Solubility and Simplicity
15.1 Radical Extensions
Some care is needed in formalising the idea of ‘solubility by radicals’. We begin from the point of view of field extensions.
Informally, a radical extension is obtained by a sequence of adjunctions of nth roots, for various n. For example, the following expression is radical:
√3
To find an extension of Q that contains this element we may adjoin in turn elements α = 3 Recall Definition 8.12, which formalises the idea of a radical extension: L : K is radical if L = K(α1, . . . , αm) where for each j = 1, . . . , m there exists njsuch that
αnjj ∈ K(α1, . . . , αj−1) ( j ≥ 1)
The elements αj form a radical sequence for L : K, and the radical degree of αj is nj.
171
FIGURE 22: Galois thought he had solved the quintic... but changed his mind.
For example, the expression (15.1) is contained in a radical extension of the form Q(α , β , γ , δ , ε ) of Q, where α3= 11, β2= 3, γ5= (7 + β )/2, δ3= 4, ε4= 1 + δ .
It is clear that any radical expression, in the sense of the introduction, is contained in some radical extension.
A polynomial should be considered soluble by radicals provided all of its zeros are radical expressions over the ground field.
Definition 15.1. Let f be a polynomial over a subfield K of C, and let Σ be the splitting field for f over K. We say that f is soluble by radicals if there exists a field M containing Σ such that M : K is a radical extension.
We emphasise that in the definition, we do not require the splitting field exten-sion Σ : K to be radical. There is a good reason for this. We want everything in the splitting field Σ to be expressible by radicals, but it is pointless to expect everything expressible by the same radicals to be inside the splitting field. Indeed,if M : K is radical and L is an intermediate field, then L : K need not be radical: see Exercise 15.6.
Note also that we require all zeros of f to be expressible by radicals. It is possible for some zeros to be expressible by radicals, while others are not—simply take a product of two polynomials, one soluble by radicals and one not. However, if an irreducible polynomial f has one zero expressible by radicals, then all the zeros must be so expressible, by a simple argument based on Corollary 5.13.
The main theorem of this chapter is:
Radical Extensions 173 Theorem 15.2. If K is a subfield of C and K ⊆ L ⊆ M ⊆ C where M : K is a radical extension, then the Galois group of L: K is soluble.
The otherwise curious word ‘soluble’ for groups arises in this context: a soluble (by radicals) polynomial has a soluble Galois group (of its splitting field over the base field).
The proof of this result is not entirely straightforward, and we must spend some time on preliminaries.
Lemma 15.3. If L : K is a radical extension in C and M is the normal closure of L: K, then M : K is radical.
Proof. Let L = K(α1, . . . , αr) with αini∈ K(α1, . . . , αi−1). Let fibe the minimal poly-nomial of αi over K. Then M ⊇ L is clearly the splitting field of ∏ri=1fi. For every zero βi jof fiin M there exists an isomorphism σ : K(αi) → K(βi j) by Corollary 5.13.
By Proposition 11.4, σ extends to a K-automorphism τ : M → M. Since αiis a mem-ber of a radical sequence for a subfield of M, so is βi j. By combining the sequences, we get a radical sequence for M.
The next two lemmas show that certain Galois groups are abelian.
Lemma 15.4. Let K be a subfield of C, and let L be the splitting field for tp− 1 over K, where p is prime. Then the Galois group of L: K is abelian.
Proof. The derivative of tp− 1 is ptp−1, which is prime to tp− 1, so by Lemma 9.13 the polynomial has no multiple zeros in L. Clearly its zeros form a group under multiplication; this group has prime order p since the zeros are distinct, so is cyclic.
Let ε be a generator of this group. Then L = K(ε) so that any K-automorphism of Lis determined by its effect on ε. Further, K-automorphisms permute the zeros of tp− 1. Hence any K-automorphism of L is of the form
αj: ε 7→ εj and is uniquely determined by this condition.
But then αiαjand αjαiboth map ε to εi j, so the Galois group is abelian.
It is possible to determine the precise structure of the above Galois group, and to remove the condition that p be prime. However, this needs extra work and is not needed at this stage. See Theorem 21.9.
Lemma 15.5. Let K be a subfield of C in which tn− 1 splits. Let a ∈ K, and let L be a splitting field for tn− a over K. Then the Galois group of L : K is abelian.
Proof. Let α be any zero of tn− a. Since tn− 1 splits in K, the general zero of tn− a is εα where ε is a zero of tn− 1 in K. Since L = K(α), any K-automorphism of L is determined by its effect on α. Given two K-automorphisms
φ : α 7→ ε α ψ : α 7→ η α
where ε and η ∈ K are zeros of tn− 1, then
φ ψ (α ) = ε η α = η ε α = ψ φ (α ) As before, the Galois group is abelian.
The main work in proving Theorem 15.2 is done in the next lemma.
Lemma 15.6. If K is a subfield of C and L : K is normal and radical, then Γ(L : K) is soluble.
Proof. Suppose that L = K(α1, . . . , αn) with αnjj ∈ K(α1, . . . , αj−1). By Proposi-tion 8.9 we may assume that nj is prime for all j. In particular there is a prime p such that α1p∈ K.
We prove the result by induction on n, using the additional hypothesis that all nj are prime. The case n = 0 is trivial, which gets the induction started.
If α1∈ K, then L = K(α2, . . . , αn) and Γ(L : K) is soluble by induction. proof is illustrated in the following diagram:
L
Observe that L : K is finite and normal, hence so is L : M, therefore Theorem 12.2 applies to L : K and to L : M.
Since tp− 1 splits in M and α1p∈ M, the proof of Lemma 15.5 implies that M(α1) is a splitting field for tp− α1pover M. Thus M(α1) : M is normal, and by Lemma 15.5 Γ(M(α1) : M) is abelian. Apply Theorem 12.2 to L : M to deduce that
Γ(M(α1) : M) ∼= Γ(L : M)/Γ(L : M(α1))
Radical Extensions 175 Now
L= M(α1)(α2, . . . , αn)
so that L : M(α1) is a normal radical extension. By induction Γ(L : M(α1)) is soluble.
Hence by Theorem 14.4(3), Γ(L : M) is soluble.
Since M is the splitting field for tp− 1 over K, the extension M : K is normal. By Lemma 15.4, Γ(M : K) is abelian. Theorem 12.2 applied to L : K yields
Γ(M : K) ∼= Γ(L : K)/Γ(L : M)
Now Theorem 14.4(3) shows that Γ(L : K) is soluble, completing the induction step.
We can now complete the proof of the main result:
Proof of Theorem 15.2. Let K0 be the fixed field of Γ(L : K), and let N : M be the normal closure of M : K0. Then
K⊆ K0⊆ L ⊆ M ⊆ N
Since M : K0is radical, Lemma 15.3 implies that N : K0is a normal radical extension.
By Lemma 15.6, Γ(N : K0) is soluble.
By Theorem 11.14, the extension L : K0is normal. By Theorem 12.2 Γ(L : K0) ∼= Γ(N : K0)/Γ(N : L)
Theorem 14.4(2) implies that Γ(L : K0) is soluble. But Γ(L : K) = Γ(L : K0), so Γ(L : K) is soluble.
The idea of this proof is simple: a radical extension is a series of extensions by nth roots; such extensions have abelian Galois groups; so the Galois group of a radical extension is made up by fitting together a sequence of abelian groups. Unfortunately there are technical problems in carrying out the proof; we need to throw in roots of unity, and we have to make various extensions normal before the Galois correspon-dence can be used. These obstacles are similar to those encountered by Abel and overcome by his Theorem on Natural Irrationalities in Section 8.8.
Now we translate back from fields to polynomials, and in doing so revert to Ga-lois’s original viewpoint.
Definition 15.7. Let f be a polynomial over a subfield K of C, with splitting field Σ over K. The Galois group of f over K is the Galois group Γ(Σ : K).
Let G be the Galois group of a polynomial f over K and let ∂ f = n. If α ∈ Σ is a zero of f , then f (α) = 0, so for any g ∈ G
f(g(α)) = g( f (α)) = 0
Hence each element g ∈ G induces a permutation g0of the set of zeros of f in Σ.
Distinct elements of G induce distinct permutations, since Σ is generated by the zeros
of f . It follows easily that the map g 7→ g0 is a group monomorphism of G into the group Snof all permutations of the zeros of f . In other words, we can think of G as a group of permutations on the zeros of f . This, in effect, was how Galois thought of the Galois group, and for many years afterwards the only groups considered by mathematicians were permutation groups and groups of transformations of variables.
Arthur Cayley was the first to propose a definition for an abstract group, although it seems that the earliest satisfactory axiom system for groups was given by Leopold Kronecker in 1870 (Huntingdon 1905).
We may restate Theorem 15.2 as:
Theorem 15.8. Let f be a polynomial over a subfield K of C. If f is soluble by radi-cals, then the Galois group of f over K is soluble.
The converse also holds: see Theorem 18.21.
Thus to find a polynomial not soluble by radicals it suffices to find one whose Galois group is not soluble. There are two main ways of doing this. One is to look at the general polynomial of degree n, which we introduced in Chapter 8 Section 8.7, but this approach has the disadvantage that it does not show that there are specific polynomials with rational coefficients that are insoluble by radicals. The alternative approach, which we now pursue, is to exhibit a specific polynomial with rational coefficients whose Galois group is not soluble. Since Galois groups are hard to cal-culate, a little low cunning is necessary, together with some knowledge of the sym-metric group.