Descent by Cyclic Isogeny - Elliptic Curves

Let

E, E^{'}

be elliptic curves over a number field

K

, and

ϕ : E \to E^{'}

an isogeny of degree

n

Suppose

E^{'} [\hat{ϕ}] ≅ ℤ ∕ n ℤ

generated by

T \in E^{'} (K)

. Then

E [ϕ] ≅ μ_{n}

as a

Gal (\bar{K} ∕ K)

-module

Theorem 16.1. Assuming that:

$f \in K (E^{'})$ and $g \in K (E)$
$div (f) = n (T) - n (0)$
$ϕ^{*} f = g^{n}$

Then

α (P) = f (P) (m o d {(K^{*})}^{n})

for all

P \in E; (K) ∖ {0, T}

Proof. Let $Q \in ϕ^{- 1} P$ . Then $δ (P) \in H^{1} (K, μ_{n})$ is represented by $σ \mapsto σ Q - Q \in E [ϕ] ≅ μ_{n}$ .

\begin{array}{l} e_{} (σ Q - Q, T) & = \frac{g (σ Q - Q + X)}{g (X)} & \begin{array}{c} for any X \in E \\ avoiding zeroes and poles of g \end{array} \\ = \frac{g (σ Q)}{g (Q)} & pick X = Q \\ = \frac{σ (g (Q))}{g (Q)} \\ = \frac{σ \sqrt[n]{f (P)}}{\sqrt[n]{f} (P)} \end{array}

( $ϕ^{*} f = g^{n}$ , $f (P) = g {(Q)}^{n}$ ).

But

\begin{array}{l} H^{1} (K, μ_{n}) & ≅ K^{*} ∕ {(K^{*})}^{n} \\ (σ \mapsto \frac{\sqrt[n]{x}}{\sqrt[n]{x}}) & \leftarrow ↤ x \end{array}

Hence $α (P) = f (P) (m o d {(K^{*})}^{n})$ . □

16.1 Descent by 2-isogeny

Proof. If $A \overset{f}{\to} B \overset{g}{\to} C$ are homomorphisms of abelian groups, then there is an exact sequence

0 ker(f) ker(gf ) ker(g)

fg coker(f) coker(gf) coker(g) 0

Since

\hat{ϕ} ϕ = {[2]}_{E}

, we get an exact sequence

0 \to \underset{≅ ℤ ∕ 2 ℤ}{\underset{⏟}{E (K) [ϕ]}} \to E (K) [2] \overset{ϕ}{\to} \underset{≅ ℤ ∕ 2 ℤ}{\underset{⏟}{E^{'} (K) [ϕ^{'}]}} \to \underset{≅ Im (α_{E^{'}})}{\underset{⏟}{\frac{E^{'} (K)}{ϕ E (K)}}} \overset{\hat{ϕ}}{\to} \frac{E (K)}{2 E (K)} \to \underset{≅ Im (α_{E})}{\underset{⏟}{\frac{E (K)}{\hat{ϕ} E^{'} (K)}}} \to 0 .

Therefore

\frac{| E (L) ∕ 2 E (K) |}{| E (K) [2] |} = \frac{| Im α_{E} | \cdot | Im α_{E^{'}} |}{4} . (1)

Mordell-Weil: $E (K) ≅ Δ \times ℤ^{r}$ , where $Δ$ is a finite group and $r = rank E (K)$ .

$\frac{E (K)}{2 E (K)} ≅ \frac{Δ}{2 Δ} \times {(\frac{ℤ}{2 ℤ})}^{r}$ .

$E (K) [2] ≅ Δ [2]$ .

Since $Δ$ is finite, we have that $\frac{Δ}{2 Δ}$ and $Δ [2]$ have the same order, and therefore

\frac{| E (K) ∕ 2 E (K) |}{| E (K) [2] |} = 2^{r} . (2)

So we are done, by using (1) and (2). □

Proof. We must show that $x, y \in K$ , $y^{2} = x (x^{2} + a x + b)$ and $v_{𝔭} (b) = 0$ then $v_{𝔭} (x) \equiv 0 (m o d 2)$ .

Case $v_{𝔭} (x) < 0$ : Lemma 9.1 gives that for some $r \geq 1$ , $v_{𝔭} (x) = - 2 r$ and $v_{𝔭} (y) = - 3 r$ , so done.

Case $v_{𝔭} (x) > 0$ : Then $v_{𝔭} (x^{2} + a x + b) = 0$ . So $v_{𝔭} (x) = v_{𝔭} (y^{2}) = 2 v_{𝔭} (y)$ . □

Lemma 16.5. Assuming that:

$b_{1} b_{2} = b$

Then

b_{1} (K^{*}) \in Im (α_{E})

if and only if

w^{2} = b_{1} u^{4} + a u^{2} v^{2} + b_{2} v^{4} (∗)

is soluble for $u, v, w \in K$ not all zero.

Proof. If $b_{1} \in {(K^{*})}^{2}$ or $b_{2} \in {(K^{*})}^{2}$ then both conditions are satisfied. So we may assume $b_{1}, b_{2} \notin {(K^{*})}^{2}$ .

Now note $b_{1} {(K^{*})}^{2} \in Im (α_{E})$ if and only if there exists $(x, y) \in E (K)$ such that $x = b_{1} t^{2}$ for some $t \in K^{*}$ . This implies

y^{2} = (b_{1} t^{2}) (b_{1}^{2} t^{2} + a b_{1} t^{2} + b)

hence

{(\frac{y}{b_{1} t})}^{2} = b_{1} t^{4} + a t^{2} + \underset{= b_{2}}{\underset{⏟}{\frac{b}{b_{1}}}} .

So ( $*$ ) has solution $(u, v, w) = (t, 1, \frac{y}{b_{1} t})$ .

Conversely, if $(u, v, w)$ is a solution to ( $*$ ) then $u v \neq 0$ and $(b_{1} {(\frac{u}{v})}^{2}, b_{1} \frac{u w}{v^{3}}) \in E (K)$ . □

Example. $E$ : $y^{2} = x^{3} - x$ ( $a = 0$ , $b = - 1$ ).

$Im (α_{E}) = ⟨ - 1 ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2}$ .

$E^{'}$ : $y^{2} = x^{3} + 4 x$ .

$Im (α_{E^{'}}) \subset ⟨ - 1, 2 ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2}$ .

\begin{array}{l} b_{1} & = - 1 & w^{2} & = - u^{4} - 4 v^{4} \\ b_{1} & = 2 & w^{2} & = 2 u^{4} + 2 v^{4} \\ b_{1} & = - 2 & w^{2} & = - 2 u^{4} - 2 v^{4} \end{array}

The first and last lines are insoluble over

ℝ

(squares are non-negative). The middle line does have a solution:

(u, v, w) = (1, 1, 2)

Therefore $Im (α_{E^{'}}) = ⟨ 2 ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2}$ .

Hence $2^{rank E (ℚ)} = \frac{2 \cdot 2}{4} = 1$ , so $rank E (ℚ) = 0$ .

So $1$ is not a congruent number.

Example. $E$ : $y^{2} = x^{3} + p x$ , $p$ a prime which is 5 modulo 8.

\begin{array}{l} b_{1} & = - 1 & w^{2} & = - u^{4} - p v^{4} \end{array}

This is insoluble over $ℝ$ , hence $Im (α_{E}) = ⟨ p ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2}$ .

$E^{'}$ : $y^{2} = x^{3} - 4 p x$ .

$Im (α_{E^{'}}) \subset ⟨ - 1, 2, p ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2}$ .

Note: $α_{E^{'}} (T^{'}) = (- 4 p) {(ℚ^{*})}^{2} = (- p) {(ℚ^{*})}^{2}$ .

\begin{array}{l} b_{1} & = 2 & w^{2} & = 2 u^{4} - 2 p v^{4} \\ b_{1} & = 2 & w^{2} & = - 2 u^{4} + 2 p v^{4} \\ b_{1} & = p & w^{2} & = p u^{4} - 4 v^{4} \end{array}

Suppose the first line is soluble. Then without loss of generality $u, v, w \in ℤ$ with $\gcd (u, v) = 1$ . If $p | u$ , then $p | w$ and then $p | v$ , contradiction. Therefore $w^{2} \equiv 2 u^{4} ⁄ \equiv 0 (m o d p)$ . So $(\frac{2}{p}) = + 1$ , which contradicts $p \equiv 5 (m o d 8)$ .

Likewise the second line has no solution since $(\frac{- 2}{p}) = - 1$ .

TODO

Example (Lindemann). $E$ : $y^{2} = x^{3} + 17 x$ .

Im (α_{E}) = ⟨ 17 ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2} .

$E^{'}$ : $y^{2} = x^{3} - 68 x$ .

Im (α_{E^{'}}) \subset ⟨ - 1, 2, 17 ⟩ \subset ℚ^{*} ∕ {(ℚ^{*})}^{2} .

$b_{1} = 2$ , $w^{2} = 2 u^{4} - 34 v^{4}$ .

Replacing $w$ by $2 w$ and dividing by $2$ gives

C : 2 w^{2} = u^{4} - 17 v^{4} .

Notation. $C (K) = {(u, v, w) \in K^{3} ∖ {0} satisfying equation above} ∕ \sim$ , where $(u, v, w) \sim (λ u, λ v, λ^{2} w)$ for all $λ \in K^{*}$ .

$C (ℚ_{2}) \neq \emptyset$ since $17 \in {(ℤ_{2}^{*})}^{4}$ .

$C (ℚ_{17}) \neq \emptyset$ since $2 \in {(ℤ_{17}^{*})}^{2}$ .

$C (ℝ) \neq \emptyset$ since $\sqrt{2} \in ℝ$ .

Therefore $C (ℚ_{v}) \neq \emptyset$ for all places $v$ of $ℚ$ .

Suppose $(u, v, w) \in C (ℚ)$ . Without loss of generality say $u, v, w \in ℤ$ , $\gcd (u,, v) = 1$ and $w > 0$ .

If $17 | w$ then $17 | u$ and then $17 | v$ , contradiction

So if $p | w$ then $p \neq 17$ and $(\frac{17}{p}) = + 1$ . Therefore $(\frac{p}{17}) = (\frac{17}{p}) = 1$ (using quadratic reciprocity). (If $p$ is odd, also $(\frac{2}{17}) = + 1$ ).

Therefore $(\frac{w}{17}) = + 1$ . But $2 w^{2} \equiv u^{4} (m o d 17)$ , so $2 \in {(𝔽_{17}^{*})}^{4} = {\pm 1, \pm 4}$ , contradiction.

So $C (ℚ) = \emptyset$ , i.e. $C$ is a counterexample to the Hasse Principle. It represents a non-trivial element of $T S (E ∕ ℚ)$ .

16.2 Birch Swinnerton Dyer Conjecture

Definition ( $L (E, s)$ ). Define

L (E, s) = \prod_{p} L_{p} (E, s)

where

L_{p} (E, s) = {\begin{matrix} {(1 - a_{p} p^{- s} + p^{1 - 2 s})}^{- 1} & if good reduction \\ {(1 - p^{- s})}^{- 1} & if split multiplicative reduction \\ {(1 + p^{- s})}^{- 1} & if nonsplit multiplicative reduction \\ 1 & if additive reduction \end{matrix}

where $# Ẽ (𝔽_{p}) = 1 + p - a_{p}$ .

Hasse’s Theorem implies

| a_{p} | \leq 2 \sqrt{p}

, which shows that

L (E, s)

converges for

Re (s) > \frac{3}{2}

Theorem (Wiles, Breuil, Conrad, Diamon, Taylor). $L (E, s)$ is the $L$ -function of a weight $2$ modular form, and hence has an analytic continuation to all of $ℂ$ (and a function equation relating $L (E, s) = L (E, 2 - s)$ ).

Conjecture (Weak Birch Swinnerton-Dyer Conjecture). ${ord}_{s = 1} L (E, s) = rank E (ℚ)$ .

Conjecture (Strong Birch Swinnerton-Dyer Conjecture). ${ord}_{s = 1} L (E, s) = rank E (ℚ)$ , which we shall call $r$ , and

\lim_{s \to 1} \frac{1}{{(s - 1)}^{r}} L (E, s) = \frac{Ω_{E} Reg E (ℚ) | T S (E ∕ ℚ) | \prod_{p} c_{p} (E)}{| E {(ℚ)}_{tors} |^{2}},

where

$c_{p}$ is given by:
$\begin{array}{l} c_{p} (E) & = Tamagawa number of E ∕ ℚ_{p} \\ = [E (ℚ_{p}) : E_{0} (ℚ_{p})] \end{array}$

If $E (ℚ) ∕ E {(ℚ)}_{tors} = ⟨ P_{1}, P_{2}, \dots, P_{r} ⟩$ , then $Reg E (ℚ) = \det {([P_{i}, P_{j}])}_{i, j = 1, \dots, r}$ , where

[P, Q] = ĥ (P + Q) - ĥ (P) - ĥ (Q) .

$Ω_{E}$ is given by:
$Ω_{E} = \int_{E (ℝ)} | \frac{d x}{2 y + a_{1} x + a_{3}} |,$

where $a_{1}, \dots, a_{6} \in ℤ$ are coefficients of a globally minimal Weierstrass equation for $E ∕ ℚ$ .

16 Descent by Cyclic Isogeny

16.1 Descent by 2-isogeny

16.2 Birch Swinnerton Dyer Conjecture