Math 124, HW #8, Page 1, December 12, 2001  Student, F.

1. We begin with the elliptic curve y^{2} + xy + y = x^{3} over ℚ and wish to find a linear change of variables transforming the elliptic curve to the form Y^{2} = X^{3} + aX + b. To do this, we write the most general linear change, x = aX + bY + c, y = dX + eY + d and then impose the “boundary” conditions. Since we’re lazy, we begin with y^{2} + xy + y − x^{3} and use Mathematica:

In[9]:=
expr0 = ExpandAll[y^2 + x*y + y - x^3 /.
    {x -> a*X + b*Y + c, y -> d*X + e*Y + f}]
Out[9]=
-c^3 + f + c*f + f^2 - 3*a*c^2*X + d*X + c*d*X + a*f*X +
  2*d*f*X - 3*a^2*c*X^2 + a*d*X^2 + d^2*X^2 - a^3*X^3 -
  3*b*c^2*Y + e*Y + c*e*Y + b*f*Y + 2*e*f*Y - 6*a*b*c*X*Y +
  b*d*X*Y + a*e*X*Y + 2*d*e*X*Y - 3*a^2*b*X^2*Y - 3*b^2*c*Y^2 +
  b*e*Y^2 + e^2*Y^2 - 3*a*b^2*X*Y^2 - b^3*Y^3

We want the coefficient of Y^{3} to be 0, so b = 0:

In[10]:= expr1 = expr0 /. b -> 0
Out[10]=
-c^3 + f + c*f + f^2 - 3*a*c^2*X + d*X + c*d*X + a*f*X +
  2*d*f*X - 3*a^2*c*X^2 + a*d*X^2 + d^2*X^2 - a^3*X^3 + e*Y +
  c*e*Y + 2*e*f*Y + a*e*X*Y + 2*d*e*X*Y + e^2*Y^2

At this point, note that a = 1 (so we have −X^{3}) and e = ±1.

In[12]:= expr2 = expr1 /. a -> 1
Out[12]=
-c^3 + f + c*f + f^2 - 3*c^2*X + d*X + c*d*X + f*X + 2*d*f*X -
  3*c*X^2 + d*X^2 + d^2*X^2 - X^3 + e*Y + c*e*Y + 2*e*f*Y +
  e*X*Y + 2*d*e*X*Y + e^2*Y^2

From the XY terms, we have e(1 + 2d) = 0, so d = −\frac{1}{2}.

In[13]:= expr3 = expr2 /. d -> -1/2
Out[13]=
-c^3 + f + c*f + f^2 - X/2 - (c*X)/2 - 3*c^2*X - X^2/4 -
  3*c*X^2 - X^3 + e*Y + c*e*Y + 2*e*f*Y + e^2*Y^2

From the X^{2} terms we have −\frac{1}{4} − 3c = 0, which implies c = −\frac{1}{12}

In[18]:= expr4 = expr3 /. c -> -(1/12)
Out[18]=
1/1728 + (11*f)/12 + f^2 - (23*X)/48 - X^3 + (11*e*Y)/12 +
  2*e*f*Y + e^2*Y^2

Now we have (from the Y term), e(2f − \frac{11}{12}) = 0, so f = \frac{−11}{24}.

In[19]:= expr5 = expr4 /. f -> -11/24
Out[19]=
-181/864 - (23*X)/48 - X^3 + e^2*Y^2

So, our transformation was: x = X − \frac{1}{12}, y = −\frac{1}{2}X ± Y − \frac{11}{24}. And, sure enough, this yields:

Y^{2} = X^{3} + \frac{23}{48}X + \frac{181}{864}

2. Let E be the elliptic curve over the finite field K = ℤ_{5} defined by y^{2} = x^{3} + x + 1.

(a) We can use brute force to find all the elements of E(K).

{ prob2() =
    local(x,y);
    for(x=0, 4,
        for(y=0,4,
            if( y^2 % 5 == (x^3 + x + 1) %5 , print([x,y]))
        )
    )
}
? prob2
[0, 1]
[0, 4]
[2, 1]
[2, 4]
[3, 1]
[3, 4]
[4, 2]
[4, 3]

The ninth point of E(K), is, of course, O.

(b) To find the group structure, we set about finding the orders of the elements.

e = ellinit([0,0,0,Mod(1,5),Mod(1,5)]);
x = [0,1]
? for(p=1,9,print(ellpow(e,x,p)))
[0, 1]
[Mod(4, 5), Mod(2, 5)]
[Mod(2, 5), Mod(1, 5)]
[Mod(3, 5), Mod(4, 5)]
[Mod(3, 5), Mod(1, 5)]
[Mod(2, 5), Mod(4, 5)]
[Mod(4, 5), Mod(3, 5)]
[Mod(0, 5), Mod(4, 5)]
[0]

Well, the point (0, 1) has order nine, so we’ve established that E(K) ≅ C_{9}. (We could have done this directly with the elorder() command.)

3. Let ⊞ be defined as in the homework.

(a) i. Identity: O no longer works as an identity. Imagine P and Q being co-linear. Then O ⊞ P =

P ⊞ O = Q ≠ P. ii. Inverses: These sort of work. If P + Q = O, then P ⊞ Q = −O = O. But unfortunately O

doesn’t really work as an identity, so… iii. Associativity: Nope. Consider y^{2} = x^{3} + 1 as in Lecture 26.

\underbrace{([−1, 0] ⊞ [0, 1])}_{[2,3]} ⊞[0, −1] = [2, 3]

[−1, 0] ⊞ \underbrace{[0, 1] ⊞ [0, −1]}_{O} = [−1, 0]

(b) If E(ℚ) is isomorphic to O, this group law works. Also if it is isomorphic to C_{2} (which would be one point on the x-axis, whose inverse is itself, and O). It also works if it’s isomorphic to C_{2} ×C_{2}. The elements of these groups correspond to points on the x axis. And there are at most three of those, so C_{2} × C_{2} is the biggest group we can have.

4. (Problem discussed with Richard.) Let g(t) be a quartic polynomial with distinct roots, and let α be a root of g(t). Let β ≠ 0 ∈ ℂ.

(a) Let

x = \frac{β}{t − α}, y = x^{2}u = \frac{β^{2}u}{(t − α)^{2}} .

We want to show that x and y give an algebraic transformation between u^{2} = g(t) and y^{2} = f(x), where

f(x) = g′(α)βx^{3} + \frac{1}{2}g′′(α)β^{2}x^{2} + \frac{1}{6}g′′′(α)β^{3}x + \frac{1}{24}g′′′′(α)β^{4}.

This is a veiled Taylor series problem. Let’s substitute:

y^{2} = f(x)
y^{2} = g′(α)βx^{3} + \frac{1}{2}g′′(α)β^{2}x^{2} + \frac{1}{6}g′′′(α)β^{3}x + \frac{1}{24}g′′′′(α)β^{4}
\frac{β^{4}u^{2}}{(t − α)^{4}} = g′(α)β \frac{β^{3}}{(t − α)^{3}} + \frac{1}{2}g′′(α)β^{2} \frac{β^{2}}{(t − α)^{2}} + \frac{1}{6}g′′′(α)β^{3} \frac{β}{(t − α)} + \frac{1}{24}g′′′′(α)β^{4}
u^{2} = g′(α)(t − α) + \frac{1}{2}g′′(α)(t − α)^{2} + \frac{1}{6}g′′′(α)(t − α)^{3} + \frac{1}{24}g′′′′(α)(t − α)^{4}

But, the right hand side is just the power series expansion around α. Since α is a root, the first term is 0 and there are only four other terms since the fifth-derivative of a fourth-degree polynomial is 0. So,

u^{2} = g(t).

(b) Suppose that g has distinct roots. We want to show that f also has distinct roots. We can write (let, without loss of generality, α = x_{1})

g(t) = (t − x_{1})(t − x_{2})(t − x_{3})(t − x_{4}).

Then, solving for t in our substitution, we have t = \frac{β}{x} + x_{1}. So,

g(t) = (\frac{β}{x} + x_{1} − x_{1}) (\frac{β}{x} + x_{1} − x_{2}) (\frac{β}{x} + x_{1} − x_{3}) (\frac{β}{x} + x_{1} − x_{4}) .

We multiply by x^{4}:

x^{4}g(t) = x^{4}u^{2} = y^{2} = f(x)
f(x) = x (\frac{β}{x} ) x (\frac{β}{x} + x_{1} − x_{2}) x (\frac{β}{x} + x_{1} − x_{3}) x (\frac{β}{x} + x_{1} − x_{4})
= β (β + xx_{1} − xx_{2}) (β + xx_{1} − xx_{3}) (β + xx_{1} − xx_{4}) .

Since the x_{i} are distinct, each product here yields a distinct root of f(x). We can conlude that u^{2} = g(t) is an elliptic curve.

5. Let 0 < β ≤ α and let C be the ellipse

\frac{x^{2}}{α^{2}} + \frac{y^{2}}{β^{2}} = 1.

(a) Because of the symmetry of an ellipse, the arc length of C is four times the arc length of y =

β\sqrt{1 − \frac{x^{2}}{α^{2}}} from x = 0 to x = α. We recall the formula for arc length S:

ds = \sqrt{dx^{2} + dy^{2}}
S = ∫ ds = ∫ \frac{dx}{dx}\sqrt{dx^{2} + dy^{2}}

S = ∫ \sqrt{1 + (\frac{dy}{dx})^{2}}dx.

Taking the derivative of y, we have:

\frac{dy}{dx} = \frac{−βx}{α\sqrt{α^{2} − x^{2}}}

So,

(\frac{dy}{dx})^{2} = \frac{β^{2}x^{2}}{α^{2}(α^{2} − x^{2})}.

Now we calculate S:

S = 4 ∫_{0}^{α} \sqrt{1 + \frac{β^{2}x^{2}}{α^{2}(α^{2} − x^{2})}}dx
S = 4 ∫_{0}^{α} \sqrt{\frac{α^{2} − x^{2} + \frac{β^{2}x^{2}}{α^{2}}}{α^{2} − x^{2}}} dx
S = 4 ∫_{0}^{α} α\sqrt{\frac{1 − \frac{x^{2}}{α^{2}} + \frac{β^{2}x^{2}}{α^{4}}}{α^{2} − x^{2}}} dx
S = 4α ∫_{0}^{α} \sqrt{\frac{1 − \frac{α^{2}−β^{2}}{α^{2}} \frac{x^{2}}{α^{2}}}{α^{2} − x^{2}}} dx

Now, let sin θ = \frac{x}{α}, which implies cos θdθ = \frac{dx}{α}, so dθ = _{\frac{dx}{α\sqrt{1− \frac{x^{2}}{α^{2}}}}} = \frac{dx}{\sqrt{α^{2}−x^{2}}} . We change the limits of integration to θ = 0 ( ⟹ x = 0) to θ = π ( ⟹ x = α) and continue:

S = 4α ∫0^{\frac{π}{2}} \sqrt{1 − \frac{α^{2} − β^{2}}{α^{2}} sin^{2} θ}dθ

Let k = ^{\frac{\sqrt{α^{2}−β^{2}}}{α}},

S = α\sqrt{1 − k^{2} sin^{2} θ}dθ.

(b) Sure enough, when α = β, the formula yields S = 4α\frac{π}{2} = 2πr, where the radius r equals α.

(c) Starting from the integral before the substitution,

S = 4α ∫_{0}^{α} \sqrt{\frac{1 − \frac{α^{2}−β^{2}}{α^{2}} \frac{x^{2}}{α^{2}}}{α^{2} − x^{2}}} dx
S = 4α ∫_{0}^{α} \sqrt{α\frac{1 − k^{2} \frac{x^{2}}{α^{2}}}{\frac{α^{2}}{α^{2}} − \frac{x^{2}}{α^{2}}}} dx

Now we substitute t = \frac{x}{α} and change the limits of integration to t = 0 and t = 1. Note that αdt = dx.

S = 4α ∫_{0}^{1} \sqrt{\frac{1 − k^{2}t^{2}}{1 − t^{2}}} dt

= 4α ∫_{0}^{1} \frac{1 − k^{2}t^{2}}{\sqrt{(1 − t^{2})(1 − k^{2}t^{2})}}dt.

(d) If α ≠ β, then our ellpise E is not a circle. So, k ≠ 0. Let g(x) = (1 − t^{2})(1 − k^{2}t^{2}). g(x) is clearly a quartic polynomial with distinct roots, x = ±1, x = ±\frac{1}{k}. We can use the equivalent result to associate g(x) with an elliptic curve. Do some algebra:

g(t) = k^{2}t^{4} − (k^{2} + 1)t^{2} + 1
g′(t) = 4k^{2}t^{3} − 2(k^{2} + 1)t
g′′(t) = 12k^{2}t^{2} − 2(k^{2} + 1)
g′′′(t) = 24k^{2}t
g′′′′(t) = 24k^{2}

Now, choosing α = 1 and β = 1, we have:

f(x) = g′(α)βx^{3} + \frac{1}{2}g′′(α)β^{2}x^{2} + \frac{1}{6}g′′′(α)β^{3}x + \frac{1}{24}g′′′′(α)β^{4}
= 2(k^{2} − 1)x^{3} + (5k^{2} − 1)x^{2} + 8k^{2}x + k^{2}.

So, the associated elliptic curve is y^{2} = 2(k^{2} − 1)x^{3} + (5k^{2} − 1)x^{2} + 8k^{2}x + k^{2}.

6. (a) y^{2} = x^{3} + ax + b. (x, y) is a point on this curve. We play with PARI a bit:

? e = ellinit([0,0,0,a,b]);
? ellpow(e,[x,y],2)
%2 = [9/(4*y^2)*x^4 + 3/(2*y^2)*a*x^2 - 2*x + 1/(4*y^2)*a^2,
-27/(8*y^3)*x^6 - 27/(8*y^3)*a*x^4 + 9/(2*y)*x^3 - 9/(8*y^3)*a^2*x^2 +
3/(2*y)*a*x + (-1/(8*y^3)*a^3 - y)]

PARI says the x-coordinate is:

x(2P) = \frac{9x^{4}}{4y^{2}} + \frac{3ax^{2}}{2y^{2}} − 2x + \frac{a^{2}}{4y^{2}}

= \frac{9x^{4} + 6ax^{2} − 8xy^{2} + a^{2}}{4y^{2}}

= \frac{9x^{4} + 6ax^{2} − 8x(x^{3} + ax + b) + a^{2}}{4y^{2}}

= \frac{x^{4} + 2ax^{2} − 8bx + a^{2}}{4y^{2}} .

Which is what we wanted.

(b) If we draw the tangent line to the point (x_{0}, y_{0}), we would have the curve:

y = (\frac{3x_{0}^{2} + a}{2y_{0}}) (x − x_{0}) + y_{0}.

We plug in the value of x checked above to find the value of y, only we let x_{0} = x and y_{0} = y:

? X = 9/(4*y^2)*x^4 + 3/(2*y^2)*a*x^2 - 2*x + 1/(4*y^2)*a^2
%3 = 9/(4*y^2)*x^4 + 3/(2*y^2)*a*x^2 - 2*x + 1/(4*y^2)*a^2
((3*x^2+a)/(2*y))*(X-x)+y
%4 = 27/(8*y^3)*x^6 + 27/(8*y^3)*a*x^4 - 9/(2*y)*x^3 + 9/(8*y^3)*a^2*x^2 - 3/(2*y)*a*x + (1/(8*y^3)*a^3 + y)

By the group law, we are interested in the negative of this, which happily coincides with what PARI gave for the y coordinate of our sum.

y(2P) = \frac{−27x^{6}}{8y^{3}} − \frac{27ax^{4}}{8y^{3}} + \frac{9x^{3}}{2y} − \frac{9a^{2}x^{2}}{8y^{3}} + \frac{3ax}{2y} − \frac{a^{3}}{8y^{3}} − y
= \frac{−27x^{6} − 27ax^{4} + 27x^{3}y^{2} − 9a^{2}x^{2} + 9axy^{2} − a^{3} − 8y^{4}}{y^{3}} = \frac{−8x^{6} − 7ax^{4} + 11bx^{3} − 8a^{2}x^{2} − 7abx − a^{3} − 8b^{2}}{y^{3}}

(c) According to the hint, we need 3P = 0, which is the same as 2P = −P. So, we can solve for the x coordinates:

x = \frac{x^{4} − 2ax^{2} − 8bx + a^{2}}{4y^{2}}
x = \frac{x^{4} − 2ax^{2} − 8bx + a^{2}}{4(x^{3} + ax + b)}
4(x^{4} + ax^{2} + bx) = x^{4} − 2ax^{2} − 8bx + a^{2}
0 = −3x^{4} − 6ax^{2} − 12bx + a^{2}

(d) Let a = 0 and b = 1. Then:

0 = −3x^{4} − 12x = −3x(x^{3} + 4).

So, x = 0 or x = \sqrt[3]{4}e^{ipi/3} or x = −\sqrt[3]{4}e^{i2pi/3} We can just plug into the curve y^{2} = x^{3} + 1 to find the points: (0, ±1) (−\sqrt[3]{4}, ±\sqrt{3}e^{iπ/2}), (\sqrt[3]{4}e^{ipi/3}, ±\sqrt{3}e^{iπ/2}), (−\sqrt[3]{4}e^{i2pi/3}, ±\sqrt{3}e^{iπ/2}) and 0.