# On simple factors of modular jacobians: endomorphism ring and simplicity of mod p reduction

Let $A_f$ be the abelian variety over $\mathbf{Q}$ arising as a $\mathbf{Q}$-simple factor of the Jacobian $J_0(N)$ of the modular curve associated to a normalized newform $f$ of weight $2$ on the congruence subgroup $\Gamma_0(N)$. There is a piece $\mathbf{T}_f$ of the Hecke ring acting on $A_f$. Such a ring is an order in the number field generated by the Fourier coefficients of $f$, whose degree equals the dimension of $A$. It is a result of Ribet (I think) that the natural map $\mathbf{T}_f\otimes\mathbf{Q}\rightarrow\textrm{End}(A)\otimes\mathbf{Q}$ is an isomorphism, where $\textrm{End}(A)$ is the ring of $\mathbf{Q}$-enndomorphisms of $A$ (I couldn't attach the index to End..).

1) Can we find an abelian variety $A_f'$ over $\mathbf{Q}$, which is $\mathbf{Q}$-isogenous to $A_f$, and such that $\textrm{End}(A_f')$ is the maximal order of $\textrm{End}(A_f')\otimes\mathbf{Q}=\textrm{End}(A_f)\otimes\mathbf{Q}$? (Notice that in the one dimensional case $\textrm{End}(A_f)$ is already maximal, being it the ring of integers)

2) Does the fact that $A_f$ has such a large endomorphism ring imply that the mod $p$ reduction of $A_f$ be simple over the prime field with $p$ elements? Here $p$ is a prime of good reduction for $A_f$.

Thanks!

• If your $A_f$ is absolutely simple, then this paper of Zywina might help with question 2: mast.queensu.ca/~zywina/papers/Splitting.pdf But I suspect that $A_f$ are usually not absolutely simple. – Barinder Banwait Oct 10 '12 at 0:04
• In the case of $A_f$, I think it's known that if $A_f$ is absolutely simple, then $A_f$ mod $p$ is simple for a set of primes $p$ of density 1. See Murty-Patankar's conjecture in Zywina's paper. – François Brunault Oct 10 '12 at 9:30
• @François: That would be true provided the Mumford-Tate conjecture was known for $A_f$, and the field $K_A^{conn}$ in Zywina's paper was $\mathbb{Q}$. I'm not sure about either of these claims. – Barinder Banwait Oct 10 '12 at 11:31
• There is a canonical $A_f$ which is defined as the quotient $J_0(N)/I_f J_0(N)$ where $I_f$ is the kernel of the map $\mathbf{T} \to K_f$ sending $T_n$ to $a_n(f)$. This is called the optimal quotient. Using Magma, you can determine the order $\mathbf{T}_f = \mathbf{Z}[a_n(f),n \geq 1]$ inside $K_f$, and you can also determine the index of $\mathbf{T}_f$ inside $\operatorname{End}(A_f)$ (for this $A_f$), so you can deduce the index of the order $\operatorname{End}(A_f)$. It is not always the maximal order, the first counterexample appears at $N=69$ I think. (...) – François Brunault Oct 21 '12 at 19:54
• (...) So optimal quotients are not always optimal from the point of view of the endomorphism ring. – François Brunault Oct 21 '12 at 19:54

I found a reference for the first question in the following PhD thesis :

The answer is yes. This is a consequence of the following general fact about abelian varieties (see Prop 2.5.4 in the thesis).

Let $A$ be an abelian variety over a field $k$. Let $R$ be an order in a number field $F$. Assume that $R$ embeds into $\operatorname{End}_k(A)$. Then there exists an abelian variety $B/k$ which is $k$-isogenous to $A$ and such that $\mathcal{O}_F$ embeds into $\operatorname{End}_k(B)$.

The idea is to take $B=A/G$ with $G=(n\mathcal{O}_F) A[n^2]$, where $n$ is the index of $R$ in $\mathcal{O}_F$.

The thesis also contains interesting examples of varieties $A_f$ with Hecke field $K_f=\mathbf{Q}(\sqrt{5})$. These are natural examples to try for Question 2 (although I have no idea how to compute the reduction of $A_f$ mod $p$).

EDIT. The answer to Question 2 is negative in general. There are newforms $f$ of weight $2$ on $\Gamma_0(N)$ such that $A_f$ splits over $\overline{\mathbf{Q}}$. This happens for example when $f$ has extra-twist. The first example appears at level $N=63$, see Table 1 p. 13 in

MR1933828 (2003i:11078) González-Jiménez, Enrique ; González, Josep. Modular curves of genus 2. Math. Comp. 72 (2003), no. 241, 397--418 (electronic).

Assume $A_f \sim E_1 \times E_2$ where everything is defined over some number field $K$. If $p$ is a prime of good reduction for $A_f$ which splits totally in $K$, then $A_f$ mod $p$ is $\mathbf{F}_p$-isogenous to a product of elliptic curves over $\mathbf{F}_p$, so it is not simple.