# Is there a largest prime p such that J_0(p) completely splits into elliptic curves

The question in the title is related to a more general question. Namely does there exist an integer $N$ such that for all curves $C/\mathbb C$ of genus $> N$ one has that not all simple isogeny factors of $J(C)$ are elliptic curves.

Now since over $\mathbb Q$ the isogeny factors of $J_0(p)$ correspond to galois orbits of newforms it seems to me that the question wether there is a largest prime $p$ such that $J_0(p)$ splits completely into elliptic curves over $\mathbb Q$ should be easier and I wonder wether the answer is already known.

I suspect that the answer is yes and that $p = 37$ is the largest prime such that $J_0(p)$ completely splits into elliptic curves. Indeed using Cremona's database of elliptic curves I verified that 37 is the biggest prime below 300000 such that $J_0(p)$ completely splits into elliptic curves.

The reason I only ask it for prime levels $p$ is that I already showed that if $p = 37$ is indeed the largest prime such that $J_0(p)$ completely splits into elliptic curves, then $N = 1200$ is the largest composite number such that $J_0(N)$ completely splits into elliptic curves.

• Your "more general question" (first sentence) is strange. $JC$ is a simple abelian variety for a general curve $C/\Bbb{C}$ of genus $g$, for every $g$. – abx Jan 27 '14 at 14:45
• I don't see why your comment makes my first question strange. It asks wether something is true for all curves of a certain genus not just the generic one, so the answer might be different. It is certainly not true that the jacobian is simple of all curves of genus $g$ if $g > 1$. – M.D. Jan 27 '14 at 15:31
• OK, sorry, I didn't realize that you are allowing the simple factor to be $J(C)$ itself. – abx Jan 27 '14 at 17:25
• Ok, I modified my question so that people in the future will not get confused in a similar way. – M.D. Jan 27 '14 at 17:32

The answer, due to Jean-Pierre Serre, can be found in an unpublished note of Henri Cohen where he characterizes the odd integers $N$ such that $J_0(N)$ is isogenous to a product of elliptic curves.

For your question, $N$ is prime, and only $p=11$, $13$, $17$, $19$ et $37$ satisfy this condition. (For $p=13$, $J_0(p)$ has dimension $0$.)

In the general case, Cohen gives the following Theorem:

Theorem. The only odd values of $N$ for which $J_0(N)$ is isogenous to a product of elliptic curves are $N\leq 21$, as well as $N = 25$, $27$, $33$, $37$, $45$, $49$, $57$, $75$, $99$ and $121$. (When $N\leq 9$, $N=13$ and $N=25$, $J_0(N)$ is zero.)

• Cool, I deduced Cohen his generalisation from Serre his result in a different way. Namely just find for each prime $p_i \leq 37$ the largest $e_i$ such that $X_0(p_i^{e_i})$ is isogenenous to a product of elliptic curves. Then any $N$ such that $J_0(N)$ splits into elliptic curves has to divide $\prod_{p\leq 37} p_i^{e_i}$ leaving a finite number of cases to check. I did this finite computation and the splitting $N$ can be found on my website. – M.D. Jan 27 '14 at 17:25

In fact we know the list of all $N$, whether even or odd, for which $J_0(N)$ is isogenous to a product of elliptic curves. See

Takuya Yamauchi, On $\mathbb Q$-simple factors of Jacobian varieties of modular curves, Yokohama Math. J. 53 (2007), no. 2, 149-160.

An alternative proof (which also corrects a minor error in Yamauchi's list) is given in section 5 of the recent paper

Noam D. Elkies, Everett W. Howe, and Christophe Ritzenthaler: Genus bounds for curves with fixed Frobenius eigenvalues, Proc. Amer. Math. Soc. 142 (2014), 71-84. arXiv: 1006.0822.

See page 82 of the online version or page 12 of the arXiv preprint.