Elliptic Curves over Number Fields: The torsion subgroup - Elliptic Curves

10 Elliptic Curves over Number Fields: The torsion subgroup

$[K : ℚ] < \infty$ , $E ∕ K$ an elliptic curve.

Notation. $𝔭$ a prime of $K$ (i.e. a prime) ideal in $O_{K}$ ).

$K_{𝔭}$ is the $𝔭$ -adic completion of $K$ , valuation ring $O_{𝔭}$ .

$k_{𝔭} = O_{K} ∕ 𝔭$ residue field.

Definition (Good reduction (prime)). $𝔭$ is a prime of good reduction for $E ∕ K$ if $E ∕ K_{𝔭}$ has good reduction.

Lemma 10.1. $E ∕ K$ has only finitely many primes of bad reduction.

Proof. Take a Weierstrass equation for $E$ with $a_{1}, \dots, a_{6} \in O_{K}$ . $E$ non-singular implies that $0 \neq Δ \in O_{K}$ .

Write $(Δ) = 𝔭^{α_{1}} \dots 𝔭_{r}^{α_{r}}$ (factorisation into prime ideals).

Let $S = {𝔭_{1}, \dots, 𝔭_{r}}$ . If $𝔭 \notin S$ then $v_{𝔭} (Δ) = 0$ . Hence $E ∕ K_{𝔭}$ has good reduction.

Therefore ${bad primes for E} \subset S$ , hence is finite. □

Remark. If $K$ has class number $1$ (e.g. $K = ℚ$ ) then we can always find a Weierstrass equation for $E$ with $a_{1}, \dots, a_{6} \in O_{K}$ which is minimal at all primes $𝔭$ .

Basic group theory: If $A$ is a finitely generatead abelian group then $A ≅ {finite group} \times ℤ^{r}$ . We call $r$ the “rank”, and ${finite subgroup}$ is the torsion subgroup.

Lemma 10.2. $E {(K)}_{tors}$ is finite.

Proof. Take any prime $𝔭$ . We saw that $E (K_{𝔭})$ has a finite index subgroup $A$ (say) with $A ≅ (O_{𝔭}, +)$ .

In particular, $A$ is torsion free

E {(K)}_{tors} ↪ E {(K_{𝔭})}_{tors} ↪ \underset{finite}{\underset{⏟}{\frac{E (K_{𝔭})}{A}}} . □

Lemma 10.3. Assuming that:

$𝔭$ is a prime of good reduction
$𝔭 ∤ n$

Then reduction modulo

𝔭

gives an injective group homomorphism

E (K) [n] ↪ Ẽ (k_{𝔭}) .

Proof. Proposition 9.5 gives that $E (K_{𝔭}) \to Ẽ (k_{𝔭})$ is a group homomorphism, with kernel $E_{1} (K_{𝔭})$ .

Corollary 8.5 and $𝔭 ∤ n$ gives that $E_{1} (K_{𝔭})$ has no $n$ -torsion. □

Example. $E ∕ ℚ$ : $y^{2} + y = x^{3} - x$ , $Δ = - 11$ .

$E$ has good reduction at all $p \neq 11$ .

$p$	$2$	$3$	$5$	$7$	$11$	$13$

$# Ẽ (𝔽_{p})$	$5$	$5$	$5$	$5$	$-$	$10$

Lemma 10.3 gives:

{\begin{matrix} # {(ℚ)}_{tors} | 5 \cdot 2^{a} & for some a \geq 0 \\ # E {(ℚ)}_{tors} | 5 \cdot 3^{b} & for some b \geq 0 \end{matrix}

Hence $# {(ℚ)}_{tors} | 5$ .

Let $T = (0, 0) \in E (ℚ)$ . Calculation gives $5 T = 0$ . Therefore $E {(ℚ)}_{tors} ≅ ℤ ∕ 5 ℤ$ .

Example. $E ∕ ℚ$ : $y^{2} + y = x^{3} + x^{2}$ , $Δ = - 43$

$p$	$2$	$3$	$5$	$7$	$11$	$13$

$# Ẽ (𝔽_{p})$	$5$	$6$	$10$	$8$	$9$	$19$

Lemma 10.3 gives:

{\begin{matrix} # E {(ℚ)}_{tors} | 5 \cdot 2^{a} & for some a \geq 0 \\ # E {(ℚ)}_{tors} | 9 \cdot 1 1^{b} & for some b \geq 0 \end{matrix}

Therefore $E {(ℚ)}_{tors} = {0}$ .

Therefore $P = (0, 0)$ is a point of infinite order.

In particular, $E (ℚ)$ is infinite.

Example. $E_{D}$ : $y^{2} = x^{3} - D^{2} x = f (x)$ . $D \in ℤ$ square-free, $Δ = 2^{6} D^{6}$ . If $p ∤ 2 D$ , then

# Ẽ_{D} (𝔽_{p}) = 1 + \sum_{x \in 𝔽_{p}} ((\frac{f (x)}{p}) + 1) .

If $p \equiv 3 (m o d 4)$ then since $f$ is odd,

(\frac{f (- x)}{p}) = (\frac{- f (x)}{p}) = (\frac{- 1}{p}) (\frac{f (x)}{p}) = - (\frac{f (x)}{p}) .

Hence

# Ẽ_{D} (𝔽_{p}) = p + 1 .

E_{D} {(ℚ)}_{tors} \supset {0, (0, 0), (\pm D, 0)} ≅ {(ℤ ∕ 2 ℤ)}^{2} .

Let $m = # E_{D} {(ℚ)}_{tors}$ .

We have $4 | m | p + 1$ forr all sufficiently large ( $p ∤ 2 m D$ ) primes $p$ with $p \equiv 3 (m o d 4)$ . Hence $m = 4$ (otherwise get contradiction to Dirichlet’s theorem on primes on arithmetic progressions).

\begin{array}{l} rank E_{D} (ℚ) \geq 1 & ⟺ \exists x, y \in ℚ, y \neq 0 \land y^{2} = x^{3} - D^{2} x \\ \overset{Lecture 1}{⟺} D is a congruent number \end{array}

Lemma 10.4. Assuming that:

$E ∕ ℚ$ is given by a Weierstrass equation with $a_{1}, \dots, a_{6} \in ℤ$
$0 \neq T = (x, y) \in E {(ℚ)}_{tors}$

Then

(i) $4 x, 8 y \in ℤ$
(ii) If $2 | a_{1}$ or $2 T \neq 0$ then $x, y \in ℤ$

Proof.

(i) The Weierstrass equation defines a formal group $Ê$ over $ℤ$ . For $r \geq 1$ ,
$Ê (p^{r} ℤ_{p}) = {(x, y) \in E (ℚ_{p}) | v_{p} (x) \leq - 2 r, v_{p} (y) \leq - 3 r} \cup {0} .$

Then Theorem 9.2 gives $Ê (p^{r} ℤ_{p}) ≅ (ℤ_{p}, +)$ if $r > \frac{1}{p - 1}$ . Hence $Ê (4 ℤ_{2})$ and $Ê (p ℤ_{p})$ for $p$ are odd are torsion free.

So if $0 \neq T = (x, y) \in E {(ℚ)}_{tors}$ then
$\begin{array}{l} v_{2} (x) & \geq - 2 \\ v_{2} (y) & \geq - 3 \\ v_{p} (x) & \geq 0 \\ v_{p} (y) & \geq 0 \end{array}$
for all odd primes $p$ .
(ii) Suppose $T \in Ê (2 ℤ_{2})$ , i.e. $v_{2} (x) = - 2$ , $v_{2} (y) = - 3$ .
Since
$\frac{Ê (2 ℤ_{2})}{Ê (4 ℤ_{2})} ≅ (𝔽_{2}, +)$

and $Ê (4 ℤ_{2})$ is torsion free, we get $2 T = 0$ .

Also $(x, y) = T = - T = (x, - y - a_{1} x - a_{3})$ . So
$\begin{array}{l} 2 y + a_{1} x + a_{3} & = 0 \\ \underset{odd}{\underset{⏟}{8 y}} + a_{1} \underset{odd}{\underset{⏟}{(4 x)}} + \underset{even}{\underset{⏟}{4 a_{3}}} & = 0 \end{array}$
Hence $a_{1}$ is odd.

So if $2 T \neq 0$ or $a_{1}$ is even then $T \notin Ê (2 ℤ_{2})$ , so $x, y \in ℤ$ . □

Example. $y^{2} + x y = x^{3} + 4 x + 1$ , $(- \frac{1}{4}, \frac{1}{8}) \in E (ℚ) [2]$ .

Theorem 10.5 (Lutz Nagell). Assuming that:

$E ∕ ℚ$ : $y^{2} = x^{3} + a x + b$ , $a, b \in ℤ$
$0 \neq T = (x, y) \in E {(ℚ)}_{tors}$

Then

x, y \in ℤ

and either

y = 0

y^{2} | (4 a^{3} + 27 b^{2})

Proof. Lemma 10.4 gives that $x, y \in ℤ$ .

If $2 T = 0$ then $y = 0$ . Otherwise $0 \neq 2 T = (x_{2}, y_{2}) \in E {(ℚ)}_{tors}$ .

Lemma 10.4 gives $x_{2}, y_{2} \in ℤ$ . But

x_{2} = {(\frac{f^{'} (x)}{2 y})}^{2} - 2 x

hence $y | f^{'} (x)$ .

$E$ non-singular gives that $f (X)$ adn $f^{'} (X)$ are coprime, so $f (X)$ and $f^{'} {(X)}^{2}$ are coprime. Therefore there exists $g, h \in ℚ [X]$ satisfying

g (X) f (X) + h (X) f^{'} {(X)}^{2} = 1 .

Doing this and clearing denominators gives

(3 X^{2} + 4 a) f^{'} {(X)}^{2} - 27 (X^{3} + a X - b) f (X) = 4 a^{3} + 27 b^{2} .

Since $y | f^{'} (x)$ and $y^{2} = f (x)$ , we get $y^{2} | (4 a^{3} + 27 b^{2})$ . □

Remark. Mazur showed that if $E ∕ ℚ$ is an elliptic curve then

E {(ℚ)}_{tors} ≅ {\begin{matrix} ℤ ∕ n ℤ & 1 \leq n \leq 12, n \neq 11 \\ ℤ ∕ 2 ℤ \times ℤ ∕ 2 n ℤ & 1 \leq n \leq 4 \end{matrix}

Moreover, all 15 possibilities occur.