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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
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
n2
or ζ
n1
). 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.

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