Course 428 — Sample Paper

Timothy Murphy

13 April 2004

Attempt 7 questions. All carry the same mark. The word ‘curve’ always means projective curve.

1. Explain informally how two points on an elliptic curve are added. Find the sum P + Q of the points P = (0, 1), Q = (1, 2) on the curve

y^{2} = x^{3} + 2x + 1

over the rationals ℚ. What is 2P? Answer: Let E be the elliptic curve. We choose any point O ∈ E as the zero point. Suppose P, Q ∈ E, the elliptic curve in question. The line P Q meets E in a third point R (which may coincice with P or Q). We set

P ∗ Q = R.

If P = Q then we take the tangent at P in place of the line P Q. Now we set

P + Q = O ∗ (P ∗ Q).

Suppose the elliptic curve is given in the standard Weierstrass form

y^{2} + c_{1}xy + c_{3}y = x^{3} + c_{2}x^{2} + c_{4}x + c_{6}.

More precisely, E is the projective curve

Y^{2}Z + c_{1}XY Z + c_{3}Y Z^{2} = X^{3} + c_{2}X^{2}Z + c_{4}XZ^{2} + c_{6}Z^{3}.

In this case we normally take O = [0, 1, 0]. If now

R = P ∗ Q = (x, y)

then

P + Q = (x, −y).

Now consider the points

P = (0, 1), Q = (1, 2)

on the elliptic curve

E(ℚ) : y^{2} = x^{3} + 2x + 1.

Suppose P Q is the line

y = mx + c.

Then

m = \frac{2 − 1}{1 − 0} = 1.

Thus the line is

y − 1 = x,

ie

y = x + 1.

This line meets the curve where

(mx + c)^{2} = x^{3} + 2x + 1.

Thus if P ∗ Q = (x_{2}, y_{2}) then

0 + 1 + x_{2} = m^{2} = 1,

ie

x_{2} = 0.

Hence

y_{2} = x_{2} + 1 = 1.

Thus

P ∗ Q = (0, 1) = P,

and so

P + Q = (0, −1).

Since the line P Q meets the curve again at P, this line is the tangent at P. Hence

P ∗ P = Q = (1, 2),

and so

2P = (1, −2).

2. Show that all cubics through 8 given points in general position in the plane pass through a 9th point.

Hence or otherwise show that addition on an elliptic curve is associative.

Answer:

(a) Let the points be P_{i} (i = 1 − 10). A cubic curve Γ has 10 coefficients:

c_{1}X^{3}+c_{2}X^{2}Y +c_{3}X^{2}Z+C_{4}XY^{2}+C_{5}XY Z+C_{6}XZ^{2}+c_{7}Y^{3}+c_{8}Y^{2}Z+c_{9}Y Z^{2}+C_{10}Z^{3} = 0.

The requirement that Γ passes through P_{i} gives 8 homogeneous linear conditions on these 10 coefficients. The solution space has dimension ≥ 10 − 8 = 2. In other words the cubics form a pencil of homogeneous dimension ≥ 1. We may suppose that no 4 of the points are collinear, and that the points do not all lie on a conic. We claim this in this case the dimension must be exactly 1. For suppose it is ≥ 2. Then we can find a cubic in the pencil passing through any further 2 points. Let us choose 2 points U, V on the line P_{7}P_{8}. Then the line ℓ = P_{7}P_{8}UV must lie entirly in the cubic, which must therefore split into

Γ = ℓC,

where C is a conic. Thus the 6 points P_{i}(i = 1 − 6) must lie on a conic. By the same argument, any 6 of the 8 given points must lie on a conic. But there is only one conic through 5 points Q_{j}(j = 1 − 5), no 4 of which are collinear. For suppose first that three of the points are collinear, say

m = Q_{1}Q_{2}Q_{3}.

Then the only conic through the 5 points is

C = mn,

where

n = Q_{4}Q_{5}.

Now suppose no three of the points are collinear; and suppose there are two conics through the 5 points. Then the conics through the points form a pencil of projective dimension ≥ 1, and we can find a conic in the pencil through any further point W.

Choose W on m = Q_{1}Q_{2}. Then the conic must degenerate into two lines,

C = mn,

and Q_{3}, Q_{4}, Q_{5} must lie on the line n, contrary to hypothesis. Thus, returning to the 8 points P_{i}, there is a unique conic through P_{1}, P_{2}, P_{3}, P_{4}, P_{5}. But as we have seen, this conic must pass through P_{6}; and by the same argument it must also pass through P_{7} and P_{8}. Hence all 8 points lie on a conic, contrary to hypothesis. Therefore the pencil is of dimension 1; and if Γ_{1}, Γ_{2} are two curves in the pencil then the general curve in the pencil is

Γ = λΓ_{1} + μΓ_{2}.

The curves Γ_{1}, Γ_{2} meet in the 8 points P_{i}. Let the curves have equations

F_{1}(X, Y, Z) = 0, F_{2}(X, Y, Z) = 0.

We can regard these as cubics in Z with coefficients in X, Y. If we form the resultant of the two cubics we obtain a homogeneous polynomial R(X, Y) of degree 9 in X, Y, whose vanishing is a condition for the two cubics to have a root in common. The 8 points P_{i} will provide 8 roots for this equation. By considering the sum of the roots, it follows that there is a 9th root in the field k we are working over. Thus the two cubics meet in a 9th point P_{9} = [X_{9}, Y_{9}, Z_{9}]. Moreover, by the argument above Y_{9}/X_{9} ∈ k; and similarly Z_{9}/X_{9} ∈ k. Hence P_{9} is defined over k.

(b) Suppose P, Q, R ∈ E. We have to show that

(P + Q) + R = P + (Q + R).

By definition,

P + Q = O ∗ (P ∗ Q),

where P ∗ Q is the point where P Q meets the curve again, and O is the point chosen as zero point. Thus we have to show that

O ∗ ((P + Q) ∗ R) = O ∗ (P ∗ (Q + R)) .

Since

U ∗ V = U ∗ W ⟺ V = W

it is sufficient to show that

(P + Q) ∗ R = P ∗ (Q + R),

ie

(O ∗ (P ∗ Q)) ∗ R = P ∗ (O ∗ (Q ∗ R)) .

Note that

U ∗ V = W ⟺ V ∗ W = U.

Thus if we set

P ∗ Q = X, Q ∗ R = Y

then

P = Q ∗ X, R = Q ∗ Y,

and our equation becomes

(O ∗ X) ∗ (Q ∗ Y) = (Q ∗ X) ∗ (O ∗ Y),

ie (since V ∗ U = U ∗ V)

(O ∗ X) ∗ (Y ∗ Q) = (O ∗ Y) ∗ (X ∗ Q).

Thus the result will follow if we show that

(P ∗ Q) ∗ (R ∗ S) = (P ∗ R) ∗ (Q ∗ S)  (†)

for any 4 points P, Q, R, S ∈ E. [Conversely, if the operation + is associative then it defines an abelian group structure on E, with

−P = O ∗ P

and

P ∗ Q = −(P + Q).

In this case,

(P ∗ Q) ∗ (R ∗ S) = P + Q + R + S = (P ∗ R) ∗ (Q ∗ S).

Thus the identity (†) holds if and only if the operation is associative.] Now let us apply the 8-point theorem to the points,

P, Q, R, S, U = P ∗ Q, V = R ∗ S, W = P ∗ R, X = Q ∗ S.

Let us define lines as follows:

ℓ = P QU, m = RSV, n = W X,
f = P RW, g = QSX, h = UV.

Then the degenerate cubics

ℓmn, fgh

pass through the 8 points, and so must have a 9th point in common. This 9th point must be where the line n meet E again, ie the point

W ∗ X = (P ∗ R) ∗ (Q ∗ S).

But by the same argument, it must be the point where the line h meet E again, ie the point

U ∗ V = (P ∗ Q) ∗ (R ∗ S).

We conclude that

(P ∗ Q) ∗ (R ∗ S) = (P ∗ R) ∗ (Q ∗ S),

as required. [This argument assumes, on the face of it, that the 9 points arising in this was are distinct. There are several ways of extending the reult to cover the special cases when some of the points coincide. Thus we could extend the definition of the pencil of cubics so that if eg P = Q then our pencil consisted of the cubics which had the same tangent at P as E. Alternatively, we could justify the general result when k = ℂ, say, by continuity. The result must then be an algebraic identity which will hold over all fields. Thirdly, we could appeal to the “Irrelevance of Algebraic Inequalities”, which states that if an identity f(x_{1}, …, x_{n}) = 0 holds subject to an inequailty g(x_{1}, …, x_{n}) ≠ 0 then it must hold in all cases. But the question is long enough as it is, and I think one can assume that no examiner would expect the student to go into this issue.]

3. Find the order of the point P = (0, 0) on the elliptic curve

y^{2} + y = x^{3} − x.

Answer: We have

(2y + 1)\frac{dy}{dx} = 3x^{2} − 1.

Thus the slope at (x, y) is

m = \frac{3x^{2} − 1}{2y + 1}.

The tangent

y = mx + c

at (x, y) meets the curve where

(mx + c)^{2} + (mx + c) = x^{3} − x.

If this meets the curve again at (x_{2}, y_{2}) then

2x + x_{2} = m^{2}.

In particular the slope at P is

m = \frac{−1}{1} = −1,

so the tangent

y = −x

meets the curve again where

x_{2} = 1,

ie at

Q = (1, −1).

Hence

Q = −2P.

The slope at Q is

m = \frac{2}{−1} = −2.

Thus the tangent at Q is

y + 1 = −(x − 1),

ie

y = −x − 2,

and this meets the curve again where

2 + x_{2} = 4,

ie

x_{2} = 2,

ie at (2, −4). Thus

−2Q = R = (2, −4).

The slope at R is

m = \frac{3 − 1}{−8 + 1} = −\frac{2}{7}.

It follows that R is of infinite order, and so therefore is P.

4. Sketch the proof of the Nagell-Lutz Theorem, that a point P = (x, y) of finite order on the elliptic curve

E(ℚ) : y^{2} + Axy + By = x^{3} + ax^{2} + bx + c,

where A, B, a, b, c ∈ ℤ, necessarily has integral coordinates x, y ∈ ℤ.

Answer: It is sufficient to show that

∥x∥_{p} ≤ 1, ∥y∥_{p} ≤ 1

for each prime p.

Let us write ∥ · ∥ for ∥ · ∥_{p}. Note that

∥x∥ > 1 ⟺ ∥y∥ > 1;

and if this is so then the terms in y^{2}, x^{3} must balance, ie

∥x∥^{3} = ∥y∥^{2}.

since otherwise y^{2} or x^{3} would dominate [ie would have larger p-adic value than any other term]. Let us suppose that this is the case.

Suppose first that p ≠ 2. In this case we can bring the equation to standard form

E(ℚ_{p}) : y^{′2} = x^{3} + a′x^{2} + b′x + c′

by “completing the square” on the left, with the change of coordinates

y′ = y + c_{1}x/2 + c_{3}/2.

The coefficients a′, b′, c′ will still be p-adic integers. Also y′ will be a p-adic integer if and only if y is a p-adic integer. So we may assume (if p ≠ 2) that the equation takes this simpler form.

Lemma. Suppose

E = E(ℚ_{p}) : y^{2} = x^{3} + ax^{2} + bx + c

is an elliptic curve with a, b, c ∈ ℤ_{p}; and suppose P = (x, y) is a point on E of finite order. Then x, y ∈ ℤ_{p}.

Remark: the result (and proof) still hold if p = 2. However, the full equation cannot in general be reduced to this simpler form if p = 2.

Proof. In homogeneous terms the equation takes the form

Y^{2}Z = X^{3} + aX^{2}Z + bXZ^{2} + cZ^{3},

where

x = X/Z, y = Y/Z.

Since y ≠ 0 we may take Y = 1, so that

Z = X^{3} + aX^{2}Z + bXZ^{2} + cZ^{3},  (∗)

with

x = X/Z, y = 1/Z,

or conversely,

X = x/y, Z = 1/y.

Note that

∥X∥ = ∥x∥/∥y∥ = ∥x∥^{−1/2} < 1;
∥Z∥ = 1/∥y∥ < 1.

By repeatedly substituting for Z in the right-hand side of (∗) we obtain a series expansion for Z in terms of X, starting

Z = X^{3} + aX^{2}(X^{3} + ⋯) + bX(X^{3} + ⋯)^{2} + cX(X^{3} + ⋯)^{3}
Z = X^{3} + O(X^{5}).

Let

E_{(p^{r})} = {P = [X, 1, Z] ∈ E(ℚ_{p}) : ∥X∥ ≤ p^{−r}, ∥Z∥ < 1}.

Lemma. E_{(p^{r})} is a subgroup of E(ℚ_{p}). Moreover, if

P_{1} = [X_{1}, 1, Z_{1}], P_{2} = [X_{2}, 1, Z_{2}] ∈ E_{(p^{r})},

and

P_{3} = P_{1} + P_{2} = [X_{3}, 1, Y_{3}]

then

X_{3} ≡ X_{1} + X_{2} mod p^{3r}.

Proof. Suppose P_{1}P_{2} is the line

Z = mX + d.

Then

m = \frac{Z_{2} − Z_{1}}{X_{2} − X_{1}} .

Subtracting the equation for P_{1} from that for P_{2},

\begin{matrix}(Z_{2} − Z_{1}) = (X_{2}^{3} − X_{1}^{3}) + a(X_{2}^{2}Z_{2} − X_{1}^{2}Z_{1}) \\ +b(X_{2}Z_{2}^{2} − X_{1}Z_{1}^{2}) + c(Z_{2}^{3} − Z_{1}^{3}). \end{matrix}

We can write this as

\begin{matrix}(Z_{2} − Z_{1}) = (X_{2}^{3} − X_{1}^{3}) + a(X_{2}^{2} − X_{1}^{2})Z_{2} − aX_{1}^{2}(Z_{2} − Z_{1}) \\ +b(X_{2} − X_{1})Z_{2} − bX_{1}(Z_{2}^{2} − Z_{1}^{2}) + c(Z_{2}^{3} − Z_{1}^{3}). \end{matrix}

Hence

\begin{matrix}m = (X_{1}^{2} + X_{1}X_{2} + X_{2}^{2}) + a(X_{1} + X_{2})Z_{2} − mX_{1}^{2} \\ +bZ_{2} − bmX_{1}(Z_{1} + Z_{2}) + cm(Z_{1}^{2} + Z_{1}Z_{2} + Z_{2}^{2}). \end{matrix}

Thus

m = \frac{U}{V} ,

where

U = X_{1}^{2} + X_{1}X_{2} + X_{2}^{2} + ⋯ ≡ 0 mod p^{2r}
V = 1 − X_{2}^{2} + ⋯ ≡ 1bmodp^{2r}.

It follows that

m ≡ 0 mod p^{2r}.

Hence

d = Z_{1} + mX_{1} ≡ 0 mod p^{3r}.

Suppose

P = [X, 1, Z] = [X/Z, 1/Z, 1] = (X/Z, 1/Z) = (x, y), say. Then

−P = (x, −y) = (X/Z, −1/Z) = [X/Z, −1/Z, 1] = [−X, 1, −Z]. In other words,

−[X, 1, Z] = [−X, 1, −Z].

In particular,

−P_{3} = [−X_{3}, 1, −Z_{3}].

Now P_{1}, P_{2}, −P_{3} lie on the line Z = mX + d. Thus, on substituting Z = mX + d in (∗) and equating the coefficients of X^{2} and X^{3},

X_{1} + X_{2} − X_{3} = −\frac{(a + 2bm + 3cm^{2})d)}{1 + am + bm^{2} + cm^{3}}
≡ 0 mod p^{3r}.

In other words,

X_{3} ≡ X_{1} + X_{2} mod p^{3r}.

In particular,

∥X_{1}∥, ∥X_{2}∥ ≤ p^{−r} ⟹ ∥X_{3}∥ ≤ p^{−r}.

Thus

P_{1}, P_{2} ∈ E_{p^{r}} ⟹ P_{1} + P_{2} ∈ E_{p^{r}},

ie E_{(p^{r})} is a subgroup.

Lemma. The only point of finite order in E_{(p)} is P = 0.

Proof. It is sufficient to show that there is no point of prime order q. For if P is of order q_{1} ⋯ q_{r} then (q_{2} ⋯ q_{r})P is of order q_{1}.

If q ≠ p this follows at once from the Lemma. For suppose P = [X, 1, Z] ∈ E_{(p)}, and suppose

∥X∥ = p^{−r}.

Then (writing X(qP) for the X-coordinate of qP)

X(qP) ≡ qX mod p^{3r}
≢ 0 mod p^{r}.

The same argument also holds if q = p, since

X(pP) ≡ pX mod p^{3r},

while

∥pX∥ = p^{−(r+1)}.

Since r + 1 < 3r it follows that

X(pP) ≢ 0 mod p^{r+1}.

We have reduced the problem to the case p = 2, where we have to return to the original equation.

The argument is similar, but more complicated.

Lemma. Suppose

E = E(ℚ_{p}) : y^{2} + Axy + By = x^{3} + ax^{2} + bx + c

is an elliptic curve with A, B, a, b, c ∈ ℤ_{p}. Then

E_{(p^{r})} = {P = [X, 1, Z] ∈ E : ∥X∥ ≤ p^{−r}, ∥Z∥ ≤ p^{−3r}}

is a subgroup of E for each r ≥ 1. Moreover, if

P_{1} = [X_{1}, 1, Z_{1}], P_{2} = [X_{2}, 1, Z_{2}] ∈ E_{(p^{r})}

and

P_{3} = P_{1} + P_{2} = [X_{3}, 1, Z_{3}]

then

X_{3} ≡ X_{1} + X_{2} mod p^{2r}.

Proof. As before, on passing to (X, Z) coordinates we can express Z as a power-series in X, though now there is a term in X^{4}:

Z = X^{3} − c_{1}X^{4} + O(X^{5}).

The computation of m and d for the line

P_{1}P_{2} : Z = mX + d

is a little more complicated, but as before

∥m∥ ≤ p^{−2r}, ∥d∥ ≤ p^{−3r}.

[In detail, the equation of the curve is now

Z + AXZ + BZ^{2} = X^{3} + aX^{2}Z + bXZ^{2} + cZ^{3}.

Subtracting the equation for P_{1} from that for P_{2},

(Z_{2} − Z_{1}) + A(X_{2}Z_{2} − X_{1}Z_{1}) + B(Z_{2}^{2} − Z_{1}^{2}) =

(X_{2}^{3} − X_{1}^{3}) + a(X_{2}^{2}Z_{2} − X_{1}^{2}Z_{1}) + b(X_{2}Z_{2}^{2} − X_{1}Z_{1}^{2}) + c(Z_{2}^{3} − Z_{1}^{3}).

We can write this as

(Z_{2} − Z_{1}) (1 + AX_{2} + B(Z_{2} + Z_{1}) − aX_{2}^{2} − b(X_{2}(Z_{2} + Z_{1}) − c(Z_{2}^{2} + Z_{2}Z_{1} + Z_{1}^{2})) = (X_{2} − X_{1}) (−AZ_{1} + (X_{2}^{2} + X_{2}X_{1} + X_{1}^{2}) + aZ_{1}(X_{2} + X_{1}) + bZ_{1}^{2}) .

Thus

m = \frac{Z_{2} − Z_{1}}{X_{2} − X_{1}} = \frac{U}{V} ,

where

U ≡ 0 mod p^{2r}, V ≡ 1 mod p^{2r};

and so

m ≡ 0 mod p^{2r}, d ≡ 0 mod p^{3r},

as before.]

Now suppose

P′_{3} = −P_{3} = [X′_{3}, 1, Z′_{3}].

Then P_{1}, P_{2}, P′_{3} lie on the line Z = mX + d. This line meets the curve where

(mX+d)+AX(mX+d)+B(mX+d)^{2} = X^{3}+aX^{2}(mX+d)+bX(mX+d)^{2}+c(mX+d)^{3}.

Thus

X_{1} + X_{2} + X′_{3} = \frac{Am + Bm^{2} − (a + 2bm + 3cm^{2})d}{1 + am + bm^{2} + cm^{3}}
≡ 0 mod p^{2r}.

[Note that the term Am means that we only have equivalence modp^{2r} rather than modp^{3r} as before.] Hence

X′_{3} ≡ −(X_{1} + X_{2}) mod p^{2r}.

In particular,

∥X_{1}∥, ∥X_{2}∥ ≡ 0 mod p^{r} ⟹ ∥X′_{3}∥ ≡ 0 mod p^{r}
⟹ ∥Z′_{3}∥ ≡ 0 mod p^{3r}

since Z_{3} = mX_{3} + d.

The formula for −[X, 1, Z] is a little more complicated than before. We know that −(x_{0}, y_{0}) = (x_{0}, y′_{0}) is the point where the line x = x_{0} meets the curve again. Considering the given equation as a quadratic in y, we see that

y_{0} + y′_{0} = −(c_{1}x + c_{3}).

Thus

−(x, y) = (x, −(y + c_{1}x + c_{3}));

or in (X, Z)-coordinates,

−[X, 1, Z] = −(X/Z, 1/Z)

= (X/Z, −(1/Z + c_{1}X/Z + c_{3}) = [X, −(1 + c_{1}X + c_{3}Z), Z]

= [−\frac{X}{1 + c_{1}X + c_{3}Z} , 1, −\frac{Z}{1 + c_{1}X + c_{3}Z} ] .

In particular,

X′_{3} = −\frac{X_{3}}{1 + c_{1}X_{3} + c_{3}Z_{3}}
= −X_{3} + c_{1}X_{3}^{2} + ⋯
≡ −X_{3} mod p^{2r}.

Thus

X_{3} ≡ X_{1} + X_{2} mod p^{2r}.

If r ≥ 2 then

X_{3} ≡ X_{1} + X_{2} mod p^{r+2};

and our argument in the simpler case above yields the following result. Lemma. Suppose

E = E(ℚ_{p}) : y^{2} + Axy + By = x^{3} + ax^{2} + bx + c,

where A, B, a, b, c ∈ ℤ_{p}. Then the only point in E_{(p^{2})} of finite order is P = 0. Finally, suppose p = 2; and suppose P = [X, 1, Z] ∈ E_{(2)} is of finite order. As before, we may assume that P is of prime order q. Let

∥X∥_{2} = 2^{−r}.

If q is odd then our previous argument holds, since

X(qP) ≡ qX mod 2^{2r}

and

∥qX∥_{2} = ∥X∥ = 2^{−r}.

Our previous argument also holds if q = 2 and r ≥ 2, since in that case

X(2P) ≡ 2X mod 2^{2r}

and

∥2X∥ = 2^{−r+1} > 2^{−2r}.

We are left with the case q = 2, r = 1. In this case either 2P = 0 or else

2P ∈ E_{(2^{2})},

in which case it follows from our previous argument that 2P is not of finite order.

5. Find all points of finite order on the elliptic curve

E(ℚ) : y^{2} = x^{3} + 17.

Answer: By Nagell-Lütz, if P = (x, y) is of finite order then x, y ∈ ℤ and either y = 0 or

y^{2} | D,

where

D = −(4b^{3} + 27c^{2}) = 27 · 17^{2},

Thus

y = 0 or y | 3 · 17.

There is no integral solution with y = 0.

If 17 | y then

17 | x^{3} ⟹ 17 | x
⟹ 17 | y^{2}
⟹ 17^{2} | y^{2}
⟹ 17^{2} | 17,

which is absurd.

Hence

y = {±1, ±3}.

If y = ±1 then

x^{3} = 16,

which has no integral solution.

If y = ±3 then

x^{3} = −8,

giving x = −2, ie the points (−2, ±3). Let P = (2, 3), so the points are ±P. It remains to determine if ±P are of finite or infinite order. The slope at (x, y) is

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

If the tangent

y = mx + c

meets the curve again at (x_{2}, y_{2}) then x, x, x_{2} are the roots of

(mx + c)^{2} = x^{3} + 17.

Thus

2x + x_{2} = m^{2}.

In particular the slope at P = (−2, 3) is

m = 126 = 2,

so the tangent is

y − 3 = 2(x + 2),

ie

y = 2x + 7;

and this meets the curve again where

−2 − 2 + x_{2} = 2^{2},

ie

x_{2} = 8.

6. Show that the elliptic curve

E : y^{2} + xy = x^{3} − x^{2} − 2x − 1

has good reduction modulo 2 and 5; and determine the groups E(𝔽_{2}) and E(𝔽_{5}). What can you deduce about the group of points of finite order on E(ℚ)?

Answer: The curve takes homogeneous form

F (X, Y, Z) ≡ Y^{2}Z + XY Z − X^{3} − X^{2}Z − 2XZ^{2} − Z^{3} = 0.

At a singular point,

∂F/partialX = Y Z − 3X^{2} − 2XZ − 2Z^{2} = 0,
∂F/partialY = 2Y Z + XZ = 0,
∂F/partialZ = Y^{2} + XY − X^{2} − 4XZ − 3Z^{2} = 0.

(a) In characteristic 2, the second equation gives

XZ = 0 ⟹ X = 0 or Z = 0.

If Z = 0 the first equation gives X = 0, and then the third equation gives Y = 0. Thus X = Y = Z = 0, which is impossible. If X = 0 then the first equation gives

Y Z = 0 ⟹ Y = 0 or Z = 0.

We have excluded Z = 0, so

X = Y = 0 ⟹ Z = 0

from the third equation, so again X = Y = Z = 0, which is impossible. We conclude that there is no singular point, ie the reduction at 2 is good.

(b) In characteristic 5, the second equation gives

Z(2Y + X) = 0 ⟹ Z = 0 or X = −2Y.

If Z = 0, then as before the first equation gives X = 0, and then the third gives Y = 0. Thus X = −2Y ⟹ Y = 2X (as −1/2 = 4/2 = 2), and the first equation gives

2X^{2} = 2Z^{2} ⟹ X = ±Z.

The third equation now gives

(4 + 2 − 1 ∓ 4 − 3)X^{2} = 0 ⟹ X = 0.

Thus X = Y = Z = 0, which is impossible. We conclude that the curve is non-singular, ie the reduction at 5 is good.

[Alternatively, one could bring the curve to reduced form since the characteristic is neither 2 nor 3. Thus the equation can be written

y^{2} − 4xy = x^{3} − x^{2} − 2x − 1,

ie

(y − 2x)^{2} = x^{3} + 3x^{2} − 2x − 1.

Writing y for y − 2x, and continuing the reduction,

y^{2} = x^{3} + 3x^{2} + 3x − 1,

ie

y^{2} = (x + 1)^{3} − 2.

Hence the discriminant

D mod 5 = −27 · (−2)^{2} ≠ 0,

ie 5 is a good prime.]

In any characteristic, the only point on the line at infinity Z = 0 is [0, 1, 0].

(a) In characteristic 2 there are just 4 finite points: (0, 0), (1, 0), (0, 1), (1, 1). Of these, (0, 1) and (1, 1) lie on the curve. Thus

E(𝔽_{2}) = ℤ/(3).

(b) In characteristic 5 we can write the equation

y^{2} − 4xy = x^{3} − x^{2} − 2x − 1,

ie

(y − 2x)^{2} = x^{3} + 3x^{2} − 2x − 1.

Setting y′ = y − 2x,

y^{′2} = x^{3} + 3x^{2} + 3x − 1,

ie

y^{′2} = x^{′3} − 2,

where x′ = x + 1.

Dropping the ′s, we have to determine the group on the curve E(𝔽_{5}) : y^{2} = x^{3} − 2.

The quadratic residues mod5 are: 0, 1, 4, ie 0, ±1. We have the following table.

x x^{3} − 2 y points
0 −2 −
1 −1 ±2 (1, ±2)
2  1 ±1 (2, ±1)
−2 0  0 (−2, 0)
−1 2  −

With O = [0, 1, 0],

∥E(𝔽_{5})∥ = 6.

It follows that

E(𝔽_{5}) = ℤ/(6).

If T ⊂ E(ℚ) is the torsion subgroup, and p is a good prime, then the map

T → E(𝔽_{p})

is an injective homomorphism. Thus in this case p = 2 gives an injective homomorphism

T → ℤ/(3).

It follows that

T = {0} or ℤ/(3).

(The prime p = 5 does not give any further information.)

7. Define a lattice L ⊂ ℂ. Show that the series

\frac{1}{z^{2}} + ∑_{ω∈L, ω≠0} (\frac{1}{(z − ω)^{2}} − \frac{1}{ω^{2}} )

defines a function φ(z) which is periodic with respect to L. Show also that φ(z) satisfies the functional equation

φ′(z)^{2} = 4φ(z)^{3} + Aφ(z) + B

for certain constants A, B. Answer:

(a) A lattice is a subgroup

L = ⟨ω_{1}, ω_{2}⟩

of the additive group ℂ generated by two non-zero complex numbers ω_{1}, ω_{2} such that

ω_{2}/ω_{1} ∉ ℝ.

[One could equally well define a lattice as a discrete subgroup of ℂ of rank 2. A discrete subgroup of ℂ is isomorphic to ℤ^{r} where r ≤ 2. In this subject we would normally exclude lattices of rank 0 (ie the group {0}) or 1 (ie the group ⟨ω⟩ consisting of multiples of some ω ∈ ℂ).]

(b) Let

φ(z) = \frac{1}{z^{2}} + ∑′ (\frac{1}{(z − ω)^{2}} − \frac{1}{ω^{2}} ) .

Then

i. The series converges absolutely for any z ∉ L; ii. the convergence is uniform in any bounded closed region ex-

cluding lattice points, and so the series defines a merormorphic function on ℂ with a double pole at each lattice point;

iii. The function is periodic with respect to L, ie

ω ∈ L ⟹ φ(z + ω) = φ(z).

To prove (i), note that

\frac{1}{(z − ω)^{2}} − \frac{1}{ω^{2}} = \frac{1}{ω(1 − z/ω)^{2}} − \frac{1}{ω^{2}}
= ω^{−2} ((1 − z/ω)^{−2} − 1)
= ω^{−2} (2z/ω + 3z^{2}/ω^{2} + ⋯)
= 2z/ω^{3} + 3z^{2}/ω^{4} + · · · .

If

ω = mω_{1} + nω_{2}

then

|ω|^{2} = ωω̄
= Q(m, n),

where Q(m, n) is a positive-definite quadratic form. It follows that

C_{1}(m^{2} + n^{2}) ≤ |ω|^{2} ≤ C_{2}(m^{2} + n^{2})

for some C_{1}, C_{2} > 0. In particular

|ω^{−r}| ≤ C(m^{2} + n^{2})^{−r/2}.

But

∑′(m^{2} + n^{2})^{−r/2}

converges for r > 2, eg by comparison with

∫ (x^{2} + y^{2})^{−r/2}dx dy.

It follows that

∑′ ω^{−r}

converges absolutely for r ≥ 3; and so the series for φ(z) converges absolutely for z ∉ L. This argument also shows that the convergence is uniform in any bounded region where say

|z − ω| ≥ ϵ > 0

for all ω ∈ L. [It is a little more difficult to prove periodicity than one might think. If one could completely separate the terms

\frac{1}{(z − ω)^{2}} and \frac{1}{ω^{2}}

it would be trivial, but unfortunatele these two series do not converge.] Suppose z ∉ L = ⟨ω_{1}, ω_{2}⟩. We regard z as fixed. It is sufficient to show that

φ(z + ω_{1}) = φ(z).

Given ϵ > 0 we can find R such that

∑_{m^{2}+n^{2}>R^{2}} |\frac{1}{(z − ω)^{2}} − \frac{1}{ω^{2}} | < ϵ

and

∑_{m^{2}+n^{2}>R^{2}} |\frac{1}{((z + ω_{1}) − ω)^{2}} − \frac{1}{ω^{2}} | < ϵ

Thus it is sufficient to consider the terms with m^{2} + n^{2} ≤ R^{2}.

But now the terms in the finite sums can be split in two. Now all the terms will cancel except for the terms

\frac{1}{(mω_{1} + nω_{2})^{2}}

when one of

m^{2} + n^{2} and (m + 1)^{2} + n^{2}

is ≤ R and the other is > R. But this implies that

|m| ≤ R + 1.

Hence

(m, n) ∈ A = {(x, y) : R^{2} − 3R < x^{2} + y^{2} < R^{2} + 3R}.

Since

|(mω_{1} + nω_{2})^{2}| ≥ C(m^{2} + n^{2}),

the discrepancy will be

< C′ ∑_{(m,n)∈A} \frac{1}{m^{2} + n^{2}} .

But this is

< C′ ∫_{(x,y)∈A′} \frac{dx dy}{x^{2} + y^{2}}

where

A′ = {(x, y) : R^{2} − 4R < x^{2} + y^{2} < R^{2} + 4R},

say. But the area of A′ is 8πR, while the value of the integrand is always ≥ 1/(R^{2} − 4R). Thus the integral is of order O(1/R) and so → 0 as R → ∞. Hence the discrepancy can be ignored, and

φ(z + ω_{1}) = φ(z).

Similarly

φ(z + ω_{2}) = φ(z),

and so

φ(z + ω) = φ(z)

for all ω ∈ L.

(c) Since

\frac{1}{(z − ω)^{2}} − \frac{1}{ω^{2}} = \frac{1}{ω^{2}} ((1 − z/ω)^{−2} − 1)

= \frac{2z}{ω^{3}} + \frac{3z^{2}}{ω^{4}} + \frac{4z^{3}}{ω^{5}} + ⋯,

in the neighbourhood of z = 0

φ(z) = \frac{1}{z^{2}} + 2G_{3}z + 3G_{4}z^{2} + ⋯,

where

G_{r} = ∑′ \frac{1}{ω^{r}}

(for r ≥ 3). If r is odd then

G_{r} = 0,

since the terms in ±ω cancel out. Thus

φ(z) = \frac{1}{z^{2}} + 3G_{4}z^{2} + 5G_{6}z^{4} + O(z^{6}).

Hence

φ′(z) = \frac{−2}{z^{3}} + 6G_{4}z + 20G_{6}z^{3} + O(z^{5}),

and so

φ′(z)^{2} = \frac{4}{z^{6}} − \frac{24G_{4}}{z_{2}} + O(1).

On the other hand,

φ(z)^{3} = \frac{1}{z^{6}} + \frac{9G_{4}}{z^{2}} + O(1).

Thus

φ′(z)^{2} − 4φ(z)^{3} = −\frac{60G_{4}}{z^{2}} + O(1).

Hence

φ′(z)^{2} − 4φ(z)^{3} + 60G_{4}φ(z) = O(1).

Thus the periodic function on the left has no poles. But such a function is bounded on a fundamental parallelogram, and so on the whole of ℂ. Hence it is constant, say

φ′(z)^{2} − 4φ(z)^{3} + 60G_{4}φ(z) = B,

ie

φ′(z)^{2} = 4φ(z)^{3} + Aφ(z) + B,

where A = 60G_{4}.

8. Find the rank of the curve

E(ℚ) : y^{2} = x^{3} − 5x.

Answer: The associated elliptic curve is

Ẽ: y^{2} = x^{3} + 20x.

The rank r of E is given by

2^{r+2} = |imχ| |imχ̃| ,

where

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

are the auxiliary homomorphisms.

We know that

{1, 5} ⊂ imχ ⊂ {±1, ±5}.

If e = −1 then ef = b = −5 ⟹ f = 5 (working always modℚ^{×2}). Thus −1 ∈ imχ if and only if the equation

u^{2} = −s^{4} + 5t^{4}

has a solution with gcd(s, t) = gcd(u, t) = 1. This equation has the obvious solution

s = 1, t = 1, u = 2.

We conclude that −1 ∈ imχ, and so

imχ = {±1, ±5}.

[The solution (s, t) = (1, 1) corresponds to a point on the curve. To see what is, recall how the auxiliary homomorphisms are defined. Any rational point on the curve is of the form (a/t^{2}, b/t^{3}), with gcd(a, t) = gcd(b, t) = 1. This lies on the curve if and only if

b^{2} = a^{3} − 5at^{4} = a(a^{2} − 5t^{4}).

Now

gcd(a, a^{2} − 5t^{4}) = gcd(a, 5t^{4}) = gcd(a, 5).

If gcd(a, 5) = 1 then

a = ±s^{2}, a^{2} − 5t^{4} = ±u^{2},

and so

u^{2} = s^{4} − 5t^{4} or u^{2} = −s^{4} + 5t^{4}.

Conversely, the solution (s, t, u) = (1, 1, 2) of u^{2} = −s^{4} + 5t4 arises from the point (a/t^{3}, b/t^{2}) = (−1, 2). The slope at this point is

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

Thus the point is of infinite order, and the rank of E(ℚ) is ≥ 1. However, it not necessary to show this in order to answer the question.] Turning to χ̃,

{1, 5} ⊂ imχ̃ ⊂ {±1, ±2, ±5, ±10}

(since 20 ≡ 5 mod ℚ^{×2}). If e = −1 then ef = b̃ = 20 ⟹ f = −20. Thus −1 ∈ imχ̃ if and only if

u^{2} = −s^{4} − 20t^{4},

which is absurd. Hence −1 ∉ imχ̃, and similarly −2, −5, −10 ∉ imχ̃. Thus

{1, 5} ⊂ imχ̃ ⊂ {1, 2, 5, 10}.

If e = 2 then ef = 20 ⟹ f = 10. Thus 2 ∈ imχ̃ if and only if the equation

u^{2} = 2s^{4} + 10t^{4}

has a solution with gcd(s, t) = gcd(u, t) = 1. Evidently u is even, say u = 2v, and

2v^{2} = s^{4} + 5t^{4}.

Since gcd(s, t) = 1, t is odd, and so

s^{4} + 5t^{4} ≡ 5 or 6 mod 8,

while

2v^{2} ≡ 0 or 2 mod 8.

Hence 2 ∉ imχ̃, and so

imχ̃ = {1, 5}.

We conclude that

2^{r+2} = 4 · 2,

ie

r = 1.

9. Find all rational points on the curve

E(ℚ) : y^{2} = x^{3} − 1.

Answer: We begin by determining the points of finite order. By Nagell-Lütz, if P = (x, y) is of finite order then x, y ∈ ℤ and

y = 0 or y^{2} | D,

where

D = −(4b^{3} + 27c^{2})
= 3^{3}.

Thus

y ∈ {0, ±1, ±3}.

If y = 0 then x = 1. Thus there is just one point of order 2, namely (1, 0).

If y = ±1 then

x^{3} = −2,

which is impossible.

Similarly, if y = ±3 then

x^{3} = 10,

which is again impossible.

Hence the only points of finite order on the curve are the point (1, 0), of order 2, and the zero point O = [0, 1, 0].

It remains to determine the rank of the curve. First we bring the root x = 1 of the cubic to 0 by the transformation x′ = x − 1. Dropping the ′s, our curve is not

E(ℚ) : y^{2} = x^{3} + 3x^{2} + 3x.

The associated elliptic curve is

Ẽ: y^{2} = x^{3} − 6x^{2} − 3x.

The rank r of E is given by

2^{r+2} = |imχ| |imχ̃| ,

where

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

are the auxiliary homomorphisms. We know that

{1, 3} ⊂ imχ ⊂ {±1, ±3}.

If e = −1 then ef = b = 3 ⟹ f = −3 (working always modℚ^{×2}). Thus −1 ∈ imχ if and only if the equation

u^{2} = −s^{4} + 3s^{2}t^{2} − 3t^{4}

has a solution with gcd(s, t) = gcd(u.t) = 1. If 3 ∤ s then s^{4} ≡ 1 mod 3 and so

u^{2} ≡ −1 mod 3

which is impossible. Hence 3 | s. But then

3 | u ⟹ 3^{2} | −3t^{4} ⟹ 3 | t,

contradicting gcd(s, t) = 1. Hence −1 ∉ imχ, and so imχ = {1, 3}.

Turning to χ̃,

{1, −3} ⊂ imχ̃ ⊂ {±1, ±3}.

If e = −1 then ef = b̃ = −3 ⟹ f = 3. Thus −1 ∈ imχ̃ if and only if the equation

u^{2} = −s^{4} + 6s^{2}t^{2} + 3t^{4}

has a solution with gcd(s, t) = gcd(u.t) = 1. As before, if 3 ∤ s then s^{4} ≡ 1 mod 3 and so

u^{2} ≡ −1 mod 3

which is impossible. Hence 3 | s. But then

3 | u ⟹ 3^{2} | 3t^{4} ⟹ 3 | t,

contradicting gcd(s, t) = 1. Hence −1 ∉ imχ̃, and so imχ̃ = {1, −3}.

We conclude that r = 0, so the only rational points on the curve are the points of finite order. Thus there is only one rational point on the curve, namely the point (1, 0) of order 2.

10. Either show that the equation

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

has no solutions in non-zero integers x, y, z; or show that the equation

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

has no solutions in non-zero integers x, y, z. Answer:

(a) We may suppose that gcd(x, y, z) = 1, which means that x, y, z are pairwise co-prime. We shall show that the equation

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

has no solution in integers with xyz ≠ 0 and gcd(x, y, z) = 1. We use the following result on “Pythagorean triples”. Lemma. Suppose

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

where x, y, z ∈ ℤ and xyz ≠ 0, gcd(x, y, z) = 1. Then one of x, y is even, and one odd. If we suppose that y is even then

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

for some u, v ∈ ℤ. Applying the Lemma to x^{4} + y^{4} = z^{2}, and supposing x is odd and y is even,

x^{2} = u^{2} − v^{2}, y^{2} = 2uv.

Clearly gcd(u, v) = 1. Also u must be odd and v even, since x^{2} ≡ 1 mod 4. So from the second equation,

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

with gcd(r, s) = 1. Hence

x^{2} = r^{4} − 4s^{4}.

Applying the Lemma again to x^{2} + 4s^{4} = r^{4},

r^{2} = a^{2} + b^{2}, 2s^{2} = 2ab,

with gcd(a, b) = 1. Hence

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

with gcd(X, Y) = 1; and so

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

Thus from the solution (x, y, z) of x^{4} + y^{4} = z^{2} we have derived a second solution (X, Y, r). Moreover, this new solution is strictly smaller than the old, in the sense that

max(|X|^{2} , |Y |^{2}) < |r|
≤ |u|
< |y|^{2}
≤ max(|x|^{2} , |y|^{2}).

This leads to a contradiction, since a strictly decreasing sequence of positive integers must terminate. Hence the equation

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

has no non-trivial solution, and the same is therefore true of

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

(b) Let

ω = e^{2π/3} = \frac{−1 + \sqrt{−3}}{2};

and let

A = ℤ[ω].

For

α = a + bω ∈ A

we set

ᾱ = a + bω^{2},
N(α) = αᾱ = a^{2} − ab + b^{2}.

Evidently

N(α) ∈ ℕ,
N(α) = 0 ⟺ α = 0, ᾱβ = ᾱβ̄,
N(αβ) = N(αβ).

We say that α ∈ A is a unit if

N(α) = 1.

It is easy to see that the only units in A are

±1, ±ω, ±ω^{2}.

We shall assume the following result. Lemma. The ring A is a unique factorisation domain. Let

π = \sqrt{−3} = 1 + 2ω.

Then π is prime, since

N(π) = 3.

Suppose

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

with gcd(x, y, z) = 1, xyz ≠ 0. If x ≡ 1 mod 3 then

x^{3} = (1 + 3u)^{3}
≡ 1 mod 3^{2}.

Similarly

x ≡ −1 mod 3 ⟹ x^{3} ≡ −1 mod 3^{2},

while

x ≡ 0 mod 3 ⟹ x^{3} ≡ 0 mod 3^{3}.

Thus

x^{3} ≡ 0, ±1 mod 3^{2}.

Since

±1 ± 1 ± 1 ≢ 0 mod 9,

It follows that just one of x, y, z is divisible by 3. We may suppose without loss of generality that

3 | z,

ie

π^{2} | z.

Thus, replacing z by −z,

x^{3} + y^{3} = π^{6}z^{3}.

We want to show that

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

has no solution with x, y, z ∈ ℤ, xyz ≠ 0. We shall prove the more general result that the equation

x^{3} + y^{3} = ϵπ^{6}z^{3}

has no solution with x, y, z ∈ A = ℤ[ω], xyz ≠ 0, and with ϵ a unit. We may suppose that

gcd(x, y, πz) = 1.

There are 3 residue classes modπ, represented by 0, ±1. Lemma. If x ≡ ±1 mod π then

x^{3} ≡ ±1 mod π^{4}.

[

Proof. If

x = ±1 + πu

then

x^{3} = (±1 + πu)^{3}
= ±1 + 3πu ± 3π^{2}u^{2} + π^{3}u^{3}
≡ ±1 − π^{3}u + π^{3}u^{3} mod π^{4}
≡ ±1 − π^{3}u(u^{2} − 1) mod π^{4}
≡ ±1 mod π^{4},

since

u ≡ 0, ±1 mod π ⟹ u^{2} − 1 ≡ 0 mod π.

] Evidently

x, y ≡ ±1 mod π.

We may suppose without loss of generality that

x ≡ 1 mod π, y ≡ −1 mod π.

We can write the equation as

(x + y)(x + ωy)(x + ω^{2}y) = ϵπ^{6}z^{3}.

Since x ≡ 1, y ≡ −1 mod π,

x + ωy ≡ 1 − ω ≡ 0 mod π,

and similarly for x + ω^{2}y. Thus

π | x + y, x + ωy, x + ω^{2}y.

On the other hand, since gcd(x, y) = 1,

gcd(x+y, x+ωy) = gcd(x+y, x+ω^{2}y) = gcd(x+ωy, x+ω^{2}y) = π.

It follows that one of (x + y), (x + ωy), (x + ω^{2}y) is divisible by π^{2} (at least), while the other two are divisible by π but not by π^{2}. On replacing y by ωy or ω^{2}y we may suppose that

π^{2} | x + y, π ∥ x + ωy, π ∥ x + ω^{2}y.

It follows that

x + y = ϵ_{1}π^{4}u^{3}, x + ωy = ϵ_{2}πv^{3}, x + ω^{2}y = ϵ_{3}πw^{3},

where ϵ_{1}, ϵ_{2}, ϵ_{3} are units, and π ∤ u, v. Now

(x + y) + ω(x + ωy) + ω^{2}(x + ω^{2}y) = 0.

Dividing by π,

ϵ′_{2}v^{3} + ϵ′_{3}w^{3} = ϵ′_{1}π^{3}u^{3},

where ϵ′_{1}, ϵ′_{2}, ϵ′_{3} are units. Dividing by ϵ′_{2} and absorbing −1 into the cubes we may suppose that ϵ′_{2} = 1 and that ϵ′_{3} ∈ {1, ω, ω^{2}}. But now

v^{3}, w^{3} ≡ ±1 mod π^{3} ⟹ ϵ′_{3} = 1.

Moreover,

v^{3} + w^{3} ≡ 1 − 1 mod π^{4} ⟹ π | u.

Setting u = πu′, the equation takes the same form

v^{3} + w^{3} = ϵ′π^{6}u^{′3}

as the equation we started from with v, w, u′ in place of x, y, z Moreover,

N(u) ≤ N(u′)^{3}
= N(u)^{3}/N(π)^{3}
= N(x + y)/N(π)^{7}
≤ N(z)/N(π).

This leads to a contradiction, since a strictly decreasing sequence of positive integers must terminate.