ELLIPTIC CURVES 111

that t − t

3

is a square of a rational number. That is, we want to solve

for the solution set E(Q) where E is the elliptic curve given by

y

2

=−t

3

+ t.

5

By the way, the “extra” solution X =−1, Y = 0toX

2

+ Y

2

=

1 corresponds to the point O at inﬁnity on this elliptic curve.

6

However, it is not much of a triangle: One of the sides is 0. In

fact, we could have excluded the solutions where X or Y is0to

begin with. But we would not have seen how the point at inﬁnity on

the elliptic curve can correspond to a “degenerate solution” to our

original problem: a triangle with one side 0, the other side ∞,in

such a way that the area is 1.

But the real question here is whether there are any nondegener-

ate rational solutions in E(Q). In a course in elementary number

theory, you can prove that the answer is “no.” So when D = 1,

the congruent number problem has a negative solution: There is

no right triangle with rational sides and area 1. For more on the

congruent number problem, see chapter 23.

Torsion and the Galois Group

It is very difﬁcult to prove that the ﬁve-step recipe that we have

given above actually deﬁnes a group law. It is not obvious that

it satisﬁes the associative law. We do have obvious inverses: the

inverse of (x

1

, y

1

)is(x

1

, −y

1

). And we do have an identity element: O.

But you will have to take associativity on faith; the algebra needed

to verify it is rather involved.

7

(Commutativity of the group law is

not difﬁcult to check from the deﬁnition.)

The group law is a big help when studying E(Q). It can help you

derive new solutions from old ones, and sometimes can even help

5

You may have noticed that this does not conform to our template y

2

= x

3

+ Ax + B for

elliptic curves. To ﬁx this, deﬁne a new variable u =−t, and get the equivalent equation

y

2

= u

3

− u.

6

Why? Well, we get this extra point if we set u =∞, as it were. That would correspond

to the point (∞, ∞) on the elliptic curve.

7

Associativity can also be proved in more elegant ways using algebraic geometry.

112 CHAPTER 9

to prove that there are no unobvious solutions—all depending on

which curve E you are studying.

There are many theorems about the set E(Z) and the group E(Q)

that can be found in the textbooks. For our purposes, in studying

representations of the absolute Galois group G of Q, we need to tell

you about a particular set of elements inside of E(C): the n-torsion.

DEFINITION: Let n be a positive integer. An element P of

E(C)ispartofthen-torsion if

n times

P + P +···+P = O.

The set of all n-torsion points is written E[n].

For example, an element P on the elliptic curve is part of the

5-torsion if P + P + P + P + P = O. This is usually written, rather

confusingly, as 5P = O. Do not confuse it with multiplying the x-

and y-coordinates of P by 5, because it means something much more

complicated. You can work out what it does mean: It will give some

Z-equation that the x-coordinate of P must satisfy, and another one

for the y-coordinate of P, for P to be 5-torsion. Therefore, the x- and

y-coordinates of P will be in Q

alg

.

In fact, the equation that the x-coordinate must solve will be a

rather messy degree-12 Z-equation, and so there will be no more

than 25 elements of the 5-torsion (for any particular elliptic curve,

of course; if you change the elliptic curve, then you will get a

different equation, and different solutions).

8

The general situation

is summarized as follows.

THEOREM 9.1: Let E be an elliptic curve, and let n be a

positive integer. All of the n-torsion points of E have

coordinates in Q

alg

, and the number of elements of E(Q

alg

) that

are n-torsion is n

2

.

8

How do we get 25 elements? The equation for the x-coordinate will be degree 12, whose

roots will give us at most 12 x-coordinates. Each of those corresponds to 2 y-coordinates,

which are ± of each other. That is 24 points. The remaining point is O.

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