˜ '

{id}

K

Definition 6.19. Let K be a ¬eld of characteristic not dividing n and which contains a

primitive n-th root of 1, ζ say. Then L/K is a simple n-Kummer extension if L = K(u) where

un = a for some a ∈ K. L/K is an (iterated ) n-Kummer extension if L = K(u1 , . . . , uk ) where

un = a1 , ..., un = ak for some elements a1 , . . . , ak ∈ K.

1 k

Note that we do not require the polynomials X n ’ aj ∈ K[X] to be irreducible in this

de¬nition.

Proposition 6.20. Let K(u)/K be a simple n-Kummer extension. Then K(u)/K is a

Galois extension and Gal(K(u)/K) is cyclic with order dividing n.

Proof. Suppose that un = a ∈ K. Then in K[X] we have

X n ’ a = (X ’ u)(X ’ ζu) · · · (X ’ ζ n’1 u).

Clearly the roots of X n ’ a are distinct and so K(u)/K is separable over K; in fact, K(u) is a

splitting ¬eld of X n ’ a over K. This means that K(u)/K is Galois.

For each ± ∈ Gal(K(u)/K) we have ±(u) = ζ r± u for some r± = 0, 1 . . . , n ’ 1. Notice that

for β ∈ Gal(K(u)/K),

β±(u) = β(ζ r± u) = ζ r± β(u) = ζ r± ζ rβ u = ζ r± +rβ u,

and so rβ± = r± + rβ . Hence the function

ρ(±) = ζ r± ,

ρ : Gal(K(u)/K) ’’ ζ ;

is a group homomorphism. As ζ is cyclic of order n, Lagrange™s Theorem implies that the

image of ρ has order dividing n. Since every element of Gal(K(u)/K) is determined by its e¬ect

on u, ρ is injective, hence | Gal(K(u)/K)| divides n. In fact, Gal(K(u)/K) is cyclic since every

subgroup of a cyclic group is cyclic.

√

Example 6.21. Let n 1 and q ∈ Q. Then Q(ζn , n q)/Q(ζn ) is a simple n-Kummer

extension.

√ √

Example 6.22. Q(i, 2)/Q(i) is a simple 4-Kummer extension for which Gal(Q(i, 2)/Q(i))

is cyclic of order 2.

74 6. A GALOIS MISCELLANY

√

Proof. We have ( 2)4 ’ 4 = 0, but

X 4 ’ 4 = (X 2 ’ 2)(X 2 + 2),

and

X 2 ’ 2 = minpolyQ(i),√2 (X).

√

The corresponding group homomorphism ρ : Gal(Q(i)( 2)/Q(i)) ’’ i has image

im ρ = {1, ’1} i.

Here is a converse to Proposition 6.20.

Proposition 6.23. Suppose that char K n and there is an element ζ ∈ K which is a

primitive n-th root of unity. If E/K is a ¬nite Galois extensions with cyclic Galois group of

order n, then there is an element a ∈ E such that E = K(a) and a is a root of a polynomial of

the form X n ’ b with b ∈ K. Hence E/K is a simple n-Kummer extension.

Proof. We have

NE/K (ζ) = ζ n = 1,

so by Hilbert™s Theorem 6.16, there is an element a ∈ E for which ζ = aσ(a)’1 . Then σ(a) =

ζ ’1 a and the elements σ k (a) = ζ ’k a for k = 0, 1, . . . , n ’ 1 are distinct, so they must be the n

conjugates of a. Also note that

X n ’ an = (X ’ a)(X ’ ζa) · · · (X ’ ζ n’1 a) = (X ’ a)(X ’ σ(a)) · · · (X ’ σ n’1 (a)),

so an ∈ K since it is ¬xed by σ. This shows that K(a) E and

n = [K(a) : K] [E : K] = n,

therefore [K(a) : K] [E : K] = n and K(a) = E.

6.5. Solvability and radical extensions

We begin by recalling some ideas about groups, see [3, 4] for further details.

Definition 6.24. A group G is solvable, soluble or soluable if there is a chain of subgroups

(called a subnormal series)

{1} = G G ··· G1 G0 = G

’1

in which Gk+1 Gk and each composition factor Gk /Gk+1 is abelian; we usually write

{1} = G G · · · G1 G0 = G.

’1

If each composition factor is a cyclic group of prime order the subnormal series is called a

composition series. A group which is not solvable is called insolvable.

Remark 6.25. For a solvable group, it is a standard result that we can always re¬ne a

subnormal series (i.e., add extra terms) to obtain a composition series. The primes as well as

the number of times each occurs are all determined by |G|, only the order of these varying for

di¬erent composition series.

Example 6.26. Let G be a ¬nite abelian group. Then G is solvable.

Example 6.27. Let G be a ¬nite p-group, where p is a prime. Then G is solvable.

In fact, for a ¬nite p-group G, there is always a normal subgroup of a p-group with index p,

so in this case we can assume each quotient Gk /Gk+1 is cyclic of order p.

Proposition 6.28. Let G be a group.

(i) If G is solvable then every subgroup H G and every quotient group G/N is solvable.

(ii) If N G and G/N are solvable then so is G.

In the opposite direction we can sometimes see that a group is insolvable. Recall that a

group is simple if it has no non-trivial proper normal subgroups.

6.5. SOLVABILITY AND RADICAL EXTENSIONS 75

Proposition 6.29. Let G be a ¬nite group. Then G is insolvable if any of the following

conditions holds:

(i) G contains a subgroup which is a non-abelian simple group (or has a quotient group

which is a non-abelian simple group).

(ii) G has a quotient group which is a non-abelian simple group.

(iii) G has a composition series in which one of the terms is a non-abelian simple group.

Example 6.30. For n 5, the alternating and symmetric groups An and Sn are insolvable.

Proof. This follows from the fact that if n 5, An is a simple group and An Sn with

quotient group Sn /An ∼ Z/2.

=

Now we explain how this relates to ¬elds and their extensions. Let K be a ¬eld and L/K a

¬nite extension. For simplicity, we assume also that char K = 0.

Definition 6.31. L/K is a radical extension of K if it has the form L = K(a1 , a2 , . . . , an )

with

adk ∈ K(a1 , a2 , . . . , ak’1 )

k

for some dk 1. Thus every element of L is expressible in terms of iterated roots of elements

of K.

Definition 6.32. If L is the splitting ¬eld of a polynomial f (X) ∈ K[X], then f (X) is

solvable by radicals over K if L is contained in a radical extension of K.

Definition 6.33. L/K is solvable if L L where L /K is a ¬nite radical Galois extension

of K.

Theorem 6.34. Let E/K be a ¬nite Galois extension. Then E/K is solvable if and only if

the group Gal(E/K) is solvable.

™

Proof. Suppose that E E where E /K is a radical Galois extension. Then

™™¦

™

Gal(E/K) is a quotient group of Gal(E /K), so it is solvable by Proposition 6.28.

Now suppose that Gal(E/K) is solvable and let n = | Gal(E/K)|. Let E be the splitting

¬eld of X n ’ 1 over E, so E contains a primitive n-th root of unity ζ and therefore it contains

a primitive d-th root of unity for every divisor d of n. Now Gal(E /E) Gal(E /K) and

by Theorem 6.7, Gal(E /E) is abelian. Also, Gal(E /K)/ Gal(E /E) ∼ Gal(E/K) which is

=

solvable, so Gal(E /K) is solvable by Proposition 6.28. We will now show that E /K is a

radical extension.

Clearly K(ζ)/K is radical. Then Gal(E /K(ζ)) Gal(E /K) is solvable. Let

{1} = G G · · · G1 G0 = Gal(E /K(ζ))

’1

be a composition series. The extension (E )G1 /K(ζ) is radical by Proposition 6.23. Similarly,

each extension (E )Gk+1 /(E )Gk is radical. Hence E /K(ζ) is radical, as is E /K.

√

Example 6.35. The Galois group of the extension Q(ζ3 , 3 2)/Q is solvable.

√

Proof. We have already studied this extension in Example 3.30 and 4.19. Clearly Q(ζ3 , 3 2)

is a radical extension of Q and

√ √

3 3

Q(ζ3 , 2) = Q(ζ3 )( 2).

√

We know that Gal(Q(ζ3 , 3 2)/Q) ∼ S3 , where we identify each element of the Galois group with

= √

a permutation of the three roots of X 3 ’ 2 in Q(ζ3 , 3 2) which we list in the order

√√ √2

3 3 3

2, 2 ζ3 , 2 ζ3 .

76 6. A GALOIS MISCELLANY

We have the following towers of sub¬elds and subgroups related under the Galois Correspon-

dence.

√

Q(ζ3 , 3 2) 5 S3

e

3 2

√A √

3 3

Q(ζ3 ) = Q(ζ3 , 2) 3 A3 = Gal(Q(ζ3 , 2)/Q(ζ3 ))

i 5

2 3

y )

{id}

Q

Here Q(ζ3 )/Q is itself a Galois extension and A3 S3 . Notice that A3 ∼ Z/3 and S3 /A3 ∼ Z/2,

= =

so we have the following composition series for S3 :

{id} A3 S3 .

It is also interesting to reverse the question and ask whether there are extensions which are

not solvable. This was a famous problem pursued for several hundred years. To ¬nd examples,

we ¬rst recall that the smallest non-abelian simple group is A5 which has order 60. We should

therefore expect to look for a polynomial of degree at least 5 to ¬nd a Galois group for a splitting

¬eld to be simple or occur as a composition factor of such a Galois group. Here is an explicit

example over Q.

Example 6.36. The splitting ¬eld of the polynomial f (X) = X 5 ’ 35X 4 + 7 ∈ Q[X] is not

solvable by radicals.

Proof. Let E = Q(f (X)) be the splitting ¬eld of f (X) over Q. Using the Eisenstein

Criterion 1.37 with p = 7, we ¬nd that f (X) is irreducible over Q. By Theorem 4.7(iii),

5 divides the order of Gal(E/Q), so this group contains an element of order 5 by Cauchy™s

Lemma.

Now observe that

f (X) = 5X 4 ’ 140X 3 = 5X 3 (X ’ 28), f (X) = 20X 4 ’ 420X 2 = 20X 2 (X ’ 21).

There are two turning points, namely a maximum at x = 0 and a minimum at x = 28. Then

f (0) = 7 > 0 > f (28) = ’4302585,

hence there are three real roots of f (X) and two non-real complex ones. Then complex conju-

gation restricts to an element of order 2 in Gal(E/Q) which interchanges the non-real roots and

¬xes the others. If we list the roots of f (X) as u1 , u2 , u3 , u4 , u5 with u1 , u2 being the non-real

roots, then the transposition (1 2) ∈ S5 corresponds to this element. Furthermore, the only

elements of S5 of order 5 are 5-cycles; by taking an appropriate power we can assume that there

is a 5-cycle of the form (1 2 3 4 5) corresponding to an element of Gal(E/Q) which we can view

as a subgroup of S5 . The next lemma shows that Gal(E/Q) ∼ S5 . =

Lemma 6.37. Let n 1. Suppose that H Sn and H contains the elements (1 2) and

(1 2 · · · n). Then H = Sn .

The proof is left as an exercise. This completes the veri¬cation of the Example.

It is worth remarking that the most extreme version of this occurs when we ask for a Galois

group which is simple. There has been a great deal of research activity on this question in the

past few decades, but apparently not all simple groups are known to occur as Galois groups of

extensions of Q or other ¬nite subextensions of C/Q. Here is an example whose Galois group

is A5 ; this is veri¬ed using Proposition 6.42.

Example 6.38. The Galois group of f (X) = X 5 +20X +16 over Q is Gal(Q(f (X))/Q) ∼ A5 .

=

6.6. GALOIS GROUPS OF EVEN AND ODD PERMUTATIONS 77

6.6. Galois groups of even and odd permutations

We have seen that for a monic separable polynomial f (X) ∈ K[X] of degree n, the Galois

group of its splitting ¬eld Ef = K(f (X)) can naturally be thought of as a subgroup of the

symmetric group Sn , where we view the latter as permuting the roots of f (X). It is reasonable

to ask when Gal(Ef /K) An rather than just Gal(Ef /K) Sn .

We ¬rst recall an interpretation of the sign of a permutation σ ∈ Sn , sgn σ = ±1. For each

pair i, j with 1 i<j n, exactly one of the inequalities σ(i) < σ(j) or σ(j) < σ(i) must

hold and the ratio (σ(j) ’ σ(i))/(j ’ i) is either positive or negative. It is easily veri¬ed that

the right-hand side of the following equation must have value ±1 and we have

σ(j) ’ σ(i)

(6.3) sgn σ = .

j’i

16i<j6n

Note that this is sometimes used as the de¬nition of sgn σ.

Suppose that f (X) factorizes over Ef as

n

f (X) = (X ’ u1 ) · · · (X ’ un ) = (X ’ ui ).

i=1

Here u1 , . . . , un ∈ Ef are the roots of f (X); as we have assumed that f (X) is separable, the ui

are distinct.

Definition 6.39. The discriminant of f (X) is

(uj ’ ui )2 ∈ Ef .

Discr(f (X)) =

16i<j6n

Notice that Discr(f (X)) = 0 since ui = uj if i = j.

Remark 6.40. There is an explicit formula for computing Discr(f (X)) is terms of its coef-

¬cients. For polynomials

p(X) = a0 + a1 X + · · · + am X m , q(X) = a0 + a1 X + · · · + am X m ,

their resultant is the (m + n) — (m + n) determinant (with n rows of ai ™s and m rows of bi ™s)

a0 a1

. . . . . . . am 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0

0 a0

a1 . . . . . . . am 0 . . . . . . . . . . . . . . . . . . . . . . . . . . 0

.. ..

.. .. .. .. .. .. .. .. .. .. .. ..

. .

. . . . . . . . . . . .

0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 a0 a1 . . . . . . . am

(6.4) Res(p(X), q(X)) = .

b0 b1 . . . . . . . bn 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0

0 b0 b1 . . . . . . . bn 0 . . . . . . . . . . . . . . . . . . . . . . . . . . 0

.. .. .. .. .. .. .. .. .. .. .. .. .. ..

. . . . . . . . . . . . . .

0 . . . . . . . . . . . . . . . . . . . . . . . . . . 0 b0 b1 . . . . . . . . . . . . bn

Then if f (X) is monic with d = deg f (X),

Discr(f (X)) = (’1)d(d’1)/2 Res(f (X), f (X)).

(6.5)

So for example,

Discr(X 3 + pX + q) = (’1)3 Res(X 3 + pX + q, 3X 2 + p)

q p 0 1 0

0 q p 0 1

= (’1) p 0 3 0 0

0 p 0 3 0

0 0 p 0 3

= ’4p3 ’ 27q 2 .

78 6. A GALOIS MISCELLANY

Here are some low degree examples of discriminants obtained using Maple.

Discr(a0 + a1 X + X 2 ) = ’4a0 + a2 .

n = 2: 1