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.)