THEOREM OF NAGELL-LUTZ: THE DECISIVE LEMMA.
IAN KIMING
Let E be an elliptic curve y 2 = x3 + ax2 + bx + c defined over Q where a, b, c ∈ Z.
Let p be a fixed prime number.
Let R be the ring consisting of rational numbers x with ordp (x) ≥ 0. The natural
map:
m
7→ (m mod p)(n mod p)−1
n
for m, n ∈ Z with p - n, induces an isomorphism of rings R/Rp ∼
= Z/p = Fp .
Similarly, one checks for ν ∈ N that the map:
m ν
· p 7→ (m mod p)(n mod p)−1
n
for m, n ∈ Z with p - n, induces an isomorphism of abelian groups
Rpν /Rpν+1 ∼
= Z/p .
For x ∈ R we denote by x̄ the reduction of x modulo p. Thus, x̄ is regarded as
an element of Fp .
For ν ∈ N we have the set
E(pν ) := fP = (x, y) ∈ E(Q) j ordp (x) ≤ −2ν , ordp (y) ≤ −3νg ∪ f0g .
If P = (x, y) ∈ E(Q) with y 6= 0 we define t(P ) := xy . Also, t(0) := 0. Notice
that, if P = (x, y) ∈ E(p) then y 6= 0 (as we would otherwise have x ∈ Z and hence
ordp (x), ordp (y) ≥ 0). Thus the map t is defined on all of E(p).
We have P ∈ E(pν ) ⇔ t(P ) ∈ Rpν .
Lemma 1. Let Pi ∈ E(p), i = 1, 2. Then t(P1 + P2 ) is defined and we have:
P1 , P2 ∈ E(pν ) ⇒ t(P1 ) + t(P2 ) − t(P1 + P2 ) ∈ Rp3ν .
Proof. As we have t(0) = 0, and t(−P ) = −t(P ), one checks immediately that we
may assume P1 , P2 , P1 + P2 6= 0.
Write then Pi = (xi , yi ), i = 1, 2. Since P1 + P2 6= 0 we may also write P1 + P2 =
(x3 , −y3 ) so that (x3 , y3 ) is the 3’rd point of intersection between E and the line
through P1 and P2 .
Assuming P1 , P2 ∈ E(pν0 ) for some ν0 ∈ N, we must show that t(P1 + P2 ) is
defined and that:
t(P1 ) + t(P2 ) − t(P1 + P2 ) ∈ Rp3ν0 .
Writing:
ordp (x1 ) = −2ν ,
ordp (y1 ) = −3ν
and
ordp (x2 ) = −2µ ,
ordp (y2 ) = −3µ
1
2
IAN KIMING
we have ν, µ ≥ ν0 and may assume without loss of generality that µ ≥ ν.
It is enough to show that t(P1 + P2 ) is defined and:
t(P1 ) + t(P2 ) − t(P1 + P2 ) ∈ Rp3ν
which we do below.
Now let us consider the following change of variables:
x̃ := p2ν x ,
(])
ỹ := p3ν y ,
z̃ := z .
e
So if (x, y, z) ∈ E(Q) then (x̃, ỹ, z̃) is a rational point on the elliptic curve E
defined by the equation:
(∗)
F (x̃, ỹ, z̃) := x̃3 + p2ν ax̃2 z̃ + p4ν bx̃z̃ 2 + p6ν cz̃ 3 − ỹ 2 z̃ = 0 .
This equation does indeed define an elliptic curve: The roots of the polynomial
x̃3 +p2ν ax̃2 +p4ν bx̃+p6ν c are p2ν ξ where ξ runs through the roots of x3 +ax2 +bx+c;
as the latter polynomial is without multiple roots, so is the former.
e Let L
e be the line through
The points (x̃i , ỹi ), i = 1, 2, are rational points on E.
these points. Since the change of variables (]) is linear we see that the 3’rd point
e and L
e is (x̃3 , ỹ3 ).
of intersection between E
e in the shape
Now, let us write the equation for L
e:
L
αx̃ + β ỹ = γ z̃ ,
where α, β, γ ∈ Q. Multiplying the equation by a suitable power of p we may
assume that α, β, γ ∈ R, and that at least one of α, β, γ is a unit in R (i.e., has
ordp = 0).
We first claim that ordp (γ) = 0. For suppose otherwise, i.e., that ordp (γ) > 0.
We then claim that µ = ν. For otherwise, we must have µ > ν.
Consider then the fact that we have:
αp3(µ−ν) x̃2 + βp3(µ−ν) ỹ2 = p3(µ−ν) γ .
In this equation, the number p3(µ−ν) ỹ2 is a unit in R whereas the terms αp3(µ−ν) x̃2
and p3(µ−ν) γ belong to Rp. So, β is not a unit, hence α is a unit. But consider
then:
αx̃1 + β ỹ1 = γ ;
in this equation the terms β ỹ1 and γ are now in Rp whereas αx̃1 is a unit, –
contradiction. So we must have µ = ν.
So the points (x̃i , ỹi ), i = 1, 2, both have coordinates that are units in R.
e either α or β is a unit in R.
Now, in the equation for the line L
e is then of form:
Suppose first that β is a unit. The equation for L
ỹ = ux̃ + v z̃
THEOREM OF NAGELL-LUTZ: THE DECISIVE LEMMA.
3
where u ∈ R and v ∈ Rp. Since (x̃i , ỹi ), i = 1, 2, 3, are the points of intersection
e and L,
e we conclude that:
between E
F (x̃, ux̃ + v z̃, z̃)
=
(x̃3 + p2ν ax̃2 z̃ + p4ν bx̃z̃ 2 + p6ν cz̃ 3 ) − (ux̃ + v z̃)2 z̃
=
(x̃ − x̃1 z̃)(x̃ − x̃2 z̃)(x̃ − x̃3 z̃) .
Since F (x̃, ux̃+v z̃, z̃) has coefficients in R, and since x̃1 , x̃2 ∈ R, we conclude that
x̃3 ∈ R. We can then consider the reduction modulo p of the above factorization:
¯1 z)(x − x̃
¯2 z)(x − x̃
¯3 z)
x3 − ūx2 z = (x − x̃
¯1 x̃
¯2 6= 0, and
as polynomials over Fp . Since x̃1 and x̃2 are units in R, we have x̃
¯
conclude that x̃3 = 0. But we also have:
¯1 x̃
¯2 + x̃
¯1 x̃
¯3 + x̃
¯2 x̃
¯3 = 0 ,
x̃
and hence a contradiction.
e as:
Similarly, if α is a unit in R, we can write the equation for L
x̃ = uỹ + v z̃
with u ∈ R, v ∈ Rp. Here u must be a unit, since otherwise the point (x̃1 , ỹ1 )
e (because x̃1 is a unit). We deduce the factorization:
could not be on L
F (uỹ + v z̃, ỹ, z̃)
=
((uỹ + v z̃)3 + p2ν a(uỹ + v z̃)2 z̃ + p4ν b(uỹ + v z̃)z̃ 2 + p6ν cz̃ 3 ) − ỹ 2 z̃
= u3 (ỹ − ỹ1 z̃)(ỹ − ỹ2 z̃)(ỹ − ỹ3 z̃) ,
and that ỹ3 ∈ R (here we use that u is a unit). Reducing this factorization mod p
we then find:
ū3 y 3 − y 2 z = ū3 (y − ỹ¯1 z)(y − ỹ¯2 z)(y − ỹ¯3 z)
as polynomials over Fp . Since now ū 6= 0, we obtain a contradiction in the same
manner as before.
This finishes the proof of the claim that ordp (γ) = 0.
So we have that γ is a unit in R. Dividing through by γ, and renaming α and
β, we may thus assume an equation of form
z̃ = αx̃ + β ỹ
e
with α, β ∈ R for the line L.
e and L
e are (x̃i , ỹi ), i = 1, 2, 3, and so their
The 3 points of intersection between E
coordinates must satisfy the equation:
0
= −ỹ 2 (αx̃ + β ỹ) + x̃3 + p2ν ax̃2 (αx̃ + β ỹ)
+p4ν bx̃(αx̃ + β ỹ)2 + p6ν c(αx̃ + β ỹ)3
= η0 x̃3 + η1 x̃2 ỹ + η2 x̃ỹ 2 + η3 ỹ 3 ,
where
η0 = 1 + p2ν aα + p4ν bα2 + p6ν cα3 ,
η1 = p2ν aβ + 2 · p4ν bαβ + 3 · p6ν cα2 β .
4
IAN KIMING
Since a, b, c, α, β ∈ R we have that η0 is a unit in R, and that η1 ∈ Rp2ν .
Now, we must have y3 6= 0: For otherwise we would deduce η0 x̃33 = 0 whence
e Hence t(P1 + P2 )
x̃3 = 0, and contradiction (the point (0, 0, 1) is not on the line L).
is defined.
Furthermore, we can consider the 3 numbers x̃i /ỹi , i = 1, 2, 3, and we find that
they are the roots of the polynomial:
η 0 w 3 + η1 w 2 + η 2 w + η3 .
Consequently,
t(P1 ) + t(P2 ) − t(P1 + P2 )
x1
x2
x3
x̃1
x̃2
x̃3
=
+
+
= pν ·
+
+
y1
y2
y3
ỹ1
ỹ2
ỹ3
ν η1
3ν
= −p ·
∈ Rp .
η0
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]