Course 428

Elliptic Curves II

Dr Timothy Murphy

Joly Theatre  Friday, 5 April 2002  16:15–17:45

Attempt 5 questions. (If you attempt more, only the best 5 will be counted.) All questions carry the same number of marks.

1. Prove Fermat’s Last Theorem either for n = 3 or for n = 4.

Answer:

n = 3 We work in the field

K = ℚ(ω),

where ω = e^{2π/3}, so that ω^{3} = 1 and 1 + ω + ω^{2} = 0. The integers in this field are the numbers

a + bω (a, b ∈ ℤ),

and the units are ±1, ±ω, ±ω^{2}. The ring of integers A = ℤ[ω], is a unique factorisation domain. Each number

α = x + yω ∈ K

has conjugate

ᾱ = x + yω^{2},

and

N(α) = αᾱ = x^{2} − xy + y^{2}.

Let

π = 1 − ω.

Then

π̄ = 1 − ω^{2} = −ω^{2}π.

Since

N(π) = 3,

it follows that 3 ramifies:

3 = ϵπ^{2},

where ϵ = −ω^{2}. The residues modπ are represented by 0, ±1. Lemma 1. If

α ≡ ±1 mod π

then

α^{3} ≡ ±1 mod π^{3}.

Proof. It is sufficient to prove the result if α ≡ 1 mod π, ie

α = 1 + πβ.

Then

α^{3} ≡ 1 + 3πβ + 3π^{2}β^{2} mod π^{3}.

Since

π^{2} | 3

the result follows.

We want to show that

x^{3} + y^{3} + z^{3} = 0

has no solution with x, y, z ∈ ℤ and xyz ≠ 0. We shall prove the more general result that for any unit ϵ ∈ A the equation

x^{3} + y^{3} + ϵz^{3} = 0

has no solution in A with xyz ≠ 0. Suppose we have such a solution. We may assume that gcd(x, y, z) = 1. Then

(x + y)(x + ωy)(x + ω^{2}y) = −ϵz^{3}.

n = 4 We have to show that the equation

x^{4} + y^{4} = z^{4}

has no non-trivial solution in ℤ (ie with xyz ≠ 0). In fact we prove the stronger result that

x^{4} + y^{4} = z^{2}

has no non-trivial solution. We may assume without loss of generality that x, y, z > 0 and that

gcd(x, y, z) = 1.

Recall that the integers x, y, z > 0 are said to form a Pythagorean triple if gcd(x, y, z) = 1 and

x^{2} + y^{2} = z^{2}.

Lemma 2. If x, y, z is Pythagorean triple then one of x, y is even and one is odd. The general Pythagorean triple with y even takes the form

x = u^{2} − v^{2}, y = 2uv, z = u^{2} + v^{2},

with u, v ≥ 0 and gcd(u, v) = 1.

In our case x^{2}, y^{2}, z is a Pythagorean triple. If y is even then

x^{2} = u^{2} − v^{2}, y^{2} = 2uv, z = u^{2} + v^{2}.

Since x is odd, one of u, v is odd and one is even. If u were even and v were odd then

x^{2} = u^{2} − v^{2} ≡ −1 mod 4,

which is impossible. Hence u is odd and v is even. Since y^{2} = 2uv and gcd(u, v) = 1 it follows that

u = s^{2}, v = 2t^{2},

with gcd(s, t) = 1. Thus

x^{2} = s^{4} − 4t^{4},

ie

x^{2} + 4t^{4} = s^{4}.

Now x, 2t^{2}, s^{2} is a Pythagorean triple. It follows that

x = a^{2} − b^{2}, 2t^{2} = 2ab, s^{2} = a^{2} + b^{2},

with gcd(a, b) = 1. Since t^{2} = ab and gcd(a, b) = 1, it follows that

a = X^{2}, b = Y^{2},

with gcd(X, Y) = 1. If we set s = Z then

X^{4} + Y^{4} = Z^{2}.

Thus one solution x, y, z of our equation leads to a second solution X, Y, Z. Also

Z^{4} = s^{4} = u^{2} < z,

since v ≠ 0. Thus each solution gives rise to a new solution with strictly smaller z, which is evidently impossible.

2. (a) Solve the equation

x^{3} + 3x − 1 = 0.

(b) Solve the equation

x^{4} + 4x − 1 = 0.

Answer:

(a) Set

x = u + v.

Then

x^{3} = u^{3} + v^{3} + 3uv(u + v).

Thus

x^{3} + 3x − 1 = u^{3} + v^{3} + 3(u + v)(uv + 1) + 1.

Let

uv + 1 = 0.

Then

u^{3} + v^{3} = −1,

while

uv = −1 ⟹ u^{3}v^{3} = −1.

Thus u^{3}, v^{3} are roots of the equation

t^{2} + t − 1 = 0.

Hence

u^{3}, v^{3} = \frac{−1 ± \sqrt{5}}{2}.

We conclude that the equation has the real solution

x = (\frac{−1 + \sqrt{5}}{2})^{1/3} + (\frac{−1 − \sqrt{5}}{2})^{1/3} ,

together with the complex solutions

x = (\frac{−1 + \sqrt{5}}{2})^{1/3} ω + (\frac{−1 − \sqrt{5}}{2})^{1/3} ω^{2},

x = (\frac{−1 + \sqrt{5}}{2})^{1/3} ω^{2} + (\frac{−1 − \sqrt{5}}{2})^{1/3} ω.

(b) We can write

x^{4} + 4x − 1 = (x^{2} + λ)^{2} − 2λx^{2} + 4x − (1 + λ^{2}).

The quadratic on the right will be a perfect square if

2^{2} = 2λ(1 + λ^{2}),

ie

λ^{3} + λ = 2.

One solution of this is λ = 1. Setting λ = 1, our equation reads:

(x^{2} + 1)^{2} = 2(x + 1)^{2}.

Thus

x^{2} + 1 = ±\sqrt{2}(x + 1),

ie

x^{2} − \sqrt{2}x + (1 − \sqrt{2}), x^{2} + \sqrt{2}x + (1 + \sqrt{2}).

Solving these two quadratic equations,

x = \frac{−\sqrt{2} ± \sqrt{2}\sqrt{2\sqrt{2} − 1}}{2},
x = \frac{\sqrt{2} ± i\sqrt{2}\sqrt{2\sqrt{2} + 1}}{2}.

3. Show that a torsion-free finitely-generated abelian group A is free, ie

A ≅ ℤ ⊕ ⋯ ⊕ ℤ.

Find a Z-basis e_{1}, …, e_{r} for the group

A = {(x, y, z) ∈ ℤ^{3} : x + 2y + 3z = 0},

ie a set of elements such that each element a ∈ A is uniquely expressible in the form

a = n_{1}e_{1} + ⋯ + n_{r}e_{r},

with n_{1}, …, n_{r} ∈ ℤ.

4. State Mordell’s Theorem, and sketch its derivation from the Weak Mordell Theorem (which states that if ℰ = ℰ (ℚ) is an elliptic curve over the rationals then the quotient-group ℰ /2ℰ is finite).

5. Determine the rank of the elliptic curve

ℰ (ℚ) : y^{2} = x^{3} − x.

Answer: Since the cubic p(x) = x^{3} − x has 3 rational roots, we have two methods of attacking the problem.

Method 1 The associated elliptic curve is

ℰ̃ : y^{2} = x^{3} + 4x.

Let the associated homomorphisms be

χ : ℰ → ℚ^{×}/ℚ^{×2}, χ̃ : ℰ̃ → ℚ^{×}/ℚ^{×2}.

Since p(x) has 3 rational roots, and b̃ = 4 is a perfect square, the rank r is given by

2^{r+2} = | im χ|| im χ̃|

Since b = −1, b̃ = 4,

im χ ⊂ {±1}, im χ̃ ⊂ {±1, ±2}.

Also

χ(0, 0) = b = −1, χ̃(0, 0) = b̃ = 4.

It follows that

im χ = {±1}.

If d ∈ im χ̃ and dd′ = b̃ = 4 then the equation

du^{4} + d′t^{4} = v^{2}

has a solution with gcd u, t = 1 = gcd v, t. If d < 0 then d′ < 0 and the equation evidently has no solution. Hence

im χ̃ ⊂ {1, 2}.

It follows that

2^{r+2} ≤ 2 · 2,

ie

r = 0.

Method 2 Let

χ_{0}, χ_{1}, χ_{−1} : ℰ → ℚ^{×}/ℚ^{×2}

be the associated homomorphisms. Thus

χ_{0}(x, y) = x mod ℚ^{×2},
χ_{1}(x, y) = x − 1 mod ℚ^{×2},
χ_{0}(x, y) = x + 1 mod ℚ^{×2},

except that

χ_{0}(0, 0) = p′(0) = −1,
χ_{1}(1, 0) = p′(1) = 2,
χ_{−1}(−1, 0) = p′(−1) = 2.

By Mordell’s Lemma,

2ℰ = ker(χ_{0} × χ_{1} × χ_{−1}),

and so

ℰ /2ℰ = im(χ_{0} × χ_{1} × χ_{−1}),

Moreover, since p(x) has 3 rational roots,

2^{r+2} = |ℰ /2ℰ |.

If (x, y) ∈ ℰ then

x = \frac{du^{2}}{t^{2}} , y = \frac{v^{2}}{t^{2}}

where d is square-free. Thus

x − 1 = \frac{du^{2} − t^{2}}{t^{2}} = \frac{ev^{2}}{t^{2}} ,
x + 1 = \frac{du^{2} + t^{2}}{t^{2}} = \frac{fw^{2}}{t^{2}} ,

where e, f are square-free; and

(χ_{0} × χ_{1} × χ_{−1})(x, y) = (d, e, f), if x ≠ 0, ±1. Moreover def is a perfect square, since

y^{2} = \frac{defu^{2}v^{2}w^{2}}{t^{6}} .

From above,

ev^{2} = du^{2} − t^{2}, fw^{2} = du^{2} + t^{2}.

It follows from this that

d > 0 ⟹ f > 0 ⟹ e > 0,
d < 0 ⟹ e < 0 ⟹ f > 0.

In particular f > 0 in all cases. Also

d | 1, e | 2, f | 2.

It follows that

im(χ_{0} × χ_{1} × χ_{−1}) ⊂ {(1, 1, 1), (−1, −1, 1), (1, 2, 2), (−1, −2, 2)}.

Hence

2^{r+2} ≤ 4 ⟹ r = 0.

6. Determine the group (ie the torsion group and rank) of the elliptic

curve

ℰ (ℚ) : y^{2} = x^{3} − 1.

Answer: First we make the constant term vanish by setting x = x′ +1. The equation becomes

ℰ : y^{2} = x′^{3} + 3x′^{2} + 3x′.

The associated elliptic curve is given by

ã = −2a, b̃ = a^{2} − 4b.

Thus the associated curve is

ℰ̃ : y^{2} = x^{3} − 6x^{2} − 3x.

If the associated homomorphisms are

χ : ℰ → ℚ^{×}/ℚ^{×2}, χ̃ : ℰ̃ → ℚ^{×}/ℚ^{×2}

then the rank r is given by

2^{r+1} = \frac{| im χ|| im χ̃|}{2},

since b̃ = −3 is not a perfect square. We have

im χ ⊂ {±1, ±3}, im χ̃ ⊂ {±1, ±3}.

Also

χ(0, 0) = 3, χ̃(0, 0) = −3.

If d | 3 and dd′ = 3 then d ∈ im χ if and only if the equation

du^{4} + 3u^{2}t^{2} + d′t^{4} = v^{2}

has a solution with gcd(u, t) = 1 = gcd(v, t). If d = −1 then d′ = 3, and the equation is

−u^{4} + 3u^{2}t^{2} + 3t^{4} = v^{2}.

Thus

−u^{4} ≡ v^{2} mod 3 ⟹ u ≡ v ≡ 0 mod 3.

Hence

9 | 3t^{4} ⟹ 3 | t,

so that gcd(t, u) > 1, contrary to hypothesis. It follows that

im χ = {1, 3}.

Similarly, if d | 3 and dd′ = 3 then d ∈ im χ̃ if and only if the equation

du^{4} − 6u^{2}t^{2} + d′t^{4} = v^{2}

has a solution with gcd(u, t) = 1 = gcd(v, t). If d = −1 then d′ = 3, and the equation is

−u^{4} − 6u^{2}t^{2} + 3t^{4} = v^{2}.

As before this implies that 3 | u, t, contrary to hypothesis. Hence

im χ̃ = {1, 3}.

It follows that

2^{r+2} = 2 · 2 ⟹ r = 0.

It remains to determine the torsion subgroup of ℰ. If (x, y) ∈ ℰ then x, y ∈ ℤ, and by Nagell-Lutz

y = 0 or y^{2} | Δ = −4b^{3} − 27c^{2} = −27.

Thus y = 0, ±1, ±3. If y = 0 then

x^{3} = 1 ⟹ x = 1.

If y = ±1 then

x^{3} = 2,

which has no rational solution. If y = ±3 then

x^{3} = 10,

which also has no rational solution. We conclude that

ℰ = {0, (1, 0)} ≅ ℤ/(2).

7. Determine the rank of the elliptic curve ℰ (ℚ) : y^{2} = x^{3} − 5x.

Answer: We observe that

P = (−1, 2) ∈ ℰ .

Let the tangent at P be

y = mx + c.

Then

m = \frac{3x^{2} − 5}{2y} = −\frac{1}{2}.

The tangent meets the curve where

(mx + c)^{2} = x^{3} − 5x.

Thus if the tangent meets the curve again at Q = (x, y) then (by considering the coefficient of x^{2})

−2 + x = m^{2}.

Thus x is non-integral, and so Q is of infinite order. Hence P is also of infinite order. In particular the rank

r ≥ 1.

The associated elliptic curve is ℰ̃ : y^{2} = x^{3} + 20x.

Since p(x) = x^{3} − 5x has just one rational root, and 20 is not a perfect square,

2^{r+2} = | im χ|| im χ̃|

where

χ : ℰ → ℚ^{×}/ℚ^{×2}, χ̃ : ℰ̃ → ℚ^{×}/ℚ^{×2}

are the auxiliary homomorphisms.

We have

im χ ⊂ {±1, ±5}, im χ̃ ⊂ {±1, ±2, ±5, ±10}.

Since

χ(0, 0) = −5, χ(−1, 2) = −1,

we have

im χ = {±1, ±5}.

If d | 20 and dd′ = 20 then d ∈ im χ̃ if and only if the equation

du^{4} + d′t^{4} = v^{2}

has a solution with t > 0 and gcd(u, t) = 1 = gcd(v, t). If d < 0 then d′ < 0 and this equation evidently has no solution. Hence

im χ̃ ⊂ {1, 2, 5, 10}.

Thus

2^{r+2} ≤ 2 · 2^{2},

ie

r ≤ 1.

Hence

r = 1.