Proof. We may assume $x\ne 0$, otherwise the result is clear. So $q(x)\ne 0$.

Let $m,n\in \mathbb{Z}$. Then

$$\begin{array}{llll}\hfill 0& \le q(mx+ny)& \hfill & \\ \hfill & =\frac{1}{2}⟨mx+ny,mx+ny⟩& \hfill & \\ \hfill & =m^{2}q(x)+mn⟨x,y⟩+n^{2}q(y)& \hfill & \\ \hfill & =q(x){\left(m+\frac{⟨x,y⟩}{2q(x)}n\right)}^2+\left(q(y)-\frac{{⟨x,y⟩}^2}{4q(x)}\right)n^2& \hfill & \end{array}$$

Take $m=-⟨x,y⟩$, $n=2q(x)\ne 0$ to deduce

$$q(y)-\frac{{⟨x,y⟩}^2}{4q(x)}\ge 0$$

hence ${⟨x,y⟩}^2\le 4q(x)q(y)$ and hence

$$|⟨x,y⟩|\le 2\sqrt{q(x)q(y)}.$$

□

Proof. Recall $\mathrm{Gal}({𝔽}_{q^r}∕{𝔽}_q)$ is cyclic of order $r$ and generated by Frobenius $x\mapsto x^q$.

Let $E$ have Weierstrass equation with coefficients $a_1,\dots ,a_6\in {𝔽}_q$ (so $a_i^q=a_i$). Define the Frobenius endomorphism

$$\begin{array}{llll}\hfill \varphi :E& \to E& \hfill & \\ \hfill (x,y)& \mapsto (x^q,y^q)& \hfill & \end{array}$$

This is an isogeny of degree $q$.

Then

$$E({𝔽}_q)=\{P\in E|\varphi (P)=P\}=\ker (1-\varphi ).$$

$$\begin{array}{llllll}\hfill {\varphi }^{\ast }\omega & ={\varphi }^{\ast }\left(\frac{dx}{y}\right)& \hfill & & \hfill & \\ \hfill & =\frac{d(x^q)}{y^q}& \hfill & & \hfill & \\ \hfill & =\frac{q{x}^{q-1}dx}{y^q}& \hfill & & \hfill & \\ \hfill & =0& \hfill & \text{(}q\equiv 0(modp)\text{)}& \hfill & & \hfill & \end{array}$$ Lemma 6.3 tells us that

$${(1-\varphi )}^{\ast }\omega =\omega -{\varphi }^{\ast }(\omega )=\omega \ne 0.$$

Hence $1-\varphi$ is separable.

By Theorem 2.8 and the fact that $1-\varphi$ is a group homomorphism, we argue as in the proof of Theorem 6.5 that

$\mathrm{deg}:\mathrm{Hom}(E,E)\to \mathbb{Z}$ is a positive definite quadratic form (Theorem 5.7, and positive definiteness is obvious since non-constant morphisms have positive degree).

Lemma 7.1 gives

Hence

$$|\#E({𝔽}_q)-1-q|\le 2\sqrt{q}.\square$$

Proof. Let $\varphi :E\to E$ be the $q$ power Frobenius map. By Corollary 7.3

$$\#E({𝔽}_q)=q+1-\mathrm{tr}(\varphi ).$$

Hence $\mathrm{tr}(\varphi )=a$, $\mathrm{deg}(\varphi )=q$.

Example Sheet 2, Q6(iii) implies ${\varphi }^2-a\varphi =q=0$, hence ${\varphi }^{n=2}-a{\varphi }^{n+1}+q{\varphi }^n=0$, so

$$\mathrm{tr}({\varphi }^{n+2})-a\mathrm{tr}({\varphi }^{n+1})+q\mathrm{tr}({\varphi }^n)=0.$$

This second order difference equation with initial conditions $\mathrm{tr}(1)=2$, $\mathrm{tr}(\varphi )=a$ has solution

$$\mathrm{tr}({\varphi }^n)={\alpha }^n+{\beta }^n,$$

where $\alpha ,\beta$ are roots of $X^2-aX+q=0$.

Again by Corollary 7.3,

$$\begin{array}{llll}\hfill \#E({𝔽}_q^n)& =q^n+1-\mathrm{tr}({\varphi }^n)& \hfill & \\ \hfill & =1+q^n-{\alpha }^n-{\beta }^n& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill Z_E(T)& =\exp \sum _{n=1}^{\infty }\left(\frac{T^n}{n}+\frac{{(qT)}^n}{n}-\frac{{(\alpha T)}^n}{n}-\frac{{(\beta T)}^n}{n}\right)& \hfill & \\ \hfill & =\frac{(1-\alpha T)(1-\beta T)}{(1-T)(1-qT)}& \hfill & \\ \hfill & =\frac{1-aT+q{T}^2}{(1-T)(1-qT)}\square & \hfill & \end{array}$$