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. 17
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e

˜ '
{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

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