Field Extensions and Automorphism Correspondence

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}. Fundamental Theorem of field-automorphism correspondence The most important instance of a fixed field E H arises when H is a subgroup of Aut(E), but we will meet a case in which it is merely a subset. In Example 3.125, we considered E = k(y1, . . . , yn), the rational function field in n variables with coefficients in a field k, and its subfield K = k(a0, . . . , an−1), where f (x) = (x −y1)(x −y2) · · · (x −yn) = a0 + a1x + · · · + an−1xn−1 + xn is the general polynomial of degree n over k. We saw that E is a splitting field of f (x) over K, for it arises from K by adjoining to it all the roots of f (x), namely, all the y’s. Now the symmetric group Sn ≤Aut(E), for every permutation of y1, . . . , yn extends to an automorphism of E, and it turns out that K = E Sn. The elements of K are usually called the symmetric functions in n variables over k.

A character6 of a group G in a field E is a (group) homomorphism σ : G →E×, where E× denotes the multiplicative group of nonzero elements of the field E. If σ ∈Aut(E), then its restriction σ|E× : E× →E× is a character in E.

A field extension E/k is a normal separable extensions if it satisfies any of the equivalent conditions in Theorem 4.34.

If E/k is a normal separable extensions and if B is an intermediate field, then a conjugate of B is an intermediate field of the form Bσ = {σ(b) : b ∈B} for some σ ∈Aut_fld(E/k). 7This result is true if finitely many transcendental elements are adjoined (remember that transcendental elements are always separable, by definition), but it may be false if infinitely many transcendental elements are adjoined.

A lattice is a partially ordered set L in which every pair of elements a, b ∈L has a greatest lower bound a ∧b and a least upper bound a ∨b.

A field extension E/k is a simple extension if there is u ∈E with E = k(u). The following theorem of E.

If f (x) = i(x −αi) ∈k[x], where k is a field, define Δ = → i< j (αi −α j), and define the discriminant to be D = D( f ) = Δ2 = i< j(αi −α j)2. It is clear that f (x) has repeated roots if and only if its discriminant D = 0. The product Δ = i< j(αi −α j) has one factor αi −α j for each distinct pair of indices (i, j) (the restriction i < j prevents a pair of indices from occurring twice). If E/k is a splitting field of f (x) and if G = Aut_fld(E/k), then each σ ∈G permutes the roots, and so Fundamental Theorem of field-automorphism correspondence σ permutes all the distinct pairs. However, it may happen that i < j while the subscripts involved in σ(αi) −σ(α j) are in reverse order. For example, suppose the roots of a cubic are α1, α2, and α3, and suppose there is σ ∈G with σ(α1) = α2, σ(α2) = α1, and σ(α3) = α3. Then σ(Δ) = ( σ(α1) −σ(α2) )( σ(α1) −σ(α3) )( σ(α2) −σ(α3) ) = (α2 −α1)(α2 −α3)(α1 −α3) = −(α1 −α2)(α2 −α3)(α1 −α3) = −Δ. In general, each term αi −α j occurs in σ(Δ) with a possible sign change. We conclude, for all σ ∈Aut_fld(E/k), that σ(Δ) = ±Δ. It is natural to consider Δ2 rather than Δ, for Δ depends not only on the roots of f (x), but also on the order in which they are listed, whereas D = Δ2 does not depend on the listing of the roots.

The resolvent cubic of f (x) = x4 + qx2 + rx + s is g(x) = (x −u)(x −v)(x −w), where u, v, w are the numbers defined in Eqs. (6).

If E is a field, then the function H ↦E H, from subsets H of Aut(E) to subfields of E, is order-reversing: If H ≤L ≤Aut(E), then E L ⊆E H.

If G = {σ1, . . . , σn} is a set of n distinct automorphisms of a field E, then [E : EG] ≥n.

If E/k is a normal separable extensions and if B is an intermediate field, that is, a subfield B with k ⊆B ⊆E, then E/B is a normal separable extensions.

But to say that θ(x) = g(x1, . . . , xn)/h(x1, . . . , xn) lies in E Sn is to say that it is unchanged by permuting its variables; that is, θ(x) is a symmetric function. . . , xn] lies in k[e1, . . . , en].

Let L and L′ be lattices, and let ϕ : L →L′ be a bijection such that both ϕ and ϕ−1 are order-reversing. Then ϕ(a ∧b) = ϕ(a) ∨ϕ(b) and ϕ(a ∨b) = ϕ(a) ∧ϕ(b).

If E/k is a normal separable extensions whose field-automorphism groups is abelian, then every intermediate field is a normal separable extensions.

If B/k is a finite separable extension, then there is u ∈B with B = k(u). In particular, if k has characteristic 0, then every finite extension B/k is a simple extension.

If f (x) ∈C[x] has degree n ≥1, then f (x) has a complex root, and hence f (x) splits: There are c, u1, . . . , un ∈C with f (x) = c(x −u1) · · · (x −un).

Let E/k be a normal separable extensions of prime degree p. If k contains a primitive pth root of unity ω, then E = k(z), where z p ∈k, and so E/k is a pure extension of type p.

Let k be a field of characteristic 0, let E/k be a normal separable extensions, and let G = Aut_fld(E/k) be a solvable group. Then E can be imbedded in a radical extension of k. Therefore, the field-automorphism groups of a polynomial over a field of characteristic 0 is a solvable group if and only if the polynomial is solvable by radicals. Remark. A counterexample in characteristic p is given in Proposition 4.56. ◀

Let p be a prime, and let k = Fp(t). The field-automorphism groups of f (x) = x p −x −t over k is cyclic of order p, but f (x) is not solvable by radicals over k.

Let k be a field with characteristic ̸= 2, let f (x) ∈k[x] be a polynomial of degree n with no repeated roots, and let D = Δ2 be its discriminant. Let E/k be a splitting field of f (x), and let G = Aut_fld(E/k) be regarded as a subgroup of Sn (as in

The resolvent cubic of f (x) = x4 + qx2 + rx + s is g(x) = x3 −2qx2 + (q2 −4s)x + r2.

Let f (x) ∈Q[x] be an irreducible quartic with field-automorphism groups G with discriminant D, and let m be the order of the field-automorphism groups of its resolvent cubic g(x). (i) If m = 6, then G ∼= S4. In this case, g(x) is irreducible and √ D is irrational. (ii) If m = 3, then G ∼= A4. In this case, g(x) is irreducible and √ D is rational. (iii) If m = 1, then G ∼= V. In this case, g(x) splits in Q[x]. (iv) If m = 2, then G ∼= D8 or G ∼= I4. In this case, g(x) has an irreducible quadratic factor.

Suppose now that k is a subfield of E and that G = Aut_fld(E/k). It is obvious that k ⊆EG, but the inclusion can be strict. For example, let E = Q( 3√ 2). If σ ∈G = Aut_fld(E/Q), then σ must fix Q, and so it permutes the roots of f (x) = x3 −2. But the other two roots of f (x) are not real, so that σ( 3√ 2) = 3√ 2. It now follows from Lemma 4.2 that σ is the identity; that is, EG = E. Note that E is not a splitting field of f (x). ◀ Our immediate goal is to determine the degree [E : EG], where G ≤Aut(E). To this end, we introduce the notion of characters.

By Proposition 4.62, the resolvent cubic is g(x) = x3 −8x + 16. The discriminant of g(x) is −4864, so that Theorem 4.60 shows that the field-automorphism groups of g(x) is S3, hence has order 6. Theorem 4.64 now shows that G ∼= S4. By Proposition 4.62, the resolvent cubic is x3 + 20x2 + 96x = x(x + 8)(x + 12). In this case, Q(u, v, w) = Q and m = 1. Therefore, G ∼= V. [This should not be a surprise if we recall Example 3.122, where we saw that f (x) arises as the irreducible polynomial of α = √ 2 + √ 3, where Q(α) = Q( √ 2, √ 3).] ◀ An interesting open question is the inverse field-automorphism problem: Which finite abstract groups G are isomorphic to Aut_fld(E/Q), where E/Q is a normal separable extensions? polynomial finite-generation proved that the symmetric groups Sn are such field-automorphism groups, and I. After the classification of the finite simple groups in the 1980s, it was shown that most simple groups are field-automorphism groups.