Pries: 405 Number Theory, Spring 2012. Homework 8. Due: Friday 4/20.

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Pries: 405 Number Theory, Spring 2012. Homework 8.

Due: Friday 4/20.

Elliptic curves continued

Read: handout from Silverman.

Complete 4 of the following problems.

1. Consider the point P = (2 , 3) on the elliptic curve E : y 2 order 6 in E (

Q

).

= x 3 + 1. Show that P has

2. Let t be a rational number such that t = 0 and t = 1 / 4. Consider the point P = ( t, t ) on the elliptic curve E : y 2 = x 3 − (2 t − 1) x 2 + t 2 x . Show that P has order 4 in E (

Q

).

Hint: you only need to compute 2 P .

3. The Nagell-Lutz theorem states the following: Let a and b be integers such that D =

− 4 a

3 − 27 b

2

= 0. If P = ( x, y ) is a point of finite order on the smooth elliptic curve

E : y 2 = x 3 + ax + b , then x and y are integers and either y = 0 or y divides D .

If p is an odd prime, use the Nagell-Lutz theorem to find all the points of finite order on y 2 = x 3 + px .

4. Let C

1 and C

2 be the cubic curves with the following equations:

C

1

: x

3

+ 2 y

3 − x − 2 y = 0 , C

2

: 2 x

3 − y

3 − 2 x + y = 0 .

(a) Find the 9 points of intersection of C

1 and C

2

.

(b) Let { (0 , 0) , P

1

, . . . , P

8

} be the nine points from part (a). Let C be an arbitrary cubic curve:

C : a

3 , 0 x

3

+ a

2 , 0 x

2

+ a

1 , 0 x + a

0 , 0

+ a

0 , 3 y

3

+ a

0 , 2 y

2

+ a

0 , 1 y + a

1 , 2 xy

2

+ a

2 , 1 x

2 y + a

1 , 1 xy = 0 .

If P

1

, . . . , P

8 are all on C , show that (0 , 0) is on C also. Hint: Concretely, each point gives a linear relation between the coefficients a i.j

and you need to show that a

0 , 0

= 0.

Remark: The problem above is related to showing that the group law on an elliptic curve is associative. It is also an example of Bezout’s theorem.

5. Let L =

Z

[ i ] = { a + bi | a, b ∈

Z

} . Show that g

3

= 0 where g

3

= 140

X w ∈ L,w =0

1 w 6

.

This is the main step in showing that

C

/L ' E : y 2 = x 3 − x . Hint: add up the terms in groups of 4, i.e., w is a + bi , i ( a + bi ), − ( a + bi ), and − i ( a + bi ).

I’m stuck on a line; I’m stuck on a curve - Peter Krebs

1

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