Polynomial Solvability by Radicals and the Degree-Five Obstruction

Let E be a field containing a subfield k. An automorphism2 of E is an isomorphism σ : E →E; we say that σ fixes k if σ(a) = a for every a ∈k. For example, consider f (x) = x2 + 1 ∈Q[x]. A splitting field of f (x) over Q is E = Q(i), and complex conjugation σ : a ↦a is an example of an automorphism of E fixing Q.

Let k be a subfield of a field E. The field-automorphism groups of E over k, denoted by Aut_fld(E/k), is the set of all those automorphisms of E that fix k. If f (x) ∈k[x], and if E = k(z1, . . . , zn) is a splitting field, then the field-automorphism groups of f (x) over k is defined to be Aut_fld(E/k). It is easy to check that Aut_fld(E/k) is a group with operation composition of functions. Note that Aut_fld(E/k) is independent of the choice of splitting field E, by Theorem 3.131. The following lemma will be used several times.

Let E/k be an algebraic extension. An irreducible polynomial p(x) is separable if it has no repeated roots. An arbitrary polynomial f (x) is separable if each of its irreducible factors has no repeated roots. An element α ∈E is called separable if either α is transcendental over k or if α is algebraic over k and its minimal polynomial irr(α, k) has no repeated roots; that is, irr(α, k) is a separable polynomial. A field extension E/k is called a separable extension if each of its elements is separable; E/k is inseparable if it is not separable.

A pure extension of type m is an extension k(u)/k, where um ∈k for some m ≥1. An extension K/k is a radical extension if there is a tower of fields k = K0 ⊆K1 ⊆· · · ⊆Kt = K in which each Ki+1/Ki is a pure extension. If um = a ∈k, then k(u) arises from k by adjoining an mth root of a. If k ⊆C, there are m different mth roots of a, namely, u, ωu, ω2u, . . . , ωm−1u, where ω = e2πi/m is a primitive mth root of unity. More generally, if k contains the mth roots of unity, then a pure extension k(u) of type m, that is, um = a ∈k, then k(u) is a splitting field of xm −a. Not every subfield k of C contains all the roots of unity; for example, 1 and −1 are the only roots of unity in Q. Since we seek formulas involving extraction of roots, it will eventually be convenient to assume that k contains appropriate roots of unity. 3Actually, a better analogy would involve polyhedra in euclidean space Rn instead of only polygons in the plane. Insolvability of the Quintic When we say that there is a formula for the roots of a polynomial f (x) analogous to the quadratic formula, we mean that there is some expression giving the roots of f (x) in terms of the coefficients of f (x). The expression may involve the field operations, constants, and extraction of roots, but it should not involve any other operations involving cosines, definite integrals, or limits, for example. We maintain that a formula as we informally described exists precisely when f (x) is solvable by radicals, which we now define.

A normal series5 of a group G is a sequence of subgroups G = G0 ≥G1 ≥G2 ≥· · · ≥Gt = {1} with each Gi+1 a normal subgroup of Gi; the factor groups of this series are the quotient groups G0/G1, G1/G2, . . . , Gn−1/Gn. 5This terminology is not quite standard. We know that normality is not transitive; that is, if H ≤K are subgroups of a group G, then H ✁K and K ✁G does not force H ✁G. A subgroup H ≤G is called a subnormal subgroup if there is a chain G = G0 ≥G1 ≥G2 ≥· · · ≥Gt = H with Gi ✁Gi−1 for all i ≥1. Normal series as defined in the text are called subnormal series by some authors; they reserve the name normal series for those series in which each Gi is a normal subgroup of the big group G. Insolvability of the Quintic In this language, Lemma 4.18 says that Aut_fld(Kt/K0) is a solvable group if Kt is a radical extension of K0 and K0 contains appropriate roots of unity.

If E is a field and H is a subset of Aut(E), then the fixed field of H is defined by E H = {a ∈E : σ(a) = a for all σ ∈H}.

Let k be a subfield of a field K, let f (x) = xn + an−1xn−1 + · · · + a1x + a0 ∈k[x], and let E = k(z1, . . . , zn) ⊆K be a splitting field. If σ : E →E is an automorphism fixing k, then σ permutes the set of roots {z1, . . . , zn} of f (x).

If f (x) ∈k[x] has degree n, then its field-automorphism groups Aut_fld(E/k) is isomorphic to a subgroup of Sn.

If k is a field of characteristic 0, then every irreducible polynomial p(x) ∈ k[x] has no repeated roots.

, that x pn −x has no repeated roots. It follows that if k ⊆E, then E/k is a separable extension, for if α ∈E, then irr(α, k) is a divisor of x pn −x.

(i) Let E/k be a splitting field of a separable polynomial f (x) ∈k[x], let ϕ : k →k′ be a field isomorphism, and let E′/k′ be a splitting field of f ∗(x) ∈k′[x] [where f ∗(x) is obtained from f (x) by applying ϕ to its coefficients]. E → E′ k ϕ k′ Then there are exactly [E : k] isomorphisms →: E →E′ that extend ϕ. (ii) If E/k is a splitting field of a separable f (x) ∈k[x], then | Aut_fld(E/k)| = [E : k].

gives σ 2 = 1E for all σ ∈Aut_fld(E/Q), and so Aut_fld(E/Q) ∼= V. Here is another way to compute G = Aut_fld(E/Q). We saw in Example 3.122 that E = Q( √ 2+ √ 3) = Q( √ 2, √ 3) is also a splitting field of g(x) = (x2 −2)(x2 −3) over Q. By

If m is a positive integer, if k is a field, and if E is a splitting field of xm −1 over k, then Aut_fld(E/k) is abelian; in fact, Aut_fld(E/k) is isomorphic to a subgroup of the multiplicative group U(Im) of all [i] with (i, m) = 1.

If p is a prime, then Aut_fld(Fpn/Fp) ∼= In, and a generator is the fixed-point-action F : u ↦u p.

Let k ⊆B ⊆E be a tower of fields, let f (x), g(x) ∈k[x], let B be a splitting field of f (x) over k, and let E be a splitting field of g(x) over k. Then Aut_fld(E/B) is a normal subgroup of Aut_fld(E/k), and Aut_fld(E/k)/ Aut_fld(E/B) ∼= Aut_fld(B/k). Insolvability of the Quintic

applies to show that there is σ ∈Aut_fld(E/k) extending τ [i.e., ρ(σ) = σ|B = τ]. The first isomorphism theorem completes the proof. • The next technical result will be needed when we apply Theorem 4.16.

(i) If B = k(α1, . . . , αn) is a finite extension of a field k, then there is a finite extension E/B that is a splitting field of some polynomial f (x) ∈k[x] (such an extension of smallest degree is called a normal4 closure of B/k). Moreover, if each αi is separable over k, then f (x) can be chosen to be a separable polynomial. (ii) If B is a radical extension of k, then the extension E/B in part (i) is a radical extension of k.

Let k be a field and let f (x) ∈k[x] be solvable by radicals, so there is a radical extension k = K0 ⊆K1 ⊆· · · ⊆Kt with Kt containing a splitting field E of f (x). If each Ki/Ki−1 is a pure extension of prime type pi, where pi ̸= char(k), and if k contains all the pith roots of unity, then the field-automorphism groups Aut_fld(E/k) is a quotient of a solvable group.

Let f (x) ∈k[x], where k is a field, and let E be a splitting field of f (x) over k. If f (x) is solvable by radicals, then its field-automorphism groups Aut_fld(E/k) is a solvable group. Remark. ◀

If n ≥5, the general polynomial of degree n f (x) = (x −y1)(x −y2) · · · (x −yn) over a field k is not solvable by radicals.

Here is an example of an inseparable extension. Let k = Fp(t) = Frac(Fp[t]), and let E = k(α), where α is a root of f (x) = x p −t; that is, α p = t. In E[x], we have f (x) = x p −t = x p −α p = (x −α)p. If we show that α /∈k, then f (x) is irreducible, by Proposition 3.126, and so f (x) = irr(α, k) is an inseparable polynomial. Therefore, E/k is an inseparable extension. It remains to show that α /∈k. Otherwise, there are g(t), h(t) ∈Fp[t] with α = g(t)/h(t). Hence, g = αh and g p = α ph p = th p, so that deg(g p) = deg(th p) = 1 + deg(h p). But p | deg(g p) and p | deg(h p), and this gives a contradiction.

In Example 4.19(ii), we showed that S4 is a solvable group. However, if n ≥5, the symmetric group Sn is not a solvable group. If, on the contrary, Sn were solvable, then so would each of its subgroups be solvable. But A5 ≤S5 ≤Sn, and A5 is not solvable because it is a nonabelian simple group. ◀