The $\mathbb{Q}$-rational cuspidal group of $J_0(N)$

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 $$\phi((d,N/d)) P_{1} - P_{d}$$.
• 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. – user196884 Nov 14 '18 at 8:18