Normat 58:3, 123–135 (2010) 123
Density of rational points on plane curves
Hugues Verdure
Institutt for Matematikk og Statistikk
Universitetet i Tromsø
9037 Tromsø
Norway
Hugues.Verdure@uit.no
Introduction
The study of Diophantine equations, that is finding integer solutions to a polyno-
mial equation with integer coefficients, is a really old branch of mathematics. We
can find examples on the subject in India (Baudhayana, 800 BC), in Alexandria
(Diophantus, 250 AC), France (Fermat, 1637). . .
While Diophantine equations are easy to formulate, solving them might be a dif-
ferent matter. The equations 3x + 8y = 7 and 3x + 9y = 7, while quite similar,
have completely different sets of solutions. The first one has infinitely many (one
of them is x = −3, y = 2), while the second one has none. Another example is the
Pythagoras/Fermat equations x
n
+y
n
= z
n
for n > 2. It is known since Pythagoras
that the equation for n = 2 has solutions (x = 3, y = 4, z = 5 is one of them), and
actually infinitely many and all known. Fermat conjectured in 1637 that the other
equations for n > 3 have no solutions, except the trivial ones, that is the one where
x, y or z are 0. He even claimed to have a proof, but the margin was not large
enough for the proof to fit in. The proof of the Taniyama-Shimura conjecture for
semistable elliptic curves by Wiles and Taylor in the early nineties, together with
works by Frey and Ribet implies that Fermat last theorem is now proved.
A very related topic is the study of rational points on algebraic curves. This is
what we propose to look at. We will give answers to the following questions: given
a plane algebraic curve, how many rational points may lie on the curve? And, in
case of infinitely many points, are they dense on the curve (that is, given any point
on the curve, does there exist a rational point on the curve infinitely near it)?
This paper is motivated by the reading of an article by Mazur [3] where he discusses
a general conjecture on rational points on algebraic varieties, conjecture that he
proves in the case of plane curves : if a plane curve has infinitely many rational
points, then they are dense on some algebraic components of the curve.
The paper is organized as follows. In the first part, we look at curves of degree
1, that is lines, in the second part, we study curves of degree 2, or conics, while in
the third part, we see the case of curves of degree 3. We always assume that the
curves are irreducible.
124 Hugues Verdure Normat 3/2010
The two first parts follow the same scheme. We find the number of rational points
that is needed for the curve to have rational coefficients. Once we have such a curve
with a rational point, we build a bijection, which is continuous both ways, between
the curve and the real line, and that sends rational points to rational numbers.
Since Q is dense in R, rational points are dense in the curve.
The third part involves more advanced mathematics. In this part, we take advan-
tage of the fact that elliptic curves have a natural structure of abelian group. We
use this to describe a mapping, from a neighbourhood of the identity element to a
neighbourhood of 0 ∈ R, that is bijective, continuous both ways and that preserves
in a way the group law. Then we show that the image of rational points by this
mapping is dense round 0. In this part, we use results and notation from [4, 5].
Density of rational points on a line
In this section, we shall see that on a line in the real plane, there are either no
rational points, or exactly one, or the rational points are dense on the line.
Assume that we have two rational points on a line L, say P = (x
P
, y
P
), and
Q = (x
Q
, y
Q
), with P 6= Q, and x
P
, x
q
, y
P
, y
Q
∈ Q. If x
P
= x
Q
, then L is the
vertical line L : x = x
P
, and the rational points are (x
P
, r), r ∈ Q, and this is of
course dense in L. If x
P
6= x
Q
, then the line L has equation
L : y =
y
P
− y
Q
x
P
− x
Q
x +
y
Q
x
P
− y
P
x
Q
x
P
− x
Q
and the coefficients are rationals. Then we have an homeomorphism between L
and R given by (x, y) 7→ x, and rational points in both L and R are in one-to-one
correspondance, which proves that rational points are dense in L.
It is also easy to give examples of lines with no or just one rational point:
L
0
: y =
√
2x + 1 has obviously no rational point, while L
1
: y =
√
2x has exactly
one, namely (0, 0).
We have thus proved the following:
Proposition 1. Let L be a line on the real plane. Then we have exactly one of the
following:
1. There are no rational points on L
2. There is exactly one rational point on L
3. An equation of L can be written with rational coefficients, and the rational
points are dense on L.
Density of rational points on a conic
A conic is an irreducible plane curve of degree 2, so it has equation
Ax
2
+ Bxy + Cy
2
+ Dx + Ey + F = 0.
Normat 3/2010 Hugues Verdure 125
Such an equation could be reducible (we don’t allow that here), in which case it
is the product of two lines, and we have essentially seen this case in the previous
section. We will also assume that it is not degenerate, that is, there are infinitely
many points on the curve (we want to eliminate cases like the empty circle, or the
circle reduced to just one point). When it is not reducible, then no 3 points on the
curve lie on a line. This is a consequence of Bezout’s theorem, but we give here a
simple proof.
Lemma 0.1. Let C : Ax
2
+ Bxy + Cy
2
+ Dx + Ey + F = 0 be a plane conic such
that there exists 3 different points on it which lie on a line. Then C is reducible.
Proof. By translating one of the three points to the origin, we can assume that the
three points are (0, 0), (x
0
, y
0
) and (tx
0
, ty
0
) for t 6= 0, 1. This also means that F =
0, Ax
2
0
+Bx
0
y
0
+Cy
2
0
+Dx
0
+Ey
0
= 0 and t
2
(Ax
2
0
+Bx
0
y
0
+Cy
2
0
)+t(Dx
0
+Ey
0
) = 0,
and in turn, (t
2
−t)(Dx
0
+Ey
0
) = 0. Since t
2
−t 6= 0, this means that Dx
0
+Ey
0
= 0,
and also Ax
2
0
+ Bx
0
y
0
+ Cy
2
0
= 0. It is then easy to see that any point of the form
(sx
0
, sy
0
) is on the curve, that is, a line of the form αx + βy = 0 is included in the
curve. We can assume that α 6= 0. Then as polynomials in the variable x, write
Ax
2
+ Bxy + Cy
2
+ Dx + Ey = (αx + βy)Q(x, y) + R(y)
where Q(x, y) is a polynomial in two variables, while R(y) is a polynomial in just y
(actually, a polynomial in two variables, but the euclidian division assures us that
the degree in x is less or equal to 0). Then, for any y, let x = −
β
α
y. It is a point on
the line, thus on the curve, which means that R(y) = 0 for all y, and therefore R
is zero, and we have proved that
Ax
2
+ Bxy + Cy
2
+ Dx + Ey = (αx + βy)Q(x, y),
hence the curve is reducible.
From now on, we assume that we are given an irreducible non-degenerate conic
of the form C : Ax
2
+ Bxy + Cy
2
+ Dx + Ey + F = 0. Then the curve is entirely
defined by 5 distinct points. Namely, we have the following lemma:
Lemma 0.2. Given 5 distinct points on the plane, with the property that no 3 of
them lie on a line, then there exists a unique conic that passes through those points.
Proof. We are given the points P
1
= (x
1
, y
1
), P
2
= (x
2
, y
2
), P
3
= (x
3
, y
3
), P
4
=
(x
4
, y
4
), and P
5
= (x
5
, y
5
). By a translation, we may assume that P
1
= (0, 0). Since
P
1
, P
2
and P
3
do not lie on a line, the quantity x
3
y
2
−x
2
y
3
6= 0, and we can define
α =
−y
3
x
3
y
2
−x
2
y
3
, β =
x
3
x
3
y
2
−x
2
y
3
, γ =
y
2
x
3
y
2
−x
2
y
3
and δ =
−x
2
x
3
y
2
−x
2
y
3
, and consider the
change of variables X = αx +βy, Y = γx + δy. This is really a change of variables,
since αδ − βγ = −1 6= 0. Moreover, since the change of variables is linear, conics
are transformed to conics, and lines into lines. Finally, this change of variables lets
P
1
unchanged, and sends the points P
2
and P
3
to (1, 0) and (0, 1) respectively. So
we might assume that the five given points are P
1
= (0, 0), P
2
= (1, 0), P
3
= (0, 1),
P
4
= (x
4
, y
4
), and P
5
= (x
5
, y
5
).
126 Hugues Verdure Normat 3/2010
We have to find coefficients A, B, C, D, E, F that satisfy
F = 0
A + D + F = 0
C + E + F = 0
Ax
2
4
+ Bx
4
y
4
+ Cy
2
4
+ Dx
4
+ Ey
4
+ F = 0
Ax
2
5
+ Bx
5
y
5
+ Cy
2
5
+ Dx
5
+ Ey
5
+ F = 0
This systems reduces to
A(x
2
4
− x
4
) + Bx
4
y
4
= −C(y
2
4
− y
4
)
A(x
2
5
− x
5
) + Bx
5
y
5
= −C(y
2
5
− y
5
)
The determinant of the system is
(x
2
4
− x
4
)x
5
y
5
− (x
2
5
− x
5
)x
4
y
4
= x
4
x
5
(y
5
(x
4
− 1) − y
4
(x
5
− 1)) .
This can’t be 0, since it would then mean x
4
= 0, or x
5
= 0 or y
5
(x
4
−1) −y
4
(x
5
−
1) = 0. But the first case is the same as saying that the points P
1
, P
3
, P
4
lie on the
same line, the second, that the points P
1
, P
3
, P
5
lie on the same line, while the third
that the points P
2
, P
4
, P
5
lie on the same line, namely the line Y =
y
4
x
4
−1
(X − 1)
(or Y =
y
5
x
5
−1
(X − 1) if x
4
= 1.) Then all the solutions are
A =
x
4
y
4
(y
2
5
− y
5
) − x
5
y
5
(y
2
4
− y
4
)
(x
2
4
− x
4
)x
5
y
5
− (x
2
5
− x
5
)x
4
y
4
C
B =
(x
2
5
− x
5
)(y
2
4
− y
4
) − (x
2
4
− x
4
)(y
2
5
− y
5
)
(x
2
4
− x
4
)x
5
y
5
− (x
2
5
− x
5
)x
4
y
4
C
We see that C = 0 leads A = B = C = D = E = F = 0 which is absurd. Then
C 6= 0, and dividing the equation of the conic by C, we can assume that C = 1,
and we have just proved that there exists a unique conic that passes through the 5
points.
We will now prove the following:
Theorem 1. Let C : Ax
2
+Bxy+Cy
2
+Dx+Ey+F = 0 be a non-degenerate conic.
If the curve has at least 5 distinct rational points on it, then there are infinitely
many rational points on it, and they are dense on the curve.
Proof. Let P
1
= (x
1
, y
1
), P
2
= (x
2
, y
2
), P
3
= (x
3
, y
3
), P
4
= (x
4
, y
4
), and P
5
=
(x
5
, y
5
) be the 5 rational points. Since it is a conic, one of the coefficients A, B, C
is not zero, say A, and we may therefore assume that A = 1. Then we have to solve
the system
Bx
1
y
1
+ Cy
2
1
+ Dx
1
+ Ey
1
+ F = −x
2
1
Bx
2
y
2
+ Cy
2
2
+ Dx
2
+ Ey
2
+ F = −x
2
2
Bx
3
y
3
+ Cy
2
3
+ Dx
3
+ Ey
3
+ F = −x
2
3
Bx
4
y
4
+ Cy
2
4
+ Dx
4
+ Ey
4
+ F = −x
2
4
Bx
5
y
5
+ Cy
2
5
+ Dx
5
+ Ey
5
+ F = −x
2
5
Normat 3/2010 Hugues Verdure 127
From the lemma, we know that this system has a unique solution, and since the
coefficients are rationals, then B, C, D, E, F have to be rational too.
If we have a rational point P = (x, y) on the conic, with x 6= x
1
(which is always
the case except for at most 2 points), then the slope of the line passing through P
1
and P is rational of course, being equal to
y−x
1
x−x
1
. This gives a map between rational
points on C except at most 2 points, and rational numbers. We show now that this
map has an inverse, namely, take a (almost any) rational number t, and consider
the line L
t
passing through the point P
1
, and with slope t. Then this line will cross
C in another rational point. This will show that the number of rational points on
the conic is infinite, but we will have to work a bit more to prove density.
The line L
t
has equation
L
t
: y = tx + (y
1
− tx
1
).
We replace y in the equation of C by this expression to find the x-coordinate of the
intersection points. This gives us
A + Bt + Ct
2
x
2
+
By
1
− Btx
1
+ 2Cty
1
− 2Ct
2
x
1
+ D + Et
x
+
Cy
2
1
+ Ct
2
x
2
1
− 2Cty
1
x
1
+ Ey
1
− Etx
1
+ F
= 0
If A + Bt + Ct
2
= 0, then we just have one intersection point, namely P
1
, but this
happens for at most 2 values of t. But if A + Bt + Ct
2
6= 0, we know that this
equation has two solutions. We know one of them, namely x
1
, and the other, say
α verifies
−x
1
− α =
By
1
−Btx
1
+2Cty
1
−2Ct
2
x
1
+D+Et
A+Bt+Ct
2
x
1
α =
Cy
2
1
+Ct
2
x
2
1
−2Cty
1
x
1
+Ey
1
−Etx
1
+F
A+Bt+Ct
2
This shows that the other intersection point is P = (α, β) with
α = −x
1
−
By
1
− Btx
1
+ 2Cty
1
− 2Ct
2
x
1
+ D + Et
A + Bt + Ct
2
and
β = tα + y
1
− tx
1
.
The good thing is that since we assume that t, x
1
, y
1
∈ Q, and we have proved
that A, B, C, D, E, F ∈ Q, then necessarily P is a rational point. It is not difficult
to show that the two maps are inverse of each other, so that we have a bijection
between the set
C(Q) − {(x, y)|x = x
1
}
and the set
Q −
t|A + Bt + Ct
2
= 0
.
Since the second one is infinite, the first one has to be infinite too, and there are
infinitely many rational points on C. As for density, since the map going from the
second set to the first set is obviously continuous, density of the rational numbers
in R implies that C(Q) is dense in C(R).
128 Hugues Verdure Normat 3/2010
Corollary 1.1. There are infinitely many Pythagorean triples.
Proof. The circle x
2
+ y
2
= 1 has at least 5 rationals points, namely (1, 0), (−1, 0),
(0, 1), (0, −1) and (
3
5
,
4
5
), the last one coming from the Pythagorean triple (3, 4, 5).
We know therefore that there are infinitely many rational points on this circle.
Now, given any such rational point, of the form (
a
c
,
b
c
), with a, b, c positive, and
relatively prime, then a
2
+ b
2
= c
2
, that is (a, b, c) is a Pythagorean triple, and
there are therefore infinitely many.
A natural question to ask is: what happens with less than 5 points? The answer
is: there exists conics with exactly 0, 1, 2, 3 or 4 rational points. We shall give an
example of each of them.
Consider the circle C
0
: (x −
√
2)
2
+ y
2
= 1. If we expand this expression, we get
−2x
√
2 + (x
2
+ 1 + y
2
) = 0. If (x, y) is a rational point on the circle with x 6= 0,
then
√
2 =
x
2
+y
2
+1
2x
∈ Q which is absurd. Then x has to be 0, but then y
2
= −1
which is also absurd. Thus the circle C
0
has no rational point.
Consider now the circle C
1
: (x −
√
2)
2
+ y
2
= 2. As we just did, we still can
show that if (x, y) is a rational point on C
1
, then necessarily x = 0. But this time,
if x = 0, there is a rational solution for y, namely, y = 0, and so C
1
has exactly one
rational point, (0, 0).
Consider now the circle C
2
: (x −
√
2)
2
+ y
2
= 3. Again, there are no rational
point (x, y) on the curve with x 6= 0. But when x = 0, we easily find two rational
points, namely (0, 1) and (0, −1), which are therefore the only two rational points
on C
2
.
If we want to have examples of conics with exactly 3 or 4 rational points, then
we have to move away from circles. The reason is that with 3 rational points on a
circle, then the center is also rational (as the intersection of the midperpendiculars),
the square of the radius also is, and this shows that the coefficients of the conic can
all be chosen rational, and we can continue the proof of the theorem from here.
Consider then the ellipse C
3
: x
2
+(
√
2+1)y
2
+
√
2y = 0. Since
√
2 is not rational,
the only way to satisfy this equation with x, y ∈ Q is when y
2
−y = 0. But if y = 0,
then x = 0, while if y = 1, then x = ±1, which shows that there are exactly 3
rational points on C
3
, namely (0, 0), (1, −1) and (1, 1).
Consider finally the ellipse C
4
: x
2
+ (
√
2 − 1)y
2
=
√
2. Since
√
2 is not rational,
the only way to satisfy the equation with x, y ∈ Q is when y
2
= 1, and then in turn,
x
2
= 1. This gives that there are exactly 4 rational points on C
4
, namely (1, 1),
(−1, 1), (1, −1) and (−1, −1).
Density of rational points on irreducible smooth cubics
We will now look at the case of plane curves of degree 3, that is, cubics.
Essentially, we needed 2 distinct points to define a line, 5 to define a conic, and
here, since we have 10 coefficients, 9 points should be sufficient to define a cubic.
In the case of lines anc conics, 2 and 5 rational points implied an infinity of such
points, and moreover, those points were dense on the curve. Can we expect that 9
Normat 3/2010 Hugues Verdure 129
rational points on a cubic curve implies infinity and density of such points?
The answer is no. Take for example the elliptic curve
y
2
+ 17xy − 120y = x
3
− 60x
2
.
This curve has exactly 15 rational points (16 if we work in the projective plane):
(0, 0), (60, −900), (−30, 180), (24, −144), (−30, 450), (60, 0), (0, 120), (−12, 288),
(240, 1800), (
15
4
,
225
8
), (240, −5760), (−12, 36), (30, −300), (−40, 400) and (30, −90).
This comes from theorems of Mordell (this curve over Q is of rank 0 as an abelian
group), and Nagell-Lutz (that describe the rational points of finite order). Actually,
Mazur proved in [2] that the structure of the torsion subgroup of the rational points
on an elliptic curve is one of few types. As a consequence, if an elliptic curve has
17 rational points, it has infinitely many.
The previous example also shows that even if the coefficients are rationals and
there exists a rational point, then there might be just a finite number of rational
points, and of course density is even further away. The method that we used for
conics doesn’t work any longer. Of course, given a fixed rational point P
1
, then
(almost) any other rational point P on the curve defines a line L through P and
P
1
with rational slope, as for the conics, and we have therefore a well defined map
C(Q) − {few points} −→ Q
P 7→ slope of (P P
1
)
But, contrary to the conics, this map has no inverse: given t ∈ Q, we can of course
define the line L
t
going through P
1
and slope t. This line intersects C at least in
P
1
. For the conics, we found exactly one other rational intersection point, found
by solving a rational polynomial of second degree with a rational root. Here, we
can’t guarantee anything. We get indeed a rational polynomial of degree 3, with
a rational root. The polynomial might have no other real root (the line has no
intersection point other than P
1
), or two real roots (the line intersects the cubic in
two other points), but these roots are generally not rational.
What happens if we have an infinity of rational points? Are they then dense on
the curve? The answer is again no, as an example will later show.
Irreducible smooth cubic with a rational point have another name, elliptic curves
(by smooth, we mean also smooth at infinity). It can be shown that by a change of
variables, they have a particular form. As mentionned earlier, the "right" way to
study these curves is to define them over the projective plane. As we will work in
different affine charts, we give the general definition. We refer to [4] for the theory
on elliptic curves.
Definition 1. An elliptic curve E over a field K of characteristic zero is a curve
in P
2
(K) given by an equation
E : Y
2
Z = X
3
+ a
4
XZ
2
+ a
6
Z
3
with −4a
3
4
− 27a
2
6
6= 0.
130 Hugues Verdure Normat 3/2010
One point here is “special”, the only point not lying in the affine plane Z = 1,
namely the point O = [0, 1, 0], usually called the point at infinity. Given a field K,
define
E(K) =
P = [x, y, z] ∈ P
2
(K), y
2
z = x
3
+ a
4
xz
2
+ a
6
z
3
.
We can define an abelian group structure on the set E(R): given two points P, Q,
draw a line (P Q) (or the tangent to the curve if P = Q). This line intersects E(R)
in exactly a third point R. Then draw a line (OR) (or the tangent if O = R).
This line intersects E(Q) in a third point S. Then define P + Q = S. The group
law is infinitely differentiable. Moreover, if P, Q ∈ E(Q), then P + Q ∈ E(Q) too,
which makes E(Q) a subgroup of E(R). Finally, since P
2
(R) is compact, and E(R)
is closed in P
2
(R), then E(R) is compact too.
Figure 1 : Addition in the affine plane
There are two types of elliptic curves defined over R: E(R) has either 1 or 2
connected components, depending on the number of real roots of X
3
+ a
4
X + a
6
(or equivalently, the number of points of order 2).
Normat 3/2010 Hugues Verdure 131
-1 1
2 3 4
-7.5
-5
-2.5
2.5
5
7.5
-1 1
2 3 4
-7.5
-5
-2.5
2.5
5
7.5
Figure 2 : E : y
2
= x
3
+ x and E : y
2
= x
3
− x with one and two components
We shall denote by E(R)
O
the component that contains the point at infinity,
called the identity component. We shall now prove the following:
Theorem 2. Let E : y
2
= x
3
+a
4
x+a
6
be an elliptic curve over Q with an infinite
number of rational points. Then E(Q) ∩ E(R)
O
is dense in E(R)
O
. Moreover, in
the case of a second connected component, then E(Q) is dense in E(R) if and only
if there exists a rational point on it.
We shall need a few lemmas to prove this result.
Lemma 2.1. Let V be any open neighborhood of O. Then there exists an open
neighborhood W of O shuch that for all P, Q ∈ W , −P ∈ W and P + Q ∈ V .
Proof. Since we are interested in what happens round O, we will work in the affine
plane Y = 1. There, the curve has equation
E : z = x
3
+ a
4
xz
2
+ a
6
z
3
and O = (0, 0). Note that we have −(x, z) = (−x, −z). Let f (x, z) = x
3
+ a
4
xz
2
+
z
3
− z. We have
∂f
∂z
= 2a
4
xz + 3a
6
z
2
− 1,
and therefore
∂f
∂z
(0, 0) = −1 6= 0. By the theorem of implicit functions, there exists
a > 0 and g : (−a, a) −→ R infinitely differentiable such that
{(x, z), f (x, z) = 0} ∩ (−a, a) × R = {(x, g(x)), x ∈ (−a, a)}.
Note that g is an odd function. Let V
0
be the image of V in the affine plane Y = 1,
and
V
00
= {x, ∃z ∈ R, (x, z) ∈ V
0
}.
132 Hugues Verdure Normat 3/2010
This is an open set of R containing 0, so we can find b > 0 such that [−b, b] ⊂
V
00
∩ (−a, a). Consider the function
ϕ : [−b, b]
2
−→ R
(x
1
, x
2
) 7−→ x ((x
1
, g(x
1
)) + (x
2
, g(x
2
)))
where + is addition of points on the curve, and x(P ) denotes the x-coordinate of
P . This function is infinitely differentiable. In particular, on the compact [−b, b]
2
,
there exists k
b
> 0 such that
|ϕ(x
1
, x
2
)| 6 k
b
||(x
1
, x
2
)||, ∀x
1
, x
2
∈ [−b, b].
Take c =
b
k
b
√
2
. Then
∀x
1
, x
2
∈ (−c, c), |ϕ(x
1
, x
2
)| 6 k
b
q
x
2
1
+ x
2
2
6
√
2k
b
c = b.
Take now W = {(x, g(x)), x ∈ (−c, c)}.
Lemma 2.2. Let V be any open neighborhood of O. Then E(Q) ∩V is infinite.
Proof. Let W be the open neighborhood from the previous lemma. For P ∈ E(R),
let
W
P
= {P + Q, Q ∈ W }
which is an open neighborhood of P since the group law and inverse are continuous.
We obviously have
E(R) =
[
P ∈E(R)
W
P
.
Since E(R) is compact, there exists a finite number of points P
1
, ··· , P
n
such that
E(R) =
n
[
i=1
W
P
i
.
Since E(Q) is infinite, there exists a point, say P
1
, such that W
P
1
∩E(Q) is infinite.
Fix Q ∈ W
P
1
∩ E(Q) and consider
θ
Q
: W
P
i
∩ E(Q) −→ V ∩ E(Q)
R 7−→ R − Q
.
This is well defined: let R ∈ W
P
1
∩ E(Q). Since E(Q) is a group, then θ
Q
(R) =
R − Q ∈ E(Q). There exists
˜
Q,
˜
R ∈ W such that Q = P
1
+
˜
Q and R = P
1
+
˜
R.
Then θ(R) = R − Q =
˜
R −
˜
Q which is in V by definition of W . θ
Q
is of course
injective, which shows that V ∩ E(Q) is infinite.
Lemma 2.3. The differential
dx
1−2a
4
xz−3a
6
z
2
is invariant, meaning that if Q ∈ E(R)
is fixed and P = (x
P
, y
P
) ∈ E(R) is variable, then
dx
P +Q
1 − 2a
4
x
P +Q
z
P +Q
− 3a
6
z
2
P +Q
=
dx
P
1 − 2a
4
x
P
z
P
− 3a
6
z
2
P
.
Normat 3/2010 Hugues Verdure 133
Proof. See [4, Proposition III.5.1], which gives an “elegant” proof, and a hint for a
straightforward but messy proof.
Lemma 2.4. There exists open neighborhoods V and U of O and 0 respectively, and
a homeomorphism L : V −→ U such that if W is the neighborhood of lemma 2.1,
then
∀P, Q ∈ W, L(P + Q) = L(P ) + L(Q).
Proof. Let a, g be as in the proof of lemma 2.1, and V = {(x, g(x)), x ∈ (−a, a)}
which is an open neighborhood of O. Since g(0) = 0, reducing a if necessary, we
may assume that 1 −2a
4
xg(x) −3a
6
g(x)
2
> for 1 > > 0 on (−a, a), so that the
integral
M(t) =
Z
t
0
dx
1 − 2a
4
xg(x) − 3a
6
g(x)
2
, t ∈ (−a, a)
is well defined. Define
M : V −→ R
P 7−→ M(x
P
)
.
Then, for P, Q in W , we know that P + Q ∈ V . Moreover, we define
L(Q) =
Z
x
Q
0
dx
R
1 − 2a
4
x
R
g(x
R
) − 3a
6
g(x
R
)
2
=
Z
x
P +Q
x
P
dx
P +R
1 − 2a
4
x
P +R
z
P +R
− 3a
6
z
2
P +R
which in turns gives
L(P ) + L(Q) =
Z
x
P +Q
0
dx
1 − 2a
4
xg(x) − 3a
6
g(x)
2
= L(P + Q).
M is obviously infinitely differentiable, and since M
0
(0) = 1, reducing the open
neighborhood if necessary, we can assume that M (and therefore L) is an homeo-
morphism from V onto its image.
Lemma 2.5. Let L, V, U, W from the previous lemma and let G = L(E(Q) ∩W ) ⊂
L(W ) . Then G = L(W ).
Proof. Let x ∈ L(W ). Assume that x > 0 for simplicity. Let > 0 small enough so
that {y ∈ U, |y − x| < } ⊂ L(W ). We shall show that {y ∈ G, |x − y| < } 6= ∅. The
set L
−1
((−, )) is an open neighborhood of O, and has therefore an infinite number
of rational points from lemma 2.2. Let P 6= O be one of them, and taking −P if
necessary, we may assume that x(P ) > 0. Let ξ = L(P ). Then 0 < ξ < . Write
x = nξ + ρ with n ∈ N and 0 6 ρ < ξ. We show by induction that nP ∈ W ∩E(Q)
and nξ = L(nP ) ∈ G. This is true for n = 1. Assume that it is true for n − 1.
Then (n − 1)P, P ∈ W ∩ E(Q), so that nP = (n − 1)P + P ∈ V ∩ E(Q), and
nξ = (n − 1)ξ + ξ = L((n − 1)P ) + L(P ) = L(nP ). Now, |L(nP ) − x| = ρ < ,
which means that nP is acually in W since L is an homeomorphism. That shows
that nξ ∈ G.
134 Hugues Verdure Normat 3/2010
We can now prove the theorem:
Proof. The set E(Q) ∩ E(R)
O
is by definition closed in E(R). By the last lemma,
there exists an open O such that O ∈ O ⊂ E(Q) ∩ E(R)
O
. Then, for any point
P ∈ E(Q) ∩ E(R)
O
, P ∈ O
P
⊂ E(Q) ∩ E(R)
O
, which shows that E(Q) ∩ E(R)
O
is also open in E(R). Since it is obviously a subset of E(R)
O
which is a connected
component of E(R), we have proved the first part of the theorem.
If there exists a second component with a rational point Q on it, then we have an
homeomorphism
E(R)
O
−→ E(R)
Q
Q 7−→ P + Q
which shows that E(Q) ∩ E(R)
Q
is dense in E(R)
Q
, and therefore E(Q) dense in
E(R).
Remark. We never used the fact that the points were in E(Q). The only thing
we used was that it was a group with an infinite number of points. We actually
proved that any infinite subgroup of E(R) is dense in the identity component. This
result is well known and an immediate consequence of standards results on infinite
subgroups of E(R) = R/Z or R/Z ×Z/2Z.
As promised earlier, here is an example of a curve with an infinite number of
rational points, but those are not dense on the curve (by the theorem, we know
that they are dense in the compenent of the point of infinity though). Consider
the curve E : y
2
= x
3
− 47088x − 2522880 in the affine plane. An application of
the Mordell-Weil theorem tells us that the group E(Q) is generated by 2 points,
namely P
1
= (240, 0) which is a point of order 2, and P
2
= (8304/25, 537408/125)
which is a point of infinite order. Both of them lie on the connected component of
O. And there are no rational point on the other component. Namely, if P, Q are on
the identity component, then P + Q is there too. By symmetry, −Q is there, and
we have a continuous map
λ
Q
: E(R)
O
−→ E(R)
R 7−→ R + Q
such that λ
Q
(−Q) = O and λ
Q
(P ) = P + Q. That means that both O and P + Q
are on the same component. Since both generators are on this component, all the
rational points have to be. The next figure plots nP
2
for n ∈ [1, 1000] (the big dots)
and the curve itself.
Normat 3/2010 Hugues Verdure 135
-200 200 400 600 800 1000
-30000
-20000
-10000
10000
20000
30000
Figure 3 : nP
2
, 1 6 n 6 1000, and E(R)
Density of rational points on curves of higher degree
We won’t try to prove anything in this section, as this is far beyond the scope of
this article. We mention that in 1983, G. Faltings [1] proved that on these curves,
the number of rational points is finite.
References
[1] G. Faltings, Endlichkeitssätze für abelsche Varietäten über Zahlkörpern. Invent.
Math. 73 (1983), 349–366; Erratum ibid. 75 (1984), 381.
[2] B. Mazur, Modular curves and the Eisenstein ideal. IHES Pub. Math. 47 (1977),
33–186.
[3] B. Mazur, The topology of rational points. Exp. Math. 1, No.1, 35–45, 1992.
[4] J.H. Silverman, The arithmetic of elliptic curves. Number 106 in Graduate texts in
mathematics, Springer-Verlag, New York, 1986.
[5] J. H. Silverman, and J. Tate, Rational Points on Elliptic Curves. Undergraduate
Texts in Mathematics, Springer-Verlag, New York, 1992
