Mathematics 202 – Seminar in Algebra 2 An “Abstract” of Galois Theory, and Solvability by Radicals April 18, 2008

The following is a compendium of the “high points” in

• the logical development of Galois Theory for fields of characteristic zero, and

• the theorem that there exist quintic polynomials f(x) ∈ ℚ[x] whose roots cannot be expressed in terms of radicals.

These are summarized in outline form, the later ones with “telegraphic” proofs showing how they follow from the earlier ones. Since the results of this subject are highly interrelated, this summary may be more helpful in piecing together what implies what than the textbook presentation or the class notes, which include lots of examples, comments, etc. so are more discursive.

(0) Terminology and Notation: For a finite extension K over F:

(a) An F -automorphism of K is a 1-1, onto, ring homomorphism σ : K → K such that σ(a) = a for all a ∈ F.

(b) Gal_{F} K is set of all F -automorphisms of K, which is a group under function composition, called the Galois group of K over F.

(c) For any subgroup H of Gal_{F} K, E_{H} = {a ∈ K : σ(a) = a, all σ ∈ H}, called the fixed field of H. (It is a field, because it is closed under differences, products, and inverses in K.)

(1) (Consequence of the “Key Corollary” of Theorem 10.7): Let F (u) contain another root v of the minimal polynomial of u over F. Then there exists an F -automorphism σ of F (u) with σ(u) = v.

(2) (The induction step of the proof of the Primitive Element Theorem) Let [K : F] < ∞ and let K be separable over F. α, β be elements of K (hence both algebraic over F). Then there exists a γ ∈ K such that F (α, β) = F (γ). (Idea of the proof: For well-chosen c ∈ F, we can find a γ of the form γ = α + cβ.)

(3) If K is any finite-dimensional extension of F of characteristic zero, then K = F (γ) for some γ ∈ K.

(4) (Lemma for the proof of Theorems 11.7 and 11.8) Let H be a subgroup of Gal_{F} K, and let E = E_{H}. Then

(a) If β = β_{1} ∈ K, and {β_{1}, …, β_{t}} is the are the distinct images of β under the elements of H, then f(x) = (x − β_{1}) ⋯ (x − β_{t}) is in E[x]. (In fact f(x) is irreducible in E[x], though we did not say this in class).

(Proof: The coefficients of g are the elementary symmetric polynomials in the β_{i}, hence are fixed by all permutations of the β_{i}, including those induced by H. Hence they are in the fixed field E_{H} = E. Let p(x) be the minimal polynomial of β_{1} over E. Then p(β_{1}) = 0, so p(σ(β_{1})) = 0 for all σ ∈ H. Hence f(x) divides p(x) in E[x]. But this implies f = p; since p is irreducible, so is f.)

(b) [K : E] = |H|.

(Proof: Apply the proof of (a) to β = u satisfying K = E(u), using the Primitive Element Theorem. We get deg p(x) = [K : E] = |H|.)

(5) (A Corollary of 4.) If K is any finite extension of F, |Gal_{F} K| divides the degree [K : F]. (This follows from 4 part b, by letting E be the fixed field of Gal_{F} K: E is a subfield of K containing F, possibly larger than F. By 4, part b, [K : E] = |Gal_{F} K|. But [K : E][E : F] = [K : F], so |Gal_{F} K| divides [K : F].)

(6) A Galois extension is a normal, separable, finite dimensional extension field K of F. If F and K have characteristic zero, this is equivalent to saying that K is the splitting field of some polynomial in f(x) ∈ F [x] (Theorem 10.15).

(7) If K is a Galois extension of F and E is any intermediate field – F ⊆ E ⊆ K, then K is also a Galois extension of E (the polynomial f ∈ F [x] is also in E[x].)

(8) (The Fundamental Theorem of Galois Theory – Theorem 11.11) Let K be a Galois extension of F.

(a) There is a one-to-one, inclusion reversing correspondence between the set of intermediate fields F ⊆ E ⊆ K and the set of subgroups Gal_{F} K ⊇ H ⊇ {id}. The correspondence is given by E ↦ Gal_{E}K and H ↦ E_{H}. Moreover, if E = E_{H}, then [E : F] = [G : H], where G = Gal_{F} K.

(b) The subfield E is normal over F if and only if its corresponding subgroup H = Gal_{E}K is a normal subgroup of Gal_{F} K. Moreover, if E is Galois over F, then Gal_{F} E is isomorphic to the quotient group Gal_{F} K/Gal_{E}K. The proof of the direct implication is based on the observation that if E is normal over F, any element σ ∈ Gal_{F} K must satisfy σ(E) ⊆ E. Therefore we have a restriction mapping

ρ : Gal_{F} K → Gal_{F} E
σ ↦ σ|_{E} ,

which is a group homomorphism. It is onto because of the “Key Corollary” of Theorem 10.7 (see 10.14). The kernel is Gal_{E}K, which is therefore a normal subgroup of Gal_{F} K. The isomorphism

Gal_{F} E ≃ Gal_{F} K/Gal_{E}K

is then a consequence of the First Isomorphism Theorem for groups.

(10) (Root Towers, I) An polynomial f(x) ∈ F [x] with F ⊂ C is said to be solvable by radicals if

there is a “root tower”

F = F_{0} ⊂ F_{1} ⊂ ⋯ ⊂ F_{n},

where α ∈ F_{n}, and for each i, F_{i} = F_{i−1}(β_{i}) where β_{i}^{n_{i}} ∈ F_{i−1} for some integer n_{i}, and F_{n} contains the splitting field of f(x).

(11) Let F be a subfield of C containing the primitive pth root of 1, denoted ζ_{p}. Let a be an element of F which is not a pth power in F. Then the splitting field K of x^{p} − a over F has degree p over F and Gal_{F} K is cyclic of order p.

(Proof: Let α be any one root of x^{p} − a, and let K = F (α). The other roots of x^{p} − a are ζ_{p}α, …, ζ_{p}^{p−1}α. Since ζ_{p} ∈ F, all these roots are in K, so K is the splitting field. By 6, K is Galois over F. If σ ∈ Gal_{F} K is not the identity, then σ(α) = ζ_{p}^{i}α for some i. It is easy to see that σ generates a subgroup of Gal_{F} K isomorphic to the cyclic group ℤ_{p}. Since K = F (α), and α is a root of a polynomial of degree p over F, [K : F] ≤ p. Hence [K : F] = p = |Gal_{F} K| and Gal_{F} K is cyclic as claimed.)

(12) Let p be a prime and let ζ_{p} be as in 11. Then

(a) ℚ(ζ_{p}) is Galois over ℚ and Gal_{ℚ}ℚ(ζ_{p}) is cyclic of order p − 1. (Proof: The minimal polynomial of ζ_{p} over ℚ is ϕ_{p}(x) = x^{p−1} +⋯+x+1. The roots of ϕ_{p} are just the powers ζ_{p}^{i} for i = 1, …, p − 1. Hence ℚ(ζ_{p}) is the splitting field of ϕ_{p} over ℚ, and we’re done by 6. The Galois group Gal_{ℚ}ℚ(ζ_{p}) is isomorphic to the multiplicative group of the field ℤ/⟨p⟩, hence is cyclic of order p − 1.)

(b) If F is any subfield of C, then F (ζ_{p}) is Galois over F and Gal_{F} F (ζ_{p}) is cyclic. (Follows since this Galois group is isomorphic to a subgroup of Gal_{ℚ}ℚ(ζ_{p}).)

(13) (Root Towers, II) If f(x) ∈ F [x] is solvable by radicals over F, then there is a root tower

F = F_{0} ⊂ F_{1} ⊂ ⋯ ⊂ F_{n}

as in 10 above (for each i, F_{i} = F_{i−1}(β_{i}) where β_{i}^{n_{i}} ∈ F_{i−1} for some integer n_{i}) such that the splitting field of f(x) is contained in F_{n}, F_{n} is Galois over F, and in addition, for each i, F_{i} is Galois over F_{i−1} with Gal_{F_{i−1}}F_{i} a cyclic group.

(Proof: Since α is expressible in terms of radicals, there is a root tower as in 10. Since \sqrt[mn]{β} = \sqrt[m]{\sqrt[n]{β}}, by inserting additional fields in the tower, we may assume that all the n_{i} are prime. Make a new root tower by first adjoining the primitive roots of unity ζ_{n_{i}} one at a time. Then adjoin the β_{i}. Each step of this new tower falls into the categories studied in 11 or 12. Hence we are done.)

(14) (Corollary of 13.) Suppose we have a root tower as in 13. Then G = Gal_{ℚ}F_{n} has a chain of

subgroups of the form:

{id} ⊂ G_{1} ⊂ G_{2} ⊂ ⋯ ⊂ G_{n} = G

where for all i, G_{i} is normal in G_{i+1}, and G_{i+1}/G_{i} is cyclic.

(Proof: Let G_{i} = Gal_{F_{i}}F_{n}. By 7, F_{n} is Galois over each of the intermediate fields F_{i}. Hence by the last part of 9,

G_{i+1}/G_{i} ≃ Gal_{F_{i+1}}F_{n}/Gal_{F_{i}}F_{n} ≃ Gal_{F_{i}}F_{i+1},

which is a cyclic group by hypothesis.)

(15) (Continuation) Since F_{n} is Galois over ℚ and α ∈ F_{n}, the splitting field K of α is contained in F_{n}, it is Galois over ℚ by 6, and Gal_{ℚ}K = Gal_{ℚ}F_{n}/Gal_{K}F_{n}. As a result, H = Gal_{ℚ}K also has chain of subgroups of the form in 14:

{e} ⊂ H_{1} ⊂ H_{2} ⊂ ⋯ ⊂ H_{n} = G

where for all i, H_{i} is normal in H_{i+1}, and H_{i+1}/H_{i} is cyclic. Groups with this property are said to be solvable groups.

(16) The symmetric group S_{5} is not solvable.

(Proof: We try to construct a chain of subgroups as in 14 or 15. Each normal subgroup must be a union of conjugacy classes in S_{n}. By examining the sizes of the conjugacy classes, we see that the only possible nontrivial normal subgroup in S_{5} is the alternating group A_{5}. But A_{5} is simple (it has no normal subgroups except the identity subgroup and A_{5}). Hence the only chains of subgroups {e} ⊂ H_{1} ⊂ H_{2} ⊂ ⋯ ⊂ H_{n} = S_{5} with H_{i} normal in H_{i+1} for all i are are {()} ⊂ A_{5} ⊂ S_{5} and {()} ⊂ S_{5}. Since A_{5} and S_{5} are not cyclic, S_{5} is not solvable.)

(17) (The “punch line”) Let f(x) = 2x^{5} − 10x + 5 ∈ ℚ[x]. By the Eisenstein Criterion with p = 5, f is irreducible. Let K be the splitting field of f. The Galois group Gal_{F} K permutes the five roots of f, so it is isomorphic to a subgroup of S_{5}. By calculus, you can see that f has exactly three real roots and one pair of complex conjugate roots. If we number the two complex roots as α_{1}, α_{2} and the three real roots as α_{3}, α_{4}, α_{5}, complex conjugation induces an automorphism of the splitting field K of f over ℚ giving the permutation (12) on the roots. Moreover, since [ℚ(α_{i}) : ℚ] = 5, 5 divides [K : F] = |Gal_{F} K|. By the Sylow theorems, Gal_{F} K must contain an element of order 5. But the only permutations of the 5 roots that have order 5 are the 5-cycles. It is easy to see that if a subgroup of S_{5} contains a 5-cycle and a 2-cycle, then it is all of S_{5}. But then by 16, Gal_{F} K is not a solvable group. So by the reasoning leading up to 15, the roots of f are not expressible by radicals(!)