Mordell’s theorem

1 Mordell’s theorem

Theorem 1.1 [Mordell] Let C be a nonsingular cubic curve with rational coefficients. Then the group Γ of rational points on C is finitely generated.

That is, there are rational points P_{1}, ..., P_{t} on C such that every rational point on C is of the form n_{1}P_{1} + ... + n_{t}P_{t} with n_{i} ∈ Z.

Viewing Γ as direct product of r copies of Z (r ≥ 0) and some cyclic groups of prime power order, we can find generators P_{1}, ..., P_{r} of infinite order and Q_{1}, ..., Q_{s} of finite order, where Q_{i} has order p_{i}^{e_{i}} for some prime p_{i}, such that the representation P = n_{1}P_{1} + ... + n_{r}P_{r} + m_{1}Q_{1} + ... + m_{s}Q_{s} is unique (n_{i} ∈ Z, m_{i} ∈ Z/p_{i}^{e_{i}}Z).

The number r is called the rank of C.

The group Γ is finite if and only if r = 0.

It is easy to find the points of finite order.

Theorem 1.2 [Nagell-Lutz] Let C be a nonsingular cubic curve with integral coefficients and equation y^{2} = x^{3} + ax^{2} + bx + c, provided with the zero point O = (0, 1, 0). Then the points of finite order on C have integral coordinates. If (x, y) has finite order, then either y = 0, or y|D, where D = −4a^{3}c + a^{2}b^{2} + 18abc − 4b^{3} − 27c^{2} is the discriminant of the curve.

In the general case where C has rational coefficients, one can use a coordinate transformation x′ = d^{2}x, y′ = d^{3}y to make the coefficients integral.

Note that there may well be points (x, y) with y|D that do not have finite order. (But the points (x, y) with y = 0 have order 2.)

The torsion group (subgroup of Γ consisting of the elements of finite order) has restricted shape: there are only 15 possibilities.

Theorem 1.3 [Mazur] The torsion group is one of Z/nZ (1 ≤ n ≤ 10 or n = 12) or Z/2mZ × Z/2Z (1 ≤ m ≤ 4).

So it is easy to find the Q_{i}. There is no known algorithm to find the P_{i}, but there are results for very many special cases.

It is unknown whether the rank r is bounded. Examples with larger r are being found every year. The current champion is the curve

y^{2} + xy + y = x^{3} − x^{2} − 20067762415575526585033208209338542750930230312178956502x+ 34481611795030556467032985690390720374855944359319180361266008296291939448732243429

with rank 28 found by Elkies (2006). For the record ranks for given torsion, see

http://web.math.hr/~duje/tors/tors.html.

2 Proof of Mordell’s theorem

After a change of coordinates we may assume the curve has the equation y^{2} = x^{3} + ax^{2} + bx + c with integral a, b, c.

Define the height of a rational number r = \frac{m}{n} (with gcd(m, n) = 1) by

H(r) = H(\frac{m}{n}) = max(|m|, |n|)

and the height of a rational point P = (x, y) on C by

H(P) = H(x).

Also define H(O) = 1. Let the logarithmic height of P be h(P) := log H(P).

The theorem is an easy consequence of four lemmas, the first three of which use the height function to describe the growth of coordinates under addition. Lemma 2.1 For any constant M, the set {P ∈ Γ | h(P) ≤ M} is finite. Lemma 2.2 Let P_{0} ∈ Γ be fixed. There is a constant κ_{0} = κ_{0}(a, b, c, P_{0}) such that h(P + P_{0}) ≤ 2h(P) + κ_{0} for all P ∈ Γ. Lemma 2.3 There is a constant κ = κ(a, b, c) such that h(2P) ≥ 4h(P) − κ for all P ∈ Γ.

The fourth lemma is the difficult part. Lemma 2.4 The subgroup 2Γ has finite index in Γ.

Now the proof of Mordell’s theorem is straightforward from these lemmas. Pick representatives Q_{1}, ..., Q_{n} of the cosets of 2Γ in Γ. Then for arbitrary P ∈ Γ we can write

P = 2P_{1} + Q_{i_{1}}

and then

P_{1} = 2P_{2} + Q_{i_{2}}
...
P_{m−1} = 2P_{m} + Q_{i_{m}}.

Let κ′ = κ′(a, b, c) be the largest of the constants κ_{0}(a, b, c, −Q_{i}). Then

h(P − Q_{i}) ≤ 2h(P) + κ′ for all P, Q_{i}

and

4h(P_{j}) ≤ h(2P_{j}) + κ = h(P_{j−1} − Q_{i_{j}}) + κ ≤ 2h(P_{j−1}) + κ + κ′

so that h(P_{m}) ≤ κ + κ′ for m sufficiently large. Now

{Q_{1}, ..., Q_{m}} ∪ {P | h(P) ≤ κ + κ′}

is a finite generating set for Γ. □

3 Proof of Lemmas 1-3

Lemma 1 is clear.

Lemma 2

For Lemma 2, first observe that the denominator of x^{3} + ax^{2} + bx + c is that of x^{3} and equals that of y^{2}, so that a point P = (x, y) of the curve satisfies x = \frac{m}{e^{2}} and y = \frac{n}{e^{3}} where m, n, e are integers with gcd(m, e) = gcd(n, e) = 1. It follows that

m ≤ H(P), e ≤ H(P)^{1/2}, n ≤ K.H(P)^{3/2},

where that last inequality is from substitution of x = \frac{m}{e^{2}} and y = \frac{n}{e^{3}} in y^{2} = x^{3} +ax^{2} +bx+c to get n^{2} = m^{3} +am^{2}e^{2} +bme^{4} +ce^{6} ≤ (1+|a|+|b|+|c|)H(P)^{3}.

Now let P = (x, y) and P_{0} = (x_{0}, y_{0}) and P + P_{0} = (ξ, η). We want to bound h(P + P_{0}) in terms of h(P). (W.l.o.g. P ≠ O, P_{0}, −P_{0}, those finitely many points are handled by increasing κ_{0} later. Now all points are finite and distinct.) The line y = λx + μ hits the curve y^{2} = x^{3} + ax^{2} + bx + c in three points with x-coordinates satisfying (λx + μ)^{2} = x^{3} + ax^{2} + bx + c and their sum is minus the coefficient of x^{2}. It follows that x + x_{0} + ξ = λ^{2} − a, where λ = \frac{y−y_{0}}{x−x_{0}} . Now

ξ = (\frac{y − y_{0}}{x − x_{0}} )^{2} − a − x_{0} − x = \frac{Ay + Bx^{2} + Cx + D}{Ex^{2} + F x + G}

for certain integers A, B, C, D, E, F, G independent of x (where y^{2} was replaced by x^{3} + ax^{2} + bx + c, cancelling the x^{3} term). Thus,

H(P + P_{0}) = H(ξ) ≤ max(|Ane + Bm^{2} + Cme^{2} + De^{4}|, |Em^{2} + F me^{2} + Ge^{4}|)

≤ max(|AK| + |B| + |C| + |D|, |E| + |F | + |G|).H(P)^{2}

and after taking logarithms

h(P + P_{0}) ≤ 2h(P) + κ_{0}.
□

Lemma 3

For Lemma 3, put P = (x, y) and 2P = (ξ, η). W.l.o.g. 2P ≠ O. As before we get 2x + ξ = λ^{2} − a, where λ = \frac{dY}{dX} (P) = \frac{3x^{2}+2ax+b}{2y}, so that

ξ = λ^{2} − a − 2x = \frac{x^{4} + ...}{4x^{3} + ...}

and numerator and denominator here have no common roots since the curve is nonsingular.

It suffices to prove the lower bound in Lemma 3.1 Let f(x), g(x) ∈ Z[x] be two polynomials without common roots (in C). Let d be the maximum of their degrees. Then, if r ∈ Q, g(r) ≠ 0 then

dh(r) − κ ≤ h(\frac{f(r)}{g(r)}) ≤ dh(r) + κ

for some constant κ depending on f, g.

Proof Since gcd(f, g) = 1 there are u, v ∈ Q(x) with u(x)f(x)+v(x)g(x) = 1.

For r = \frac{m}{n} (with gcd(m, n) = 1) let F (r) = n^{d}f(r) and G(r) = n^{d}g(r) so that F (r) and G(r) are integers. Now u(r)F (r) + v(r)G(r) = n^{d}.

Let A be the l.c.m. of the denominators of the coefficients of u, v and let e be the maximum of their degrees. Then An^{e}u(r) and An^{e}v(r) are integers, and hence gcd(F (r), G(r))|An^{d+e}. On the other hand, if say f(x) = a_{0}x^{d} + ... + a_{d} has degree d, then F (r) = a_{0}m^{d} + ... + a_{d}n^{d} and gcd(n, F (r))|a_{0} and gcd(F (r), G(r))|Aa_{0}^{d+e}.

Put R := Aa_{0}^{d+e}. Now

H(\frac{f(r)}{g(r)}) = H(\frac{F (r)}{G(r)}) ≥ \frac{1}{R} max(|F (r)|, |G(r)|)

gives

\frac{H(\frac{f(r)}{g(r)})}{H(r)^{d}} ≥ \frac{max(|F (r)|, |G(r)|)}{R max(|m|^{d}, |n|^{d})} = \frac{max(|f(r)|, |g(r)|)}{R max(|r|^{d}, 1)}.

The righthand side is bounded below by a positive constant C (since there is a finite nonzero limit when r tends to infinity, and a nonzero minimum on a compact piece since f and g do not vanish simultaneously. So

H(\frac{f(r)}{g(r)}) ≥ C.H(r)^{d}

and

h(\frac{f(r)}{g(r)}) ≥ dh(r) − κ

as desired. The other inequality is easier (and not needed). □

4 2Γ has finite index in Γ

Remains to prove Lemma 4. Since that is difficult, we only do a special case, namely that where x^{3} + ax^{2} + bx + c has a rational root x_{0}, that is, where there is a rational point (x_{0}, 0) of order 2. Change coordinates so that this point becomes (0, 0). Now the equation is y^{2} = x^{3} + ax^{2} + bx, that is, c = 0.

The discriminant becomes D = b^{2}(a^{2} − 4b), and since the curve is nonsingular, this is nonzero.

Play with two curves: C defined by y^{2} = x^{3} + ax^{2} + bx and C̄ defined by y^{2} = x^{3} + āx^{2} + b̄x, where ā = −2a and b̄ = a^{2} − 4b.

Now ¯ ā = 4a and ¯ b̄ = ā^{2} − 4b̄ = 16b so that ¯ C̄ becomes the curve y^{2} = x^{3} + 4ax^{2} + 16bx, and (x, y) ∈ ¯ C̄ iff (\frac{1}{4}x, \frac{1}{8}y) ∈ C.

Define ϕ : C → C̄ by (x, y) ↦ (x̄, ȳ) with x̄ = x+a+ \frac{b}{x} = \frac{y^{2}}{x^{2}} and ȳ = y(1− \frac{b}{x^{2}} ) for x ≠ 0, and map both (0, 0) and O to Ō. Then ϕ is a group homomorphism with kernel {(0, 0), O}.

Define ψ : C̄ → C as the composition of ϕ̄ and (x, y) ↦ (\frac{1}{4}x, \frac{1}{8}y). Then ψ is a group homomorphism with kernel {(0, 0), Ō}.

The composition of ϕ and ψ is the map P ↦ 2P on C.

All these statements follow by straightforward computation.

The desired result that 2Γ has finite index in Γ will follow from the two facts that ϕΓ has finite index in Γ̄, and ψΓ̄ has finite index in Γ. By symmetry it suffices to show one of these, say the latter.

We need a description of ϕΓ. We have

(i) Ō ∈ ϕΓ.

(ii) (0, 0) ∈ ϕΓ iff b̄ = a^{2} − 4b is a square.

(iii) (x̄, ȳ) ∈ ϕΓ for x̄ ≠ 0 iff x̄ is a square in Q.

(Indeed, (i) is clear. We have x̄ = \frac{y^{2}}{x^{2}} , so x̄ is a square, and x̄ = 0 iff y = 0, that is, x(x^{2} + ax + b) = 0 for some rational point (x, y) with x ≠ 0 on C, that is, if a^{2} − 4b is a square. Finally, if x̄ = r^{2}, then the point (x, y) with x = \frac{1}{2}(r^{2} − a + \frac{ȳ}{r}) and y = xr lies on C and maps to (x̄, ȳ).)

Let Q^{∗} be the multiplicative group of the nonzero rationals, and Q^{∗2} the subgroup of squares. Define a map α : Γ → Q^{∗}/Q^{∗2} by P = (x, y) ↦ x for x ≠ 0, (0, 0) ↦ b, O ↦ 1.

Now α is a group homomorphism: First of all, it maps the unit element O to the unit element 1. Suppose P_{1}+P_{2}+P_{3} = O. We show that α(P_{1})α(P_{2})α(P_{3}) = 1. (And that suffices to prove that α is a homomorphism.) The points P_{1}, P_{2}, P_{3} lie on a line y = λx + μ and x_{1}, x_{2}, x_{3} are roots of (λx + μ)^{2} = x^{3} + ax^{2} + bx. The product of the roots is minus the constant term, that is, is μ^{2}, so that α(P_{1})α(P_{2})α(P_{3}) = x_{1}x_{2}x_{3} = μ^{2} = 1 in Q^{∗}/Q^{∗2}. If P_{1} = (0, 0) then μ = 0 and x_{2}, x_{3} are roots of λ^{2}x = x^{2} + ax + b and α(P_{1})α(P_{2})α(P_{3}) = bx_{2}x_{3} = b^{2} = 1 in Q^{∗}/Q^{∗2}. If P_{1} = O then P_{2} = −P_{3} and x_{2} = x_{3} and α(P_{1})α(P_{2})α(P_{3}) = 1.x_{2}x_{3} = 1 in Q^{∗}/Q^{∗2}. □

Next, the image of α is finite (and is contained in the set of divisors of b): Let P = (x, y) = (\frac{m}{e^{2}} , \frac{n}{e^{3}} ) be a point of C. Then α(P) = \frac{m}{e^{2}} = m in Q^{∗}/Q^{∗2}. From n^{2} = m(m^{2} + ame^{2} + be^{4}) we see that each prime divisor p of m occurs to some even power in m, unless it also occurs (to an odd power) in m^{2} + ame^{2} + be^{4} and hence in be^{4}, and hence in b, since gcd(m, e) = 1.

Next, from the description of the image of ϕ (applied to ψ) it is clear that the kernel of α is precisely the image of ψ. Consequently, α induces an isomorphism from Γ/ψ(Γ̄) to a subgroup of Q^{∗}/Q^{∗2} contained in the subgroup of divisors of b. In particular, Γ/ψ(Γ̄) is finite. □