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.)

  • $\begingroup$ In the paper, the $\mathbb{Q}$-rational cuspidal group of $J_1(2p)$, Takagi showed my question is true for $J_1(2p)$. Here is a link : projecteuclid.org/euclid.jmsj/1414090242 $\endgroup$
    – user116980
    Nov 15, 2018 at 8:39
  • $\begingroup$ Given the lengthy argument Takagi needs for $J_1(2p)$ I am not convinced that "it is easy to see" for $J_0(N)$ and that there is actually a significant argument missing in Lings paper. $\endgroup$ Nov 30, 2022 at 1:01
  • $\begingroup$ This seems to be quite challenging and deep. The most recent paper on this subject is by Hwajong Yoo in the Journal of Number Theory (in the January 2023 issue) and is titled "The rational cuspidal divisor class group of $X_{0}(N)$." This paper states as Conjecture 1.3 that $\mathcal{C}(N)_{\mathbb{Q}}$ is generated by degree $0$ rational cuspidal divisors, and mentions that this was a question asked by Ribet. (The notation in the paper is a bit different than in this question.) $\endgroup$ Nov 30, 2022 at 13:39

1 Answer 1


I have thought about this question as well in the past. And must say I do not find it "easy" to see as well. In a paper together with Etropolski, Hoeij, Morrow, Zureick-Brown we needed something similar but then for $X_1(N)$, and we solved it by explicitly computing $C(N)$ and the action of galois on it using sage for the small values of $N$ for which we needed the result, see proposition 4.11 in that article. If it is indeed easy to see (also for the equivalent question for $X_1(N)$) then I must say I it could have saved us some work. At least these explicit computations showed that the equivalent claim for $X_1(N)$ and $N \leq 55$ is also true.

When I was working on that article I tried looking at it in the following way: $L \subset \mathbb Q(\zeta_N)$ be the smallest galois extension over which all cusps are defined. And let $G := Gal(L/\mathbb Q)$ furthermore let $H$ be the kernel of $\mathsf{Cusp}^0 \to \mathcal{C}(N)$. Then we have an exact sequence $$0 \to H \to \mathsf{Cusp}^0 \to \mathcal{C}(N) \to 0$$ Taking the group cohomology of the above exact sequence gives $$0 \to H^G \to \mathsf{Cusp}^{0,G} \to \mathcal{C}(N)^G \to H^1(G,H) \to H^1(G,\mathsf{Cusp}^{0}).$$

I was trying to solve the question by either showing $H^1(G,H)=0$ or $H^1(G,H) \to H^1(G,\mathsf{Cusp}^{0})$ being injective. But failed in that.

Note it can't be some abstract statement about curves and finite sets of points on them over cyclotomic fields such that all their differences are torsion in the jacobian.

For example if one takes $E$ to be the elliptic curve defined by $y^2 + xy + y = x^3 - x^2 - 91x - 310$ and $\text{"Cusp"} := \infty\mathbb Z + (-5, i + 2)\mathbb Z + (-5, -i + 2)\mathbb Z$ the free rank 3 module generated by three points over $\mathbb Q(i)$ then the image of $\text{"Cusp"}^0$ in $J(E)(\mathbb Q(i)) \cong E(\mathbb Q(i))$ is isomorphic to $\mathbb Z/4\mathbb Z$ and generated by $(-5, i + 2)-\infty$ however $2(-5, -i + 2)-2\infty \sim (-21/4, 17/8) - \infty \in J(E)(\mathbb Q)$ is a point of order 2. While on the other hand the galois invariant elements in $\text{"Cusp"}^0$ are generated by $(-5, -i + 2)+(-5, i + 2) - 2\infty$. Whose image in $J(E)(\mathbb Q)$ is trivial.

In the last paragraph I put $\mathsf{Cusp}$ in parenthesis since the curve $E$ is not a modular curve and the three points on it are not actual cusps. However since the situation there is very similar it does show that some intrinsic argument pertaining to the particularity of modular curves and cusps is needed, and we can't just use an abstract argument using torsion points in jacobians of curves over cyclotomic fields.


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