A Variant of Theorem 5.1.3 with Simpler Hypothesis

Proposition 5.4.1   Suppose $A=A_f\subset J_0(N)$ is a newform abelian variety and $q$ is a prime that exactly divides $N$ . Suppose $\mathfrak{m}\subset\mathbb{T}(N)$ is a non-Eisenstein maximal ideal of residue characteristic $\ell$ and that $\ell\nmid m_A$ , where $m_A$ is the modular degree of $A$ . Then $\Phi_{A,q}(\overline{\mathbb{F}}_q)[\mathfrak{m}]=0$ .

Proof. The component group of $\Phi_{J_0(N),q}(\overline{\mathbb{F}}_q)$ is Eisenstein by [Rib87], so

$$\Phi_{J_0(N),q}(\overline{\mathbb{F}}_q)[\mathfrak{m}]=0.$$

By Lemma 4.2.2, the image of $\Phi_{J_0(N),q}(\overline{\mathbb{F}}_q)$ in $\Phi_{A^{\vee},q}(\overline{\mathbb{F}}_q)$ has no  $\mathfrak{m}$ torsion. By the main theorem of [CS01], the cokernel $\Phi_{J_0(N),q}(\overline{\mathbb{F}}_q)$ in $\Phi_{A^{\vee},q}(\overline{\mathbb{F}}_q)$ has order that divides $m_A$ . Since $\ell\nmid m_A$ , it follows that the cokernel also has no  $\mathfrak{m}$ torsion. Thus Lemma 4.2.2 implies that $\Phi_{A^{\vee},q}(\overline{\mathbb{F}}_q)[\mathfrak{m}]=0$ . Finally, the modular polarization $A\to A^{\vee}$ has degree $m_A$ , which is coprime to $\ell$ , so the induced map $\Phi_{A,q}(\overline{\mathbb{F}}_q)\to \Phi_{A^{\vee},q}(\overline{\mathbb{F}}_q)$ is an isomorphism on $\ell$ primary parts. In particular, that $\Phi_{A^{\vee},q}(\overline{\mathbb{F}}_q)[\mathfrak{m}]=0$ implies that $\Phi_{A,q}(\overline{\mathbb{F}}_q)[\mathfrak{m}]=0$ . $\qedsymbol$

If $E$ is a semistable elliptic curve over  $\mathbb{Q}$ with discriminant $\Delta$ , then we see using Tate curves that $\overline{c}_p = {\mathrm{ord}}_p(\Delta).$

Theorem 5.4.2   Suppose $A=A_f\subset J_0(N)$ is a newform abelian variety with $L(A_{/\mathbb{Q}},1)\neq 0$ and $N$ square free, and let $\ell$ be a prime. Suppose that $p\nmid N$ is a prime, and that there is an elliptic curve $E$ of conductor $pN$ such that:

1. [Rank] The Mordell-Weil rank of $E(\mathbb{Q})$ is positive.

2. [Divisibility]We have $a_p(E) = -1$ , $$\ell \mid \overline{c}_{E,p}$$ , and
   $$\ell\nmid 2 \cdot N \cdot p \cdot c_{E,p} \cdot \prod_{q\mid N} \overline{c}_{E,q}.$$

3. [Irreducibility] The mod $\ell$ representation $\overline{\rho}_{E,\ell}$ is irreducible.

4. [Noncongruence]The representation $\overline{\rho}_{E,\ell}$ is not isomorphic to any representation $\overline{\rho}_{g,\lambda}$ where $g\in S_2(\Gamma_0(N))$ is a newform of level dividing $N$ that is not conjugate to $f$ .

Then there is an element of order $\ell$ in ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(\mathbb{Q}, A_f)$ that is not visible in $J_0(N)$ but is strongly visible in $J_0(pN)$ . More precisely, there is an inclusion

$$E(\mathbb{Q})/\ell E(\mathbb{Q}) \hookrightarrow {\mathrm{Ker}}({... ...wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(\mathbb{Q}, A_f))[\lambda],$$

where $C\subset J_0(pN)$ is isogenous to $A_f\times E$ , the homomorphism $A_f \to C$ has degree a power of $2$ , and $\lambda$ is the maximal ideal of $\mathbb{T}(N)$ corresponding to $\overline{\rho}_{E,\ell}$ .

Proof. The divisibility assumptions of Hypothesis 2 on the $\overline{c}_{E,q}$ imply that the Serre level of $\overline{\rho}_{E,\ell}$ is $N$ and since $\ell\nmid N$ , the Serre weight is $2$ (see [RS01, Thm. 2.10]). Since $\ell$ is odd, Ribet's level lowering theorem [Rib91] implies that there is some newform $h =\sum b_n q^n\in S_2(\Gamma_0(N))$ and a maximal ideal $\lambda$ over $\ell$ such that $a_q(E) \equiv b_q \pmod{\lambda}$ for all primes $q\neq p$ . By our non-congruence hypothesis, the only possibility is that $h$ is a $G_{\mathbb{Q}}$ -conjugate of $f$ . Since we can replace $f$ by any Galois conjugate of $f$ without changing $A_f$ , we may assume that $f=h$ . Also $a_p(f) \equiv -(p+1)\pmod{\lambda}$ , as explained in [Rib83, pg. 506].

Hypothesis 3 implies that $\lambda$ is not Eisenstein, and by assumption $\ell\nmid m_A$ , so Proposition 5.4.1 implies that $\Phi_{A,q}(\overline{\mathbb{F}}_q)[\lambda]=0$ for each $q\mid N$ .

The theorem now follows from Theorem 5.1.3. $\qedsymbol$

Remark 5.4.3   The condition $a_p(E) = -1$ is redundant. Indeed, we have $\overline{c}_{E,p} \neq c_{E,p}$ since $\overline{c}_{E,p}$ is divisible by $\ell$ and $c_{E,p}$ is not. By studying the action of Frobenius on the component group at $p$ one can show that this implies that $E$ has nonsplit multiplicative reduction, so $a_p(E) = -1$ .

Remark 5.4.4   The non-congruence hypothesis of Theorem 5.4.2 can be verified using modular symbols as follows. Let $W \subset H_1(X_0(N),\mathbb{Z})_{{\mathrm{new}}}$ be the saturated submodule of $H_1(X_0(N),\mathbb{Z})$ that corresponds to all newforms in $S_2(\Gamma_0(N))$ that are not Galois conjugate to $f$ . Let $\overline{W} = W\otimes \mathbb{F}_{\ell}$ . We require that the intersection of the kernels of $T_q\vert _{\overline{W}} - a_q(E)$ , for $q\neq p$ , has dimension 0 .