An endomorphism that maps to zero in the p-adic tate module

Question

Let $E$ be an ordinary elliptic curve defined over say, $\mathbb F_p$ with endomorphism ring $\mathcal O_K$ where $[K:Q] = 2$. Consider the representation on the one dimensional p-adic Tate module: $$T: \mathcal O_K\otimes \mathbb Q_p \to \mathbb Q_p = End(T_p(E)).$$

For dimension reasons, this map has to be non injective. On the other hand, the map $\mathcal O_K\otimes \mathbb Q \to T_p(E)$ is injective.

Therefore, we have a sequence of elements $f_n \in \mathcal O_K\otimes \mathbb Q_p$ that form a cauchy sequence with respect to the p-adic topology where the limit is non-zero but it's image under the representation is $0$.

In particular, if we let $\sigma$ denote the dual of the Frobenius $\phi$, then we see that we can choose $f_n = a_n\sigma^n$ where $a_n$ is an endomorphism.

The cauchy condition is then effectively that $a_{n+1}\sigma - a_n$ is divisible by a high power of $\phi^n$.

Question: What is an explicit sequence that realizes the non injectivity?

Answer 1

For $p$ odd prime and an elliptic curve $E/\Bbb{F}_p$ with $E[p] \ne O$

• The Frobenius $\phi(x,y) = (x^p,y^p)$ is inseparable of degree $p$, with $\phi^* $ its dual isogeny $[p] = \phi^* \phi$ is inseparable thus $\#E[p]=\deg_s([p])$ can't be $p^2$ so it must be $= p$ which implies $\# E[p^n]=\deg_s([p^n]) = p^n$, whence $E[p^n]$ is cyclic $= \langle Q_n \rangle,[p] Q_n= Q_{n-1}$.

• $\Bbb{Z}_p \cong End(E[p^\infty]) $ through the isomorphism $b \mapsto (mQ_n \mapsto [b m \bmod p^n] Q_n)$

• The ring homomorphism $End_{\overline{\Bbb{F}}_p}(E) \to End(E[p^\infty])$ is injective since isogenies have finite kernel, whence $End_{\overline{\Bbb{F}}_p}(E)$ is a commutative subring of $\Bbb{Z}_p$

• $[\#E(\Bbb{F}_p)] = [\deg(\phi-1)] = (\phi^*-1)(\phi-1)= [p+1-tr(\phi)]$, $[tr(\phi)] = \phi+\phi^* \in End_{\overline{\Bbb{F}}_p}(E)\cap \Bbb{Z}$

  Let $H(T)=T^2-tr(\phi) T+p\in \Bbb{Z}[T]$. Since $H(\phi)=0\in End_{\overline{\Bbb{F}}_p}(E)$ and $\deg(\phi) = p$ then $\phi$ is not $\in \Bbb{Q}\cap End_{\overline{\Bbb{F}}_p}(E)$ whence $H(T)$ is $\phi$'s $\Bbb{Q}$-minimal polynomial and (since any endomorphism $u$ is a root of $T^2-tr(u)T + \deg(u)$) then $End_{\overline{\Bbb{F}}_p}(E)$ is a subring of $O_K, K= \Bbb{Q}[T]/(H(T))$

• The map $f : \Bbb{Z}[\phi] \to End(E[p^\infty]) \to \Bbb{Z}_p$ is a ring homomorphism and it has no integer in its kernel thus it is injective.

• For some $s \in 0,1$ $$f(\phi) = \frac{tr(\phi) + (-1)^s tr(\phi)\sqrt{1 - \frac{4p}{tr(\phi)^2}}}{2} \in \Bbb{Z}_p$$ (with Hensel lemma $\sqrt{1 - \frac{4p}{tr(\phi)}}$ is in $\Bbb{Z}_p$ since $[p] =\phi^*\phi \implies p \nmid tr(\phi)$)

• $End_{\overline{\Bbb{F}}_p}(E) \otimes_\Bbb{Z} \Bbb{Q}_p = \Bbb{Q}_p+\phi \Bbb{Q}_p $ and the homomorphism $\Bbb{Q}_p+\phi \Bbb{Q}_p \to \Bbb{Q}_p, \phi \mapsto f(\phi)$ has kernel $\{ a+b\phi,a=- f(\phi)b\in \Bbb{Q}_p \}$

• $\lim_{n \to \infty} \phi^{n!}-1$ is in the kernel but I don't see why $\lim_{n \to \infty} \phi^n$ should converge in $\Bbb{Q}_p+\phi \Bbb{Q}_p$

Answer 2

Adapted from the other answer:

The sequence $f_n = \phi^{n!}-1$ is Cauchy since the difference of consecutive terms is $\mathbb \phi^{n!}(\phi^n - 1)$. The $\phi^{n!}$ kills a large subgroup of $\mu_{p^\infty}$ part while $\phi^n-1$ kills a large subgroup of the etale torsion contained in $E[\mathbb F_{q^n}]$. Therefore, in all, it kills a large subgroup of $E[p^\infty]$.

For the same reason, the limit acts as $0$ on the Tate module.

On the other hand, the limit is not zero in $\mathbb O_K\otimes \mathbb Q_p$ because $\phi^{n!}-1$ acts as $-1$ on the $\mu_{p^\infty}$ part and therefore is not divisible by $p^n$.