We are nearing our journey’s end. From here forward, we

will give examples of Galois representations, reciprocity

laws, and their applications to Diophantine equations. In

this chapter, we discuss representations of the absolute

Galois group G to GL(1, F

p

) and GL(2, F

p

). We explain

the ﬁrst case in detail. The second example depends on

elliptic curves, and we must content ourselves with a brief

description and a lengthy example. The example gives

us an opportunity to put together some of the facts we

have learned about elliptic curves, factoring polynomials

modulo q,andFrob

q

. We will see how they interact to

enable us to prove a reciprocity law concerning the 2-

torsion points on an elliptic curve.

Roots of Unity

We start with the polynomial x

n

− 1. We know what the roots of

this polynomial are, using de Moivre’s Theorem: They are the n

complex numbers ζ , ζ

2

, ζ

3

, ..., ζ

n

, where ζ = cos

2π

n

+ i sin

2π

n

.It

is important to know that these n numbers are all unequal, that

they all solve the same polynomial equation x

n

− 1 = 0, and that

ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 201

ζ

n

= 1. One consequence of these facts is that a ≡ b (mod n)ifand

only if ζ

a

= ζ

b

.

1

We look now at what the elements of the absolute Galois group

G can do to these numbers. A key point is that when we know

what σ , an element of G, does to ζ , we know what it does to every

root of the polynomial x

n

− 1. For example, suppose that σ (ζ ) = ζ

3

.

Then σ (ζ

2

) = σ(ζ · ζ ) = σ (ζ ) · σ (ζ ) = ζ

3

· ζ

3

= ζ

6

= ζ

3·2

, and similarly

σ (ζ

k

) = ζ

3k

for any positive integer k.

So if we know σ (ζ ), then we know everything that we need to

know about how σ permutes the roots. Because σ (1) = 1, σ (ζ ) must

be a root not equal to 1 (i.e., ζ , ζ

2

, ..., ζ

n−2

or ζ

n−1

). We can write

σ (ζ ) = ζ

a

for some a between 1 and n − 1. To distinguish elements

of G, we label them by the exponent a, so that σ

a

(ζ ) = ζ

a

. Of course,

there will be many σ ’s in G with the same effect on ζ , and we will

write σ

a

for any one of them that takes ζ to ζ

a

.

Which numbers a can occur? Suppose there is some number

d > 1 so that both a and n are multiples of d. In that case, σ

a

is

expelled from the G-club, because it will not be a permutation of

the roots. For example, let n = 10, so ζ is a “primitive” tenth root

of unity.

2

We claim that σ

2

is not a permutation. What goes wrong?

σ

2

(ζ ) = ζ

2

, and σ

2

(ζ

6

) = ζ

12

= ζ

10+2

= ζ

10

ζ

2

= 1 · ζ

2

= ζ

2

. Because σ

2

sends both ζ and ζ

6

to the same number ζ

2

, it follows that σ

2

is not

a one-to-one correspondence. In general:

THEOREM 18.1: If a and n have a common factor other than

1, then there is no σ

a

in G. But if a and n have no common

factor other than 1, then there are σ

a

’s in G.

Now we will take the case where n = p, a prime number bigger

than 2. (This restriction to odd primes is reasonable, because x

2

− 1

has only the boring rational roots 1 and −1.) We can now think of

the exponent a as a nonzero element of F

p

, in other words as an

1

To review what a ≡ b means, look back at chapter 4.

2

The n roots of x

n

− 1 form a cyclic group under multiplication, and to say that ζ is

“primitive” means that ζ generates the entire group. See page 40 for a review of cyclic

groups.

Start Free Trial

No credit card required