202 CHAPTER 18

element of F

×

p

.

3

Remember that now our deﬁning equation for ζ is

ζ

p

= 1.

The Galois group G(x

p

− 1) has exactly p − 1 elements:

σ

1

, ..., σ

p−1

. On the other hand, any element of Q(x

p

− 1) is

of the form c

1

ζ + c

2

ζ

2

+···+c

p−1

ζ

p−1

for some rational numbers

c

1

, ..., c

p−1

. If you want, you can completely work out how G acts

on the ﬁeld Q(x

p

− 1).

EXERCISE: Given a number a not divisible by p, show that

there is a permutation π of {1, 2, ..., p − 1} such that

σ

a

(c

1

ζ + c

2

ζ

2

+···+c

p−1

ζ

p−1

) = c

1

ζ

a

+ c

2

ζ

2a

+···+c

p−1

ζ

(p−1)a

= c

π(1)

ζ + c

π(2)

ζ

2

+···+c

π(p−1)

ζ

p−1

so that the c

i

’s get permuted by this permutation π (where π

depends on a).

How Frob

q

Acts on Roots of Unity

We next pick some prime q other than p, and we try to identify

Frob

q

as one of these permutations of the roots of x

p

− 1.

THEOREM 18.2: If q = p, then q is unramiﬁed in Q(x

p

− 1),

and Frob

q

is a σ

q

. In other words, if ζ is a primitive pth root of

unity, then Frob

q

(ζ ) = ζ

q

.

We can explain why this theorem is plausible, even though we

cannot quite prove it here. We start with a numerical example of

the easiest case: p = 3. The solutions of the equation x

3

− 1 = 0 are

x = 1, x =

−1+

√

−3

2

, and x =

−1−

√

−3

2

. One way to get these solutions is to

apply de Moivre’s Theorem; another is to factor x

3

− 1 = (x − 1)(x

2

+

x + 1) and apply the quadratic formula to solve x

2

+ x + 1 = 0. To

avoid writing these numbers over and over, it is customary to write

ω =

−1+

√

−3

2

. You can check that ω

2

=

−1−

√

−3

2

, so the three solutions to

the equation are 1, ω, and ω

2

. Remember that ω

3

= 1.

4

3

See page 39 for a discussion of F

×

p

.

4

Thus, when p = 3, we write ω rather than ζ .

ONE- AND TWO-DIMENSIONAL REPRESENTATIONS 203

Let q be any prime other than 3, and let us see what Frob

q

(ω)

is. The two key facts to remember are that Frob

q

(ω) must be a root

of the same minimal polynomial as ω, which is x

2

+ x + 1, and that

q is a divisor of N(Frob

q

(ω) − ω

q

). The two roots of x

2

+ x + 1 are ω

and ω

2

, so we must have Frob

q

(ω) = ω or Frob

q

(ω) = ω

2

.Itisalso

easy to see that if Frob

q

(ω) = ω

q

, then N(Frob

q

(ω) − ω

q

) = N(0) = 0,

and q does divide 0. So setting Frob

q

(ω) = ω

q

does not contradict

either one of the requirements for Frob

q

(ω) that we stated back

on page 185. (Remember that ω

q

must be ω or ω

2

, because ω

3

= 1;

therefore, powers of ω just keep repeating over and over again.)

The other possibility is that ω

q

= ω

j

, where j is 1 or 2, and

Frob

q

(ω) = ω

k

, where k is 3 − j. (This is a fancy way of writing that

k is also 1 or 2, and k = j.) Because ω

3

= 1, we can compute that

with these assumptions, N(Frob

q

(ω) − ω

q

) = N(ω

3−j

− ω

j

), where j is

1or2.Ifj = 1, you get N(ω

2

− ω) = N(−

√

−3) = 3, and if j = 2, you

get N(ω − ω

2

) = N(

√

−3) = 3. Either way, there is no way for the

prime q to divide N(ω

3−j

− ω

j

). So we need to have

Frob

q

(ω) = ω

q

. (18.3)

This same argument works if 3 is replaced by any odd prime p.

The last step is a bit trickier in general: You need to show that if

j + k is not a multiple of p, then N(ζ

p−k

− ζ

j

)isp, which can be done

with some algebraic identities.

Now we explore one particular case of Theorem 18.2. We look at

what happens when q ≡ 1 (mod p). This means that Frob

q

(ζ ) = ζ .

Why? We can rewrite the congruence q ≡ 1 (mod p) as the equation

q = 1 + kp,soζ

q

= ζ

1+kp

= (ζ

1

)(ζ

kp

) = ζ · (ζ

p

)

k

= ζ · (1)

k

= ζ . In other

words, Frob

q

will be the identity permutation. Now, remember what

we said in chapter 16 about thinking of Frob

q

as a permutation of

the roots of an irreducible polynomial. When q ≡ 1 (mod p), Frob

q

is a permutation that does not change any of the roots, so each of

the cycles in the permutation has size 1. And, by Theorem 16.1, this

means that x

p

− 1 must factor into p factors of degree 1 in F

q

.

For example, let p = 5 and q = 11. Then by what we wrote in

the previous paragraph, because 11 ≡ 1 (mod 5), x

5

− 1 must factor

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