3Solutions to polynomial equations
II Galois Theory
3.3 Radical extensions
We are going to put these together and look at radical extensions, which allows
us to characterize what it means to “solve a polynomial with radicals”.
Definition (Radical extension). A field extension
L/K
is radical if there is
some further extension E/L and with a sequence
K = E
0
⊆ E
1
⊆ ··· ⊆ E
r
= E,
such that each
E
i
⊆ E
i+1
is a cyclotomic or Kummer extension, i.e.
E
i+1
is a
splitting field of t
n
− λ
i+1
over E
i
for some λ
i+1
∈ E
i
.
Informally, we say
E
i+1
is obtained by adding the roots “
n
p
λ
i+1
” to
E
i
.
Hence we interpret a radical extension as an extension that only adds radicals.
Definition (Solubility by radicals). Let
K
be a field, and
f ∈ K
[
t
].
f
. We say
f is soluble by radicals if the splitting field of f is a radical extension of K.
This means that f can be solved by radicals of the form
n
√
λ
i
.
Let’s go back to our first lecture and describe what we’ve done in the language
we’ve developed in the course.
Example. As we have shown in lecture 1, any polynomial
f ∈ Q
[
t
] of degree at
most 4 can be solved by radicals.
For example, assume
deg f
= 3. So
f
=
t
3
+
at
2
+
bt
+
c
. Let
L
be the
splitting field of
f
. Recall we reduced the problem of “solving”
f
to the case
a
= 0 by the substitution
x 7→ x −
a
3
. Then we found our
β, γ ∈ C
such that
each root
α
i
can be written as a linear combination of
β
and
γ
(and
µ
), i.e.
L ⊆ Q(β, γ, µ).
Then we showed that
{β
3
, γ
3
} =
(
−27c ±
p
(27c)
2
+ 4 × 27b
3
2
)
.
We now let
λ =
p
(27c)
2
+ 4 × 27b
3
.
Then we have the extensions
Q ⊆ Q(λ) ⊆ Q(λ, µ) ⊆ Q(λ, µ, β),
and also
Q ⊆ L ⊆ Q(λ, µ, β).
Note that the first extension
Q ⊆ Q
(
λ
) is a Kummer extension since it is a
splitting field of
t
2
−λ
2
. Then
Q
(
λ
)
⊆ Q
(
λ, µ
) is the third cyclotomic extension.
Finally,
Q
(
λ, µ
)
⊆ Q
(
λ, µβ
) is a Kummer extension, the splitting field of
t
3
−β
3
.
So Q ⊆ L is a radical extension.
Let’s go back to the definition of a radical extension. We said
L/K
is radical
if there is a further extension
E/L
that satisfies certain nice properties. It would
be great if
E/K
is actually a Galois extensions. To show this, we first need a
technical lemma.
Lemma. Let
L/K
be a Galois extension,
char K
= 0,
γ ∈ L
and
F
the splitting
field of
t
n
−γ
over
L
. Then there exists a further extension
E/F
such that
E/L
is radical and E/K is Galois.
Here we have the inclusions
K ⊆ L ⊆ F ⊆ E,
where
K, L
and
F
are given and
E
is what we need to find. The idea of the proof
is that we just add in the “missing roots” to obtain
E
so that
E/K
is Galois,
and doing so only requires performing cyclotomic and Kummer extensions.
Proof.
Since we know that
L/K
is Galois, we would rather work in
K
than in
L
.
However, our
γ
is in
L
, not
K
. Hence we will employ a trick we’ve used before,
where we introduce a new polynomial
f
, and show that its coefficients are fixed
by
Gal
(
L/K
), and hence in
K
. Then we can look at the splitting field of
f
or
its close relatives.
Let
f =
Y
ϕ∈Gal(L/K)
(t
n
− ϕ(γ)).
Each
ϕ ∈ Gal
(
L/K
) induces a homomorphism
L
[
t
]
→ L
[
t
]. Since each
ϕ ∈ Gal
(
L/K
) just rotates the roots of
f
around, we know that this induced
homomorphism fixes
f
. Since all automorphisms in
Gal
(
L/K
) fix the coefficients
of f , the coefficients must all be in K. So f ∈ K[t].
Now since
L/K
is Galois, we know that
L/K
is normal. So
L
is the splitting
field of some
g ∈ K
[
t
]. Let
E
be the splitting field of
fg
over
K
. Then
K ⊆ E
is normal. Since the characteristic is zero, this is automatically separable. So
the extension K ⊆ E is Galois.
We have to show that
L ⊆ E
is a radical extension. We pick our fields as
follows:
– E
0
= L
– E
1
= splitting field of t
n
− 1 over E
0
– E
2
= splitting field of t
n
− γ over E
1
– E
3
= splitting field of t
n
− ϕ
1
(γ) over E
2
– . . .
– E
r
= E,
where we enumerate Gal(L/K) as {id, ϕ
1
, ϕ
2
, ···}.
We then have the sequence of extensions
L = E
0
⊆ E
1
⊆ E
2
⊆ ··· ⊆ E
r
Here
E
0
⊆ E
1
is a cyclotomic extension, and
E
1
⊆ E
2
,
E
2
⊆ E
3
etc. are
Kummer extensions since they contain enough roots of unity and are cyclic. By
construction, F ⊆ E
2
. So F ⊆ E.
Theorem. Suppose
L/K
is a radical extension and
char K
= 0. Then there is
an extension E/L such that E/K is Galois and there is a sequence
K = E
0
⊆ E
1
⊆ ··· ⊆ E,
where E
i
⊆ E
i+1
is cyclotomic or Kummer.
Proof. Note that this is equivalent to proving the following statement: Let
K = L
0
⊆ L
1
⊆ ···L
s
be a sequence of cyclotomic or Kummer extensions. Then there exists an
extension
L
s
⊆ E
such that
K ⊆ E
is Galois and can be written as a sequence
of cyclotomic or Kummer extensions.
We perform induction on s. The s = 0 case is trivial.
If
s >
0, then by induction, there is an extension
M/L
s−1
such that
M/K
is
Galois and is a sequence of cyclotomic and Kummer extensions. Now
L
s
is a
splitting field of
t
n
−γ
over
L
s−1
for some
γ ∈ L
s−1
. Let
F
be the splitting field
of
t
n
− γ
over
M
. Then by the lemma and its proof, there exists an extension
E/M
that is a sequence of cyclotomic or Kummer extensions, and
E/K
is Galois.
K
L
s−1
L
s
= L
s−1
(
n
√
γ)
M
F = M(
n
√
γ)
E
However, we already know that
M/K
is a sequence of cyclotomic and Kummer
extensions. So
E/K
is a sequence of cyclotomic and Kummer extension. So
done.