Math 124, HW #10, Page 1, December 12, 2001  Student, F.

Problem 2

We are interested in computing E(ℚ)_{tor} for the eliptic curve:

y^{2} + xy = x^{3} − 8369487776175x + 9319575518172005625.

. First, we want to “transform” this curve to a different form. We complete the square by adding \frac{x^{2}}{4} to the right. This yields:

(y + \frac{1}{2}x)^{2} = x^{3} + \frac{x^{2}}{4} − 8369487776175x + 9319575518172005625.

Now, let y′ = y + \frac{1}{2}x. We use some algebra helpers:

? foo(x)=x^3+(x^2)/4-a*x+b
? foo(x-c)
%9 = x^3 + (-3*c + 1/4)*x^2 + (-a + (3*c^2 - 1/2*c))*x +
     (c*a + (b + (-c^3 + 1/4*c^2)))

So, let x′ = x − \frac{1}{12}. Then, we have:

Y^{2} = X^{3} − \frac{401735413256401}{48}x + 8052113850303732744601/864

Now we need to make another substitution to get integer coefficients. Let Y′ = (864)^{3}Y, and X′ = (864)^{2}X. So we have:

864^{−6}Y′ = 864^{−6}X′^{3} − \frac{401735413256401}{48 · 864^{2}}x + 8052113850303732744601/864
Y′ = X′^{3} − 4663949607051700990574592X′ + 3876846619145478710717898068248756224

At this point, we have a form that we can apply the Nagell-Lutz theorem. First, we calculate the discriminant:

Δ = −16(4a^{3} + 27b^{2})
Δ = 8023697806944174419538713292813885598725893795748881000860483584000000

From here on, we’ll just explain the process. First, let y = 0 and solve the resulting equation to find the points of order 2. Then, Consider all y^{2} | Δ. Then try integer x’s. This is algorithmic, since there are a

finite number of cases we need to try. y < \sqrt{Δ} and the equation will yield an upper bound for x since the x^{3} term will dominate. We wouldn’t want to do this with such big numbers in real life, though. There are many many cases to check.

PARI can do this anyway, though, using a different algorithm:

? ?elltors
elltors(e,{flag=0}): torsion subgroup of elliptic curve e: order, structure,
generators. If flag = 0, use Doud's algorithm; if flag = 1, use Lutz-Nagell.
? e=ellinit([1,0,0,-8369487776175,9319575518172005625]);
? elltors(e);
? elltors(e)
%32 = [12, [6, 2], [[2307180, 1512439725], [1669206, -834603]]]

As we can see, the torsion group is of order 12 and isomorphic to C_{6}×C_{2} with generators (2307180, 1512439725) and (1669206, −834603). It’s very reassuring that our “equivalent” elliptic curves have the same structure torsion subgroups:

? e2=ellinit([0,0,0,-401735413256401/48,8052113850303732744601/864]);
? elltors(e2)
%39 = [12, [6, 2], [[27686161/12, 1513593315], [20030473/12, 0]]]
? e3=ellinit([0,0,0,-4663949607051700990574592,
        3876846619145478710717898068248756224]);
? elltors(e3)
%44 = [12, [6, 2], [[1722300703488, 976226130956943360], [1246055664384, 0]]]

The “elltors” function does not use Lutz-Nagell by default.

Problem 3

1. I wrote a quick PARI program to count the number of points on this curve:

{ prob3() =
    local(p,y,order);
    forprime(P=5,30,
     order = 1; \\ point at infinity
        for(x=0,P-1,
            for(y=0,P-1,
                if( (y^2 - 1 -x^3) % P == 0, order++)
            )
        );
        print("p = ", P, " has order ", order);
    )
}

The output is:

p = 5 has order 6
p = 7 has order 12
p = 11 has order 12
p = 13 has order 12
p = 17 has order 18
p = 19 has order 12
p = 23 has order 24
p = 29 has order 3

2. We notice that if p ≅ 2 (mod 3), then the order of the group is p + 1. (We see this for 5, 11, 17, 23.)

3. Claim: If p ≅ 2 (mod 3), then the order of the group of points over the finite field ℤ/pℤ is p + 1. Proof: First, we wish to show that the map f : ℤ_{p} → ℤ_{p}, f(x) ↦ x^{3} is bijective. Our proof will resemble the standard proof of Fermat’s Little Theorem. Write p = 3k + 2. Recall that x^{p−1} = 1. We can write:

1 · 2 · . . . p = 1^{3k+2} · 2^{3k+2} · . . . · p^{3k+2}.

We multiply by ∏_{x=1} px^{−2} on both sides, and we have:

∏_{x=1} px^{−1} = ∏_{x=1} px^{3k}

But since every non-zero element in ℤ_{p} has an inverse, so we can just rewrite,

∏_{x=1} px = ∏_{x=1} px^{3k} = ∏_{x=1} p (x^{3})^{k} .

So the set S = {x^{3k}} is a complete set of residues. Now we can finish the At this point, let us prove that f is injective. Suppose x^{3} = y^{3} but x ≠ y. Then S can’t be a complete set of residues, since it would have at most p − 1 elements—x^{3k} = y^{3k}. Since f is a map between finite fields, injectivity implies surjectivity, so f is bijective. The map g(x) ↦ x + 1 is clearly bijective, so the composition g ∘ f is bijective.

Now we look at y^{2} = x^{3} + 1. We’ve shown that {x^{3} + 1 | x ∈ 0 … p − 1} is a complete set of residues. So, which elements of ℤ_{p}^{×} are squares? As we’ve shown on a previous problem set, half of them! And, for each, there are two solutions. Suppose z ∈ ℤ_{p}^{×}, then if y^{2} ≅ z, then (−y)^{2} = (p − y)^{2} = z. So, from here we have \frac{p−1}{2} · 2 = p − 1 points. We also have the point (−1, 0) and the point at infinity, a sum total of p + 1 points.

Problem 4: Lutz-Nugell, again

Let p be a prime and let E be the curve defined by the equation y^{2} = x^{3} + px = x^{2}(x + p). We calculate the discriminant Δ = −16(4p^{3}) = −(4p)^{3}. If y = 0, then x = 0. Now, suppose y ≠ 0. Recalling that y^{2} | Δ, y can only be 1, 2, 4, p, 2p, and 4p. We can disprove each of these cases by looking at integer solutions of y^{2} = x^{2}(x + p) = mn, m, n ∈ Integers—y = 1: nope, no solutions because both m and n must be 1, but this cannot be. y = 2: mn = 4, so either m = 1 and n = 4 or m = n = 2. Again, nothing works. y = 4: similarly, nothing works. y = p: In this case, we have mn = p^{2}. Nope, doesn’t work. y = 2p: still doesn’t work. y = 4p: doesn’t work, sorry.

So there are two points: (0, 0) and O, the point at infinity.

Problem 5: Finitely generated abelian groups

1. Let E be an elliptic curve over the real numbers. We wish to show that E(ℝ) is not a finitely generated abelian group. It depends on the coefficients, but suppose we have an interval of non-zero length(a, b) such that if x ∈ (a, b), then there exists a real solution (x, y) on the elliptic curve. Examining the graph of any elliptic curve, we can always find such an interval. Suppose the group is abelian and finitely generated by n points. Then, for every x ∈ (a, b) we could associate an element of ℤ^{n} representing the poewrs of the generators. But, then there are a countable number of real numbers in an interval. This is a contradiction—we know that there are an uncountable number of real numbers in any interval.

2. Let E be an elliptic curve over the field ℤ_{p}. We wish to show that E(ℤ_{p}) is a finitely generated abelian group. It has at most p^{2} elements (the ordered pairs (a, b)), so it is certainly finitely generated. Furthermore, the way we defined addition, xy = yx, so the group is abelian.