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1
the fact that the reduction map π1 : Z[ζn ] ’’ Fpd1 is injective on the powers of ζn , we ¬nd that
1
r r r r
p11 p11 p11 p11
= 0 in Z[ζn ]. This shows that minpolyQ,ζ
f (ζn )= 0 and so f (ζn ) (ζn ) = 0 and so ζn is
n
conjugate to ζn .
6.2. CYCLOTOMIC EXTENSIONS 69

r
p1
Repeating this argument starting with ζn1 and using the prime p2 we ¬nd that
r r
p 1p 2
minpolyQ,ζ (ζn1 2 ) =0
n
r r
p 1p 2
and so ζn1 2 is conjugate to ζn . Continuing in this fashion, for each j = 1, . . . , m we have
r r
rr r
p11 p22 ···pj j p11 ···pj j
minpolyQ,ζn (ζn ) = 0, and thus ζn
is conjugate to ζn . When j = m, this shows
t ) = 0. Hence ζ t is conjugate to ζ in the extension Q(ζ )/Q.
that minpolyQ,ζn (ζn n n
n

Theorem 6.3. For n > 2, consider the cyclotomic extension Q(ζn )/Q where ζn = e2πi/n .
Then Q(ζn )R = Q(ζn ). Furthermore,
()
Q(ζn )R = Q(ζn ) = Q(ζn + ζ n ) = Q(cos(2π/n)),
and
•(n)
[Q(cos(2π/n)) : Q] = .
2
Proof. Recall that
Gal(Q(ζn )/Q) ∼ Z/n— ,
=
r
where the residue class of r acts by sending ζn to ζn . Complex conjugation corresponds to the
residue class of ’1 ≡ n ’ 1 (mod n). Making use of the elementary identities
1
eθi = cos θ + sin θ i, cos θ = (eθi + e’θi ),
2
we obtain
1 1 ’1
cos(2π/n) = (ζn + ζ n ) = (ζn + ζn ).
2 2
Complex conjugation ¬xes each of the real numbers cos(2πk/n) for k = 1, 2, . . . , n ’ 1. The
residue class of r acts by sending cos(2π/n) to cos(2πr/n); it is elementary to show that
cos(2πr/n) = cos(2π/n) unless r ≡ 1 (mod n). Hence
( ) = {id, ( )} = Gal(Q(cos(2π/n))/Q).
Thus we have
Q(ζn ) ( ) = Q(cos(2π/n)),
and so [Q(cos(2π/n)) : Q] = •(n)/2. Notice that ζn is a root of the polynomial
X 2 ’ 2 cos(2π/n)X + 1 ∈ Q(cos(2π/n))[X],
so we also have
minpolyQ(cos(2π/n)),ζn (X) = X 2 ’ 2 cos(2π/n)X + 1.
(6.1)
Example 6.4. We have
[Q(ζ24 ) : Q] = •(24) = 8
and
Gal(Q(ζ24 )/Q) ∼ Z/2 — Z/2 — Z/2.
=
Proof. By Theorem 1.42 we have [Q(ζ24 ) : Q] = 8. Also,
√ √
1 3 1 3
6 3 8
ζ24 = i, ζ24 = + i, ζ24 = ’ + i,
2 2 2 2
√√
and all of these numbers are in Q(ζ24 ), hence Q( 2, 3, i) Q(ζ24 ). It is easy to check that
√√
[Q( 2, 3, i) : Q] = 8,
which implies that √√
Q(ζ24 ) = Q( 2, 3, i).
Using this we ¬nd that
Gal(Q(ζ24 )/Q) ∼ Z/2 — Z/2 — Z/2.
=
70 6. A GALOIS MISCELLANY

We also have cos(2π/24) = cos(π/12) ∈ Q(ζ24 ). Since

3
cos(2π/12) = cos(π/6) = ,
2
we have √
3
2 cos2 (π/12) ’ 1 =
2
and so
3
4 cos4 (π/12) ’ 4 cos2 (π/12) + 1 = ,
4
giving
16 cos4 (π/12) ’ 16 cos2 (π/12) + 1 = 0.
Then
16X 4 ’ 16X 2 + 1 = 16 minpolyQ,cos(π/12) (X).
Note that case (i) of Kaplansky™s Theorem 4.22 applies to the polynomial minpolyQ,cos(π/12) (X).
For this example, Gal(Q(ζ24 )/Q) has 23 ’ 1 = 7 subgroups of each of the orders 2 and 4; it
is an interesting exercise to ¬nd them together with their ¬xed sub¬elds.
Remark 6.5. The minimal polynomial for cos(π/12) can also be found as follows. We have
¦24 (ζ24 ) = 0, hence since
¦24 (X) = X 8 ’ X 4 + 1,
we obtain
8 4
ζ24 ’ ζ24 + 1 = 0.
’4
Then after multiplying by ζ24 we have
’4
4
ζ24 ’ 1 + ζ24 = 0,
giving
’4
4
(ζ24 + ζ24 ) ’ 1 = 0.
Now
’1 ’4 ’2
(ζ24 + ζ24 )4 = (ζ24 + ζ24 ) + 4(ζ24 + ζ24 ) + 6,
4 2

hence
’4 ’1 ’2
ζ24 + ζ24 = (ζ24 + ζ24 )4 ’ 4(ζ24 + ζ24 ) ’ 6.
4 2

Similarly,
’1 ’2
(ζ24 + ζ24 )2 = ζ24 + ζ24 + 2,
2

so
’2 ’1
ζ24 + ζ24 = (ζ24 + ζ24 )2 ’ 2.
2

Combining these we have
’1 ’1
(ζ24 + ζ24 )4 ’ 4(ζ24 + ζ24 )2 + 1 = 0,
and so
16 cos4 (π/12) ’ 16 cos2 (π/12) + 1 = 0.
This method will work for any n where •(n) is even, i.e., when n > 2.
Remark 6.6. The polynomial that expresses cos nθ as a polynomial in cos θ is the n-th
Chebsyhev polynomial of the ¬rst kind Tn (X) ∈ Z[X]. Here are the ¬rst few of these polynomi-
als:
T2 (X) = 2X 2 ’ 1, T3 (X) = 4X 3 ’ 3X,
T4 (X) = 8X 4 ’ 8X 2 + 1, T5 (X) = 16X 5 ’ 20X 3 + 5X,
T6 (X) = 32X 6 ’ 48X 4 + 18X 2 ’ 1, T7 (X) = 64X 7 ’ 112X 5 + 56X 3 ’ 7X.
These form a system of orthogonal polynomials which can be computed in Maple using the
command orthopoly[T](n,X).
6.3. ARTIN™S THEOREM ON LINEAR INDEPENDENCE OF CHARACTERS 71

Now let K be a ¬eld with characteristic char K n. The polynomial ¦n (X) has integer
coe¬cients, so we can view it as an element of K[X] since either Q K or Fp K and we
can reduce the coe¬cients modulo p. In either case it can happen that ¦n (X) factors in K[X].
However, we can still describe the splitting ¬eld of X n ’ 1 over K and its Galois group.
Theorem 6.7. If char K n, then the splitting ¬eld of X n ’ 1 over K is K(ζ), where ζ ∈ K
is a primitive n-th root of unity and Gal(K(ζ)/K) is abelian with order dividing •(n).
Proof. Working in K, we know that ¦n (ζ) = 0, hence the roots of minpolyK,ζ (X) ∈ K[X]
are primitive roots of 1. So X n ’ 1 splits over K(ζ) and each element ± ∈ Gal(K(ζ)/K) has
the action ±(ζ) = ζ r± , where gcd(r± , n) = 1. Hence Gal(K(ζ)/K) is isomorphic to a subgroup
of Gal(Q(ζn )/Q) ∼ (Z/n)— , in particular it is abelian.
=
Remark 6.8. When p = char K > 0, this Galois group only depends on the largest sub¬eld
of K which is algebraic over Fp . For example, if K = Fpd (T ) then the value of d is the crucial
factor. The precise outcome can be determined with the aid of Proposition 5.21.
Example 6.9. We have the following splitting ¬elds and Galois groups.
(i) The splitting ¬eld of X 4 ’ 1 over F3 (T ) is F9 (T ) and
Gal(F9 (T )/F3 (T )) ∼ (Z/4)— ∼ Z/2.
= =
(ii) By Proposition 5.20, X 4 ’ 1 splits over F5 (T ) and the Galois group Gal(F5 (T )/F5 (T ))
is trivial.
Proof. (i) By Proposition 5.20, X 4 ’ 1 is separable over F3 (T ) and has irreducible factors
(X ’ 1), (X + 1) and (X 2 + 1). The splitting ¬eld of (X 2 + 1) over F3 is F9 = F3 (ζ), where
ζ 2 + 1 = 0, so (X 2 + 1) splits over F9 (T ). Also,
Gal(F9 /F3 ) ∼ (Z/4)— ∼ Z/2,
= =
with generator σ satisfying σ(ζ) = ζ ’1 = ’ζ. This generator clearly extends to an automor-
phism of F9 (T ) which ¬xes T .
(ii) By Proposition 5.20, X 4 ’ 1 splits over F5 .

6.3. Artin™s Theorem on linear independence of characters
Let G be a group and K a ¬eld.
Definition 6.10. A group homomorphism χ : G ’’ K — is called a character of G with
values in K.
Example 6.11. Given any ring homomorphism • : R ’’ K we obtain a character of R—
in K by restricting • to a map χ• : R— ’’ K — .
Example 6.12. Given an automorphism ± : K ’’ K, χ± : K — ’’ K — is a character of
K — in K.
Example 6.13. Let E/K be a Galois extension and σ ∈ Gal(E/K). Then χσ : E — ’’ E —
is a character.
Definition 6.14. Let χ1 , . . . , χn be characters of a group G in a ¬eld K. Then χ1 , . . . , χn
are linearly independent if for t1 , . . . , tn ∈ K,
t1 χ1 + · · · + tn χn = 0 =’ t1 = · · · = tn = 0.
If χ1 , . . . , χn are not linearly independent they are linearly dependent.
In this de¬nition, the functional equation means that for all g ∈ G,
t1 χ1 (g) + · · · + tn χn (g) = 0.
Theorem 6.15 (Artin™s Theorem). Let χ1 , . . . , χn be distinct characters of a group G in a
¬eld K. Then χ1 , . . . , χn are linearly independent.
72 6. A GALOIS MISCELLANY

Proof. We proceed by induction on n. For n = 1 the result is easily veri¬ed. For the
inductive assumption, suppose that it holds for any n k.
Let χ1 , . . . , χk+1 be a set of k + 1 distinct characters for which there are t1 , . . . , tk+1 ∈ K
not all zero and such that
(6.2) t1 χ1 + · · · + tk+1 χk+1 = 0.
If one of the ti is zero, say tr = 0, then χ1 , . . . , χr’1 , χr+1 , . . . , χk+1 is linearly dependent,
contradicting the inductive assumption. Hence all of the ti must be non-zero. As χ1 = χ2 , there
must be an element g0 ∈ G for which χ1 (g0 ) = χ2 (g0 ). So for all g ∈ G, Equation (6.2) applied
to g0 g yields
t1 χ1 (g0 g) + · · · + tk+1 χk+1 (g0 g) = 0,
and therefore since χj (g0 g) = χj (g0 )χj (g), we see that
t1 χ1 (g0 )χ1 + · · · + tk+1 χk+1 (g0 )χk+1 = 0.
Multiplying Equation (6.2) by χ1 (g0 ) and subtracting gives
t2 (χ2 (g0 ) ’ χ1 (g0 ))χ2 + t3 (χ3 (g0 ) ’ χ1 (g0 ))χ3 + · · · + tk+1 χk+1 = 0,
in which the coe¬cient t2 (χ2 (g0 )’χ1 (g0 )) is not zero. Hence χ2 , . . . , χk+1 is linearly dependent,
again contradicting the inductive assumption. So χ1 , . . . , χk+1 is linearly independent, which
demonstrates the inductive step.
Suppose that E/K is a ¬nite Galois extension with cyclic Galois group Gal(E/K) = σ of
order n. For each u ∈ E — , the element uσ(u) · · · σ n’1 (u) ∈ E satis¬es
σ(uσ(u) · · · σ n’1 (u)) = σ(u) · · · σ n’1 (u)σ n (u) = σ(u) · · · σ n’1 (u)u,
hence in uσ(u) · · · σ n’1 (u) ∈ E σ = K. Using this we can de¬ne a group homomorphism
NE/K : E — ’’ K — ; NE/K (u) = uσ(u) · · · σ n’1 (u).
NE/K is called the norm mapping for E/K and generalizes the norm mapping for ¬nite ¬elds
of Section 5.3.
There is another homomorphism
δE/K : E — ’’ E — ; δE/K (u) = uσ(u)’1 .
Notice that for u ∈ E — ,
NE/K (δE/K (u)) = (uσ(u)’1 )(σ(u)σ 2 (u)’1 · · · σ n’1 (u)σ n (u)’1 ) = 1,
since σ n (u) = u. So im δE/K ker NE/K . Our next result is an important generalization of
Proposition 5.27.
Theorem 6.16 (Hilbert™s Theorem 90). Let E/K be a ¬nite Galois extension with cyclic
Galois group Gal(E/K) = σ of order n. Then im δE/K = ker NE/K . Explicitly, if u ∈ E — and
u σ(u) · · · σ n’1 (u) = 1, then there is a v ∈ E — such that u = vσ(v)’1 .
Proof. Let u ∈ ker NE/K .
The characters σ k : E — ’’ E — with k = 0, 1, . . . , n’1 are distinct and linearly independent
by Artin™s Theorem 6.15. Consider the function
id +uσ + uσ(u)σ 2 + · · · + uσ(u) · · · σ n’2 (u)σ n’1 : E — ’’ E.
This cannot be identically zero, so for some w ∈ E, the element
v = w + uσ(w) + uσ(u)σ 2 (w) + · · · + uσ(u) · · · σ n’2 (u)σ n’1 (w)
is non-zero. Notice that
uσ(v) = uσ(w) + uσ(u)σ 2 (w) + uσ(u)σ 2 (u)σ 3 (w) + · · · + uσ(u)σ 2 (u) · · · σ n’1 (u)σ n (w) = v,
since
uσ(u)σ 2 (u) · · · σ n’1 (u)σ n (w) = w.
Thus we have u = vσ(v)’1 as required.
6.4. SIMPLE RADICAL EXTENSIONS 73

6.4. Simple radical extensions
In this section we will investigate splitting ¬elds of polynomials of the form X n ’ a, where
char K n. We call these simple radical extensions and later in De¬nition 6.31 we introduce a
more general notion of radical extension.
Proposition 6.17. Let f (X) = X n ’ a ∈ K[X] be irreducible and separable over K. Then
the splitting ¬eld of f (X) over K has the form E = K(u, ζ), where u is a root of f (X) and ζ
is a primitive n-th root of 1.
Corollary 6.18. If K contains a primitive n-th root of unity ζ then the splitting ¬eld of
f (X) = X n ’ a over K has the form E = K(u), where u is a root of f (X). The Galois group
Gal(K(f (X))/K) is cyclic of order n with a generator σ for which σ(u) = ζu.
In the more general situation of Proposition 6.17,
{id} Gal(K(ζ)/K) Gal(K(f (X))/K(ζ)),
where Gal(K(ζ)/K) and Gal(K(ζ)/K))/ Gal(K(ζ, u)/K(ζ) are abelian and in fact cyclic. The
Galois Correspondence identi¬es the following towers of sub¬elds and subgroups.
K(ζ, u) Gal(K(ζ, u)/K)
` 8


K(ζ) Gal(K(ζ)/K)
9

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