$p$-Torsion of Twisted Powers

Let $p$ and $\ell$ be as in Conjecture 4.1. In order to apply Theorem 3.1, it is necessary to know that $p$ does not divide the orders of certain groups. In this section, we use that $a_\ell(E) \not\equiv 2\pmod{p}$ to deduce that certain groups do not have any $p$ torsion. The key idea is that the condition on $a_\ell(E)$ implies that $+1$ is not an eigenvalue of $\Frob_\ell$ on the $p$-adic Tate module attached to $E$.

First, we recall that certain torsion points on the closed fiber of a Néron model lift to the generic fiber. Let $K$ be a finite extension of  $\mathbb{Q}_\ell$ with ring of integers $\O$ and residue class field $k$.

Lemma 5.1   Let $A$ be an abelian variety over $K$ with Néron model $\mathcal{A}$ over $\O$. Then for every integer $n$ not divisible by $\ell$, there is an isomorphism

$$A(K)[n] \xrightarrow{ \cong } \mathcal{A}(k)[n].$$

Proof. This is a standard fact, whose proof we recall for the convenience of the reader. Let $A^{1}(K)$ denote the kernel of the natural reduction map $r:A(K)\rightarrow \mathcal{A}(k)$. Because $A^{1}(K)$ is a formal group, it is pro-$p$, so $[n]:A^{1}(K)\rightarrow {}A^{1}(K)$ is an isomorphism. Since $\mathcal{A}$ is smooth over $\O$, Hensel's lemma (see BLR) implies that the reduction map is surjective, so the following sequence is exact:

$$0\rightarrow A^1(K) \rightarrow A(K) \rightarrow \mathcal{A}(k) \rightarrow 0.$$

The snake lemma applied to the multiplication by $n$ diagram attached to this exact sequence yields the following exact sequence:

$$0\rightarrow 0\rightarrow A(K)[n]\rightarrow \mathcal{A}(k)[n] \r... ...rightarrow A(K)/n A(K) \rightarrow \mathcal{A}(k)/n\mathcal{A}(k)\rightarrow 0,$$

which proves the proposition. $\qedsymbol$

Let $E$ be an elliptic curve over  $\mathbb{Q}$ with associated newform $f = \sum a_n q^n$, and fix a prime $p$ that is rigid for $E$. Suppose $K$ is the extension of  $\mathbb{Q}$ corresponding to a surjective Dirichlet character $\chi: (\mathbb{Z}/\ell\mathbb{Z})^* \rightarrow\!\!\!\!\rightarrow \boldsymbol{\mu}_p$ of prime conductor; then $K$ is the subfield of $\mathbb{Q}(\boldsymbol{\mu}_\ell)$ fixed by $\ker(\chi)$, so it is of degree $p$, is totally ramified at $\ell$, and is unramified outside $\ell$. Let  $A=\ker(\tr : \Res_{K/\mathbb{Q}} E_K \rightarrow E)$. We next compute the Tamagawa number $c_{A,\ell}=\char93 \Phi_{A,\ell}(\mathbb{F}_\ell)$ of $A$ at $\ell$ and the $p$-torsion of several abelian varieties.

Proposition 5.2   Let $E$, $\chi$, $K$, and $A$ be as above and suppose that $\ell\nmid N_E$ and $a_\ell \not\equiv 2\pmod{p}$. Then the following groups have no nontrivial $p$-torsion:

$$A(K), \quad A(\mathbb{Q}_\ell),\quad R(\mathbb{Q}_\ell),\quad (R/E)(\mathbb{Q}_\ell),$$   and$$\quad \Phi_{A,\ell}(\mathbb{F}_\ell).$$

Proof. The reason the $p$-torsion vanishes in all these cases is that the condition $a_\ell \not\equiv 2\pmod{p}$ implies in each case that $\Frob_\ell$ has no $+1$ eigenvalue. The details are as follows.

We first show that $R(\mathbb{Q}_\ell)[p]=\{0\}$, where $R=\Res_{K/\mathbb{Q}}E_K$. By definition,

$$R(\mathbb{Q}_\ell) = E_K(\mathbb{Q}_\ell\otimes _\mathbb{Q}K) = E(K_v)\times \cdots \times E(K_v)$$   ($p$ copies)$$,$$

where $K_v$ is the completion of $K$ at the unique prime of $K$ lying over $\ell$. The action of $\Frob_\ell\in \Gal(\mathbb{Q}_\ell^{\ur}/\mathbb{Q}_\ell)$ on $E[p](\mathbb{Q}_\ell^{\ur})=E[p](\overline{\mathbb{Q}}_\ell)$ has characteristic polynomial $F(x) = x^2-a_\ell x + \ell \in \mathbb{F}_p[x]$. Since $a_\ell \not\equiv 2\pmod{p}$ and $\ell\equiv 1\pmod{p}$, it follows that $\Frob_\ell$ does not have $+1$ as an eigenvalue, so $E(\mathbb{Q}_\ell)[p]=\{0\}$. If $z\in E(K_v)[p]$, then the field $L=\mathbb{Q}_\ell(z)$ is an unramified subfield of the totally ramified field $K_v$, so $z\in E(\mathbb{Q}_\ell)[p]=\{0\}$. Thus $E(K_v)[p]=\{0\}$, which implies that $E(K)[p]=\{0\}$ and $R(\mathbb{Q}_\ell)[p]=\{0\}$. Since $R_K/E_K \cong E_K \times \cdots \times E_K$ ($p-1$ times), we see that

$$(R/E)(\mathbb{Q}_\ell)[p]\subset (R/E)(K_v)[p] = (E(K_v)\times \cdots \times E(K_v))[p] = \{0\}.$$

Finally, we turn to the component group $\Phi_{A,\ell}$. Let $\mathcal{A}$ denote the Néron model of $A$. By Lang's Lemma the natural map $\mathcal{A}(\mathbb{F}_\ell) \rightarrow \Phi_{A,\ell}(\mathbb{F}_\ell)$ is surjective. Thus if $\Phi_{A,\ell}(\mathbb{F}_\ell)[p]\neq \{0\}$, then $\mathcal{A}(\mathbb{F}_\ell)[p]\neq \{0\}$. However, by Lemma 5.1 and observation of the previous paragraph,

$$\mathcal{A}(\mathbb{F}_\ell)[p] = A(\mathbb{Q}_\ell)[p]\subset R(\mathbb{Q}_\ell)[p]=\{0\},$$

so $\Phi_{A,\ell}(\mathbb{F}_\ell)[p]=\{0\}$, as claimed. $\qedsymbol$