# Separable extensions II

What we know is a drop, what we don’t know is an ocean, Isaac Newton

# Introduction

Basically, the fundamental theorem of Galois theory says that you can tell a lot about a field extension by looking at its Galois group. More specifically, it says that for a finite and Galois field extension K/F there is a one-to-one correspondence between its intermediate fields and the subgroups of its Galois group. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., [K : F] = |Gal(K : F)|.

Notice that Galois’s fundamental theorem does not hold for all extensions. The easiest nontrivial extensions are the algebraic extensions, and more specifically, simple extensions (a field extension which is generated by the adjunction of a single element, called a primitive element), i.e., K = F(a). Let a ∈ K be algebraic over F, let f(x) be its irreducible polynomial, then F(a) = F[a] ≋ F[x]/⟨f(x)⟩, [F[a] : F] = deg f(x) = d, and a basis of F(a) as an F-vector space is {1, a, a2, ···, ad-1}, where d = deg(f), that is, a simple extension's degree equals the degree of the primitive element's minimal polynomial.

Fact. Suppose α is algebraic over F with minimal polynomial f(x) and K = F(α) ⇒ ∀ σ ∈ Aut(K/F)=Gal(K/F), σ(α) is also a root in f(x). Conversely, if β is any other root of f(x) in K ⇒ ∃! τ ∈ Aut(K/F)=Gal(K/F) with τ(α) = β. Therefore, |Gal(K/F)| is equal to the number of roots of f in K, and is (in particular) finite and at most [K : F].

If α1, α2, ···, αs are the distinct roots of f in K. ∀σ ∈ Gal(K/F), σ(αi) = αj and we know that |Gal(K/F)| = s. Futhermore, [K : F] = deg(p), so [K : F] = |Gal(K/F)| if and only if deg(p) is equal to s, the number of distinct roots of the minimal polynomial of α over F.

Certainly, s ≤ deg(p), but they may not be equal because the irreducible polynomial might not split completely in K, e.g., F = ℚ, α = $\sqrt[3]{2}$, p(x) = x3 -2 which has only one root in K = ℚ($\sqrt[3]{2}$), so [ℚ($\sqrt[3]{2}$) : ℚ] = 3 = deg(x3 -2) > 1 = Gal(K/Q). An algebraic extension is called normal if the minimal polynomial of every element split in K. Secondly, some of the roots of p in K may be repeated. An algebraic extension is called separable if the minimal polynomial of every element has no repeated roots in its splitting field. If we take any finite extension K : F that is both normal and separable, then it is indeed true that |Gal(K : F)| = [K : F].

# Definition

Let F be a field. An irreducible polynomial f ∈ F[x] is separable if it has distinct roots (or it has no repeated roots) in any splitting field, i.e., the number of distinct roots of f equals the degree of the polynomial and each root of f has multiplicity 1, f = k(x - α1)(x - α2)···(x - αn) where the roots αi are all distinct.

A nonzero arbitrary polynomial g(x) ∈ F[x] is called separable if all irreducible factors are separable over F, e.g., (x -2)2 ∈ ℚ[x] is separable because its only irreducible factor (x -2) is indeed separable, (x2-1) is separable over ℚ, as is (x -3)7.

This definition is contested. In this definition, separability depends on the field F, and any polynomial over a perfect field is indeed separable. It is convenient for Galois Theory, but many prefer another definition. Definition: An arbitrary polynomial over a given field F is separable if its roots are distinct in an algebraic closure of F.

Let K/F be a finite extension. An element α ∈ K is separable over F (A finite extension is always algebraic, and thus α is algebraic) if its irreducible polynomial over F is separable. A finite extension K/F is separable if every element of K is separable over F. K/F is said to be inseparable if it is not separable, that is, there is at least one element of K that is not separable over F.

Recall. Zeros of an irreducible polynomial. The derivate of a polynomial f(x) = a0 + a1x + ··· anxn is well defined for any field F in F[x]. It is defined as f’(x) = a1 + 2a2x + ··· + nanxn-1 ∈ F[x]. We have the basic properties as the derivate of calculus: (af + bg)’(x) = af’(x) + bg’(x), (fg)’(x) = f’(x)g(x) + f(x)g’(x), and (f(g(x)))’ = f’(g(x))g’(x).

Lemma. Let f(x) and g(x) be polynomials over a field F. (Fields and Galois Theory. Morandi. P., Springer.)

• If f has no repeated roots in any splitting field, then f is separable over F.

Proof: If f has no repeated roots in any splitting field, then neither does any irreducible factor of f ⇒ f is separable over F.

• If g divides f (g | f) and if f is separable over F ⇒ g is separable.

Proof: If g divides f, then the irreducible factors of g are also irreducible factors of f ⇒ [f is separable, i.e., no irreducible factor of f has a repeated root] g is separable.

• If f1,···, fn are separable polynomials over F, then the product f1···fn is separable over F.

Proof: The set of irreducible factors of the fi ∀i, 1 ≤ i ≤ n (all of them with no repeated roots) is precisely the set of irreducible factor of the polynomial f1···fn, so f1···fn is separable over F.

• If f(X) ∈ F[X] is separable over F, then f is separable over any extension field of F. An element, say α, in an extension E that is separable over F is also separable over E.

Proof.

Let E/F be an arbitrary extension of F. Let f(x) be separable over F and g(x) in E[x] a factor of f(x) ⇒ f(x) = g(x)h(x) in E[X].

Since f(x) is separable, we can write 1 = f(x)u(x) + f’(x)v(x) for some polynomials u(x) and v(x) in F[x] ⇒ [f(x) = g(x)h(x)] 1 = (g(x)h(x))u(x) + (g(x)h’(x) + g’(x)h(x))v(x) = g(x)(h(x))u(x) + h’(x)v(x)) + g’(x)(h(x)v(x)) ⇒ There is a polynomial-linear combination of g and g’ equals 1, so g is separable.

Suppose α ∈ E, E is an arbitrary extension of F, α separable over F. Since its minimal polynomial in E[x], min(α, E), divides its minimal polynomial in F[x], min(α, F) ⇒ [min(α, E) | min(α, F), min(α, F) separable] ⇒ separability of α over F implies separability of α over E∎

Proposition. Let F be any field.

1. An irreducible polynomial f ∈ F[x] is separable iff the greatest common factor of f and f' is 1, it is expressed or written as (f, f') = 1, f is relatively prime to its derivate.
2. If F has characteristic zero, F is a finite field or perfect, then every polynomial in F[x] is separable.
3. If F has characteristic zero, F is a finite field or perfect, then any finite extension K/F is separable.

# In positive characteristic fields, there exist inseparable irreducible polynomial

If F has characteristic zero, F is a finite field or perfect, then every polynomial in F[x] is separable. If F has characteristic p (p > 0), there can exist inseparable irreducible polynomial. Let F be the field of rational functions with coefficients in the field with p (=2) elements = $\mathbb{F_2}(t)=${$\frac{f(t)}{g(t)}|f(t),g(t)∈\mathbb{F_2}(t), g(t)≠0$}.

f(x) = x2 -t ∈ $\mathbb{F_2}(t)$, f is irreducible over F because it has degree 2 and it has no roots in F [Reducibility test for polynomials with degrees two or three. Let F be a field. If f(x) ∈ F[x] and deg(f(x)) = 2 or 3, then f(x) is reducible over F iff f has a zero in F.]. Notice that a root of f in F is a rational function a(t)/b(t) ∈ F such that (a(t)/b(t))2 - t = 0 ⇒ (a(t)/b(t))2 = t ⇒ a(t)2 = b(t)2t. This is not possible because the degree of the left hand side has even degree (being the square of a polynomial), but the polynomial of the right hand side has odd degree, so we have a contradiction ⊥

Consider the splitting field K of f(x) over F ([K : F] = 2, deg(x2-t)=2, x2-t is irreducible over F, and it does exist such a splitting field, $K = F(\sqrt{t})$

Since char(K) = char(F) = char($\mathbb{F_2}$) = 2 ⇒ $2(\sqrt{t})=0 ⇒ \sqrt{t} + \sqrt{t} = 0 ⇒ \sqrt{t} = - \sqrt{t}$ in K, therefore f(x) has only one root in K, and it does not have distinct roots (deg = 2, but it has only one root) or, in other words, it has a multiple root ⇒ f(x)∈$\mathbb{F_2}(t)$ is not separable.

The extension K/F is normal (it is the splitting field of a polynomial), but not Galois, because Gal(K/F) has only one element, the identity that maps $\sqrt{t}$ to itself (it needs to send $\sqrt{t}$ to another root of the polynomial, but there’s a single root, namely $\sqrt{t}$): 1 = |Gal(K/F)| < [K : F] = 2.

To sum up, the minimum irreducible polynomial of $\sqrt{t}$ over $\mathbb{F_2}(t)$ is f(x) = x2 -t. (f, f’) = f ≠ 1 (because f’ = 2x = 0), f splits into $(x -\sqrt{t})^2$, so it’s clearly inseparable. Futhermore, [$\mathbb{F_2}(\sqrt{t}):\mathbb{F_2}(t)$] = 2 and Aut($\mathbb{F_2}(\sqrt{t})/\mathbb{F_2}(t)$) = {id} as $\sqrt{t}$ is the only root of x2 -t -irreducible-, so it is not Galois: 1 = |Aut($\mathbb{F_2}(\sqrt{t})/\mathbb{F_2}(t)$)| < [$\mathbb{F_2}(\sqrt{t}):\mathbb{F_2}(t)$] = 2.

# Examples

• In $K = \mathbb{F_3}, f ∈ \mathbb{F_3}$[x], let f(x) = x6 + x5 + x4 + 2x3 +2x2 + x +2. f’ = 5x4 +4x3 +6x2 + 4x + 1 = 2x4 + x3 +x + 1.

We compute gcd(f, f’) using the Euclidean Algorithm:

1. f = (2x2 +x)f’ + 2x2 + 2.
2. f’ = (x2 + 2x + 2)(2x2 + 2) + 0.

Therefore (f, f’) = 2x2 +2, which is the same x2 +1 up to scaling (monic gcd) ⇒ [An irreducible polynomial f ∈ F[x] is separable iff (f, f') = 1] f is not separable and the roots of x2 + 1 are multiple root of f. In fact, f(x) = (x2 +1)2(x2 +x + 2).

• In ℚ[x], f(x) = x4 -3x2 = 0 ⇒ [x2 = y] y2 -3y = 0 ⇒ y = 3 and 0 ⇒ [x2 = y] $x=±\sqrt{3},0$. There are three possible roots in an algebraic closure of ℚ, say $\bar \mathbb{Q}$, but deg(f) = 4, so it is not separable (it depends on the definition 😄). In our previous definition, f(x) = x2(x2 -3) is separable because its two irreducible factors x and x2-3 are separable.

• In $K = \mathbb{F_2}, f ∈ \mathbb{F_2}$[x], let f(x) = x3 -1. f(x) = (x -1)(x2 +x +1). Thus, f is separable iff x2 +x +1 is separable. Besides, x2 + x +1 is irreducible over $\mathbb{F_2}$ because it has no roots (02 +0 +1 = 1 ≠ 0, 12 + 1 + 1 = 1 ≠ 0) and deg(x2 +x + 1) = 2. Since $K = \mathbb{F_p}$ is a perfect field for every prime p ⇒ irreducible polynomials over $\mathbb{F_p}$ are always separable.

Let E/F be an algebraic extension of fields of characteristic p. The separable closure of F in E is S = {α ∈ E | α is separable over F}. It is an algebraic separable extension of F, and the separable degree of E/F is simply [S : F], the degree of the field extension S/F. [E : F] = [E : S][S : F]. Fact about separability for finite extensions: E/F is separable if and only if [E: F] = [E : F]S

Separability of towers. Let L/F be a finite extension and let F ⊆ K ⊆ L be an intermediate field. Then, L/F is separable if and only if L/K and K/F are separable.

Proof.

⇒) First, let’s assume that L/F is separable. ∀α ∈ L, α is separable over F. In particular, this is true for any α ∈ K, so K/F is separable.

Let α ∈ L, let mK and mF be the minimum polynomials of α over K and F respectively. Within K[x] we can use the division algorithm, mF = qmK + r (deg(r) < deg(mK)) ⇒ r(α) = mF(α) -q(α)mK(α) = 0 -0 = 0 ⊥ (minimality of mK -we are within K[x]-) ⇒ r = 0 ⇒ mF = qmK ⇒ mK divides mF and mF is separable over F (K/F is separable) ⇒ [If g divides f (g | f) and f is separable over F ⇒ g is separable.] mK is separable over K ⇒ L/K is separable.

⇐) [L : F] = [L : K][K : F]

[L : F]S = [L : K]S[K : F]S = [E/F is separable if and only if [E: F] = [E : F]S, L/K and K/F are separable by assumption] [L : K][K : F] ⇒ L/F is separable.