The Invariant Differential - Elliptic Curves

Definition (Space of differentials). The space of differentials $Ω_{C}$ is the $K (C)$ -vector space generated by $d f$ for $f \in K (C)$ , subject to the relations

(i) $d (f + g) = d f + d g$
(ii) $d (f g) = f d g + g d f$
(iii) $d a = 0 \forall a \in K$

Let

0 \neq ω \in Ω_{C}

. Let

P \in C

be a smooth point, and

t \in K (C)

a uniformiser at

P

Then

ω = f d t

for some

f \in K {(C)}^{*}

. We define

{ord}_{p} (ω) = {ord}_{p} (f)

(independent of choice of

t

Fact: Suppose

f \in K {(C)}^{*}

{ord}_{p} (f) = n \neq 0

. If

char K ∤ n

, then

{ord}_{p} (d f) = n - 1

Lemma 6.1. Assuming that:

$char K \neq 2$
$E$ : $y^{2} = (x - e_{1}) (x - e_{2}) (x - e_{3})$ , $e_{1}, e_{2}, e_{3} \in K$ distinct

Then

ω = \frac{d x}{y}

is a differential on

E

with no zeroes or poles

⟹ g (E) = 1

. In particular, the

K

-vector space of regular differentials on

E

is 1-dimensional, spanned by

ω

Proof. Let $T_{i} = (e_{i}, 0)$ , $E [2] = {0, T_{1}, T_{2}, T_{3}}$ .

div (y) = (T_{1}) + (T_{2}) + (T_{3}) - 3 (0) . (1)

For $P \in E ∖ {0}$ ,

div (x - x_{p}) = (P) + (- P) - 2 (0) .

If $P \in E ∖ E [2]$ , then ${ord}_{p} (x - x_{p}) = 1 ⟹ {ord}_{p} (d x) = 0$ .

If $P \in E ∖ E [2]$ then ${ord}_{p} (x - x_{p}) = 1$ $⟹ {ord}_{p} (d x) = 0$ .

If $P = T_{i}$ then ${ord}_{p} (x - x_{p}) = 2$ $⟹ {ord}_{p} (d x) = 1$ .

If $P = 0$ then ${ord}_{p} (x) = - 2$ $⟹ {ord}_{p} (d x) = - 3$ .

Therefore

div (d x) = (T_{1}) + (T_{2}) + (T_{3}) - 3 (0) . (2)

(1) and (2) implies $div (\frac{d x}{y}) = 0$ . □

Lemma 6.2. Assuming that:

$P \in E$ ,
$\begin{array}{l} τ_{P} : E & \to E \\ X & \mapsto P + X \end{array}$
$ω = \frac{d x}{y}$ as above

Then

τ_{P}^{*} ω = ω

(we say

ω

is the invariant differential).

Proof. $τ_{P}^{*} ω$ is a regular differential on $E$ . So $τ_{P}^{*} ω = λ_{p} ω$ .

The map $E \to ℙ^{1}$ , $P \mapsto λ_{P}$ is a morphism of smooth projective curves but not surjective (misses $0$ and $\infty$ ). Therefore it is constant (by Theorem 2.8), i.e. there exists $λ \in K^{*}$ such that $τ_{P}^{*} ω = λ ω$ for all $P \in E$ .

Taking $P = 0_{E}$ shows $λ = 1$ . □

Remark. If $K = ℂ$ , $ℂ ∕ Λ \tilde{\to} E (ℂ)$ , $z \mapsto (℘ (z), ℘^{'} (z))$ ,

\frac{d x}{y} = \frac{℘^{'} (z) d z}{℘^{'} (z)} = d z .

(invariant under $z \mapsto z + (const)$ ).

Lemma 6.3. Assuming that:

$ϕ, ψ \in Hom (E_{1}, E_{2})$
$ω$ an invariant differential on $E_{2}$

Then

{(ϕ + ψ)}^{*} ω = ϕ^{*} ω + ψ^{*} ω

Proof. Write $E = E_{2}$

\begin{array}{l} E \times E & \to E \\ μ : (P, Q) & \mapsto P + Q \\ {p r}_{1} : (P, Q) & \mapsto P \\ {p r}_{2} : (P, Q) & \mapsto Q \end{array}

Fact: $Ω_{E \times E}$ is a 2-dimensional $K (E \times E)$ -vector space with basis ${p r}_{1}^{*} ω$ and ${p r}_{2}^{*} ω$ .

Therefore

μ^{*} ω = f {p r}_{1}^{*} ω + g {p r}_{2}^{*} ω (1)

for some $f, g \in K (E \times E)$ .

For fixed $Q \in E$ , let

\begin{array}{l} ι_{Q} : E & \to E \times E \\ P & \mapsto (P, Q) \end{array}

Applying $ι_{Q}^{*}$ to (1) gives

\begin{array}{l} {(μ ι_{Q})}^{*} ω & = (ι_{Q}^{*} f) {({p r}_{1} ι_{Q})}^{*} ω + (ι_{Q}^{*} g) {({p r}_{2} ι_{Q})}^{*} ω \\ \underset{= ω}{\underset{⏟}{τ_{Q}^{*} ω}} & = \underset{by Lemma 6.2}{\underset{⏟}{(ι_{Q}^{*} f)}} ω + 0 \end{array}

Therefore $ι_{Q}^{*} f = 1$ for all $Q \in E$ , so $f (P, Q) = 1$ for all $P, Q \in E$ .

Similarly $g (P, Q) = 1$ for all $P, Q \in E$ .

(1) ⟹ μ^{*} ω = {p r}_{1}^{*} ω + {p r}_{2}^{*} ω .

Now pull back by

\begin{array}{l} E_{1} & \to E \times E \\ P & \mapsto (ϕ (P), ψ (P)) \end{array}

to get

{(ϕ + ψ)}^{*} ω = ϕ^{*} ω + ψ^{*} ω . □

Lemma 6.4. Assuming that:

$ϕ : C_{1} \to C_{2}$ a non-constant morphism

Then

ϕ

is separable if and only if

ϕ^{*} : Ω_{C_{1}} \to Ω_{C_{2}}

is non-zero

Example. $𝔾_{m} = 𝔸^{1} ∖ {0}$ (multiplicative group).

\begin{array}{l} ϕ : 𝔾_{m} & \to 𝔾_{m} \\ x & \mapsto x^{n} \end{array}

( $n \geq 2$ integer).

$ϕ^{*} (d x) = d (x^{n}) = n x^{n - 1} d x$ . So if $char K ∤ n$ then $ϕ$ is separable.

Theorem 2.8 implies $# ϕ^{- 1} (Q) = \deg ϕ$ for all but finitely many $Q \in 𝔾_{m}$ .

But $ϕ$ is a group homomorphism, so

# ϕ^{- 1} (Q) = # \ker (Q)

for all $Q \in 𝔾_{m}$ . Thus $# \ker ϕ = \deg ϕ = n$ and hence $K (= \bar{K})$ contains exactly $n$ $n$ -th roots of unity.

Proof. Lemma 6.3 + induction gives ${[n]}^{*} ω = n ω$ .

$char K ∤ n$ implies $[n]$ is separable. So $# {[n]}^{- 1} (Q) = \deg [n]$ for all but finitely many $Q \in E$ . But $[n]$ is a group homomorphism, so ${[n]}^{- 1} (Q) = # E [n]$ for all $Q \in E$ .

Putting these two statements together gives

# E [n] = \deg [n] \overset{Corollary 5.9}{=} n^{2} .

Group theory (structure theorem) gives that

E [n] ≅ ℤ ∕ d_{1} ℤ \times \dots \times ℤ ∕ d_{t} ℤ

for some $1 < d_{1} | d_{2} | \dots | d_{t} | n$ .

Let $p$ be a prime with $p | d_{1}$ . Then $E [p] ≅ {(ℤ ∕ p ℤ)}^{t}$ . But $# E [p] = p^{2}$ , so $t = 2$ . But $# E [p] = p^{2}$ , so $t = 2$ , i.e. $E [n] ≅ ℤ ∕ d_{1} ℤ \times ℤ ∕ d_{2} ℤ$ . Since $d_{1} | d_{2} | n$ and $d_{1} d_{2} = n^{2}$ , we get $d_{1} = d_{2} = n$ .

Thus $E [n] ≅ {(ℤ ∕ n ℤ)}^{2}$ . □

Remark. If $char K = p$ then $[p]$ is inseparable. It can be shown that:

\begin{array}{l} either E [p^{r}] & ≅ ℤ ∕ p^{r} ℤ \forall r \geq 1 & “ordinary” or E [p] & = 0 & “supersingular” \end{array}

Do not use this remark on Example Sheet 2!

6 The Invariant Differential