Isogenies - Elliptic Curves - Cambridge III notes

5 Isogenies

Let $E_{1}, E_{2}$ be elliptic curves.

Definition (Isogeny). An isogeny $ϕ : E_{1} \to E_{2}$ is a nonconstant morphism with $ϕ (0_{E_{1}}) = 0_{E_{2}}$ (by Theorem 2.8, a morphism is nonconstant if and only if surjective on $\bar{K}$ -points).

We say $E_{1}$ and $E_{2}$ are isogenous in this case.

Definition. $Hom (E_{1}, E_{2}) = {isogenies E_{1} \to E_{2}} \cup {0}$ .

This is an abelian group under

(ϕ + ψ) (P) = ϕ (P) + ψ (P) .

If $E_{1} \overset{ϕ}{\to} E_{2} \overset{ψ}{\to} E_{3}$ are isogenies then $ψ \circ ϕ$ is an isogeny.

Tower Law implies $\deg (ψ ϕ) = \deg (ϕ) \deg (ψ)$ .

Proposition 5.1. Assuming that:

$0 \neq n \in ℤ$

Then

[n] : E \to E

is an isogeny.

Proof. $[n]$ is a morphism by Theorem 4.4. We must show $[n] \neq [0]$ .

Assume $char K \neq 2$ .

Case $n = 2$ : Lemma 4.5 implies $E [2] \neq E$ implies $[2] \neq [0]$ .

Case $n$ odd: Lemma 4.5 implies $\exists 0 \neq T \in E [2]$ . Then $n T = T \neq 0$ which gives $[n] \neq [0]$ .

Now use $[m n] = [m] \circ [n]$ .

If $char K = 2$ , then could rpelace Lemma 4.5 with an explicit lemma about $3$ -torsion points. □

Corollary 5.2. $Hom (E_{1}, E_{2})$ is a torsion free $ℤ$ -module.

Theorem 5.3. Assuming that:

$ϕ : E_{1} \to E_{2}$ an isogeny

Then

ϕ (P + Q) = ϕ (P) + ϕ (Q)

for all

P, Q \in E_{1}

Proof (sketch). $ϕ$ incudes

\begin{array}{l} ϕ_{*} : {Div}^{0} (E_{1}) & \to {Div}^{0} (E_{2}) \\ \sum_{P \in E_{1}} n_{P} P & \mapsto \sum_{P \in E_{1}} n_{P} ϕ (P) \end{array}

Recall $ϕ^{*} : K (E_{2}) ↪ K (E_{1})$ .

Fact: If

f \in \bar{K} {(E_{1})}^{*}

then

div (N_{K (E_{1}) ∕ K (E_{2})} f) = ϕ_{*} (div f) .

So $ϕ_{*}$ sends principal divisors to principal divisors.

Since $ϕ (0_{E_{1}}) = 0_{E_{2}}$ , the following diagram commutes:

P E1 E2 Q

ϕ∼=∼=ϕ∗ [(P )− (0E1)] Pic0(E1) Pic0(E2) [(Q )− (0E2)]

ϕ_{*}

a group homomorphism implies

ϕ

is a group homomorphism. □

Lemma 5.4. Assuming that:

$ϕ : E_{1} \to E_{2}$ an isogeny

Then there exists a morphism

ξ

making the following diagram commute:

(

x_{i} =

x

-coordinate on a Weierstrass equation for

E_{i}

). Moreover if

ξ (t) = \frac{r (t)}{s (t)}

r, s \in K [t]

coprime, then

\deg ϕ = \deg ξ = \max (\deg (r), \deg (s)) .

Proof. For $i = 1, 2$ . $K (E_{i}) ∕ K (x_{i})$ is a degree $2$ Galois extension, with Galois group generated by ${[- 1]}^{*}$ .

Theorem 5.3 implies that $ϕ \circ [- 1] = [- 1] \circ ϕ$ .

So if $f \in K (x_{2})$ then

{[- 1]}^{*} ϕ^{*} f = ϕ^{*} {[- 1]}^{*} f = ϕ^{*} f .

Therefore $ϕ^{*} f \in K (x_{1})$ .

K (E1) = K (x1,y1)

K (x1)

K (E2) = K (x2,y2)

2dd2eeggKϕξ (x2)

In particular,

ϕ^{*} = ξ (x_{1})

for some rational function

ξ

Tower Law implies $\deg ϕ = \deg ξ$ .

Now $K (x_{2}) ↪ K (x_{1})$ , $x_{2} \mapsto ξ (x_{1}) = \frac{r (x_{1})}{s (x_{1})}$ , $r, s \in K [t]$ coprime.

We claim the minimal polynomial of $x_{1}$ over $K (x_{2})$ is

F (t) = r (t) - s (t) x_{2} \in K (x_{2}) [t] .

Check:

$F (x_{1}) = 0$ : true by the definition of our embedding.
$F$ is irreducible in $K [x_{2}, t]$ (since $r, s$ coprime), hence irreducible in $K (x_{2}) [t]$ by Gauss’s lemma.

Therefore

\deg ξ = [K (x_{1}) : K (x_{2})] = \deg F = \max (\deg r, \deg s) . □

Lemma 5.5. $\deg [2] = 4$ .

Proof. Assume $char K \neq 2, 3$ (the result is true even in the case of $char K \in {2, 3}$ , but we will only prove the simpler case).

$E$ : $y^{2} = x^{3} + a x + b = f (x)$ .

If $P = (x, y) \in E$ , then

\begin{array}{l} x (2 P) & = {(\frac{3 x^{2} + a}{2 y})}^{2} - 2 x \\ = \frac{{(3 x^{2} + a)}^{2} - 8 x f (x)}{4 f (x)} \\ = \frac{x^{4} + \dots}{f (x)} \end{array}

The numerator and denominator are coprime. Indeed, otherwise there exists $𝜃 \in \bar{K}$ with $f (𝜃) = f^{'} (𝜃) = 0$ and hence $f$ has a multiple root (contradiction).

Now Lemma 5.4 implies $\deg [2] = \max (4, 3) = 4$ . □

Definition (Quadratic form in an abelian group). Let $A$ be an abelian group. We say $q : A \to ℤ$ is a quadratic form if

(i) $q (n x) = n^{2} q (x)$ for all $n \in ℤ$ , $x \in A$ .
(ii) $(x, y) \mapsto q (x + y) - q (x) - q (y)$ is $ℤ$ -bilinear.

Lemma 5.6. $q : A \to ℤ$ is a quadratic form if and only if it satisfies the parallelogram law:

q (x + y) + q (x - y) = 2 q (x) = 2 q (y) \forall x, y \in A .

Proof.

$\Rightarrow$ Let $⟨ x, y ⟩ = q (x + y) - q (x) - q (y)$ . Then
$⟨ x, x ⟩ = q (2 x) - 2 q (x) = 2 q (x)$

(the equality in the last step is using property (i) with $n = 2$ ).

But by (ii),
$⟨ x + y, x + y ⟩ + ⟨ x - y, x - y ⟩ = 2 ⟨ x, x ⟩ + 2 ⟨ y, y ⟩ .$

Hence (upon dividing by $2$ ),

$q (x + y) + q (x - y) = 2 q (x) + 2 q (y) .$
$\Leftarrow$ See Example Sheet 2. □

Theorem 5.7. $\deg : Hom (E_{1}, E_{2}) \to ℤ$ is a quadratic form.

Note. We define $\deg (0) = 0$ .

For the proof we assume $char K \neq 2, 3$ . Write $E_{2}$ : $y^{2} = x^{3} + a x + b$ .

Let $P, Q \in E_{2}$ with $P, Q, P + Q, P - Q \neq 0$ . Let $x_{1}, \dots, x_{4}$ be the $x$ -coordinates of the $4$ points.

Lemma 5.8. There exists $W_{0}, W_{1}, W_{2} \in ℤ [a, b] [x_{1}, x_{2}]$ of degree in $x_{1}$ and of degree $\leq 2$ in $x_{2}$ such that

(1 : x_{3} + x_{4} : x_{3} x_{4}) = (W_{0} : W_{1} : W_{2}) .

Proof. Two methods.

(1) Direct calculation: $\begin{array}{l} W_{0} & = {(x_{1} - x_{2})}^{2} \\ W_{1} & = \dots \\ W_{2} & = \dots \end{array}$
see formula sheet.
(2) Let $y = λ x + ν$ be the line through $P$ and $Q$ . $\begin{array}{l} x^{3} + a x + b - {(λ x + ν)}^{2} & = (x - x_{1}) (x - x_{2}) (x - x_{3}) \\ = x^{3} - s_{1} x^{2} + x_{2} x - s_{3} \end{array}$
where $s_{i}$ is the $i$ -th elementary symmetric polynomial in $x_{1}, x_{2}, x_{3}$ .

Comparing coefficients:
$\begin{array}{l} λ^{2} & = s_{1} \\ - 2 λ ν & = s_{2} - a \\ ν^{2} & = s_{3} + b \end{array}$
Eliminating $λ$ and $ν$ gives
$\underset{= F (x_{1}, x_{2}, x_{3})}{\underset{⏟}{{(s_{2} - a)}^{2} - 4 s_{1} (s_{3} + b) = 0}}$

where $F$ has degree $\leq 2$ in $x_{i}$ .

$x_{3}$ is a root of the quadratic
$W (t) = F (x_{1}, x_{2}, t) .$

Repeating for the line through $P$ and $- Q$ shows that $x_{4}$ is the other root. Therefore
$W_{0} (t - x_{3}) (t - x_{4}) = W_{0} t^{2} - W_{1} t + W_{2},$

hence
$(1 : x_{3} + x_{4} : x_{3} x_{4}) = (W_{0} : W_{1} : W_{2}) . □$

We show that if $ϕ, ψ \in Hom (E_{1}, E_{2})$ , then

\deg (ϕ + ψ) + \deg (ϕ - ψ) \leq 2 \deg (ϕ) + 2 \deg (ψ) .

We may assume $ϕ, ψ, ϕ + ψ, ϕ - ψ \neq 0$ (otherwise trivial, or use $\deg [- 1] = 1$ , $\deg [2] = 4$ ).

\begin{array}{l} ϕ & : (x, y) \mapsto (ξ_{1} (x), \dots) \\ ψ & : (x, y) \mapsto (ξ_{2} (x), \dots) \\ ϕ + ψ & : (x, y) \mapsto (ξ_{3} (x), \dots) \\ ϕ - ψ & : (x, y) \mapsto (ξ_{4} (x), \dots) \end{array}

Lemma 5.8 implies

(1 : ξ_{3} + ξ_{4} : ξ_{3} ξ_{4}) = ({(ξ_{1} - ξ_{2})}^{2} : \dots) .

Put $ξ_{i} = \frac{r_{i}}{s_{i}}$ , $r_{i}, s_{i} \in K [t]$ coprime.

\underset{coprime}{\underset{⏟}{(s_{3} s_{4} : r_{3} s_{4} + r_{4} s_{4} : r_{3} r_{4})}} = ({(r_{1} s_{2} - r_{2} s_{1})}^{2} : \dots) . (∗)

Therefore

\begin{array}{l} \deg (ϕ + ψ) + \deg (ϕ - ψ) & = \max (\deg (r_{3}), \deg (s_{3})) + \max (\deg (r_{4}), \deg (s_{4})) \\ = \max (\deg (s_{3} s_{4}), \deg (r_{3} s_{4} + r_{3} s_{3}), \deg (r_{3} r_{4})) \\ \leq 2 \max (\deg (r_{1}), \deg (s_{1})) + 2 \max (\deg (r_{2}), \deg (s_{2})) \\ = 2 \deg (ϕ) + 2 \deg (ψ) & (1) \end{array}

Now replace $ϕ, ψ$ by $ϕ + ψ, ϕ - ψ$ to get

\begin{array}{l} \deg (2 ϕ) + \deg (2 ψ) & \leq 2 \deg (ϕ + ψ) + 2 \deg (ϕ - ψ) \\ 4 \deg (ϕ) + 4 \deg (ψ) & \leq 2 \deg (ϕ + ψ) + 2 \deg (ϕ - ψ) \\ 2 \deg (ϕ) + 2 \deg (ψ) & \leq \deg (ϕ + ψ) + \deg (ϕ - ψ) & (2) \end{array}

(1) and (2) give that $\deg$ satisfies the parallelogram law, hence $\deg$ is a quadratic form.

This proves Theorem 5.7.

Corollary 5.9. $\deg (n ϕ) = n^{2} \deg (ϕ)$ for all $n \in ℤ$ , $ϕ \in Hom (E_{1}, E_{2})$ . In particular, $\deg [n] = n^{2}$ .

Example 5.10. Let $E ∕ K$ be an elliptic curve. Suppose $char K \neq 2$ . Let $0 \neq T \in E (K) [2]$ .

Without loss of generality $E$ : $y^{2} = x (x^{2} + a x + b)$ , $a, b \in K$ , $b (a^{2} - 4 b) \neq 0$ , and $T = (0, 0)$ .

If $P = (x, y)$ and $P^{'} = P + T = (x^{'}, y^{'})$ then

\begin{array}{l} x^{'} & = {(\frac{y}{x})}^{2} - a - x \\ = \frac{x^{2} + a x + b}{x} - a - x \\ = \frac{b}{x} \\ y^{'} & = (\frac{y}{x}) x^{'} \\ = - \frac{b y}{x} \end{array}

Let

\begin{array}{l} ξ & = x + x^{'} + a = {(\frac{y}{x})}^{2} \\ η & = y + y^{'} = \frac{y}{x} (x - \frac{b}{x}) \\ η^{2} & = {(\frac{y}{x})}^{2} [{(x + \frac{b}{x})}^{2} - 4 b] \\ = ξ [{(ξ - a)}^{2} - 4 b] \\ = ξ (ξ^{2} - 2 a ξ + a^{2} - 4 b) \end{array}

Let $E^{'}$ : $y^{2} = x (x^{2} + a^{'} x + b^{'})$ , where $a^{'} = - 2 a$ , $b^{'} = a^{2} - 4 b$ . There is an isogeny

\begin{array}{l} ϕ : E & \to E^{'} \\ (x, y) & \mapsto ({(\frac{y}{x})}^{2} : \frac{y (x^{2} - b)}{x^{2}} : 1) \\ - 2 - 3 0 {ord}_{0_{E}} (∙) \\ 1 0 3 + 3 \\ 0_{E} & \mapsto (0 : 1 : 0) = 0_{E^{'}} \end{array}

To compute the degree: ${(\frac{y}{x})}^{2} = \frac{x^{2} + a x + b}{x}$ (coprime numerator and denominator as $b \neq 0$ ). Lemma 5.4 gives $\deg ϕ = 2$ .

We say $ϕ$ is a $2$ -isogeny.

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