Seminar on Elliptic Curves and the
Weil conjectures
The Invariant Differential
University of Regensburg
Faculty of Mathematics
Andreas Pangerl
Regensburg, 01.06.2016
Chapter 5
The Invariant Differential
Let E/K be an elliptic curve given by the Weierstrass equation
E : y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 .
As we have seen in Proposition 1.5 of [1, Chapter 3], the so-called invariant
differential
dx
∈ ΩE
ω=
2y + a1 x + a3
associated with the Weierstrass equation neither has zeros nor poles, that
is div(ω) = 0. In the following we want to show that it is invariant under
translation by points on the elliptic curve.
For Q ∈ E we consider the isomorphism translation-by-Q
τQ : E −→ E
P 7−→ P + Q
and the induced map τQ∗ : ΩE −→ ΩE on the K(E)-vector space of differentials ΩE . Then we have
Proposition 1. Let E and ω be as above and let Q ∈ E. Then
τQ∗ ω = ω.
Proof. One can prove the assertion that the equality
dx(P + Q)
dx(P )
=
2y(P + Q) + a1 x(P + Q) + a3
2y(P ) + a1 x(P ) + a3
holds by using the addition formula of (III.2.3c) and standard differentiation
rules. However, there is a much more elegant way of showing this.
By (II.4.2a) ΩE is a one-dimensional K(E)-vector space. Thus, there is a
rational function aQ ∈ K(E)∗ , depending a priori on Q, such that
τQ∗ ω = aQ ω.
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5. The Invariant Differential
2
Note that aQ 6= 0, because τQ is an isomorphism. We now can compute
div(aQ ) = div(τQ∗ ω) − div(ω)
= τQ∗ div(ω) − div(ω)
= 0,
since div(ω) = 0. Therefore, by (II.3.1) aQ must be constant, i.e., aQ ∈ K ∗ .
Finally, we consider the map
f : E −→ P1
Q 7−→ [aQ , 1].
Now, f is a rational map because it is determined by the aQ ’s, which can be
expressed as a rational function depending on the coordinates of Q. But it is
not surjective, because it misses the points [0, 1] and [1, 0] and therefore we
can conclude by (II.2.1) and (II.2.3) that f is constant. Thus, aQ does not
depend on Q and we find its value by noting that
aQ = aO = 1
for all Q ∈ E.
This proves our assertion.
With Proposition 1 we now can show
Theorem 2. Let E and E 0 be elliptic curves, let ω be an invariant differential
on E and let
φ, ψ : E 0 −→ E
be two isogenies. Then
(φ + ψ)∗ ω = φ∗ ω + ψ ∗ ω.
(5.1)
Remark 3. The two ”plus signs” in (5.1) represent completely different
operations. The first is addition in Hom(E 0 , E), which is essentially addition
using the group law on E. The second is the addition in the vector space of
differentials ΩE 0 .
Proof. For φ = [0] or ψ = [0] it is clear. If φ + ψ = [0], we can use the fact
that
ψ ∗ = (−φ)∗ = φ∗ ◦ [−1]∗
and it suffices to show that
[−1]∗ ω = −ω.
The negation formula from (III.2.3a)
[−1](x, y) = (x, −y − a1 x − a3 ),
5. The Invariant Differential
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allows us to calculate
∗
[−1]
dx
2y + a1 x + a3
=
dx
dx
=−
2(−y − a1 x − a3 ) + a1 x + a3
2y + a1 x + a3
and the statement follows. So we now can assume that φ, ψ and φ + ψ are
non-zero.
Let (x1 , y1 ) and (x2 , y2 ) be ”independent” Weierstrass coordinates for
E, meaning that they satisfy the given Weierstrass equation but no other
algebraic relation. Moreover, let
(x3 , y3 ) = (x1 , y1 ) + (x2 , y2 ),
so by the addition formula (III.2.3c) on E, x3 and y3 are rational combinations of x1 , y1 , x2 and y2 . Furthermore, for any (x, y) let ω(x, y) denote the
corresponding invariant differential
ω(x, y) =
dx
.
2y + a1 x + a3
Then, using the addition formula and the standard rules of differentiation,
we can express ω(x3 , y3 ) in terms of ω(x1 , y1 ) and ω(x2 , y2 ). More precisely,
we have
ω(x3 , y3 ) = f (x1 , y1 , x2 , y2 )ω(x1 , y1 ) + g(x1 , y1 , x2 , y2 )ω(x2 , y2 ),
(5.2)
with f, g ∈ K(x1 , y1 , x2 , y2 ).
We now want to show that both f and g are identically 1 by using
Proposition 1. Suppose we assign fixed values to x2 and y2 by choosing some
Q ∈ E and setting
x2 = x(Q) and y2 = y(Q).
Then dx2 = dx(Q) = 0, so ω(x2 , y2 ) = 0 and with Proposition 1 we get
ω(x3 , y3 ) = ω(τQ (x1 , y1 )) = τQ∗ ω(x1 , y1 ) = ω(x1 , y1 ).
Substituting this in (5.2), we find that
f (x1 , y1 , x(Q), y(Q)) ≡ 1
as a rational function in K(x1 , y1 ). Furthermore, this is true for every Q ∈ E
and it follows that f is identically 1. By reversing the roles of x1 , y1 and
x2 , y2 , we see that the same is true for g.
5. The Invariant Differential
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To recapitulate, we have shown that if
(x3 , y3 ) = (x1 , y1 ) + (x2 , y2 )
(+ is addition in E),
ω(x3 , y3 ) = ω(x1 , y1 ) + ω(x2 , y2 )
(+ is addition in ΩE ).
then
Now let (x0 , y 0 ) be Weierstrass coordinates on E 0 , and set
(x1 , y1 ) = φ(x0 , y 0 ),
(x2 , y2 ) = ψ(x0 , y 0 ) and
(x3 , y3 ) = (φ + ψ)(x0 , y 0 ).
Then we have
ω((φ + ψ)(x0 , y 0 )) = ω(φ(x0 , y 0 )) + ω(ψ(x0 , y 0 )).
In other words (φ + ψ)∗ ω = φ∗ ω + ψ ∗ ω.
By Proposition 1 and Theorem 2 we now get the following useful results
virtually for free.
Corollary 4. Let ω be an invariant differential on an elliptic curve E and
let m ∈ Z. Then
[m]∗ ω = mω.
Proof. The assertion is clearly true for m = 0, 1. By applying Theorem 2
with E = E 0 , φ = [m] and ψ = [1] we obtain
[m + 1]∗ ω = [m]∗ ω + ω
and the desired result now follows by induction.
Next, we give a new proof of (III.4.2a), stating that the map [m] for
m ∈ Z \ {0} is non-constant. In doing so, we need (II.4.2c), that is, a nonconstant morphism φ : E1 −→ E2 is separable if and only if the induced map
φ∗ : ΩE2 −→ ΩE1 is non-zero.
Corollary 5. Let E/K be an elliptic curve and let m ∈ Z with m 6= 0.
Assume either that char(K) = 0 or that m is prime to char(K). Then the
multiplication-by-m map on E is a finite, separable endomorphism.
Proof. Let ω be an invariant differential on E. Then, Corollary 4 and our
assumption on m implies that
[m]∗ ω = mω 6= 0,
so [m] 6= [0]. Hence, [m] is finite and because [m]∗ is non-zero, (II.4.2c) tells
us that [m] is separable.
5. The Invariant Differential
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As another application of Theorem 2 and Corollary 4, we examine when
a linear combination involving the Frobenius endomorphism we have seen in
(III.4.6) is separable.
Corollary 6. Let E be an elliptic curve defined over a finite field Fq of characteristic p > 0, let φ : E −→ E be the q th -power Frobenius endomorphism
and let m, n ∈ Z. Then the map
m + nφ : E −→ E
is separable if and only if p - m.
In particular, the map 1 − φ is separable.
Proof. Let ω be an invariant differential on E. By (II.4.2c), we know that a
map ψ : E −→ E is inseparable if and only if ψ ∗ ω = 0.
Applying this to the map ψ = m + nφ and using Theorem 2 as well as
Corollary 4, we get
(m + nφ)∗ ω = mω + nφ∗ ω.
But φ∗ ω = 0 because φ is inseparable, so
(m + nφ)∗ ω = mω.
Now, mω = 0 if and only if p | m, which gives us the desired result.
Lastly, we also can prove that the endomorphism ring of an elliptic curve
defined over a field of characteristic zero is commutative.
Corollary 7. Let E/K be an elliptic curve and let ω be a non-zero invariant
differential on E. We define a map from End(E) to K in the following way:
End(E) −→ K,
φ 7→ aφ
such that φ∗ ω = aφ ω.
(i) The map φ 7→ aφ is a ring homomorphism.
(ii) The kernel of φ 7→ aφ is the set of inseparable endomorphisms of E.
(iii) If char(K) = 0, then End(E) is a commutative ring.
Proof. As in the proof of Proposition 1, the fact that ΩE is a one-dimensional
K(E)-vector space implies that φ∗ ω = aφ ω for some function aφ ∈ K(E).
We claim that aφ ∈ K. This is clear if aφ = 0, while if aφ =
6 0 we use the fact
that div(ω) = 0 to compute
div(aφ ) = div(φ∗ ω) − div(ω) = φ∗ div(ω) − div(ω) = 0.
Hence, by (II.3.1) we conclude aφ ∈ K ∗ .
(i) By Theorem 2 we have
aφ+ψ ω = (φ + ψ)∗ ω = φ∗ ω + ψ ∗ ω = aφ ω + aψ ω = (aφ + aψ )ω,
that is aφ+ψ = aφ + aψ . Similarly,
aφ◦ψ ω = (φ ◦ ψ)∗ ω = ψ ∗ (φ∗ ω) = ψ ∗ (aφ ω) = aφ ψ ∗ (ω) = aφ aψ ω,
which proves that aφ◦ψ = aφ aψ .
(ii) We have
aφ = 0 ⇐⇒ φ∗ ω = 0 ⇐⇒ φ is inseparable by (II.4.2c).
(iii) If char(K) = 0, then every non-constant endomorphism is separable, so
(ii) says that End(E) injects into K. Hence End(E) is commutative.
References
[1] Joseph H. Silverman. The Arithmetic of Elliptic Curves. Graduate Texts
in Mathematics, Springer-Verlag, New York, 1986.
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