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