204 CHAPTER 18

into ﬁve factors in F

11

. In fact, you can check that

x

5

− 1 ≡ (x − 1)(x − 3)(x − 4)(x − 5)(x − 9) (mod 11). (18.4)

The congruence (18.4) means that if you multiply out the ﬁve

factors on the right and then view each coefﬁcient of the result

modulo 11, you will get x

5

− 1. Check: The product of those ﬁve

factors is x

5

− 22x

4

+ 176x

3

− 638x

2

+ 1, 023x − 540, and 22, 176,

638, and 1,023 are all multiples of 11, while −540 =−49 × 11 − 1 ≡

−1 (mod 11).

On the other hand, if you take q = 7 (for example), you will not

be able to ﬁnd ﬁve integers b

1

, ..., b

5

so that

x

5

− 1 ≡ (x − b

1

)(x − b

2

)(x − b

3

)(x − b

4

)(x − b

5

) (mod 7).

One-Dimensional Galois Representations

We can now construct some one-dimensional representations of G.

The number a in the equation σ

a

(ζ ) = ζ

a

can be thought of as a

1-by-1 matrix, that is, an element of GL(1, F

p

). The fact that a has a

multiplicative inverse modulo p is exactly what makes a an element

of GL(1, F

p

), rather than just an element of F

p

.

We describe this one-dimensional Galois representation very

carefully. After all, it is the ﬁrst Galois representation we have

been able to understand completely in this book. We call it φ.If

γ is any element in the absolute Galois group G, φ(γ ) is going to

be some element in GL(1, F

p

), in other words, φ(γ ) ≡ a (mod p), for

some integer a not divisible by p. What is φ(γ )?

First of all, φ(γ ) will only depend on the restriction of γ to the

ﬁeld Q(f ) where f (x) = x

p

− 1. We call the Galois group of Q(f ),

which we normally denote by G(f ), H for short. Then φ(γ ) depends

only on r

H

(γ ). (Look back at chapter 14 for the details of the

restriction morphism.)

We saw in our recent discussion that every element of H is of the

form σ

a

for some a not divisible by p.Sor

H

(γ ) = σ

a

for some a not

divisible by p. We simply deﬁne φ(γ ) to be that a.

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