3Solutions to polynomial equations
II Galois Theory
3.2 Kummer extensions
We shall now consider a more general case, and study the splitting field of
t
n
− λ ∈ K
[
t
]. As we have previously seen, we will need to make use of the
n
th
primitive roots of unity.
The definition of a Kummer extension will involve a bit more that it being
the splitting field of
t
n
− λ
. So before we reach the definition, we first studying
some properties of an arbitrary splitting field of
t
n
−λ
, and use this to motivate
the definition of a Kummer extension.
Definition (Cyclic extension). We say a Galois extension
L/K
is cyclic is
Gal(L/K) is a cyclic group.
Theorem. Let
K
be a field,
λ ∈ K
non-zero,
n ∈ N
,
char K
= 0 or 0
< char K ∤
n. Let L be the splitting field of t
n
− λ. Then
(i) L contains an nth primitive root of unity, say µ.
(ii) L/K
(
µ
) is a cyclic (and in particular Galois) extension with degree [
L
:
K(µ)] | n.
(iii) [L : K(µ)] = n if and only if t
n
− λ is irreducible in K(µ)[t].
Proof.
(i)
Under our assumptions,
t
n
− λ
and (
t
n
− λ
)
′
=
nt
n−1
have no common
roots in L. So t
n
− λ has distinct roots in L, say α
1
, ··· , α
n
∈ L.
It then follows by direct computation that
α
1
α
−1
1
, α
2
α
−1
1
, ··· , α
n
α
−1
1
are
distinct roots of unity, i.e. roots of
t
n
−
1. Then one of these, say
µ
must
be an nth primitive root of unity.
(ii)
We know
L/K
(
µ
) is a Galois extension because it is the splitting field of
the separable polynomial t
n
− λ.
To understand the Galois group, we need to know how this field exactly
looks like. We let
α
be any root of
t
n
−λ
. Then the set of all roots can be
written as
{α, µα, µ
2
α, ··· , µ
n−1
α}
Then
L = K(α
1
, ··· , α
n
) = K(µ, α) = K(µ)(α).
Thus, any element of
Gal
(
L/K
(
µ
)) is uniquely determined by what it sends
α
to, and any homomorphism must send
α
to one of the other roots of
t
n
− λ, namely µ
i
α for some i.
Define a homomorphism
σ
:
Gal
(
L/K
(
µ
))
→ Z/nZ
that sends
ϕ
to the
corresponding i (as an element of Z/nZ, so that it is well-defined).
It is easy to see that σ is an injective group homomorphism. So we know
Gal
(
L/K
(
µ
)) is isomorphic to a subgroup of
Z/nZ
. Since the subgroup of
any cyclic group is cyclic, we know that
Gal
(
L/K
(
µ
)) is cyclic, and its size
is a factor of
n
by Lagrange’s theorem. Since
|Gal
(
L/K
(
µ
))
|
= [
L
:
K
(
µ
)]
by definition of a Galois extension, it follows that [L : K(µ)] divides n.
(iii)
We know that [
L
:
K
(
µ
)] = [
K
(
µ, α
) :
K
(
µ
)] =
deg q
α
. So [
L
:
K
(
µ
)] =
n
if and only if
deg q
α
=
n
. Since
q
α
is a factor of
t
n
− λ
,
deg q
α
=
n
if and
only if
q
α
=
t
n
−λ
. This is true if and only if
t
n
−λ
is irreducible
K
(
µ
)[
t
].
So done.
Example. Consider
t
4
+ 2
∈ Q
[
t
]. Let
µ
=
√
−1
, which is a 4th primitive root
of unity. Now
t
4
+ 2 = (t − α)(t + α)(t − µα)(t + µα),
where
α
=
4
√
−2
is one of the roots of
t
4
+ 2. Then we have the field extension
Q ⊆ Q(µ) ⊆ Q(µ, α), where Q(µ, α) is a splitting field of t
4
+ 2.
Since
√
−2 ∈ Q
(
µ
), we know that
t
4
+ 2 is irreducible in
Q
(
µ
)[
t
] by looking at
the factorization above. So by our theorem,
Q
(
µ
)
⊆ Q
(
µ, α
) is a cyclic extension
of degree exactly 4.
Definition (Kummer extension). Let
K
be a field,
λ ∈ K
non-zero,
n ∈ N
,
char K
= 0 or 0
< char K ∤ n
. Suppose
K
contains an
n
th primitive root of
unity, and
L
is a splitting field of
t
n
− λ
. If
deg
[
L
:
K
] =
n
, we say
L/K
is a
Kummer extension.
Note that we used to have extensions
K ⊆ K
(
µ
)
⊆ L
. But if
K
already
contains a primitive root of unity, then
K
=
K
(
µ
). So we are left with the cyclic
extension K ⊆ L.
To following technical lemma will be useful:
Lemma. Assume
L/K
is a field extension. Then
Hom
K
(
L, L
) is linearly in-
dependent. More concretely, let
λ
1
, ··· , λ
n
∈ L
and
ϕ
1
, ··· , ϕ
n
∈ Hom
K
(
L, L
)
distinct. Suppose for all α ∈ L, we have
λ
1
ϕ
1
(α) + ··· + λ
n
ϕ
n
(α) = 0.
Then λ
i
= 0 for all i.
Proof. We perform induction on n.
Suppose we have some λ
i
∈ L and ϕ
i
∈ Hom
K
(L, L) such that
λ
1
ϕ
1
(α) + ··· + λ
n
ϕ
n
(α) = 0.
The
n
= 1 case is trivial, since
λ
1
ϕ
1
= 0 implies
λ
1
= 0 (the zero homomorphism
does not fix K).
Otherwise, since the homomorphisms are distinct, pick
β ∈ L
such that
ϕ
1
(β) = ϕ
n
(β). Then we know that
λ
1
ϕ
1
(αβ) + ··· + λ
n
ϕ
n
(αβ) = 0
for all α ∈ L. Since ϕ
i
are homomorphisms, we can write this as
λ
1
ϕ
1
(α)ϕ
1
(β) + ··· + λ
n
ϕ
n
(α)ϕ
n
(β) = 0.
On the other hand, by just multiplying the original equation by ϕ
n
(β), we get
λ
1
ϕ
1
(α)ϕ
n
(β) + ··· + λ
n
ϕ
n
(α)ϕ
n
(β) = 0.
Subtracting the equations gives
λ
1
ϕ
1
(α)(ϕ
1
(β) − ϕ
n
(β)) + ··· + λ
n−1
ϕ
n−1
(α)(ϕ
n−1
(β) − ϕ
n
(β)) = 0
for all
α ∈ L
. By induction,
λ
i
(
ϕ
i
(
β
)
− ϕ
n
(
β
)) = 0 for all 1
≤ i ≤ n −
1. In
particular, since ϕ
1
(β) − ϕ
n
(β) = 0, we have λ
1
= 0. Then we are left with
λ
2
ϕ
2
(α) + ··· + λ
n
ϕ
n
(α) = 0.
Then by induction again, we know that all coefficients are zero.
Theorem. Let
K
be a field,
n ∈ N
,
char K
= 0 or 0
< char K ∤ n
. Suppose
K
contains an
n
th primitive root of unity, and
L/K
is a cyclic extension of degree
[L : K] = n. Then L/K is a Kummer extension.
This is a rather useful result. If we look at the splitting field of a polynomial
t
n
− λ
, even if the ground field includes the right roots of unity, a priori, this
doesn’t have to be a Kummer extension if it doesn’t have degree
n
. But we
previously showed that the extension must be cyclic. And so this theorem shows
that it is still a Kummer extension of some sort.
This is perhaps not too surprising. For example, if, say,
n
= 4 and
λ
is
secretly a square, then the splitting field of
t
4
− λ
is just the splitting field of
t
2
−
√
λ.
Proof.
Our objective here is to find a clever
λ ∈ K
such that
L
is the splitting
field of t
n
− λ. To do so, we will have to hunt for a root β of t
n
− λ in L.
Pick
ϕ
a generator of
Gal
(
L/K
). We know that if
β
were a root of
t
n
− λ
,
then
ϕ
(
β
) =
µ
−1
β
for some primitive
n
th root of unity
µ
. Thus, we want to find
an element that satisfies such a property.
By the previous lemma, we can find some α ∈ L such that
β = α + µϕ(α) + µ
2
ϕ
2
(α) + ··· + µ
n−1
ϕ
n−1
(α) = 0.
Then, noting that
ϕ
n
is the identity and
ϕ
fixes
µ ∈ K
, we see that
β
trivially
satisfies
ϕ(β) = ϕ(α) + µϕ
2
α + ··· + µ
n−1
ϕ
n
(α) = µ
−1
β,
In particular, we know that ϕ(β) ∈ K(β).
Now pick
λ
=
β
n
. Then
ϕ
(
β
n
) =
µ
−n
β
n
=
β
n
. So
ϕ
fixes
β
n
. Since
ϕ
generates Gal(L/K), we know all automorphisms of L/K fixes β
n
. So β
n
∈ K.
Now the roots of
t
n
− λ
are
β, µβ, ··· , µ
n−1
β
. Since these are all in
β
, we
know K(β) is the splitting field of t
n
− λ.
Finally, to show that
K
(
β
) =
L
, we observe that
id, ϕ|
K(β)
, . . . ,
ϕ
n
|
K(β)
are
distinct elements of
Aut
K
(
K
(
β
)) since they do different things to
β
. Recall our
previous theorem that
[K(β) : K] ≥ |Aut
K
(K(β))|.
So we know that n = [L : K] = [K(β) : K]. So L = K(β). So done.
Example. Consider
t
3
−
2
∈ Q
[
t
], and
µ
a third primitive root of unity. Then
we have the extension
Q ⊆ Q
(
µ
)
⊆ Q
(
µ,
3
√
2
). Then
Q ⊆ Q
(
µ
) is a cyclotomic
extension of degree 2, and
Q
(
µ
)
⊆ Q
(
µ,
3
√
2
) is a Kummer extension of degree 3.