ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 211

curve—O, P, Q, and R—so in all there must be an even number

of points on the curve.

Another Example

We now try a slightly more complicated example. Let E be the

elliptic curve y

2

= x

3

− 1, so that A = 0 and B =−1. We can apply

the theorem for any prime q other than 2 and 3. The three points in

E[2] (other than O)areP = (1, 0), Q = (ω, 0), and R = (ω

2

, 0), where

ω is the number

−1+

√

−3

2

that showed up above in our discussion of

the solutions of x

3

− 1 = 0.

EXERCISE: Show that P + Q = R.

Now choose your favorite prime q (other than 2 and 3, which are

ramiﬁed), and consider how Frob

q

shifts these points around. We

know that Frob

q

(P) = P, because the coordinates of P are in Z, and

elements of the absolute Galois group G always ﬁx elements of Z.

What about Frob

q

(Q)? The two possibilities are Frob

q

(Q) = Q and

Frob

q

(Q) = R.

We take q = 5. We saw above in equation (18.3) that Frob

5

(ω) =

ω

5

= ω

2

, and so Frob

5

(Q) = R. Because R = P + Q, we see, using the

prescription on page 206 as usual, that

ψ(Frob

5

) =

11

01

.

Therefore, χ

ψ

(Frob

5

) = 1 + 1 ≡ 0 (mod 2).

What about Frob

7

? Repeat the same computation, and you will

see that Frob

7

(ω) = ω. Therefore, Frob

7

(Q) = Q, and so

ψ(Frob

7

) =

10

01

.

This means that χ

ψ

(Frob

7

) = 1 + 1 ≡ 0 (mod 2).

In fact, you can see that these two examples show the two

possibilities for ψ(Frob

q

), and in either case χ

ψ

(Frob

q

) ≡ 0 (mod 2).

Again, from Theorem 18.5, we conclude that #E(F

q

) is always even.

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