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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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