0(N)$, which is called the cuspidal group of $J_0(N)$.

By Manin and Drinfeld, the group $\mathcal{C}(N)$ is finite. Let $\mathcal{C}(N)_\mathbb{Q}$ be the $\mathbb{Q}$-rational cuspidal group of $J_0(N)$, which is defined by the subgroup of $\mathcal{C}(N)$ consisting of the elements fixed by the action of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\mathbb{Q}$.

Here is my question:

Is the group $\mathcal{C}(N)_\mathbb{Q}$ generated by the images of the degree-0 $\mathbb{Q}$-rational cuspidal divisors?

(Here, by the degree-0 $\mathbb{Q}$-rational cuspidal divisors we mean the degree-0 cuspidal divisors which are fixed by the action of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.) A priori, the group generated by the images of the degree-0 $\mathbb{Q}$-rational cuspidal divisors is only a subgroup of $\mathcal{C}(N)_\mathbb{Q}$.

In the paper by Ling, "On the $\mathbb{Q}$_rational cuspidal subgroup and the component group of $J_0(p^r)$" published in Israel Journal of Mathematics 99 (1997), 29--54, he says that

it is easy to see that the $\mathbb{Q}$-rational cuspidal subgroup $\mathcal{C}(N)_\mathbb{Q}$ of $J_0(N)$ is generated by divisors coming from divisors of the kind $\phi((d, N/d))P_1-(P_d)$ as $d$ runs through the positive divisors of $N$.

(This is on page 34.) Can anyone prove this statement? (This is equivalent to my question.)


I more or less make the same claim as Ling in a paper that I wrote. I think the key thing that one needs to know is how the Galois group acts on the cusps of $X_{0}(N)$, and for this I found the book "Arithmetic on modular curves" by Glenn Stevens to be helpful. In particular, Theorem 1.3.1 from that book says that if $d | N$ and $\begin{bmatrix} c \\ d \end{bmatrix}$ represents a cusps of $X_{0}(N)$ and $\tau_{s} \in Gal(\mathbb{Q}(\zeta_{N})/\mathbb{Q})$ sends $\zeta_{N} = e^{2 \pi i / N}$ to $\zeta_{N}^{s}$, then $$ \tau_{s}\left(\begin{bmatrix} c \\ d \end{bmatrix}\right) = \begin{bmatrix} c \\ s' d \end{bmatrix} $$ where $s s' \equiv 1 \pmod{N}$. From this, it takes a little bit of thought to see that $Gal(\mathbb{Q}(\zeta_{N})/\mathbb{Q})$ acts transitively on all $\phi((d,N/d))$ cusps with "denominator" $d$. Therefore the divisor $P_{d}$ (from Ling's paper) is a single Galois orbit, and from this it follows that any element of $\mathcal{C}(N)_{\mathbb{Q}}$ is an integer linear combination of the divisors $\phi((d,N/d)) P_{1} - P_{d}$.

I feel like there ought to be a simpler pure thought reason why the answer to your first question is yes, but I can't quite see it right now.

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  • $\begingroup$ Ling's claim is twofold: one is that the divisors $\phi((d, N/d))P_1-(P_d)$ span the $\mathbb{Q}$-rational cuspidal divisors, which you showed already. The other is that my answer is yes, this is why I asked above. $\endgroup$ – user196884 Nov 14 '18 at 8:18

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