VIGRE Number Theory Working Group “Mazur Seminar” Talk 1

Universal Families and Ruling Out Small Primes

Bryden Cais

September 16, 2003

1 Introduction

The purpose of this talk is to give the general flavor of the background (i.e. pre Mazur) material on the possibilities for torsion subgroups of elliptic curves over ℚ. To that end, there are three sections: in the first, we show how to write down a one parameter family of elliptic curves defined over ℚ having a rational point of of order N for 3 ≤ N ≤ 10 and N = 12. (The case N = 2 can also be treated but is somewhat different). We then give some motivation as to why it is these values of N that occur and why A Ogg. conjectured that these are the only possible values. In the second section, we will show that no elliptic curve over ℚ can have a rational point of order 11. We do this by showing that the modular curve X_{1}(11), which classifies pairs (E, P) of elliptic curves E and points P of E having order 11, has no rational points other than cusps. In the third section, we tackle the same problem for points of order 35. This is substantially more subtle than the previous case as the genus of X_{1}(35) is large.

2 Familes of elliptic curves

In this section, following [8], we construct a one-parameter family of elliptic curves over ℚ with a rational point of exact order 5. In the course of this construction, we will explicitly see that the modular curve X_{1}(5) has genus 0. This construction can be done similarly for a point of exact order N when 3 ≤ N ≤ 10 or N = 12.

Let E be an elliptic curve with a point P of order 5. Embed E in ℙ^{2} with coordinates X, Y, Z in the usual way, with x = X/Z and y = Y/Z. By performing a change of coordinates if necessary, we may suppose that P = (0, 0, 1), O = (0, 1, 0), and the tangent line to E at P is the line y = 0. This forces E to have Weierstrass form

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2}.

Observe that a_{2} is a unit since otherwise P would be an inflection point, and hence have order 3. We therefore make the change of variables x ↦ λ^{2}x, y ↦ λ^{3}y (which fixes fixing P, O) where λ = a_{3}/a_{2} to bring the (affine) equation of E to the form

y^{2} + αxy + βy = x^{3} + βx^{2}.  (1)

Observe that this form of the Weierstrass equation for E is uniquely determined by E. Indeed, the only coordinate changes preserving the Weirstrass form of the equation of E and fixing P, O are of the form (x, y) ↦ (u^{2}x, u^{3}y); but these do not preserve the form (1) unless u = 1.

Now, it is not difficult to determine that on the curve (1),

2P = (−β, β(α − 1))  3P = (1 − α, α − β − 1).

Moreover, P has order 5 if and only if 3P = −2P, i.e. if and only if α = 1 + β. Therefore, E has Weierstrass equation

E_{β} : y^{2} + (1 + β)xy + βy = x^{3} + βx^{2}.  (2)

The discriminant of E_{β} is

Δ(β) = −β^{5}(β^{2} + 11β − 1),  (3)

which is nonzero except for β ∈ \{0, \frac{−11±5\sqrt{5}}{2}\} and the point at infinity on the projective line ℙ^{1}. Hence, for any field and any β ∈ K − {β : Δ(β) = 0} we obtain an elliptic curve E_{β} defined over K and unique up to isomorphism, with a point P = (0, 0) of exact order 5. In particular, for K = ℚ, we have obtained a one parameter family of elliptic curves defined over ℚ with a rational point of order 5.

We thus have the identification of Y_{1}(5) ≃ ℙ^{1} − {∞, {Δ = 0}} and hence X_{1}(5) ≃ ℙ^{1} (which shows that X_{1}(5) has genus 0).

The situation is almost identical for elliptic curves with a point of exact order N for 3 ≤ N ≤ 10 and N = 12 (N = 2 must be treated differently). Indeed, these are precisely the cases when X_{1}(N) has genus zero (see Table 1). Since X_{1}(N) over ℚ always has a rational point (namely any cusp) this gives the identification X_{1}(N)(ℚ) ≃ ℙ^{1}(ℚ) in the above cases; i.e. in each of these cases, we can obtain a one parameter family of elliptic curves defined over ℚ having an exact point of order N. That these are the only instances when X_{1}(N) has genus 0 (together with calculations for other N that we shall give a flavor of below) lead A. Ogg to conjecture exactly what groups can occur as the torsion subgroup of an elliptic curve over ℚ.

Table 1

N Genus of X_{1}(N) Genus of X_{0}(N)
2  0  0
3  0  0
4  0  0
5  0  0
6  0  0
7  0  0
8  0  0
9  0  0
10  0  0
11  1  1
12  0  0
13  2  0
14  1  1
15  1  1
16  2  0
17  5  1
18  2  0
20  3  1
21  5  1
24  5  1
25  12  0
27  13  1
35  25  3
49  69  1

3 Elliptic curves with a point of order 11

In this section we will show that no elliptic curve over ℚ can have a rational point of order 11. We will accomplish this by deriving a model for the modular curve X_{1}(11) over ℚ and will subsequently show that X_{1}(11)(ℚ) contains only cusps.

Let E be an elliptic curve with a rational point P of order 11. As usual, we shall suppose that E ⊂ ℙ^{2} is a nonsingular cubic curve having the point at infinity O = (0, 1, 0) as a flex and that the group structure on E is such that 3 points add up to 0 if and only if they are collinear. Since P has order 11, we see that no three of the points O, P, 3P, 4P are collinear. It follows that we may choose our coordinate system in ℙ^{2} so that

P = (0, 0, 1)  3P = (1, 0, 0)
4P = (1, 1, 1)  5P = (x_{1}, x_{2}, x_{3})

with x_{1}, x_{2}, x_{3} ∈ ℚ. Observe that these requirements specify E uniquely: indeed, the four points O, P, 3P, 4P are in “general” position, and there are no projective automorphisms fixing these four points other than the identity. This rigidifies our problem.

We now explicitly determine the points −P, −3P, −4P, −5P in terms of x_{1}, x_{2}, x_{3}. By our description of the group law on E we see that −5P, O, 5P are collinear, as are −5P, P, 4P. It follows that −5P is the unique point of intersection of the lines L_{1} joining O, 5P and L_{2} joining P, 4P. We easily determine the parameterizations

L_{1} :α(x_{1}, x_{2} − x_{1}, x_{3}) + (0, x_{1}, 0)
L_{2} :β(1, 1, 0) + (0, 0, 1),

where we are using the fact that x_{1} ≠ 0 (since otherwise 5P, P, O would be collinear and 6P = 0). We find that L_{1}, L_{2} intersect at the point where β = x_{1}/x_{3}, that is, the point (x_{1}, x_{1}, x_{3}). It follows that −5P = (x_{1}, x_{1}, x_{3}). In this manner, we produce the following table:

P = (0, 0, 1)  −P = (0, x_{2}, x_{3} − x_{1})
3P = (1, 0, 0)  −3P = (x_{3} − x_{1}, x_{3} − x_{2} − x_{1}, 0)
4P = (1, 1, 1)  −4P = (1, 0, 1)
5P = (x_{1}, x_{2}, x_{3})  −5P = (x_{1}, x_{1}, x_{3}).

Finally, since 2P is the point of intersection of the lines through −5P, 3P and p, −3P, we determine that

2P = (x_{1}(x_{3} − x_{1}), x_{1}(x_{3} − x_{2} − x_{1}), x_{3}(x_{3} − x_{2} − x_{1})).

Now the condition that 11P = O is precisely the condition that 2P, 4P, 5P are collinear, i.e. that the determinant

|\begin{matrix}1 & 1 & 1 \\ x_{1} & x_{2} & x_{3} \\ x_{1}(x_{3} − x_{1}) & x_{1}(x_{3} − x_{2} − x_{1}) & x_{3}(x_{3} − x_{2} − x_{1}) \end{matrix}| = x_{3}(x_{2} − x_{1})(x_{3} − x_{2} − x_{1}) + x_{1}^{2}(x_{3} − x_{1})

is zero. Since x_{1} ≠ 0, we let x_{2}/x_{1} = y and x_{3}/x_{1} = x and find that a rational point of order 11 on E gives a rational point on the affine curve (x − 1)^{2} = xy(x − y).

Now the birational transformations

v = 1 − x + xy  x = \frac{(v − 1)^{2}}{u − v + 1}
u = xy(y − 1)  y = \frac{u}{v − 1}

transform the curve (x − 1)^{2} = xy(x − y) into the elliptic curve M

u^{2} + u = v^{3} − v^{2},

which gives a minimal model for X_{1}(11).

We have thus seen that any elliptic curve E/ℚ with a rational point of order 11 gives rise to a rational point of u^{2} + u = v^{3} − v^{2}. We now show that this elliptic curve has rank 0, and that the points of finite order correspond to degenerate cases above.

Using the Nagell-Lutz method for finding torsion points, we compute that

M_{tors}(ℚ) = {O, (0, 0), (0, 1), (−1, 0), (−1, 1)} ≃ ℤ/5.

These 5 points correspond to

(x_{1}, x_{2}, x_{3}) ∈ {(0, 1, 0), (0, 1, 1), (0, 0, 1), (1, 1, 1), (1, 0, 1)}.

But in each of these cases, one sees that the point 5P = (x_{1}, x_{2}, x_{3}) on our original elliptic curve is equal to one of O, P, −P, 4P, −4P and hence does not have order 11. It therefore remains to show that M(ℚ) has rank 0.

Using the program PARI, we can compute that

L_{M}(1) = 0.2538418608559106843377589237 …,

where L_{M} is the L function attached to the elliptic curve M. From work of Kolyvagin and Gross-Zagier, we know that if L_{M} is nonvanishing at s = 1 (i.e. has analytic rank 0) than M(ℚ) has rank 0. We shall actually prove that M(ℚ) has rank 0 without appealing to such machinery, but it’s at least good to know in advance that our work will not be in vain.

We first make the change of variables u + 1/2 = s/2^{3} and v = t/2^{2} so that M is put in the form

M : s^{2} = t^{3} − 4t^{2} + 16 := f(t).

Now suppose that M has a rational point. It is easy to see that this point has the form (m/e^{2}, n/e^{3}) for integers m, n, e with e > 0 and (m, e) = 1. We first claim that we can take m, n to be odd. Indeed, let M(𝔽_{2}) denote the reduction of M modulo 2. It is easy to see that #M(𝔽_{2}) = 3. It follows that the image of [3]M(ℚ) in M(𝔽_{2}) is zero (since reduction modulo 2 is a group homomorphism) and hence that [3]M(ℚ) lies in the kernel of the reduction map; i.e. the set of all points in M(ℚ) with the form (m/e^{2}, n/e^{3}) and e even, m, n odd. Since M_{tors}(ℚ) is a group of order 5, we have the claim.

We now show that multiplication by 2 on M(ℚ) is a group isomorphism (i.e. we perform a “2-descent”).

Let λ_{1}, λ_{2}, λ_{3} denote the three roots of f(t) is some (henceforth fixed) algebraic closure K, with λ_{1} real. Set L = ℚ(λ_{1}). We state the following (easily verified) facts about L:

1. L has discriminant D_{L} = −2^{2} · 11.

2. The polynomial f has discrminant D_{f} = −2^{8} · 11 3. L has class number 1.

4. An integral basis for the ring of integers O of L is {1, π, π^{2}}, where π = λ_{1}/2.

5. The index I of ℤ[λ_{1}] in O is 8. This follows from items 1 and 2, and the formula I^{2} = D_{f}/D_{L}.

6. The unit group of L has rank 1, generated by the fundamental unit π − 1

7. The prime 2 = P_{2}^{3} is totally ramified in L, with P_{2} ⊂ O the unique prime above 2.

8. The prime 11 factors as P_{11}Q_{11}^{2} in L for prime ideals P_{11}, Q_{11} ⊂ O.

Now, in K we have the factorization

n^{2} = (m − λ_{1}e^{2})(m − λ_{2}e^{2})(m − λ_{3}e^{2}),

and we set γ = (m − λ_{2}e^{2})(m − λ_{3}e^{2}). Observe that γ ∈ O ⊂ L. If P is a prime ideal of O, and A is any O ideal, we let v_{P} (A) denote the order of A at P. Suppose that v_{P} (m − λ_{1}e^{2}) is odd. Then it is clear that v_{P} (γ) > 0. We certainly have v_{P} (e) = 0 (since otherwise v_{P} (m) > 0 and (e, m) ≠ 1). Writing

f(x) = (x − λ_{1})f′(λ_{1}) + O((x − λ_{1})^{2})

we have

(m − λ_{1}e^{2})γ = (m − λ_{1}e^{2})f′(λ_{1}) + O((m − λ_{1}e^{2})^{2}),

from which it follows that v_{P} (f′(λ_{1})) ≥ min{v_{P} (m − λ_{1}e^{2}), v_{P} (γ)}. We conclude that the ideal (γ, m − λ_{1}e^{2}) divides N_{L/ℚ}(f′(λ_{1})) = −D_{f}.

We therefore have (m − λ_{1}e^{2}) = AB^{2} for ideals A, B ⊂ O with A squarefree and A dividing D_{f}. From our list above, we see that A is divisible only by primes above 11 or 2. We show that in fact, neither of these possibilities can occur. Indeed, since n^{2} = N_{L/ℚ}(m−λ_{1}e^{2}) = N_{L/ℚ}(A)N_{L/ℚ}(B)^{2} we see that N_{L/ℚ}(A) is a square. It follows that if 11 divides N_{L/ℚ}(A) then P_{11}Q_{11} divides A and hence (m−λ_{1}e^{2}); therefore P_{11}^{2}Q_{11}^{2} divides (m−λ_{1}e^{2})^{2} so that 11|(m−λ_{1}e^{2})^{2}. Letting α = (m−λ_{1}e^{2})^{2}/11 we have α ∈ O and 11α ∈ ℤ[λ_{1}]. Since 11 does not divide e, we have α ∉ ℤ[λ_{1}]. It follows that 11 divides the index I of ℤ[λ_{1}] in O. But this is clearly false as the index is 8.

Since 2 = P_{2}^{3} is totally ramified in L, if 2|N_{L/ℚ}(A) then we must have v_{P_{2}}(A) = 1 (since A is squarefree). But then we see that N_{L/ℚ}(A) is exactly divisible by 2 and is hence not a square.

We have thus shown that (m − λ_{1}e^{2}) is the square of some ideal B ⊂ O. Using the fact that O has trivial ideal class group, we may write (m − λ_{1}e^{2}) = ±ϵ^{r}η^{2} for some η ∈ O and r ∈ {0, 1}, where ϵ = π −1 is the fundamental unit of L. Now since π = λ_{1}/2 satisfies x^{3} −2x^{2} +2, we see that N_{L/ℚ}(π) = 2, i.e. π = P_{2} (up to a unit). Since λ_{1} = 2π we see that 2, λ are congruent to 0 modulo π^{2}. Since we have put ourselves in the situation where m is odd, we have m − λ_{1}e^{2} ≡ 1 mod π^{2}. There are exactly four residue classes modulo π^{2} (namely 0, 1, π, 1 + π), so for any η ∈ O we have η^{2} ≡ 1 or 0 modulo π^{2}. Hence, ±ϵη^{2} is congruent to (π − 1) or 0 modulo π^{2}. Since m − λ_{1}e^{2} ≡ 1 mod π^{2} it follows that m − λ_{1}e^{2} = ±η^{2} for some η ∈ O. However, since complex conjugation interchanges the roots λ_{2}, λ_{3} and fixes λ_{1}, we see that

(m − e^{2}λ_{2})(m − e^{2}λ_{3}) = |(m − e^{2}λ_{2})|^{2}

where | · | is the usual absolute value on ℂ. Hence (m − e^{2}λ_{2})(m − e^{2}λ_{3}) ∈ ℝ is positive. Since n^{2} = (m − e^{2}λ_{1})(m − e^{2}λ_{2})(m − e^{2}λ_{3}) is also positive, we have m − e^{2}λ_{1} > 0 so that m − e^{2}λ_{1} = η^{2} for some η ∈ L. Let L = L_{1} and let L_{2}, L_{3} denote the galois conjugates of L_{1} in K. We thus see that (m − λ_{i}e^{2}) ∈ L_{i} is a square. for each i.

But now the homomorphism ϕ_{i} : M(ℚ) ⟶ L_{i}^{×}/L_{i}^{×}^{2} given by (m/e^{2}, n/e^{3}) ↦ (m/e^{2} − λ_{i})L_{i}^{×}^{2} has trivial image. It follows that the standard “Mordell Weil” homomorphism M(ℚ) ⟶ L_{1}^{×}/L_{1}^{×}^{2}× L_{2}^{×}/L_{2}^{×}^{2} × L_{3}^{×}/L_{3}^{×}^{2} has trivial image. Since the kernel is precisely 2M(ℚ), we have shown that M(ℚ) = 2M(ℚ) and hence that M has rank 0. It follows that there are no elliptic curves E/ℚ with a rational point of order 11.

4 Elliptic curves with a point of order 35

.

When the curve X_{1}(N) has large genus it is unreasonable to expect to be able to show that X_{1}(N) has no rational points other than cusps by a direct argument. The strategy for curves of high genus like this is to show that they admit a rational map to some other curve, of lower genus, which has no rational points (other than the “expected” ones coming from the cusps of X_{1}(N)). An obvious candidate is X_{0}(N). In terms of moduli spaces, the natural map X_{1}(N) ⟶ X_{0}(N) takes a pair (E, P) ∈ X_{1}(N) and sends it to the pair (E, ⟨P ⟩) ∈ X_{0}(N), where ⟨P ⟩ is the subgroup generated by P. On the complex analytic side, the natural inclusion of groups Γ_{1}(N) ↪ Γ_{0}(N) induces a holomorphic map of Riemann Surfaces Γ_{1}(N)/ℌ^{∗} ⟶ Γ_{0}(N)/ℌ^{∗}. This is a Galois covering with galois group Γ_{0}(N)/ ± Γ_{1}(N) ≃ (ℤ/Nℤ)^{×}/{±1}.

Of course, this strategy may fail miserably: Indeed, it can happen (as in the cases N = 13, 16, 18, 25) that X_{1}(N) has no non-cuspidal rational points, while X_{0}(N) has infinitely many rational points. In the cases listed above, the genus of X_{0}(N) is zero, and since X_{0}(N) always has a rational point (namely, any cusp) we have the identification X_{0}(N)(ℚ) ≃ ℙ^{1}(ℚ). In moduli theoretic terms, there are (one parameter) families of elliptic curves with a “rational” subgroup of order N for N = 13, 16, 18, 25. Here, a rational subgroup of order N is a subgroup of E[N](ℚ̄) which is fixed (as a group but not necessarily point-wise) by the action of Gal(ℚ̄/ℚ). thus, in these cases, one needs different methods to show that X_{1}(N) has no non-cuspidal rational points.

This strategy can also be rather complicated to implement. For N = 17, 20, 21, 24, 27 the curve X_{0}(N) has genus 1, and we can hope to “get lucky” and be able to show by a direct argument that X_{0}(N) has no rational points other than the ones coming from the cusps of X_{1}(N). This works, for example, in the case N = 17 [7]. However, for N = 35, the curves X_{1}(N) and X_{0}(N) have genus 25 and 3 respectively. Thus, even if we are “lucky” and X_{0}(35) has no non-cuspidal rational points, it may in practice be very hard to show this by directly working with a set of equations for X_{0}(35) in some projective space.

In this section, we will construct a quotient curve of X_{0}(35) having genus 1, and show directly that this quotient has no rational points other than those coming from the cusps of X_{0}(35) (which in turn must come from the cusps of X_{1}(35)). It will follow that there are no elliptic curves over ℚ possessing a rational subgroup of order 35, and hence, no elliptic curves having a rational point of order 35.

We will now proceed to determine a model for X_{0}(35) and the above mentioned quotient curve. For any n|N we let W_{n} be the Atkin-Lehner involution as defined in [4]. Let Ω^{1}(X) denote, as usual, the space of holomorphic differentials on X. Recall [4, pg. 9] that we have the natural identification

Ω^{1}(X_{0}(N)/W_{n}) ≃ {f ∈ S_{2}(N) : f|_{W_{n}} = f},  (4)

where a weight 2 cusp form f, invariant under W_{n} is mapped to the differential ω given locally by f(z)dz. Moreover, we know that the genus g of a curve X is given by the dimension (as a complex vector space) of Ω^{1}(X). Consulting Table 5 of [1], we see that S_{2}(35) has dimension 3, and that we can find a basis of S_{2}(35) consisting of eigenforms for the W_{n} with n|35. One sees that S_{2}(35) splits into a one dimensional space of cusp forms with eigenvalue +1 with respect to W_{5} and eigenvalue −1 for W_{7}, and a two dimensional space of cusp forms having eigenvalue −1 for W_{5} and +1 for W_{7}. Since W_{35} = W_{5}W_{7}, it follows that no cusp form on Γ_{0}(35) has eigenvalue 1 for W_{35}. From the identification (4), we see that the dimension (over ℂ) of Ω^{1}(X_{0}(35)/W_{35}) is zero, hence we have an isomorphism X_{0}(35)/W_{35} ≃ ℙ^{1}. Since W_{35} is an involution, this shows that X_{0}(35) admits a degree 2 map to ℙ^{1} and is hence a hyperelliptic curve of genus 3. It follows that we have en embedding X_{0}(35) → ℙ^{2}. If we give ℙ^{2} homogenous coordinates (X, Y, Z), then we can realize the image of X_{0}(35) ∩ {Z ≠ 0} under this embedding as the locus y^{2} = p(x), where p is a polynomial of degree 2g + 2 = 8 and x = X/Z, y = Y/Z. Moreover, we also see that the dimension of Ω^{1}(X_{0}(35)/W_{5}) is one, and hence that X_{0}(35)/W_{5} is an elliptic curve. It is this quotient that we will determine has no “unexpected” rational points. It is worth pointing out that one can explicitly determine the action of W_{5} on X_{0}(35) in terms of pairs (E, C) of elliptic curves and rational subgroups C of order 35: indeed, we know that C contains a subgroup of order 5, call it D, and a subgroup or order 7, say F. Let ϕ : E → E/D be the quotient map and let F′ denote the image of F under ϕ. Finally, let D′ be the kernel of the dual map ϕ̃ and let C′ be the composition of D′ and F′. Then C′ has order 35 and in the quotient, X_{0}(35)/W_{5}, the pair (E, C) is mapped to the pair (E/D, C′) [5].

To determine the polynomial p(x) we follow [4]. We know that the map

X_{0}(35) ⟶ X_{0}(35)/W_{35} ≃ P^{1}

is given (on affine pieces) by (x, y) ↦ x (equivalently, the involution W_{35} is the unique “hyperelliptic” involution of X_{0}(35) given by y ↦ −y). Thus, x ∈ ℂ(X_{0}(35)) has exactly 2 distinct poles (distinctness follows from the fact that the map is ramified only over the roots of p, which cannot include ∞). Observe that the divisor of poles of y is g +1 = 4 times the divisor of poles of x. Hence the differentials

\frac{dx}{y}  \frac{xdx}{y}  \frac{x^{2}dx}{y}

are holomorphic and have distinct divisors. It follows that they span Ω^{1}(X_{0}(35)). Moreover, we can require that x have a simple pole at ∞ so that the orders of vanishing of \frac{dx}{y}, \frac{xdx}{y}, \frac{x^{2}dx}{y} at ∞ are, respectively, 3,2,1. Now S_{2}(35) has basis f_{1}, f_{2}, f_{3}, where the order of vanishing of f_{j} at ∞ is j. Equivalently, letting q be a local parameter at ∞ in the Riemann surface X_{0}(35)(ℂ), we may write f_{j} = q^{j} + O(q^{j+1}). It follows from the identification (4) that we have (at least up to scalar multiplication)

\frac{dx}{y} = f_{3} and \frac{xdx}{y} = f_{2}.

From this, we easily obtain

x = \frac{f_{2}}{f_{3}} and y = \frac{dx}{f_{3}},

with x determined up to a constant. Using Stein’s tables [9], we find

f_{2} = q^{2} − 3q^{4} − q^{5} + 2q^{6} + q^{7} + O(q^{8})
f_{3} = q^{3} − 2q^{4} − q^{5} + 2q^{6} + q^{7} + O(q^{8}),

and hence, using the fact that up to a factor of 2πi we have dx = q \frac{dx}{dq}dt (for a local parameter t),

x = q^{−1} + 2 + 2q + 3q^{2} + 5q^{3} + 6q^{4} + 10q^{5} + 12q^{6} + 18q^{7} + O(q^{8})
y = −q^{−4} − 2q^{−3} − 3q^{−2} + 17 + 68q + 208q^{2} + 524q^{3} + 1221q^{4} + 2626q^{5} + 5385q^{6} + O(q^{7}).

Using these q expansions and elementary linear algebra, we find that

y^{2} − x^{8} + 12x^{7} − 50x^{6} + 108x^{5} − 131x^{4} + 76x^{3} + 10x^{2} − 44x + 19

is a function on X_{0}(35) with a zero of order at least 20 at the cusp ∞. It follows from general principles that this function is identically zero. Replacing x by x + 1 gives the equation

y^{2} = x^{8} − 4x^{7} − 6x^{6} − 4x^{5} − 9x^{4} + 4x^{3} − 6x^{2} + 4x + 1  (5)
= (x^{2} + x − 1)(x^{6} − 5x^{5} − 9x^{3} − 5x − 1),  (6)

which is the affine piece of X_{0}(35).

Now, since W_{5} is an involution, we know that the map X_{0}(35) ⟶ X_{0}(35)/W_{5} has degree 2. Moreover, we have seen that the genus of X_{0}(35)/W_{5} is 1 while the genus of X_{0}(35) is 3. By Hurwitz’s formula, we find that X_{0}(35) ⟶ X_{0}(35)/W_{5} is ramified at four points. On physically examining the possible involutions of (6), we see that W_{5} must correspond to the involution x ↦ −1/x, y ↦ y/x^{4} since this is the only involution of (6) that ramifies at exactly four points (those corresponding to x = ±i). It follows that the function field of X_{0}(35)/W_{5} is generated by the functions u = x − 1/x and v = y(1 + 1/x^{4}). Straightforward calculations with (6) now show that X_{0}(35)/W_{5} is given (at least in affine coordinates) by

v^{2} = (u + 1)(u^{2} + 2)^{2}(u^{3} − 5u^{2} + 3u − 19.)

We know that this is an elliptic curve, and hence can put it in standard Weierstrass form by an appropriate birational transformation. We find [5, pg. 221] that the transformation

z = \frac{7v}{2(u^{2} + 2)(u + 1)^{2}} − \frac{1}{2}
w = \frac{u − 6}{u + 1}

leads to

z^{2} + z = w^{3} + w^{2} + 9w + 1,  (7)

or equivalently,

t^{2} = s^{3} + 4s^{2} + 16 · 9s + 16 · 5.  (8)

We can use the method of Nagell-Lutz to determine the torsion group of (7). We find that the full torsion group is isomorphic to ℤ/3ℤ and is given explicitly by the points {(1, 3), (1, −4), O}. these points correspond to the values w = 1, ∞, which in turn correspond to u = ∞, −1. Since u = x − 1/x, we observe that u = ∞ comes from x = 0, ∞ and u = −1 does not come from a rational value of x. Since x = 0, ∞ are cusps of X_{0}(35), we see that none of the torsion points of X_{0}(35)/W_{5} come from non-cuspidal rational points of X_{0}(35).

We can again appeal to PARI to compute that

L_{X_{0}(35)/W_{5}}(1) = 0.7029112391349055151579958867 …,

so that (by BSD for modular elliptic curves of analytic rank 0) the rank of X_{0}(35)/W_{5}(ℚ) is zero.

Of course, It can also be shown by the methods of the preceeding section that the elliptic curve (8) has rank 0. We will not carry this out here.

At any rate, we see that X_{0}(35) has no rational points other than cusps. Thus, there are no elliptic curves E defined over ℚ with a rational point of order 35.

References

[1] Birch, B.J. and W. Kyuk, eds. Modular Functions of One Variable IV, Springer-Verlag LNM 476 (1975).

[2] Billing, G., and K. Mahler. On exceptional points on cubic curves. J. London Math. Soc. 15, 32–43 (1940).

[3] Connell, I. Points of order 11 on Elliptic Curves. Nieuw Archief voor Wisk. 13, 3, 257–288 (1995).

[4] Galbraith, S. Equations for Modular Curves. http://www.isg.rhul.ac.uk/∼sdg/thesis.html (1996).

[5] Kubert, D. Universal Bounds on the Torsion of Elliptic Curves. Proc. London Math. Soc. (3) 33, 193–237 (1976).

[6] Mazur, B., and J. Tate. Points of Order 13 on Elliptic Curves. Inventiones math. 22, 41–49 (1973).

[7] Ogg, A. Rational Points of Finite Order on Elliptic Curves. Inventiones math. 12, 105–111 (1971).

[8] Silverberg, A. Open Questions in Arithmetic Algebraic Geometry. In “Arithmetic Algebraic Geometry,” (B. Conrad and K. Rubin, eds). American Math. Soc. IAS/Park City, Vol. 9 (2001).

[9] Stein, W. The Modular Forms Database: Tables.

http://modular.fas.harvard.edu/Tables/basis cuspforms/basis gamma0.html.