108 CHAPTER 9

place according to the rules of the ﬁeld R.) Let

x

3

= λ

2

− x

1

− x

2

.Lety

3

=−(λx

3

+ ν). Then P + Q has

coordinates (x

3

, y

3

).

5. If x

1

= x

2

and y

1

= y

2

= 0, then compute the number

λ =

3x

2

1

+A

2y

1

. Compute ν as before, and then use the same

formula for x

3

and y

3

as in the previous case.

A Much-Needed Example

This description is somewhat mysterious.

4

In fact, it is not obvious

(unless you looked at the footnote) that (x

3

, y

3

) even solves the

same equation y

2

= x

3

+ Ax + B that we started with. We take

some examples in E(Q) with the particular curve y

2

= x

3

+ 1, which

means that A = 0 and B = 1. Remember that (0, 1), (0, −1), and

(2, 3) are points on this curve. The ﬁrst case of the group law

tells us that O + (0, 1) = (0, 1), and the second case tells us that

(0, 1) + O = (0, 1). The third case tells us that (0, 1) + (0, −1) = O.

Now things get messier. We compute (0, 1) + (2, 3). We start by

computing λ =

3−1

2−0

=

2

2

= 1. Next, ν = 1 − 1 · 0 = 1. Then x

3

= 1

2

−

0 − 2 =−1, and y

3

=−(1 ·−1 + 1) = 0. So we just worked out that

(0, 1) + (2, 3) = (0, −1).

Finally, we compute (2, 3) + (2, 3). Using the last case, we com-

pute that λ =

3·2

2

+0

2·3

=

12

6

= 2. Next, ν = 3 − 2 · 2 =−1. Now, x

3

= 4 −

2 − 2 = 0 and y

3

=−(2 · 0 + (−1)) = 1. So (2, 3) + (2, 3) = (0, 1).

EXERCISE: Let E be the elliptic curve y

2

= x

3

+ 17. Compute

(−1, 4) + (2, 5) and (2, 5) + (2, 5).

SOLUTION: To compute (−1, 4) + (2, 5), we ﬁrst compute that

λ =

1

3

, and then ν =

13

3

, and then (−1, 4) + (2, 5) = (−

8

9

, −

109

27

).

To compute (2, 5) + (2, 5), we ﬁrst compute that λ =

6

5

, and

then ν =

13

5

, and then (2, 5) + (2, 5) = (−

64

25

,

59

125

).

4

Geometrically, what is going on is this: We take P and Q and connect them with a line.

This line will intersect the elliptic curve in exactly three points: P, Q, and a third point T.

We negate the y coordinate of T and the resulting point is P + Q. This is what underlies

formula (4). The other formulas have similar geometric interpretations. The fact that

this geometric construction deﬁnes a group law is amazing and not so easy to prove.

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