5
$\begingroup$

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.)

$\endgroup$
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.