Abstract Algebra 2 (MATH 4140/5140)
Radical Extensions That Are Galois
To be able to apply the FTGT to solvability of polynomial equations by radicals and to
constructibility by straightedge and compass, we need radical extensions — or in particular,
square root extensions — that are also Galois extensions.
Recall: Let K be a field, and let L be an extension field of K.
Definition. L is a radical extension of K if there exists a radical tower
T : K = K0 ≤ K1 ≤ · · · ≤ Kr = L where for every j = 1, 2, . . . , r,
m
Kj = Kj−1 (αj ) and αj j ∈ Kj−1 for some αj ∈ Kj and mj ≥ 2
(i.e., each αj is an mj -th root of an element of Kj−1 ).
(∗)
Notation: I(T ) := {mj : 1 ≤ j ≤ r} (= the set of indices of the roots adjoined in T ).
Theorem. Let K be a field of characteristic 0, and let K ≤ L be a radical extension with
radical tower T . Then L has an extension field L such that
(i) L is a Galois extension of K,
(ii) L is a radical extension of K with a radical tower T satisfying I(T ) = I(T ), and
(iii) L contains the m-th roots of unity for every m ∈ I(T ).
The proof the Theorem relies on the following Lemma about ‘joining’ radical extensions.
Lemma. Let K, L, L0 be subfields of an extension field F of K such that K ≤ L and K ≤ L0
are radical extensions with radical towers T and T 0 , respectively. Then, for the smallest
subfield L∗ of F extending both L and L0 we have that L∗ is a radical extension of K witnessed
by a radical tower T ∗ that is a continuation of T 0 and satisfies I(T ∗ ) = I(T ) ∪ I(T 0 ).
Proof. Let F, K, L, L0 , L∗ , T , T 0 be as in the theorem, and assume T is as displayed in (??).
Consider the following tower (a continuation of T 0 ):
T ∗ : K ≤ · · · ≤ L0 =: L00 ≤ L00 (α1 ) =: L01 ≤ L01 (α2 ) =: L02 ≤ · · · ≤ Lr−1 (αr ) =: L0r .
|
{z
}
T0
Kj ≤ L0j for all j = 0, 1, . . . , r. Reason:
.
T ∗ is a radical tower with I(T ∗ ) = I(T ) ∪ I(T 0 ). Reason:
.
1.
2.
L0r = L0 (α1 , . . . , αr ) = L∗ . Reasons. 1.:
; 2.:
.
Proof of the Theorem. Let K, L and T be as in the Theorem, with T as in (??).
L = K(γ) for some γ ∈ L. Reason:
.
Q
m
Let fγ ∈ K[x] be the minimal polynomial of γ over K, let h := fγ m∈I(T ) (x − 1), and let
L be a splitting field for h over K. Since γ is a root of h, we can choose L so that γ ∈ L,
hence L = K(γ) ≤ L. Let γ = γ1 , γ2 , . . . , γn denote the roots of fγ in L, and let ζm be a
primitive m-th root of unity in L for each m ∈ I(T ).
L has properties (i) and (iii). Reason:
.
1
2
L = K({γ1 , γ2 , . . . , γn } ∪ {ζm : m ∈ I(T )}). Reason:
Hence, the smallest subfield of L that contains all fields
.
K(γ` ) (` = 1, 2, . . . , n) and K(ζm ) (m ∈ I(T ))
is L itself. Here, K(γ) = K(γ1 ) is a radical extension of K with radical tower T , and
all K(ζm ) (m ∈ I(T )) are simple radical extensions of K with index set {m} ⊆ I(T ).
Therefore, a repeated application of the Lemma will imply that L has property (ii) if we can
show that all K(γ` ) (` = 2, . . . , n) are also radical extensions of K with radical towers T`
satisfying I(T` ) = I(T ). So, let us choose an ` (2 ≤ ` ≤ n).
L has a K-automorphism ϕ` such that ϕ` (γ) = γ` . Reason:
The image of T under ϕ` is the tower
.
Tj : K = ϕ` (K) = ϕ` (K0 ) ≤ ϕ` (K1 ) ≤ · · · ≤ ϕ` (Kr ) = ϕ` (L) = ϕ` (K(γ)) = K(γ` ).
T` is a radical tower such that I(T` ) = I(T ), because for every j = 1, 2, . . . , r,
ϕ` (Kj ) =
.
The special case I(T ) = {2} of the Theorem yields the following result for square root
extensions.
Corollary 1. Let K be a field of characteristic 0. For every square root extension K ≤ L
there exists an extension field L of L such that L is a Galois extension of K as well as a
square root extension of K.
This corollary yields stronger necessary conditions for constructibility than the ones we
proved earlier1. We will only state and prove the new, stronger necessary conditions for
constructibility for the complex plane approach.
Corollary 2. Let B be a set of points in the complex plane C such that 0, 1 ∈ B, and let EB
denote the subfield of C generated by the complex numbers in B and their conjugates. If a
point z ∈ C is constructible from B, then
(1) z is algebraic over EB , and
(2)∗ for the minimal polynomial f ∈ EB [x] of z over EB and for any splitting field L for
f over EB , we have that [L : EB ] is a power of 2.
Note. [L : EB ] does not depend on which splitting field L we choose. Thus, we may assume
that z ∈ L, and hence EB (z) ≤ L. Therefore, deg(f ) = [EB (z) : EB ] | [L : EB ], so condition
(2)∗ is indeed stronger than the earlier necessary condition that deg(f ) is a power of 2.
Proof of Corollary 2. Assume z ∈ C is constructible from B. We proved earlier that (1)
must hold. Let f , L be as in (2)∗ .
EB has a square root extension M such that z ∈ M and the extension EB ≤ M is Galois.
Reason:
.
M contains a splitting field L for f over EB . Reason:
.
[M : EB ] is a power of 2, and hence so is [L : EB ]. Reason:
.
1See
Corollaries 1 and 2 on the handout ‘Necessary Conditions for Constructibility’.