Courbes elliptiques et formes modulaires/Elliptic Curves and Modular Forms.
MAT 6630
Homework 1. Due January 31, 2017
To get full credit solve 4 of the following problems (you are welcome to attempt them all).
The answers may be submitted in English or French.
1. A natural number n is said to be congruent if it is equal to the area of a right triangle
with rational sides. If we set a, b, c to be the lengths of the sides of the triangle, then n
is congruent if and only if the equations
a2 + b 2 = c 2
n=
ab
2
(1)
have solutions in Q.
Prove that for n any rational number there is a bijection between the solutions of (1)
with a, b, c in Q, with a 6= 0 and a 6= c, and the solutions of
Y 2 = X 3 − n2 X
(2)
with X, Y ∈ Q, Y 6= 0.
Hint: Take
nb
2n2
(a, b, c) → −
,
a+c a+c
(X, Y ) →
n2 − X 2 2Xn n2 + X 2
,−
,
Y
Y
Y
.
Solution: Notice that
2n2
a2 b 2
nb
ab2
a(c2 − a2 ) a2 (c2 − a2 )
,
,
,
−
= −
= −
a+c a+c
2(a + c) 2(a + c)
2(a + c)
2(a + c)
a(c − a) a2 (c − a)
= −
,
.
2
2
Then Y 6= 0 iff a 6= c and a 6= 0.
!
2
a(c
−
a)
a(c
−
a)
−
− n2
X(X 2 − n2 ) = −
2
2
a(c − a)
=−
2
a2 (c − a)2 a2 b2
−
4
4
=−
a3 (c − a) 2
c + a2 − 2ac − b2
8
a4 (c − a)2
a3 (c − a)
2
=−
2a − 2ac =
= Y 2.
8
4
then the first function is well defined.
On the other hand, in
2
n2 − X 2 2Xn n2 + X 2
,−
,
Y
Y
Y
,
2
a = 0 iff n −X
= 0 iff X 2 = n2 but that implies that Y = 0. On the other hand,
Y
2
2
2
2
a = c iff n −X
= n +X
iff X 2 = 0, but that implies that Y = 0.
Y
Y
2
2
a +b =
ab
1
=
2
2
n2 − X 2
Y
2
n2 − X 2
Y
2Xn
+ −
Y
2
=
n2 + X 2
Y
2
= c2 .
2Xn
nX(X 2 − n2 )
−
=
= n.
Y
Y2
Thus, the second function is also well-defined.
Now we check that the functions are mutual inverses:
 2
ab 2 a(c−a) 2 
a(c−a)
a(c−a)
ab
ab 2
+ − 2
2 − 2
2
2
 2 − − 2

,
−
,


2
2
a (c−a)
a (c−a)
a2 (c−a)
2
=
2
b2 − (c − a)2
b2 + (c − a)2
, b,
2(c − a)
2(c − a)
=
2
(c2 − a2 ) − (c − a)2
(c2 − a2 ) + (c − a)2
, b,
2(c − a)
2(c − a)
= (a, b, c) .
 
−
n2 −X 2
Y
n2 +X 2
Y
−
n2 −X 2
Y
2
=
X 2 − n2
Y
X2
Y
,
n2 −X 2
Y
2 n2 +X 2
Y
2
−
n2 −X 2
Y


.
2
2 2 !
X − n2
X
,
= (X, Y )
Y
Y
Since the functions are inverse to each other and well-defined, they are bijections.
2. Find a necessary and sufficient conditions for the line L : Y = cX +d to be an inflectional
tangent to the affine curve C : Y 2 = X 3 + aX + b, i.e. to meet C at a point P with
I(P, L ∩ C) = 3. Hence find a general formula for the elliptic curves C in canonical form
having a rational point of order 3.
3. Verify Bezout’s theorem for the curves Y 2 = X 3 − X 2 and Y = X 2 + X.
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Solution: First we look for intersection points in the affine plane:
C : y 2 = x3 − x2 ,
D : y = x2 + x.
Replacing the second equation into the first,
(x2 + x)2 = x3 − x2 ⇒ x2 (x2 + x + 2) = 0.
We obtain the points (0, 0), (α± , −2), where α± =
√
−1± −7
.
2
On the other hand, in the infinity line,
0 = x3 ,
0 = x2 .
We obtain the point (0 : 1 : 0).
Now we compute
I((0, 0), C ∩D) = I(y 2 −x3 +x2 , y −x2 −x) = I(y 2 −x3 +x2 −y(y −x2 −x), y −x2 −x)
= I(−x3 + x2 + yx2 + yx, y − x2 − x) = I(x, y − x2 − x) + I(−x2 + x + yx + y, y − x2 − x)
= I(x, y)+I(−x2 +x+yx+y −(y −x2 −x), y−x2 −x) = I(x, y)+I(2x+yx, y−x2 −x)
= I(x, y) + I(x, y − x2 − x) + I(2 + y, y − x2 − x) = 2I(x, y) + I(2 + y, y − x2 − x) = 2
since y + 2 is nonzero in (0, 0).
For α = α± , we have to analize
I((α, −2), C ∩ D) = I((y − 2)2 − (x + α)3 + (x + α)2 , y − 2 − (x + α)2 − (x + α)).
As before, we obtain,
2I(x + α, y − 2) + I(2 + (y − 2), (y − 2) − (x + α)2 − (x + α))
= I(y, y − x2 − 2αx − x) = I(y, −x2 − 2αx − x) = I(y, x) + I(y, −x − 2α − 1) = 1
since −x − 2α − 1 is nonzero in (0, 0).
Finally,
I(O, C ∩D) = I(z −x3 +x2 z, z −x2 −xz) = I(z −x3 +x2 z −x(z −x2 −xz), z −x2 −xz)
= I(z + 2x2 z − xz, z − x2 − xz) = I(z, z − x2 − xz) + I(1 + 2x2 − x, z − x2 − xz)
= I(z, −x2 ) = 2
Then we get 2 + 1 + 1 + 2 = 6 = deg C deg D, consistent with Bezout’s theorem.
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4. Prove that the equation
(X 2 − 2)(X 2 − 17)(X 2 − 34) = 0
(3)
has a solution in R, in Qp for all primes p, but it does not have solutions in Q.
Solution: It is clear that
(X 2 − 2)(X 2 − 17)(X 2 − 34) = 0
√
√
√
has solutions in R (and they are ± 2, ± 17, ± 34). It is also clear that there are
no solutions in Q (the solutions in R are not in Q).
34
,
or
equals 1. Indeed,
Now for p 6= 2, 17, we have that at least one of p2 , 17
p
p
17
if the first two symbols are −1, then the third must be 1, since 34
= p2
.
p
p
If ap = 1 (with a = 2, 17 or 34), then the equation X 2 − a has a nonzero solution
modulo p. The derivative is 2X which is different from zero. Then by Hensel’s
lemma, the solution extends to Zp and therefore to Qp .
If p = 17, we have X = 6 is a solution for X 2 − 2 modulo 17. We check the derivative
is 2X|6 = 12, different from zero modulo 17. By Hensel’s lemma, it extends to a
solution in Z17 and therefore Q17 .
If p = 2, we have X = 1 is a solution for X 2 − 17 modulo 8. We check the derivative
is 2X|1 = 2, different from zero modulo 4, and we can apply Hensel’s lemma again
to obtain a solution in Z2 and therefore in Q2 .
5. (a) Show that for 0 6= x ∈ Q,
Y
|x|p =
p
1
|x|
where the product is taken over all primes p = 2, 3, 5, . . .
(b) If x ∈ Q and |x|p ≤ 1 for every prime p, show that x ∈ Z.
(c) Show that there are no solutions to X 4 = 2 in any of the fields Qp with p = 2, 3, 5.
What happens with X 4 = 4?
Solution:
(a) Let x =
a
b
∈ Q. We can factor a and b and get a prime factorization for x:
x = ±pe11 . . . perr
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where the exponents can be either positive or negative. Then we note
1
i=j
pi
|pi |pj =
1 i 6= j
Therefore,
|x|pi = |pei i |pi =
Y
|x|p =
p
pe11
1
pei i
1
1
=
.
e
. . . pr r
|x|
(b) If x = 0, then x ∈ Z. If x 6= 0, we can write it as a product of primes as in the
previous point. Then |x|pi = p1ei ≤ 1 iff ei ≥ 0. If this happens for every prime, then
i
x ∈ Z.
P
i
(c) Write X = 21k ∞
i=0 ai 2 . Then
!4
∞
X
ai 2i
= 24k+1 .
i=0
By looking at the powers of 2 dividing in each side, we get that ai = 0 when i < k.
We divide by 24k , we obtain
∞
X
!4
ai 2i−k
= 2.
i=k
Looking modulo 2, ak = 0. But then 24 divides the left term and we obtain a
contradiction. Thus, there are no solutions in Q2 to X 4 = 2.
A similar idea applies to X 4 = 4, leading to no solutions in Q2 .
P
i
For 3, we write X = 31k ∞
i=0 ai 3 . Again
!4
∞
X
ai 3i
= 34k 2.
i=0
By comparing powers of 3 dividing in both sides, we get that ai = 0 for i < k. We
divide by 3k and can reduce to the case
!4
∞
X
ai 3i
=2
i=0
Looking modulo 3, a40 ≡ 2 mod 3, but that is impossible, since
is no solution in Q3 .
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2
3
= −1. Thus, there
On the other hand, for X 4 − 4, we have that 1 is a solution modulo 3. The derivative
is 4X 3 |1 = 4, and we can apply Hensel’s lemma to extend the solution modulo 3 to
a solution in Z3 and therefore in Q3 .
The case of 5 is similar. Here ±2 is not a square modulo 5, and that implies that
both 2 and 4 are not squares modulo 5, so neither equation has solutions in Q5 .
6. Consider the plane affine curve C : Y 2 = X 3 + p. Prove that the point (0, 0) on the
reduced curve over Fp does not lift to Z2p . Why does this not violate Hensel’s lemma?
Solution: In Fp , the curve is given by Y 2 = X 3 . Then (0, 0) is clearly a solution.
If we had a solution in Zp , we should have a solution (a, b) in Z/p2 Z that is congruent
to (0, 0) modulo p. Then (a, b) = (a1 p, b1 p). We have
b21 p2 ≡ a31 p3 + p mod p2 .
But that implies
0 ≡ p mod p2 ,
a contradiction.
This does not contradict Hensel’s lemma because the derivatives of f = Y 2 − X 3 − p
are
∂f = −3X 2 (0,0) = 0.
∂x (0,0)
∂f = 2Y |(0,0) = 0.
∂y (0,0)
Therefore the hypotheses are not met.
7. Consider the projective variety V : Y 2 Z = X 3 + 17Z 3 , and let P1 = (x1 , y1 ), and
P2 = (x2 , y2 ) be distinct points of V ∩ U2 . . Let L be the line through P1 and P2 .
(a) Show that V ∩ L = {P1 , P2 , P3 } and express P3 = (x3 , y3 ) in terms of P1 and P2 .
(If L is tangent to V , then P3 may be equal to P1 or P2 ).
(b) Calculate P3 for P1 = (−1, 4) and P2 = (2, 5).
(c) Show that if P1 , P2 ∈ V (Q), then P3 ∈ V (Q).
Solution: (a) By Bezout’s theorem, there are three points of intersection. Let us
assume that P1 + P2 6= O. Because the points are different, the line through them is
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given by
(y − y1 ) =
(y2 − y1 )
(x − x1 ) = m(x − x1 ).
(x2 − x1 )
Now we plug this into the equation
(y1 + m(x − x1 ))2 = x3 + 17.
From there, we see that the sum of the roots is m2 . Therefore,
2
y2 − y1
y 2 + y12 − 2y1 y2 − (x32 − x21 x2 − x1 x22 + x31 )
x3 =
− x1 − x2 = 2
x2 − x1
(x2 − x1 )2
34 − 2y1 y2 + x21 x2 + x1 x22
x3 =
(x2 − x1 )2
y3 = y1 +
(y2 − y1 )
(x3 − x1 )
(x2 − x1 )
(b) We apply the formulas
34 − 2 · 4 · 5 + (−1)2 2 + (−1)22
8
=− ,
2
(2 − (−1))
9
8
109
(5 − 4)
− − (−1) =
.
y3 = 4 +
(2 − (−1))
9
27
x3 =
(c) Since x3 , y3 are rational functions of x1 , x2 , y1 , y2 , then they must be rational if
x1 , x2 , y1 , y2 are rational.
8. Consider the projective variety V : X 2 + Y 2 = Z 2 . Show that the rational map
V →k
(X : Y : Z) →
X +Z
Y
is regular in (−1 : 0 : 1). Compute its value.
Remark: In fact the rational map V → P1 (k) given by (X : Y : Z) → (X + Z : Y ) is
regular.
Y
Solution: Notice that in V , Z 2 − X 2 = Y 2 , from where X+Z
= Z−X
. But the
Y
righthand side can be evaluated at (−1 : 0 : 1) and it yields the value 0.
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