Elliptic Curves over Finite Fields - Elliptic Curves

7 Elliptic Curves over Finite Fields

Lemma 7.1. Assuming that:

$A$ an abelian group
$q : A \to ℤ$ a positive definit quadratic form

Then

| \underset{= ⟨ x, y ⟩}{\underset{⏟}{q (x + y) - q (x) - q (y)}} | \leq 2 \sqrt{q (x) q (y)} \forall x, y \in A .

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

Let $m, n \in ℤ$ . Then

\begin{array}{l} 0 & \leq q (m x + n y) \\ = \frac{1}{2} ⟨ m x + n y, m x + n y ⟩ \\ = m^{2} q (x) + m n ⟨ x, y ⟩ + n^{2} q (y) \\ = q (x) {(m + \frac{⟨ x, y ⟩}{2 q (x)} n)}^{2} + (q (y) - \frac{{⟨ x, y ⟩}^{2}}{4 q (x)}) n^{2} \end{array}

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

q (y) - \frac{{⟨ x, y ⟩}^{2}}{4 q (x)} \geq 0

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

| ⟨ x, y ⟩ | \leq 2 \sqrt{q (x) q (y)} .

□

Theorem 7.2 (Hasse’s Theorem). Assuming that:

$E ∕ 𝔽_{q}$ an elliptic curve

Then

| # E (𝔽_{q}) - (q + 1) | \leq 2 \sqrt{q} .

Proof. Recall $Gal (𝔽_{q^{r}} ∕ 𝔽_{q})$ 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}{l} ϕ : E & \to E \\ (x, y) & \mapsto (x^{q}, y^{q}) \end{array}

This is an isogeny of degree $q$ .

Then

E (𝔽_{q}) = {P \in E | ϕ (P) = P} = \ker (1 - ϕ) .

\begin{array}{l} ϕ^{*} ω & = ϕ^{*} (\frac{d x}{y}) \\ = \frac{d (x^{q})}{y^{q}} \\ = \frac{q x^{q - 1} d x}{y^{q}} \\ = 0 & (q \equiv 0 (m o d p)) \end{array}

Lemma 6.3 tells us that

{(1 - ϕ)}^{*} ω = ω - ϕ^{*} (ω) = ω \neq 0 .

Hence $1 - ϕ$ is separable.

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

\underset{= # E (𝔽_{q})}{\underset{⏟}{# \ker (1 - ϕ)}} = \deg (1 - ϕ) .

$\deg : Hom (E, E) \to ℤ$ 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

| \deg (1 - ϕ) - 1 - \deg ϕ | \leq 2 \sqrt{\deg ϕ} .

Hence

| # E (𝔽_{q}) - 1 - q | \leq 2 \sqrt{q} . □

Definition. For $ϕ, ψ \in End (E) = Hom (E, E)$ , we put $⟨ ϕ, ψ ⟩ = \deg (ϕ + ψ) - \deg ϕ - \deg ψ$ and $tr (ϕ) = ⟨ ϕ, 1 ⟩$ .

Corollary 7.3. Assuming that:

$E ∕ 𝔽_{q}$ an elliptic curve
$ϕ \in End (E)$ the $q$ -power Frobenius

Then

# E (𝔽_{q}) = q + 1 - tr (ϕ)

and

| tr (ϕ) | \leq 2 \sqrt{q}

7.1 Zeta functions

For $K$ a number field, let

ζ_{K} (s) = \sum_{𝔞 \subset O_{K}} \frac{1}{{(N 𝔞)}^{s}} = \prod_{\begin{array}{c} 𝔭 \subset O_{K} \\ prime \end{array}} {(1 - \frac{1}{{(N 𝔭)}^{s}})}^{- 1} .

For $K$ a function field, i.e. $K = 𝔽_{q} (C)$ where $C ∕ 𝔽_{q}$ is a smooth projective curve,

ζ_{K} (s) = \prod_{x \in | C |} {(1 - \frac{1}{{(N x)}^{s}})}^{- 1},

where

| C | = {closed points on C}

(closed points are orbits for action of $Gal (\bar{𝔽_{q}} ∕ 𝔽_{q})$ on $C (\bar{𝔽_{q}})$ ) and $N x = q^{\deg x}$ , $\deg x$ is the size of orbit.

We have $ζ_{K} (s) = F (q^{- s})$ for some $F \in ℚ [[T]]$ ,

\begin{array}{l} F (T) & = \prod_{x \in | C |} {(1 - T^{\deg x})}^{- 1} \\ \log F (T) & = \sum_{x \in | C |} \sum_{m = 1}^{\infty} \frac{1}{m} T^{m \deg x} & - \log (1 - x) = \sum_{m = 1}^{\infty} \frac{x^{m}}{m} \\ T \frac{d}{d T} (\log F (T)) & = \sum_{x \in | C |} \sum_{m = 1}^{\infty} \deg x T^{m \deg x} \\ = \sum_{n = 1}^{\infty} (\sum_{\begin{array}{c} x \in | C | \\ \deg x | n \end{array}} \deg x) T^{n} & n = m \deg x \\ = \sum_{n = 1}^{\infty} # C (𝔽_{q^{n}}) T^{n} \end{array}

Therefore

F (T) = \exp (\sum_{n = 1}^{\infty} \frac{# C (𝔽_{q^{n}})}{n} T^{n}) .

Definition (Zeta function). The zeta function $Z_{C} (T)$ of a smooth projective curve $C ∕ 𝔽_{q}$ is defined by

Z_{C} (T) = \exp (\sum_{n = 1}^{\infty} \frac{# C (𝔽_{q^{n}})}{n} T^{n}) .

Theorem 7.4. Assuming that:

$E ∕ 𝔽_{q}$ an elliptic curve with $# E (𝔽_{q}) = q + 1 - a$

Then

Z_{E} (T) = \frac{1 - a T + q T^{2}}{(1 - T) (1 - q T)} .

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

# E (𝔽_{q}) = q + 1 - tr (ϕ) .

Hence $tr (ϕ) = a$ , $\deg (ϕ) = q$ .

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

tr (ϕ^{n + 2}) - a tr (ϕ^{n + 1}) + q tr (ϕ^{n}) = 0 .

This second order difference equation with initial conditions $tr (1) = 2$ , $tr (ϕ) = a$ has solution

tr (ϕ^{n}) = α^{n} + β^{n},

where $α, β$ are roots of $X^{2} - a X + q = 0$ .

Again by Corollary 7.3,

\begin{array}{l} # E (𝔽_{q}^{n}) & = q^{n} + 1 - tr (ϕ^{n}) \\ = 1 + q^{n} - α^{n} - β^{n} \end{array}

Therefore

\begin{array}{l} Z_{E} (T) & = \exp \sum_{n = 1}^{\infty} (\frac{T^{n}}{n} + \frac{{(q T)}^{n}}{n} - \frac{{(α T)}^{n}}{n} - \frac{{(β T)}^{n}}{n}) \\ = \frac{(1 - α T) (1 - β T)}{(1 - T) (1 - q T)} \\ = \frac{1 - a T + q T^{2}}{(1 - T) (1 - q T)} □ \end{array}

Remark. Hasse’s Theorem tells us that $| a | \leq 2 \sqrt{q}$ . $α = \bar{β}$ , and so $| α | = | β | = \sqrt{q}$ $(*)$ .

Let $K = 𝔽_{q} (E)$ . $ζ_{K} (s) = 0$ , so $Z_{E} (q^{- s}) = 0$ , so $q^{s} = α$ or $β$ . Then $q^{Re (s)} = | α |$ or $| β |$ , so by $(*)$ , $Re (s) = \frac{1}{2}$ .

“This is an analog of the Riemann hypothesis.”