# Silverman AEC exercise V.5.10(e)

I'm working my way through exercise V.5.10, but get stuck at (e).

Let $E/\mathbb{F}_q$ be an elliptic curve, let $\phi:E\rightarrow E$ be the $q^{\text{th}}$-power Frobenius endomorphism, and let $p=\operatorname{char}(\mathbb{F}_q)$. Let $p^i$ be the largest power of $p$ such that $p^{2i}\mid q$. Prove that $$\operatorname{tr}(\phi)\equiv0\pmod{p}\iff \operatorname{tr}(\phi)\equiv0\pmod{p^i}.$$

The $\Leftarrow$ direction is clear, so the $\Rightarrow$ is the hard part. We know that $E$ must be supersingular by earlier parts, yet it's not clear to me how to use those earlier solutions to help solve this. Any hints/suggestions?

Let $E/\mathbb{F}_q$ be an elliptic curve, and assume that $\operatorname{tr}(\phi)=0\pmod{p}$. Then by (a), we know that $E$ is supersingular. By Theorem V.3.1(a), we know that $j(E)\in\mathbb{F}_{p^2}$. Therefore there exists some elliptic curve $C/\mathbb{F}_{p^2}$ such that $E\cong C$. In particular, $\#E(\mathbb{F}_{q^2})=\#C(\mathbb{F}_{q^2})$. Let's write $q=p^{2m+r}$ for $r\in\left\{0,1\right\}$.
Now consider the recurrence relations from Exercise 5.13, applied to $C$. That is, write $c_i=p^{2i}+1-\#C(\mathbb{F}_{p^{2i}})$. Then using the recurrence relation, it is easily shown that $c_i\equiv0\pmod{p^i}$. In particular, $c_{2m+r}\equiv0\pmod {p^{2m+r}}$, i.e. $c_{2m+r}\equiv0\pmod{q}$.
Again consider the recurrence relation from Exercise 5.13, but now applied to $E$. Write $a_i=q^i+1-\#E(\mathbb{F}_{q^i})$, so that $a_2=a_1^2-2q$. But since $\#E(\mathbb{F}_{q^2})=\#C(\mathbb{F}_{({p^2})^{2m+r}})$, we know that $a_2=c_{2m+r}$. Therefore by the above, $a_2\equiv0\pmod{q}$, and therefore $a_1^2\equiv0\pmod{q}$.
Combining with Hasse's theorem, this leaves $a_1\in\left\{0,\pm\sqrt{q},\pm\sqrt{2q},\pm\sqrt{3q},\pm2\sqrt{q}\right\}$. As $a_1=\operatorname{tr}(\phi)$, the result now follows.