Separable extensions

From now on we consider only fields of characteristic zero, i.e. extensions of the rational field. As we shall see "An algebraic field extension is Galois if and only if it is normal and separable" and this latter condition is always attained in zero characteristic field. We recall that a polynomial is separable if it has no multiple roots.

8.8.1 Definition. An algebraic extension K of a field F is said separable extension if each element of K is separable over F, i.e. its minimal polynomial is separable.  □

8.8.2 Proposition. Let a and b two algebraic elements over a field F of characteristic zero. Then there exists an element α ∈ F(a,b) such that F(a,b) = F(α).

Proof. Let f(x) and g(x) in F[x] be the minimal polynomials of a and b respectively. Let K be an extension in which both polynomials split, and let

the roots in K of f(x) and g(x) respectively. We know from Proposition 8.3.6 that over a field of characteristic 0, every irreducible polynomial is "separable" -- its roots in any splitting field are all distinct, hence since f(x) and g(x) are irreducible, all the a_i (and b_j) are distinct. For every i = 1, ..., n and every j = 2, ..., m, the equation in λ

admits only one root in K, given by

F being of characteristic 0, it possesses infinitely many elements. We pick an element γ ∈ F which does not satisfy any of the above n(m − 1) equations. Set α = a + γb. We shall prove that

It suffices to prove that F(α) ⊇ F(a,b) and to this end it is sufficient to prove that b ∈ F(α). Consider the polynomial

The element b is a common root of g(x) and h(x), thus x − b is a common factor in a proper extension of F(α) of both g(x) and h(x). It is more precisely their gcd: indeed h(x) cannot have (in a given extension) other roots of g(x), because we chose γ such that α − γb_j ≠ a_i for every i. We cannot have either gcd(g(x), h(x) = (x − b)^t with t > 1, because g(x) has only distinct roots. Thu (x − b) is the greatest common factors between g(x) and h(x) and hence belongs to F(α)[x], since F(α) is the common field of coefficients of g(x) and h(x); b ∈ F(α).  □

We have thus proved the following

8.8.3 Primitive Element Theorem. Every K finite extension of a field F of characteristic zero is simple, i.e. there exists α ∈ K called primitive element such that K = F(α).  □

Exercises

1. Prove that ℚ(√2, √3) = ℚ(√2 + √3).

2. Let K a finite extension of a finite field F. Prove that K is a simple extension.

3. Let F a finite field. Prove that the ring F[x] contains irreducible polynomials over F of arbitrary degree.

Solutions

1. It suffices to show that √2 and √3 can be written as combination of coefficients in ℚ of the basis 1, √2 + √3, (√2 + √3)², (√2 + √3)³. Indeed √2 = −2/3(√2 + √3) + 1/2 (√2 + √3)³, from which we obtain also that √3 = (√2 + √3) − √2 = 11/2 (√2 + √3) − 1/2 (√2 + √3)³.  ■

2. Consider K ⊃ F. We know that the multiplicative subgroup K* of a finite field K is cyclic (see Theorem 8.3.11). If a a generator of K*, then certainly K = F[a]. K* = ⟨b⟩ for some b, so K* = K\{0} = {b¹, b², b³, ..., b^{(|K| − 1)} = 1}. And F(b) includes 0. So we have all the elements of K now. ■

3. Let |F| = p^m and let K the splitting field of the polynomial x^pmn − x. Since the degree of this polynomial It results F ⊆ K (it suffices to show that ∀a ∈ F then a^pm = a which in turn satisfies the relation p^mn = a as well, i.e. every element of F is a root). The field K is finite with p^mn elements. By the result obtained in the previous exercise, K = F(b) for a given b ∈ K. Let p(x) the minimal polynomial of b over F. If p(x) has degree n then n = [F(b) : F] by theorem 8.2.13 (d).  ■