Try
The roots of X³−2 are a, aω and aω², where a³=2 and ω²+ω+1=0. The tower Q ⊆ Q(ω) ⊆ Q(ω,a) contains all three roots. Its first step can be written ω=(−1+√−3)/2; its second step adjoins a cube root. Notice that the real field Q(a) alone misses two roots. A formula for one root and a splitting field are different objects.
A radical tower over a field F is a finite tower F=E₀ ⊆ E₁ ⊆ … ⊆ Eᵣ, with Eᵢ=Eᵢ₋₁(αᵢ) and αᵢ raised to a positive integer nᵢ belonging to Eᵢ₋₁. We allow complex intermediate values. A nonzero polynomial is solvable by radicals over F when all its roots lie in some radical tower over F; the tower need not equal its splitting field. Throughout this lesson F has characteristic zero, and all fields lie in one algebraic closure. These hypotheses matter.
Predict
Suppose αⁿ=a belongs to the base field K. Must K(α)/K be Galois? Predict the difference between K=Q and K=Q(ω) when n=3 and a=2. The missing ingredient is the collection μₙ of all nth roots of unity. If K contains μₙ, all roots of Xⁿ−a are already ζα with ζ in K.
Review solvable groups, the Galois correspondence and cyclotomic fields. We will prove both directions of the radical-solvability criterion, including why additional roots of unity and normal closures are necessary.
Experiment
Use the following table as a worked example. Fill in the last column yourself before reading it. Products of automorphisms act from right to left.
| Field step | Degree | Automorphisms over the preceding field |
|---|
| Q ⊆ Q(ω) | 2 | Identity and conjugation ω↦ω² |
| Q(ω) ⊆ Q(ω,a) | 3 | a↦a, aω, aω²; ω fixed |
| Q ⊆ Q(ω,a) | 6 | S₃, with normal C₃ and quotient C₂ |
The degree-three claim follows from total degree six and [Q(ω):Q]=2, as established in the complete S₃ example. The chain S₃ ⊇ C₃ ⊇ 1 has cyclic factors of orders two and three. Reverse the subgroup inclusions to recover the field tower. This is the shape of the general proof, but the example alone does not prove the theorem.
Now replace a by a nonzero α with αⁿ in K and assume μₙ⊆K. For every K-automorphism σ of K(α), record σ(α)/α. Check that these ratios are nth roots of unity. Why is the ratio for στ the product of the two ratios? Identify exactly where fixing μₙ is used.
Understand
A radical step with roots of unity has cyclic Galois group
Let α≠0, αⁿ=a∈K and μₙ⊆K. The field E=K(α) is the splitting field of Xⁿ−a: every root is αζ for ζ∈μₙ. Characteristic zero makes the polynomial separable, so E/K is Galois. The map σ↦σ(α)/α embeds Gal(E/K) into μₙ. It is injective because α generates E. It is a homomorphism because σ fixes each root of unity: if τ(α)=ζα, then στ(α)=ζσ(α). The group μₙ is cyclic, so the Galois group is cyclic, of order dividing n. Do not assert its order is always n; Xⁿ−a may be reducible. If α=0, the extension is trivial.
A cyclic step can be generated by a radical
Conversely, let E/K be cyclic Galois of degree n, let σ generate its group, and suppose K contains a primitive nth root of unity ζ. For b∈E form the resolvent
R(b)=b+ζ⁻¹σ(b)+ζ⁻²σ²(b)+…+ζ⁻⁽ⁿ⁻¹⁾σⁿ⁻¹(b).
Here σʲ means j repeated applications; the expression is a sum in E, not a new formal variable. The distinct automorphisms 1,σ,…,σⁿ⁻¹ are linearly independent as functions with coefficients in E, by the independence proof in the Galois-correspondence lesson. Therefore R is not the zero function: there is a b with α=R(b)≠0. Reindex the sum, using σⁿ=1 and ζⁿ=1, to obtain σ(α)=ζα. Hence αⁿ is fixed by σ and belongs to K. Also σʲ(α)=ζʲα gives n distinct images. The minimal polynomial of α therefore has degree at least n and at most [E:K]=n. Thus E=K(α). This proves the converse without assuming that an arbitrary choice of b works. Degree one is harmless: E=K already.
From radical formulas to a solvable group
Let a radical tower E₀=F ⊆ … ⊆ Eᵣ contain all roots of a polynomial f. We construct a finite Galois extension Nᵣ/F containing Eᵣ whose group is solvable. This is stronger than merely taking the normal closure and hoping its group stays solvable.
Let m be the product of the tower exponents nᵢ (take m=1 for an empty tower), choose a primitive mth root ζ, and start with N₀=F(ζ). This is finite Galois and abelian over F. Indeed an automorphism sends ζ to ζᵘ for a unit u modulo m; restriction gives an injective homomorphism into the abelian group (Z/mZ)×. Every μₙᵢ is now present.
Inductively assume Nᵢ₋₁/F is finite Galois with solvable group and contains Eᵢ₋₁. Write αᵢⁿ=a∈Eᵢ₋₁, with n=nᵢ, and first handle a≠0. In a common algebraic closure adjoin to Nᵢ₋₁ one nth root of each σ(a), for every σ∈Gal(Nᵢ₋₁/F). Call the resulting field Nᵢ, choosing αᵢ for the identity conjugate. Because μₙ is already present, each binomial Xⁿ−σ(a) splits there.
The polynomial P(X)=∏σ (Xⁿ−σ(a)) has coefficients fixed by Gal(Nᵢ₋₁/F), since that group permutes its factors. Thus P∈F[X]. Combine P with a polynomial over F whose splitting field is Nᵢ₋₁. Their product has splitting field Nᵢ, so Nᵢ/F is finite Galois. Repeated factors cause no problem in characteristic zero: the extension is still separable.
Restriction onto Gal(Nᵢ₋₁/F) is surjective by the Galois correspondence. Its kernel Gal(Nᵢ/Nᵢ₋₁) embeds into a product of copies of μₙ: each automorphism multiplies each selected radical by an nth root of unity, and the selected radicals generate the field. These ratios compose multiplicatively because the base fixes μₙ. The kernel is therefore abelian. A group with solvable normal subgroup and solvable quotient is solvable, by the extension proof in the groups lesson. This proves the inductive claim. If a=0, simply take Nᵢ=Nᵢ₋₁.
Let L be the splitting field of f over F. It lies in Nᵣ. Since L/F is Galois, restriction identifies Gal(L/F) with a quotient of Gal(Nᵣ/F). Quotients of solvable groups are solvable. This completes the first direction, including radical towers that are not themselves normal.
From a solvable group to radical formulas
Now let L/F be the finite splitting field of f and assume G=Gal(L/F) is solvable. Put n=|G|, K=F(ζₙ) and M=LK, the smallest field containing both L and K. The first extension K/F is a radical step because ζₙⁿ=1. Both L/F and K/F are finite Galois, so M/F is finite Galois; in particular M/K is Galois.
Restriction embeds H=Gal(M/K) into G: fixing both L and K fixes M. Thus H is solvable and its order divides n. A finite solvable group has a chain H=H₀ ⊇ H₁ ⊇ … ⊇ Hₛ=1 in which Hⱼ₊₁ is normal in Hⱼ and each quotient is cyclic of prime order. To justify the refinement, start with an abelian-factor chain. In each finite abelian quotient choose a maximal proper subgroup; its quotient is simple abelian, hence cyclic of prime order. Continue within that subgroup, whose order decreases, and take preimages. Normality is required in the preceding group, not necessarily in all of H.
Reverse this chain under the Galois correspondence. It produces K=D₀ ⊆ D₁ ⊆ … ⊆ Dₛ=M, where Dⱼ is the subfield of M fixed by every element of Hⱼ. Each Dⱼ₊₁/Dⱼ is cyclic Galois of prime degree p dividing |H| and hence n. Every Dⱼ contains K and therefore a primitive pth root of unity. The cyclic-step resolvent proof supplies a generator β with βᵖ∈Dⱼ. Each field step is consequently radical. Prepend F⊆K: the resulting radical tower contains M and therefore every root of f.
We have proved: over a characteristic-zero field, a polynomial is solvable by radicals exactly when the Galois group of its splitting field is solvable. The proof establishes existence of radical expressions; it does not promise a short expression or a numerical algorithm with good stability.
Apply
For a changed example take f=X³−5. Build its splitting field, prove its degree, identify its group and write a radical tower. Then diagnose three claims: “every degree-five polynomial is unsolvable,” “a radical extension is automatically Galois,” and “any b works in the resolvent.” Supply a counterexample or a precise repair for each.
Check the changed polynomial and the three repairs
For b=∛5, Eisenstein at 5 gives [Q(b):Q]=3. The real field Q(b) excludes ω, so adjoining ω has degree two. The six automorphisms act faithfully on three roots and give S₃, which is solvable. The tower Q⊆Q(ω)⊆Q(ω,b) is radical, and contains b,bω,bω². The polynomial X⁵−2 is a degree-five example solvable by radicals: adjoin ζ₅ and then the fifth root 2^(1/5). Its roots are ζ₅ʲ2^(1/5). The field Q(∛2) is a radical extension but is not normal. For a nontrivial cyclic extension, b=1 gives R(1)=0 because the roots-of-unity coefficients sum to zero; independence proves the existence of a successful b, not success for every b.
Finally suppose a polynomial’s splitting-field group has been independently proved to be S₅ or A₅. Explain why the nonsolvability proof in the groups lesson rules out a radical formula for all its roots. Merely knowing its degree is five does not identify its Galois group. This lesson proves the conditional obstruction; it does not certify a particular quintic’s group by inspection.
Continue
Reconstruct both directions without the text. Mark where characteristic zero, roots of unity, normality, surjectivity of restriction and automorphism independence enter. Then try the saved proof tasks and the mixed assessment. Writing a complete argument and checking a changed example provide stronger evidence than recognizing a group name. Representations, modules and independently assessed course projects remain further parts of the undergraduate algebra pathway.
