Weierstrass Equations - Elliptic Curves

In this section, we drop the assumption that

K = \bar{K}

, but we instead assume that

K

is a perfect field.

Fact: If

D \in Div (E)

is defined over

K

(i.e. fixed by

Gal (\bar{K} ∕ K)

) then

L (D)

has a basis in

K (E)

(not just

\bar{K} (E)

Proof of Theorem 3.1. $L (2 . 0_{E}) \subset L (3 . 0_{E})$ . Pick basis $1, x$ for $L (2 . 0_{E})$ and $1, x, y$ for $L (3 . 0_{E})$ .

Note: ${ord}_{0_{E}} (x) = - 2$ , ${ord}_{0_{E}} (y) = - 3$ .

The 7 elements $1, x, y, x^{2}, x y, x^{3}, y^{2}$ in the 6-dimensional space $L (6 . 0_{E})$ must satisfy a dependence relation.

Leaving out $x^{3}$ or $y^{2}$ gives a basis for $L (6 . 0_{E})$ since each term has a different order pole at $0_{E}$ .

Therefore the coefficients of $x^{3}$ and $y^{2}$ are non-zero. Rescaling $x$ and $y$ (if necessary) we get

y^{2} + a_{1} x y + a_{3} y = x^{3} + a_{2} x^{2} + a_{4} x + a_{6}

for some $a_{i} \in K$ .

Let $E^{'}$ be the curve defined by this equation (or rather its projective closure). There is a morphism

\begin{array}{l} ϕ : E & \to E^{'} \subset ℙ^{2} \\ P & \mapsto (x (P) : y (P) : 1) = (\frac{x}{y} (P) : 1 : \frac{1}{y} (P)) \\ 0_{E} & \mapsto (0 : 1 : 0) \end{array}

Then

\begin{array}{l} [K (E) : K (x)] & = \deg (E \overset{x}{\to} ℙ^{1}) \\ = {ord}_{0_{E}} (\frac{1}{x}) \\ = 2 [K (E) : K (y)] & = \deg (E \overset{y}{\to} ℙ^{1}) \\ = {ord}_{0_{E}} (\frac{1}{y}) \\ = 3 \end{array}

This gives us a diagram of field extensions:

By the Tower Law (since

2, 3

are coprime), we get that

[K (E) : K (x, y)] = 1

. Hence

K (E) = K (x, y) = ϕ^{*} K (E^{'})

, so

\deg ϕ = 1

, so

ϕ

is birational.

If $E^{'}$ is singular then $E$ and $E^{'}$ are rational, contradiction.

So $E^{'}$ is smooth. Then Remark 2.9 implies that $ϕ^{- 1}$ is a morphism. So $ϕ$ is an isomorphism. □

Proposition 3.2. Assuming that:

$K$ is perfect
$E$ , $E^{'}$ are elliptic curves over $K$
$E, E^{'}$ are in Weierstrass form

Then

E ≅ E^{'}

over

K

if and only if the equations are related by a change of variables

\begin{array}{l} x & = u^{2} x^{'} + r \\ y & = u^{3} y^{'} + u^{2} s x^{'} + t \end{array}

where $u, r, s, t \in K$ with $u \neq 0$ .

Proof.

$\Leftarrow$ Obvious.
$\Rightarrow$ $⟨ 1, x ⟩ = L (2 . 0_{E}) = ⟨ 1, x^{'} ⟩$ hence $x = λ x^{'} + r$ for some $λ, r \in K$ with $λ \neq 0$ .
$⟨ 1, x, y ⟩ = L (3 . 0_{E}) = ⟨ 1, x^{'}, y^{'} ⟩$ implies $y = μ y^{'} + σ x^{'} + t$ for some $μ, σ, t \in K$ with $μ \neq 0$ .

Looking at coefficients of $x^{3}$ and $y^{2}$ , we get $λ^{3} = μ^{2}$ . So $λ = u^{2}$ , $μ = u^{3}$ for some $0 \neq u \in K$ .

Put $s = \frac{σ}{u^{2}}$ . □

A Weierstrass form equation defines an elliptic curve if and only if it defines a smooth curve, which happens if and only if

Δ (a_{1}, \dots, a_{6}) \neq 0

, where

Δ \in ℤ [a_{1}, \dots, a_{6}]

is a certain polynomial.

char K \neq 2, 3

, we may reduce to the case

E : y^{2} = x^{3} + a x + b

, with discriminant

Δ = - 16 (4 a^{3} + 27 b^{2})

Corollary 3.3. Assuming that:

$char K \neq 2, 3$ and $K$ is perfect
elliptic curves $E : y^{2} = x^{3} + a x + b$ , $E^{'} : y^{2} = x^{3} + a^{'} x + b^{'}$

Then

E

and

E^{'}

are isomorphic over

K

if and only if

a^{'} = u^{4} a

b^{'} = u^{6} b

for some

u \in K^{*}

Proof. $E$ and $E^{'}$ are related by a substitution as in Proposition 3.2 with $r = s = t = 0$ . □

Proof.

\begin{array}{l} E ≅ E^{'} & ⟺ {\begin{matrix} a^{'} = u^{4} a \\ b^{'} = u^{6} b \end{matrix} for some u \in K^{*} \\ ⟹ (a^{3} : b^{2}) = ({(a^{'})}^{3} : {(b^{'})}^{2}) \\ ⟺ j (E) = j (E^{'}) \end{array}

and the converse holds if $K = \bar{K}$ (to go backwards on the $⟹$ step, we only need to take some kind of $n$ -th root). □

3 Weierstrass Equations