3Solutions to polynomial equations
II Galois Theory
3.4 Solubility of groups, extensions and polynomials
Let
f ∈ K
[
t
]. We defined the notion of solubility of
f
in terms of radical
extensions. However, can we decide whether
f
is soluble or not without resorting
to the definition? In particular, is it possible to decide whether its soluble by
just looking at
Gal
(
L/K
), where
L
is the splitting field of
f
over
K
? It would
be great if we could do so, since groups are easier to understand than fields.
The answer is yes. It turns out the solubility of
f
corresponds to the solubility
of
Gal
(
L/K
). Of course, we will have to first define what it means for a group
to be soluble. After that, we will find examples of polynomials
f
of degree at
least 5 such that
Gal
(
L/K
) is not soluble. In other words, there are polynomials
that cannot be solved by radicals.
Definition (Soluble group). A finite group
G
is soluble if there exists a sequence
of subgroups
G
r
= {1} ◁ ··· ◁ G
1
◁ G
0
= G,
where G
i+1
is normal in G
i
and G
i
/G
i+1
is cyclic.
Example. Any finite abelian group is solvable by the structure theorem of finite
abelian groups:
G
∼
=
Z
⟨n
1
⟩
× ··· ×
Z
⟨n
r
⟩
.
Example. Let
S
n
be the symmetric group of permutations of
n
letters. We
know that G
3
is soluble since
{1} ◁ A
3
◁ S
3
,
where S
3
/A
3
∼
=
Z/⟨2⟩ and A
3
/{0}
∼
=
Z/⟨3⟩.
S
4
is also soluble by
{1} ◁ {e, (1 2)(3 4)} ◁ {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ◁ A
4
◁ S
4
.
We can show that the quotients are
Z/⟨
2
⟩
,
Z/⟨
3
⟩
,
Z/⟨
2
⟩
and
Z/⟨
2
⟩
respectively.
How about
S
n
for higher
n
? It turns out they are no longer soluble for
n ≥
5.
To prove this, we first need a lemma.
Lemma. Let G be a finite group. Then
(i) If G is soluble, then any subgroup of G is soluble.
(ii)
If
A ◁ G
is a normal subgroup, then
G
is soluble if and only if
A
and
G/A
are both soluble.
Proof.
(i) If G is soluble, then by definition, there is a sequence
G
r
= {1} ◁ ··· ◁ G
1
◁ G
0
= G,
such that G
i+1
is normal in G
i
and G
i
/G
i+1
is cyclic.
Let
H
i
=
H ∩G
i
. Note that
H
i+1
is just the kernel of the obvious homo-
morphism
H
i
→ G
i
/G
i+1
. So
H
i+1
◁ H
i
. Also, by the first isomorphism
theorem, this gives an injective homomorphism
H
i
/H
i+1
→ G
i
/G
i+1
. So
H
i
/H
i+1
is cyclic. So H is soluble.
(ii)
(
⇒
) By (i), we know that
A
is solvable. To show the quotient is soluble,
by assumption, we have the sequence
G
r
= {1} ◁ ··· ◁ G
1
◁ G
0
= G,
such that
G
i+1
is normal in
G
i
and
G
i
/G
i+1
is cyclic. We construct the
sequence for the quotient in the obvious way. We want to define
E
i
as the
quotient
G
i
/A
, but since
A
is not necessarily a subgroup of
E
, we instead
define
E
i
to be the image of quotient map
G
i
→ G/A
. Then we have a
sequence
E
r
= {1} ◁ ··· ◁ E
0
= G/A.
The quotient map induces a surjective homomorphism
G
i
/G
i+1
→
E
i
/E
i+1
, showing that E
i
/E
i+1
are cyclic.
(⇐) From the assumptions, we get the sequences
A
m
= {1} ◁ ··· ◁ A
0
= A
F
n
= A ◁ ··· ◁ F
0
= G
where each quotient is cyclic. So we get a sequence
A
m
= {1} ◁ A
1
◁ ··· ◁ A
0
= F
n
◁ F
n−1
◁ ··· ◁ F
0
= G,
and each quotient is cyclic. So done.
Example. We want to show that
S
n
is not soluble if
n ≥
5. It is a well-known
fact that
A
n
is a simple non-abelian group, i.e. it has no non-trivial subgroup.
So A
n
is not soluble. So S
n
is not soluble.
The key observation in Galois theory is that solubility of polynomials is
related to solubility of the Galois group.
Definition (Soluble extension). A finite field extension
L/K
is soluble if there
is some extension L ⊆ E such that K ⊆ E is Galois and Gal(E/K) is soluble.
Note that this definition is rather like the definition of a radical extension,
since we do not require the extension itself to be “nice”, but just for there to be
a further extension that is nice. In fact, we will soon see they are the same.
Lemma. Let
L/K
be a Galois extension. Then
L/K
is soluble if and only if
Gal(L/K) is soluble.
This means that the whole purpose of extending to
E
is just to make it a
Galois group, and it isn’t used to introduce extra solubility.
Proof. (⇐) is clear from definition.
(
⇒
) By definition, there is some
E ⊆ L
such that
E/K
is Galois and
Gal
(
E/K
) is soluble. By the fundamental theorem of Galois theory,
Gal
(
L/K
) is
a quotient of
Gal
(
E/K
). So by our previous lemma,
Gal
(
L/K
) is also soluble.
We now come to the main result of the lecture:
Theorem. Let
K
be a field with
char K
= 0, and
L/K
is a radical extension.
Then L/K is a soluble extension.
Proof.
We have already shown that if we have a radical extension
L/K
, then
there is a finite extension
K ⊆ E
such that
K ⊆ E
is a Galois extension, and
there is a sequence of cyclotomic or Kummer extensions
E
0
= K ⊆ E
1
⊆ ··· ⊆ E
r
= E.
Let
G
i
=
Gal
(
E/E
i
). By the fundamental theorem of Galois theory, inclusion of
subfields induces an inclusion of subgroups
G
0
= Gal(E/K) ≥ G
1
≥ ··· ≥ G
r
= {1}.
In fact,
G
i
▷ G
i+1
because
E
i
⊆ E
i+1
are Galois (since cyclotomic and Kummer
extensions are). So in fact we have
G
0
= Gal(E/K) ▷ G
1
▷ ··· ▷ G
r
= {1}.
Finally, note that by the fundamental theorem of Galois theory,
G
i
/G
i+1
= Gal(E
i+1
/E
i
).
We also know that the Galois groups of cyclotomic and Kummer extensions are
abelian. Since abelian groups are soluble, our previous lemma implies that
L/K
is soluble.
In fact, we will later show that the converse is also true. So an extension is
soluble if and only if it is radical.
Corollary. Let
K
be a field with
char K
= 0, and
f ∈ K
[
t
]. If
f
can be solved
by radicals, then
Gal
(
L/K
) is soluble, where
L
is the splitting field of
f
over
K
.
Again, we will later show that the converse is also true. However, to construct
polynomials that cannot be solved by radicals, this suffices. In fact, this corollary
is all we really need.
Proof.
We have seen that
L/K
is a Galois extension. By assumption,
L/K
is
thus a radical extension. By the theorem,
L/K
is also a soluble extension. So
Gal(L/K) is soluble.
To find an
f ∈ Q
[
t
] which cannot be solved by radicals, it suffices to find an
f
such that the Galois group is not soluble. We don’t know many non-soluble
groups so far. So in fact, we will find an f such that Gal(L/Q) = S
5
.
To do so, we want to relate Galois groups to permutation groups.
Lemma. Let
K
be a field,
f ∈ K
[
t
] of degree
n
with no repeated roots. Let
L
be the splitting field of
f
over
K
. Then
L/K
is Galois and there exist an
injective group homomorphism
Gal(L/K) → S
n
.
Proof.
Let
Root
f
(
L
) =
{α
1
, ··· , α
n
}
. Let
P
α
i
be the minimal polynomial of
α
i
over
K
. Then
P
α
i
| f
implies that
P
α
i
is separable and splits over
L
. So
L/K
is
Galois.
Now each
ϕ ∈ Gal
(
L/K
) permutes the
α
i
, which gives a map
Gal
(
L/K
)
→ S
n
.
It is easy to show this is an injective group homomorphism.
Note that there is not a unique or naturally-defined injective group homo-
morphism to
S
n
. This homomorphism, obviously, depends on how we decide to
number our roots.
Example. Let
f
= (
t
2
−
2)(
t
2
−
3)
∈ Q
[
t
]. Let
L
be the splitting field of
f
over
Q. Then the roots are
Root
f
(L) = {
√
2, −
√
2,
√
3, −
√
3}.
We label these roots as
α
1
, α
2
, α
3
, α
4
in order. Now note that
L
=
Q
(
√
2,
√
3
),
and thus [
L
:
Q
] = 4. Hence
|Gal
(
L/Q
)
|
= 4 as well. We can let
Gal
(
L/Q
) =
{id, φ, ψ, λ}, where
id(
√
2) =
√
2 id(
√
3) =
√
3
φ(
√
2) = −
√
2 φ(
√
3) =
√
3
ψ(
√
2) =
√
2 ψ(
√
3) = −
√
3
λ(
√
2) = −
√
2 λ(
√
3) = −
√
3
Then the image of Gal(L/Q) → S
4
is given by
{e, (1 2), (3 4), (1 2)(3 4)}.
What we really want to know is if there are polynomials in which this map is
in fact an isomorphism, i.e. the Galois group is the symmetric group. If so, then
we can use this to produce a polynomial that is not soluble by polynomials.
To find this, we first note a group-theoretic fact.
Lemma. Let p be a prime, and σ ∈ S
p
have order p. Then σ is a p-cycle.
Proof. By IA Groups, we can decompose σ into a product of disjoint cycles:
σ = σ
1
···σ
r
.
Let σ
i
have order m
i
> 1. Again by IA Groups, we know that
p = order of σ = lcm(m
1
, ··· , m
r
).
Since
p
is a prime number, we know that
p
=
m
i
for all
i
. Hence we must have
r
= 1, since the cycles are disjoint and there are only
p
elements. So
σ
=
σ
1
.
Hence σ is indeed an p cycle.
We will use these to find an example where the Galois group is the symmetric
group. The conditions for this to happen are slightly awkward, but the necessity
of these will become apparent in the proof.
Theorem. Let
f ∈ Q
[
t
] be irreducible and
deg f
=
p
prime. Let
L ⊆ C
be the
splitting field of f over Q. Let
Root
f
(L) = {α
1
, α
2
, ··· , α
p−2
, α
p−1
, α
p
}.
Suppose that
α
1
, α
2
, ··· , α
p−2
are all real numbers, but
α
p−1
and
α
p
are not.
In particular,
α
p−1
=
¯α
p
. Then the homomorphism
β
:
Gal
(
L/Q
)
→ S
n
is an
isomorphism.
Proof.
From IA groups, we know that the cycles (1 2
··· p
) and (
p −
1
p
)
generate the whole of
S
n
. So we show that these two are both in the image of
β
.
As
f
is irreducible, we know that
f
=
P
α
1
, the minimal polynomial of
α
1
over Q. Then
p = deg P
α
i
= [Q(α
1
) : Q].
By the tower law, this divides [
L
:
Q
], which is equal to
|Gal
(
L/Q
)
|
since the
extension is Galois. Since
p
divides the order of
Gal
(
L/Q
), by Cauchy’s theorem
of groups, there must be an element of
Gal
(
L/Q
) that is of order
p
. This maps
to an element σ ∈ im β of order exactly p. So σ is a p-cycle.
On the other hand, the isomorphism
C → C
given by
z 7→ ¯z
restricted to
L
gives an automorphism in
Gal
(
L/Q
). This simply permutes
α
p−1
and
α
p
, since
it fixes the real numbers and
α
p−1
and
α
p
must be complex conjugate pairs. So
τ = (p − 1 p) ∈ im β.
Now for every 1
≤ i < p
, we know that
σ
i
again has order
p
, and hence
is a
p
-cycle. So if we change the labels of the roots
α
1
, ··· , α
p
and replace
σ
with
σ
i
, and then waffle something about combinatorics, we can assume
σ = (1 2 ··· p − 1 p). So done.
Example. Let t
5
− 4t + 2 ∈ Q[t]. Let L be the splitting field of f over Q.
First note that
deg f
= 5 is a prime. Also, by Eisenstein’s criterion,
f
is
irreducible. By finding the local maximum and minimum points, we find that
f
has exactly three real roots. So by the theorem, there is an isomorphism
Gal(L/Q) → S
5
. In other words, Gal(L/Q)
∼
=
S
5
.
We know S
5
is not a soluble group. So f cannot be solved by radicals.
After spending 19 lectures, we have found an example of a polynomial that
cannot be solved by radicals. Finally.
Note that there are, of course, many examples of
f ∈ Q
[
t
] irreducible of
degree at least 5 that can be solved by radicals, such as f = t
5
− 2.