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Fearless Symmetry by Robert Gross, Avner Ash

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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
p1
. On the other hand, any element of Q(x
p
1) is
of the form c
1
ζ + c
2
ζ
2
+···+c
p1
ζ
p1
for some rational numbers
c
1
, ..., c
p1
. 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
p1
ζ
p1
) = c
1
ζ
a
+ c
2
ζ
2a
+···+c
p1
ζ
(p1)a
= c
π(1)
ζ + c
π(2)
ζ
2
+···+c
π(p1)
ζ
p1
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
3j
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
3j
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
pk
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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