When Frobenius map equal to multiplication-by-m map Here is a homework, the result brought me some trouble.
Let $p = 7$, and consider the finite field ${ \mathbb{F}}_{p^{2}}$ , which we may represent explicitly as
 $${ \mathbb{F}}_{p^{2}}\simeq   {\mathbb{F}}_{p}[i]/(i^{2} +1)=\{a+bi:a,b\in{ \mathbb{F}}_{p}\}.$$
Now consider the elliptic curve $E/{ \mathbb{F}}_{p^{2}}$
 defined by
$$y^{2} = x^{3} + (1 + i)x.$$
The group of ${ \mathbb{F}}_{p^{2}}$
-rational points on $E$ is isomorphic to $\mathbb{Z}/6\mathbb{Z }\oplus \mathbb{Z}/6\mathbb{Z }$ and is generated by the affine points
$$P_{1} =(i,i), P_{2} =(i+2,2i),$$
Now consider the Frobenius endomorphism $\pi_{E}$ of $E$:
1.Prove that $\pi_{E} = 7$ by showing that $\pi_{E} -7$  is the zero homomorphism(remember, it is not enough to show that$E({ \mathbb{F}}_{p})$
 is in the kernel of $\pi_{E} -7$, you need to prove that $E(\overline{{ \mathbb{F}}}_{p})$
is in the kernel of $\pi_{E} -7$ ).
My Question: 
$(i)$I have no idea how to check  $E(\overline{{ \mathbb{F}}}_{p})$
is in the kernel of $\pi_{E} -7$ ; Is there some useful method to check the kernel of some endomorphism(As  $E[m]$ is so simple and cute).
(ii) Can we decide the group structure $E(\overline{{ \mathbb{F}}}_{p})$ in this exercise?
2.Then find an endomorphism $\alpha$ that satisfies $\alpha^{2} = -1$(give $\alpha$ explicitly).
My Question: 
(iii) How to check $\alpha\neq m$ for some integer $m$?(Actually, in this exercise $End(E)$ is an order in a quaternion algebra)
Here is my method:(I don't think it is a general way)
Assume $\alpha=m$,
$m^{2}P_{j}=\alpha^{2}P_{j}=-P_{j},~j=1,2.$
We will have $6|(m^{2}+1)$, but there is no such integer satisfies this condition which contradict our assumption.
Thank you for any help.
:)
 A: 1) Of course in principle you could write down rational maps for the map $\pi-7$ and check that this is $0$ on every point of $E$. Note that you have to consider $E(\overline{\mathbb F_{7^2}})$ since the base field is $\mathbb F_{7^2}$, and not $E(\overline{\mathbb F_7})$ as you wrote.
But the quickest way is using the fact that if $l\neq 7$ is a prime than you have an injective map $\varphi\colon\text{End}(E)\otimes \mathbb Z_l\to \text{End}(T_l(E))$ where $T_l$ is the $l$-adic Tate module of $E$. Such a map sends an endomorphism $\phi$ to the same map seen as an endomorphism of the Tate module. Since $T_l(E)\simeq \mathbb Z_l\times \mathbb Z_l$, then you can define the characteristic polynomial of the Frobenius as the characteristic polynomial of $\varphi(\pi)$. One can show that such a polynomial is independent of $l$ and in fact equals $x^2-a_{7^2}x+7^2$ where $a_{7^2}=7^2+1-\#E(\mathbb F_{7^2})=14$. So this characteristic polynomial equals in your case $x^2-14x+49=(x-7)^2$. Now every endomorphism satisfies its characteristic polynomial, so you must have that $(\varphi(\pi)-7)^2=0$ as an endomorphism of $T_l(E)$. But since $\varphi$ is injective, this means that $(\pi-7)^2=0$ as an endomorphism of $E$, and this implies that $\pi-7=0$ because $\text{End}(E)$ has no zero divisors.
I don't know if you can decide the group structure of $E(\overline{\mathbb F_{7^2}})$, but if $K$ is an extension of $\mathbb F_{7^2}$ of degree $n$, then $\#E(K)=1+|K|-2\cdot 7^n=(7^n-1)^2$.
2) If $\alpha$ satisfies $\alpha^2=-1$, then it is automatical that $\alpha$ is not multiplication by $m$, because if you compose multiplication by $m$ with itself you get multiplication by $m^2$, which is obviously different from $-1$.
I think that in your case there are only 2 choices for $\alpha$, namely the maps sending $x\mapsto -x$ and $y\mapsto \pm iy$.
A: There is a technique for proving that two endomorphisms $\alpha$ and $\beta$ are equal.
Lemma7.12
In the previous question, we can choose $\alpha=\pi_{E}$, $\beta=7$ and $n=15$.
I am not so prefer this method as we have to compute $\alpha_{n}$ and $\beta_{n}$ which need us to know precisely the structure of $E[n]$(also basis ($P_{1},~P_{2}$)). It is not so easy to compute for large n. 
Some more refinement would be appreciated.
