Proof of Theorem 3.11

We continue to use the notation and hypotheses of Section 4.1 (so $\ell^2 \nmid N$ ) and assume in addition that $A$ is a newform quotient, and that $\ell \nmid {m_{\scriptscriptstyle{A}}}$ . We have to show that then $\ell \nmid {c_{\scriptscriptstyle{A}}}$ . Just as in the previous proof, it suffices to check that the image of $H^0(A_{{{\bf {Z}}_{(\ell)}}},\Omega^1_{A/{{{\bf {Z}}_{(\ell)}}}})$ in ${{{\bf {Z}}_{(\ell)}}}[[q]]$ is saturated. Since $A$ is a newform quotient, if $\ell \mid N$ , then $W_\ell$ acts as a scalar on $A$ and on  $S_2(\Gamma_0(N);{{{\bf {Z}}_{(\ell)}}})[I]$ . So again, using Lemma 4.2, it suffices to show that the map $H^0({A_{{{\bf {Z}}_{(\ell)}}}},\Omega^1_{{A/{{{\b... ...{Z}}_{(\ell)}}}},\Omega^1_{{J/{{{\bf {Z}}_{(\ell)}}}}})\otimes \mathbf{F}_\ell$ is injective.

The composition of pullback and pushforward in the following diagram is multiplication by the modular exponent of $A$ :

$$\xymatrix{ & {H^0(J_{{{{\bf{Z}}_{(\e... ...m_A} & & {H^0(A_{{{{\bf{Z}}_{(\ell)}}}}, \Omega^1_{A/{{{\bf{Z}}_{(\ell)}}}})} }$$

Since $m_A \in {{\bf {Z}}_{(\ell)}}^{\times}$ , the map $\pi^*$ is a section to the map $\pi_*$ up to a unit and hence its reduction modulo $\ell$ is injective, which is what was left to be shown. Let tex2html_wrap_inline$&pi#pi;_*$ and tex2html_wrap_inline$&pi#pi;^*$denote the maps obtained by tensoring the diagram above with tex2html_wrap_inline$F_&ell#ell;$. Then tex2html_wrap_inline$&pi#pi;_*&cir#circ;&pi#pi;^*$ is multiplication by an integer coprime to tex2html_wrap_inline$&ell#ell;$ from the finite dimension tex2html_wrap_inline$F_&ell#ell;$-vector space tex2html_wrap_inline$H^0(A_Z_(&ell#ell;), &Omega#Omega;^1_A/Z_(&ell#ell;))&otimes#otimes;F_&ell#ell;$ to itself, hence an isomorphism. In particular, tex2html_wrap_inline$&pi#pi;^*$ is injective, which is what was left to show.

theorem_type[rmk][lem][][definition][][] Adam Joyce observed that one can also obtain injectivity of tex2html_wrap_inline$&pi#pi;^*$ as a consequence of Prop. 7.5.3(a) of [BLR90].