Formal Groups - Elliptic Curves - Cambridge III notes

Lemma 8.1 (Hensel’s Lemma). Assuming that:

$R$ is complete with respect to an ideal $I$
$F \in R [X]$ , $s \geq 1$
$a \in R$ satisfies $F (a) \equiv 0 (m o d I^{s})$ , $F^{'} (a) \in R^{\times}$

Then there exists a unique

b \in R

such that

F (b) = 0

and

b \equiv 0 (m o d I^{s})

Proof. Let $u \in R^{\times}$ with $F (a) = u (m o d I)$ (e.g. we could take $u = F^{'} (a)$ ). Replacing $F (X)$ by $\frac{F (X + a)}{u}$ , we may assume $a = 0$ , $F^{'} (0) \equiv 1 (m o d I)$ .

We put $x_{0} = 0$ ,

x_{n + 1} = x_{n} - F (x_{n}) (1)

Easy induction gives

x_{n} \equiv 0 (m o d I^{s}) \forall n (2)

Let

F (X) - F (Y) = (X - Y) (F^{'} (0) + X G (X, Y) + Y H (X, Y)) (3)

for some $G, H \in R [X, Y]$ .

Claim: $x_{n + 1} \equiv x_{n} (m o d I^{n + s})$ for all $n \geq 0$ .

Proof: By induction on $n$ . Case $n = 0$ is true.

Suppose $x_{n} \equiv x_{n = 1} (m o d I^{n + s - 1})$ . Then

F (x_{n}) - F (x_{n - 1}) = (x_{n} - x_{n - 1}) (1 + c) (m o d I^{n + s})

for some $c \in I$ . Hence

F (x_{n}) - F (x_{n - 1}) = x_{n} - x_{n - 1} (m o d I^{n + s})

Hence

x_{n} - F (x_{n}) \equiv x_{n - 1} - F (x_{n - 1}) (m o d I^{n + s})

and hence $x_{n + 1} \equiv x_{n} (m o d I^{n + s})$ .

This proves the claim.

Therefore ${(x_{n})}_{n \geq 0}$ is Cauchy. Since $R$ is complete, we have $x_{n} \to b$ as $n \to \infty$ for some $b \in R$ .

Taking limit $n \to \infty$ in (1) gives $b = b - F (b)$ , hence $F (b) = 0$ .

Taking limit $n \to \infty$ in (2) gives $b \equiv 0 (m o d I^{s})$ .

Uniqueness is proved using (3) and the “useful remark” (if $x \in I$ then $1 - x \in R^{\times}$ ). □

Hence there exists a unique

w (t) \in R = ℤ [a_{1}, \dots, a_{6}] [[t]]

such that

Remark. Taking $u = 1$ in the proof of Lemma 8.1, $w (t) = \lim_{n \to \infty} w_{n} (t)$ where $w_{0} (t) = 0$ , $w_{n + 1} (t) = f (t, w_{n} (t))$ .

In fact,

w (t) = t^{3} (1 + A_{1} t + A_{2} t^{2} + \dots)

where $A_{1} = a_{1}$ , $A_{2} = a_{1}^{2} + a_{2}$ , $A_{3} = a_{1}^{3} + 2 a_{1} a_{2} + 2 a_{3}$ , …

Lemma 8.2. Assuming that:

$R$ an integral domain which is complete with respect to an ideal $I$
$a_{1}, \dots, a_{6} \in R$
$K = Frac (R)$

Then

Ê (I) : = {(t, w) \in E (K) | t, w \in I}

is a subgroup of $E (K)$ .

Proof. Taking $(t, w) = (0, 0)$ show $0_{E} \in Ê (I)$ .

So it suffices to show that $P_{1}, P_{2} \in Ê (I)$ then

\underset{= P_{3}}{\underset{⏟}{- P_{1} - P_{2}}} \in Ê (I) .

$P_{1}, P_{2} \in Ê (I)$ implies $t_{1}, t_{2} \in I$ , $w_{1} = w (t_{1}) \in I$ , $w_{2} = w (t_{2}) \in I$ .

\begin{array}{l} w (t) & = \sum_{n = 2}^{\infty} A_{n - 2} t^{n + 1} & (A_{0} = 1) \\ λ & = {\begin{matrix} \frac{w (t_{2}) - w (t_{1})}{t_{2} - t_{1}} & t_{1} \neq t_{2} \\ w^{'} (t_{1}) & t_{1} = t_{2} \end{matrix} \\ = \sum_{n = 2}^{\infty} A_{n - 2} (t_{1}^{n} + t_{1}^{n - 1} t_{2} + \dots + t_{2}^{n}) \in I \\ ν & = w_{1} - λ t_{1} \in I \end{array}

Substituting $w = λ t + ν$ into $w = f (t, w)$ gives

\begin{array}{l} λ t + ν & = t^{3} + a_{1} t (λ t + ν) + a_{2} t^{2} (λ t + ν) + a_{3} {(λ t + ν)}^{2} + a_{4} t {(λ t + ν)}^{2} + a_{6} {(λ t + ν)}^{3} \\ A & = (coefficient of t^{3}) = 1 + a_{2} λ + a_{4} λ^{2} + a_{6} λ^{3} \\ B & = (coefficient of t^{2}) = a_{1} λ + a_{2} ν + a_{3} λ^{2} + 2 a_{4} λ ν + 3 a_{6} λ^{2} ν \end{array}

We have $A \in R^{\times}$ and $B \in I$ . Hence $t_{3} = - \frac{B}{A} - t_{1} - t_{2} \in I$ , $w_{3} = λ t_{3} + ν \in I$ . □

Taking

R = ℤ [a_{1}, \dots, a_{6}] [[t]]

I = (t)

, then Lemma 8.2 gives that there exists

ι \in ℤ [a_{1}, \dots, a_{6}] [[t]]

with

ι (0) = 0

and

Taking

R = ℤ [a_{1}, \dots, a_{6}] [[t_{1}, t_{2}]]

I = (t_{1}, t_{2})

, Lemma 8.2 gives that there exists

F \in ℤ [a_{1}, \dots, a_{6}] [[t_{1}, t_{2}]]

with

F (0, 0) = 0

and

Definition (Formal group). Let $R$ be a ring. A formal group over $R$ is a power series $F (X, Y) \in R [[X, Y]]$ satisfying

(i)
$F (X, Y) = F (Y, X)$
(ii)
$F (X, 0) = X$ and $F (0, Y) = Y$
(iii)
$F (X, F (Y, Z)) = F (F (X, Y), Z)$

This looks like it would only define a monoid, but in fact inverses are guaranteed to exist in this context.

Definition (Morphism / isomorphic (formal groups)). Let $F$ and $G$ be formal groups over $R$ given by power series $F$ and $G$ .

(i)
A morphism $f : F \to G$ is a power series $f \in R [[T]]$ such that $f (0) = 0$ satisfying
$f (F (X, Y)) = G (f (x), f (Y)) .$
(ii)
$F ≅ G$ if there exists $F \overset{f}{\to} G$ and $G \overset{g}{\to} F$ morphisms such that $f (g (X)) = g (f (X)) = X$ .

Theorem 8.3 (All formal groups are isomorphic). Assuming that:

$char R = 0$

Then any formal group

F

over

R

is isomorphic to

{\hat{𝔾}}_{a}

over

R \otimes ℚ

. More precisely

(i)
There is a unique power series
$\log (T) = T + \frac{a_{2}}{2} T^{2} + \frac{a_{3}}{3} T^{3} + \dots$

with $a_{i} \in R$ such that
$\log (F (X, Y)) = \log (X) + \log (Y) . (∗)$

(ii)

There is a unique power series

\exp (T) = T + \frac{b_{2}}{2!} + \frac{b_{3}}{3!} T^{3} + \dots

with $b_{i} \in R$ such that

\exp (\log (T)) = \log (\exp (T)) = T .

Proof.

(i) Notation $F_{1} (X, Y) = \frac{\partial F}{\partial x} (X, Y)$ .
Uniqueness: Let
$p (T) = \frac{d}{d T} \log (T) = 1 + a - 2 T + a_{3} T^{2} + \dots .$

Differentiating ( $*$ ) with respect to $X$ gives
$p (F (X, Y)) F_{1} (X, Y) = p (X) + 0 .$

Putting $X = 0$ gives $p (Y) F_{1} (0, Y) = 1$ , hence $p (Y) = F_{1} {(0, Y)}^{- 1}$ , so $p$ is uniquely determined by $F$ and hence $\log$ is too.

Existence: Let
$p (T) = F_{1} {(0, T)}^{- 1} = 1 = a_{2} T + a_{3} T^{2} + \dots$

(say). Let
$\log (T) = T + \frac{a_{2}}{2} T^{2} + \frac{a_{3}}{3} T^{3} + \dots .$

Calculate
$\begin{array}{l} F (F (X, Y), Z) & = F (X, F (Y, Z)) \\ F_{1} (F (X, Y), Z) F_{1} (X, Y) & = F_{1} (X, F (Y, Z)) & \frac{d}{d x} \\ F_{1} (Y, Z) F_{1} (0, Y) & = F_{1} (0, F (Y, Z)) & put X = 0 \\ ⟹ F_{1} (Y, Z) p {(Y)}^{- 1} & = p {(F (Y, Z))}^{- 1} \\ ⟹ F_{1} (Y, Z) p (F (Y, Z)) & = p (Y) \\ \log (F (Y, Z)) & = \log (Y) + h (Z) & integrate w.r.t. Y \end{array}$
for some power series $h$ .

Symmetry $Y \leftrightarrow Z$ gives $h (Z) = \log (Z)$ .

This proves existence.
(ii) We prove a lemma first. □

Proof. We construct polynomials $g_{n} (T) \in R [T]$ such that

\begin{array}{l} f (g_{n} (T)) & \equiv T (m o d T^{n + 1}) \\ g_{n + 1} (T) & \equiv g_{n} (T) (m o d T^{n + 1}) \end{array}

Then $g (T) = \lim_{n \to \infty} g_{n} (T)$ satisfies $f (g (T)) = T$ .

To start the induction, we set $g_{1} (T) = a^{- 1} T$ . Now suppose $n \geq 2$ and $g_{n - 1} (T)$ exists. Then

f (g_{n - 1} (T)) \equiv T + b T^{n} (m o d T^{n + 1}) .

We put $g_{n} (T) = g_{n - 1} (T) + λ T^{n}$ for some $λ \in R$ to be chosen later.

Then

\begin{array}{l} f (g_{n} (T)) & = f (g_{n - 1} (T) + λ T^{n}) \\ \equiv f (g_{n - 1} (T)) + λ a T^{n} (m o d T^{n + 1}) \\ \equiv T + (b + λ a) T^{n} (m o d T^{n + 1}) \end{array}

We take $λ = - \frac{b}{a}$ ( $a \in R^{\times}, b \in R ⟹ λ \in R$ ).

This completes the induction step.

We get $g (T) = a^{- 1} T + \dots \in R [[T]]$ such that $f (g (T)) = T$ . Applying the same construction to $g$ gives $h (T) = a T + \dots \in R [[T]]$ such that $g (h (T)) = T$ .

Now note $f (T) = f (g (h (T))) = h (T)$ . □

Notation. Let $F$ (e.g. ${\hat{𝔾}}_{a}$ , ${\hat{𝔾}}_{m}$ , $Ê$ ) be a formal group given by a power series $F \in R [[X, Y]]$ .

Suppose $R$ is a ring complete with respect to ideal $I$ . For $x, y \in I$ , put

x \oplus_{F} y = F (x, y) \in I .

Then $F (I) : = (I, \oplus_{F})$ is an abelian group.

Examples:

${\hat{𝔾}}_{a} (I) = (I, +)$
${\hat{𝔾}}_{m} (I) = (1 + I, \times)$
$Ê (I) = subgroup of E (K) in Lemma 8.2$

Corollary 8.5. Assuming that:

$F$ a formal group over $R$
$n \in ℤ$ such that $n \in R^{\times}$

Then

(i) $[n] : F \to F$ is an isomorphism of formal groups
(ii) If $R$ is complete with respect to ideal $I$ then $F (I) \overset{\times n}{\to} F (I)$ is an isomorphism of groups. In particular, $F (I)$ has no $n$ -torsion.

Proof. We have

\begin{array}{l} [1] (T) & = T \\ [n] (T) & = F ([n - 1] (T), T) \end{array}

(for $n < 0$ use $[- 1] (T) = ι (T)$ ).

Since

F (X, Y) = X + Y + X Y (\dots)

we get

[2] (T) = F (T, T) 2 T + \dots \in R [[T]]

and by induction we get

[n] (T) = n T + \dots \in R [[T]] .

Lemma 8.4 shows that if $n \in R^{\times}$ then $[n]$ is an isomorphism. This proves (i), and (ii) follows. □

8 Formal Groups