MA 426
MATHEMATICS DEPARTMENT
FOURTH YEAR UNDERGRADUATE EXAMS – APRIL 2014
Course Title: ELLIPTIC CURVES
Model Solution No: 1
a) Strong Bézout’s Theorem: Assume that the polynomials defining C and C 0
have no non-zero common factor. Let P1 , . . . , Pk be the points of intersection of
C(K) and C 0 (K). For each i, let ni be the multiplicity of Pi . Then we have
n1 + · · · + nk = nn0 .
Marks: 2
b) The line L has degree 1 by definition, and E is a cubic (so has degree 3). We saw
that E is always irreducible, hence never contains a line. Therefore, by the Strong
Bézout Theorem L ∩ E(K) has 3 points of intersection counted with multiplicities.
Marks: 2
c) Definition of group law: Let P, Q ∈ E(K), and L = P Q the line that goes
through them. Let R be the third point of intersection of L with E(K). By
definition P + Q is the third point of intersection of the line through R and ∞.
The line through P Q is the same as the line QP , so they both intersect E(K) at
the same third point of intersection R. Hence, P + Q = Q + P , and the group law
on E(K) is commutative.
Marks: 6 = 3+3
d) Since the line P1 P2 is non-vertical, it has an equation of the form y = mx + c,
where m 6= 0. The x-coordinates of the points of intersection of this line with
E(K) satisfy the cubic equation
(mx + c)2 = x3 + ax + b ⇐⇒ x3 − m2 x2 + (a − 2mc)x + b − c2 = 0.
The roots of this cubic are x1 , x2 , x3 and x1 + x2 + x3 = m2 ; and by definition of
the group law, we have −y3 = mx3 + c.
The points P1 , P2 are defined over K when x1 , x2 , y1 , y2 ∈ K. In that case, the
slope m and the y-intercept c are both in K. Hence x3 , y3 ∈ K, and P3 is defined
over K.
We just proved that P1 , P2 ∈ E(K) =⇒ P1 + P2 ∈ E(K). (Note that the
line P1 P2 being vertical means that P1 + P2 = 0 ∈ E(K). Also, P1 = 0 or
P2 = 0 =⇒ P1 + P2 ∈ E(K).) Morever P ∈ E(K) =⇒ −P ∈ E(K). So, E(K)
is a subgroup of E(K).
Marks: 10: 6+4
MA 426
2
−2
e) The slope of the tangent line at P is m = 3(2)
= 53 . So the equation of the
2(3)
tangent line at P is y − 3 = 35 (x − 2), which is the same as y = 53 x − 31 . Let
R = (xR , yR ) be the third point of intersection of this line with E. Then, we have
. From this, we get yR = 53 (− 11
) − 31 = − 64
.
xR = m2 − 2xP = ( 35 )2 − 2(2) = − 11
9
9
27
11 64
Therefore, Q = 2P = −R = (− 9 , 27 ).
Marks: 8
f) By definition of the law on E, −P = (x, −y). Now, P is a 2-torsion point if and only
if P + P = 0. This is equivalent to saying that P = (x, y) = (x, −y) = −P . This is
possible if and only if y = 0 and x satisfies the cubic polynomial X 3 +AX +B. This
polynomial has at most three roots counted with multiplicities. This accounted
for three 2-torsion points, to which we must add the zero element 0. Hence, the
curve E has at most four 2-torsion points counted with multiplicities.
√
√
For E : Y 2 = X 3 − 5X, the cubic polynomial factors as X(X
−
5)(X
−
5). So,
√
the set of 2-torsion points on E is E(Q)[2] = {0, (0, 0), (± 5, 0)}. There are two
that are defined over Q; namely, 0 and (0, 0).
Marks: 12 = 8 + 4
MA 426
MATHEMATICS DEPARTMENT
FOURTH YEAR UNDERGRADUATE EXAMS – APRIL 2014
Course Title: ELLIPTIC CURVES
Model Solution No: 2
a) By definition, for all P = (x, y) ∈ E(Fq ), we have
n
n
φqn (P ) = (xq , y q ) = φnq (x, y) = φnq (P ).
By the separability criterion, we have
cφnq −1 = cnφq − c1 = 0 − 1 = −1,
since φq is purely inseparable. Hence, φqn − 1 is separable.
Marks: 3 = 1 + 2
b) By definition, E(Fqn ) = ker(φqn − 1). Since φqn − 1 is separable, we have
#E(Fqn ) = # ker(φqn − 1) = deg(φqn − 1).
Marks: 4 = 2 + 2
c) We have α2 + β 2 = (α + β)2 − 2αβ = a2q − 2q, where aq is the trace of φq . Hence
α2 + β 2 is an integer, and Q(x) ∈ Z[x]. We have
Q(x) = (x2 − α2 )(x2 − β 2 ) = x4 − (α2 + β 2 )x2 + q 2 = (x − α)(x − β)(x + α)(x + β)
= (x2 − aq x + q)h(x),
where h(x) = (x + α)(x + β). Hence, Q(x) is divisible by (x2 − aq x + q), the
characteristic polynomial of φq . Since both x2 − aq x + q, Q(x) ∈ Z[x] are monic,
h(x) ∈ Z[x].
Marks: 4
d) Since x2 − aq x + q divides Q(x), we see that Q(φq ) = 0. This together with the fact
that φq2 = φ2q imply that the characteristic polynomial of φq2 is x2 −(α2 +β 2 )x+q 2 .
Therefore, we have
#E(Fq2 ) = q 2 + 1 − α2 − β 2 .
Marks: 5 = 2 + 3
e) We have E(F5 ) = {0, (1, 1), (1, 4), (2, 1), (2, 4)}; so #E(F5 ) = 5. This means that
a = 5 + 1 − 5 = 1. So the characteristic
polynomial of√ Frobenius is x2 − x + 5. The
√
1+ −19
roots of this polynomial are α =
and β = 1− 2−19 . So,
2
#E(F25 ) = 52 + 1 − α2 − β 2 = 35.
Marks: 4
MA 426
MATHEMATICS DEPARTMENT
FOURTH YEAR UNDERGRADUATE EXAMS – APRIL 2014
Course Title: ELLIPTIC CURVES
Model Solution No: 3
a) Write P = (x, y) with x, y ∈ Q. It is enough to show that for all prime p,
vp (x) < 0 ⇐⇒ vp (y) < 0 ⇐⇒ vp (x) = −2r,
vp (y) = −3r
(r > 0).
If vp (x) = −k < 0 then vp (x3 ) = −3k, vp (Ax) = vp (A) + vp (x) ≥ −k and
vp (B) ≥ 0 (using integrality of A, B). So vp (x3 + Ax + B) = −3k, but this also
equals vp (y 2 ) = 2vp (y) so is even. This means that k = 2r with r ∈ Z positive and
vp (x) = −2r, vp (y) = −3k/2 = −3r.
Conversely, if vp (x) ≥ 0 then vp (x3 + Ax + B) ≥ 0 so vp (y) ≥ 0.
Marks: 3
b) By definition rp (0) = 0. Let P = (x, y) ∈ E(Q) be non-zero. By using a), we
can write this point in projective coordinates as P = [ac : b : c3 ] (note that the
coordinates are coprime). So rp (P ) = [ac : b̄ : c̄3 ], where x̄ is the residue of x ∈ Z
modulo p.
Obviously, 0 ∈ ker(rp ). So, let P = (x, y) ∈ E(Q) be non-zero. Then, we have
rp (P ) = 0 ⇐⇒ [ac : b̄ : c̄3 ] = [0 : 1 : 0] ⇐⇒ p | c.
So, ker(rp ) = {0} ∪ {(x, y) ∈ E(Q) : vp (x) < 0}.
Marks: 4 = 2 + 2
c) The first part of the Lutz-Nagell Theorem states that torsion points in E(Q) are
integral. Therefore, such points are not in ker(rp ). Thus ker(rp ) ∩ E(Q)tors = {0}.
Marks: 4 = 2 + 2
d) The point P = (−1, 0) belongs to E(Q). Therefore the order of E(Q)tors is a
multiple of 2.
The discriminant of the curve E is ∆ = −(4(2)3 + 27(3)2 ) = −275 = −52 · 11. So
we may bound #E(Q)tors using any of the primes p 6= 5, 11.
For p = 3, E(F3 ) = {0, (0, 0), (1, 0), (2, 0)} has order 4. So the order of E(Q)tors is
either 2 or 4.
For p = 7, E(F7 ) = {0, (2, 1), (2, 6), (3, 1), (3, 6), (6, 0)} has order 6. So #E(Q)tors =
gcd(4, 6) = 2.
Marks: 7 = 2 + 1 + 4
e) By looking for points of small height, we find that Q = (3, 6) belongs to E(Q).
The order of Q is necessarily infinite.
Marks: 2
MA 426
MATHEMATICS DEPARTMENT
FOURTH YEAR UNDERGRADUATE EXAMS – APRIL 2014
Course Title: ELLIPTIC CURVES
Model Solution No: 4
a) The descent map δ : E(Q) → Q× /(Q× )2 is defined by

if P = 0,
 (Q× )2
B(Q× )2 if P = (0, 0),
δ(P ) :=

x(Q× )2 if P = (x, y) 6= (0, 0).
The image of δ consists of the classes b(Q× )2 such that
(i) b | B is squarefree, and
(ii) the equation
bl4 + Al2 m2 + (B/b)m4 = n2
has a solution (l, m, n) ∈ Z3 , with m > 0 and gcd(l, m) = gcd(n, m) = 1.
Since there are only finitely many (squarefree) divisors of B, im(δ) is finite.
Marks: 4 = 2 + 1 + 1
b) Mordell-Weil Theorem: Let E be an elliptic curve defined over Q. Then the
group E(Q) is finitely generated, i.e.,
E(Q) ∼
= Zr ⊕ T,
where r ≥ 0 is the rank of E(Q), and T is finite.
Marks: 2
c) To find the rank r, we need to work with both E and the curve E 0 : Y 2 =
X(X 2 + A0 X + B 0 ), where A0 = −2A and B 0 = A2 − 4B. In this specific example,
E 0 : Y 2 = X(X 2 + 6X + 29).
For the curve E, im(δ) lies in Q(S, 2), where S = {5}. By definition, δ(0, 0) =
−5(Q× )2 . Since X 2 − 3X − 5 is indefinite, we need to consider either of b = −1 or
5 in order find the image of δ. For b = −1, we need to solve the equation
−l4 − 3l2 m2 + 5m4 = n2 ,
in l, m, n ∈ Z (m > 0). This has the solution (1, 1, 1), which yields the point
P = (−1, 1) on E. Hence im(δ) = {(±Q× )2 , ±5(Q× )2 }.
For the curve E 0 , im(δ 0 ) lies in Q(S 0 , 2), where S 0 = {29}. By definition, δ(0, 0) =
29(Q× )2 . In this case, X 2 + 6X + 29 is definite. So im(δ 0 ) = {(Q× )2 , 29(Q× )2 }.
MA 426
Since B 0 = 29 is not a square, we have #E(Q)/2E(Q) = 2 · 4/2 = 4.
It is given that E(Q)tors ' Z/2Z. So E(Q) ' Zr × Z/2Z by the Mordell-Weil
Theorem. Hence, we have
E(Q)/2E(Q) ' (Z/2Z)r+1 = (Z/2Z)2 ,
which impies that r = 1.
Marks: 14 = 7 + 3 + 4
MA 426
MATHEMATICS DEPARTMENT
FOURTH YEAR UNDERGRADUATE EXAMS – APRIL 2014
Course Title: ELLIPTIC CURVES
Model Solution No: 5
a) The congruence number curve for n = 2 is the curve E : Y 2 = X 3 − 4X. We need
to show that E(Q) has rank r = 0. It will then follow from the lectures that 2 is
not a congruence number.
For the curve E, im(δ) lies in Q(S, 2), where S = {2}. By definition, we have
δ(0, 0) = −(Q× )2 .
Since X 2 − 4 is indefinite, we need to consider b = 2 or b = −2 in order to find
im(δ). For b = 2, we get the equation −2l4 + 2m4 = n2 . Taking (l, m, n) = (1, 1, 0),
we get the point (−2, 0). So, im(δ) = {±(Q× )2 , ±2(Q× )2 }.
For the other curve E 0 : Y 2 = X(X 2 + 16) to be considered, the image of δ 0 lies in
Q(S 0 , 2), where S 0 = {2}. By definition, δ(0, 0) = (Q× )2 . In this case, X 2 + 16 is
definite. To find im(δ 0 ), we need to let b = 2, and solve the equation
2l4 + 8m4 = 2(l4 + 4m4 ) = n2 ,
in l, m, n ∈ Z such that m > 0 and gcd(l, m) = gcd(n, m) = 1. This is impossible
since l and m are coprime. Hence, im(δ 0 ) = {(Q× )2 }. Since B 0 = 16 is a square, it
follows that #E(Q)/2E(Q) = 4.
By the Lutz-Nagell Theorem E(Q)tors ' (Z/2Z)2 . Let r be the rank of E(Q), so
that E(Q) ' Zr × (Z/2Z)2 by the Mordell-Weil Theorem. Then, we have
E(Q)/2E(Q) ' (Z/2Z)r+2 ' (Z/2Z)2 ,
which impies that r = 0.
Marks: 15 = 3 + 12
b) Let n ≥ 1 be a natural number, and write n = u2 n0 where u ∈ Z and n0 (is
squarefree). Then, n is a congruence number if and only if n0 is a congruence
number. Indeed, assume that there exist rational numbers 0 < a < b < c such
that a2 + b2 = c2 and n0 = ab/2. Then, it is easy to see that ua, ub, uc solve the
congruence number problem for n. The converse is also true. This together with
a) imply that 50 is not a congruence number.
Marks: 5