ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 209

The 2-Torsion

We conclude this chapter with a relatively simple case where we

can actually prove Theorem 18.5, using various facts stated earlier

in the book. Namely, we will look at the Galois representation

ψ coming from an elliptic curve where we set p = 2. We will be

able to compute ψ(Frob

q

) explicitly (which is much more difﬁcult if

p > 2) and test the theorem: For q any prime except for divisors of

2(4A

3

+ 27B

2

), we must have the number of points on E(F

q

) odd or

even depending on whether χ

ψ

(Frob

q

) is odd or even.

Pick an elliptic curve E deﬁned by the equation y

2

= x

3

+ Ax + B.

We are letting p = 2 in the above theorem, meaning that we want

to identify the points on E that satisfy the equation 2P = O.

7

It

happens that these points are particularly easy to describe. In

addition to O itself, they are the points P = (a,0), Q = (b, 0), and

R = (c, 0) where a, b, and c all solve the equation x

3

+ Ax + B = 0.

(This is a cubic equation, so it will have three solutions. One of the

consequences of the restriction that 4A

3

+ 27B

2

= 0 is that the three

solutions will be unequal, so there really are three different points

in our list.) It will always happen that P + Q = R.

Be forewarned that some of the analysis gets quite complicated,

and you can skip it if you just want to go on to the next chapter. But

if you plow through it, you should get an idea of how many of the

ideas in this chapter and chapter 16 ﬁt together beautifully.

An Example

We take a particularly easy example to get started understanding

the theorem. Let E be the elliptic curve deﬁned by the equation

y

2

= x

3

− x (in other words, we are taking A =−1 and B = 0). To

ﬁnd the points in E[2], we need to solve the equation x

3

− x = 0.

This equation has three solutions: x = 0, x = 1, and x =−1. Cor-

responding to those solutions there are three points on the el-

liptic curve: P = (0, 0), Q = (1, 0), and R = (−1, 0). If you use the

7

Remember that this means P + P = O, where the plus sign means adding points

according to the recipe given in chapter 9.

210 CHAPTER 18

complicated group law on E, you can check that P + P = O, Q + Q =

O, and R + R = O. In fact, you can also check that P + Q = R.

Now, let q be any odd prime. We want to understand how Frob

q

permutes these three points. In this case, we can easily understand

what happens: Frob

q

is an element of the absolute Galois group G,

and the absolute Galois group leaves ﬁxed the rational numbers

0, 1, and −1. In other words, Frob

q

(P) = P, Frob

q

(Q) = Q, and

Frob

q

(R) = R.

8

In this case, the matrix ψ(Frob

q

) is well-deﬁned, and using the

prescription on page 206, we see that

ψ(Frob

q

) =

10

01

.

The trace of this matrix is 2. Therefore, χ

ψ

(Frob

q

) = 2, and Theo-

rem 18.5 is the assertion that

2 ≡ q + 1 − #E(F

q

) (mod 2).

Because 2 ≡ 0 (mod 2), this is a fancy way of saying that the

number q + 1 − #E(F

q

) is even. Because q is an odd prime, q + 1

is even, so in the end we have to understand why the number of

points in E(F

q

) is even.

How can we arrange the points on E(F

q

) to make them easy to

count? Sort them into three classes:

1. O.

2. P, Q, and R.

3. The rest of the solutions of y

2

≡ x

3

− x (mod q).

We can see that there is one element in the ﬁrst class and three

elements in the second class. What about the third class? The other

solutions of y

2

≡ x

3

− x (mod q) will have y ≡ 0 (mod q). If a typical

point in this class is (j, k), then (ptj, −k) will also be in this class and

will be different from (j, k), because k ≡ 0 (mod q).

Thus, because we can pair them up, there are an even number

of points in the third class. There are four other points on the

8

The third equation in fact is implied by the other two, and the fact that P + Q = R.

Why? If σ is any element of the absolute Galois group G, then σ (P + Q) = σ (P) + σ (Q)if

P and Q are any points on the elliptic curve E = (Q

alg

).

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