The Eisenstein Nature of Component Groups

The theorem below, which generalizes some of the results of [13] and [15], was conjectured by the second author after computing many component groups of quotients of $J_0(p)$ using the results of this paper. M. Emerton read an early version of this paper and subsequently announced a proof of the theorem below (see [6]).

Theorem 7.2 (Emerton)   Let $p$ be a prime and let $f_1,\ldots,f_n$ be a set of representatives for the Galois-conjugacy classes of newforms in $S_2(\Gamma_0(p))$. Let $A_1,\ldots, A_n$ be the optimal quotients associated to $f_1,\ldots,f_n$, respectively. Then for each $i$, $i=1,\ldots,n$, we have

$$\char93 A_i(\mathbf{Q})_{\tor}=\char93 \Phi_{A_i}(\overline{\mathbf{F}}_p)=\char93 \Phi_{A_i}(\mathbf{F}_p).$$

Furthermore,

$$\char93 \Phi_{J_0(p)}(\overline{\mathbf{F}}_p)= \prod_{i=1}^n \char93 \Phi_{A_i}(\overline{\mathbf{F}}_p).$$

Before Emerton proved the above assertion, the second author verified it using the algorithm of this paper for all $p\leq 757$, and, up to a power of $2$, for all $p< 2000$.

Remark 7.3   It is tempting to guess that, e.g., the natural map

$$\Phi_{J_0(113)}(\overline{\mathbf{F}}_{113})\rightarrow \prod_{i=1}^4 \Phi_{A_i}(\overline{\mathbf{F}}_{113})$$

is an isomorphism, but this is incorrect. Two of the $\Phi_{A_i}(\overline{\mathbf{F}}_{113})$ have order $2$, so the product $\prod_{i=1}^4 \Phi_{A_i}(\overline{\mathbf{F}}_{113})$ is not a cyclic group. However, Mazur proved that the groups $\Phi_{J_0(p)}(\overline{\mathbf{F}}_p)$ are cyclic for all primes $p$.