Roots without repetition
A polynomial is separable when, over a splitting field, it has no repeated roots. The detector is the formal derivative: f has a repeated root iff gcd(f, f′) ≠ 1. For an *irreducible* f this can fail only if f′ = 0 identically — which over a characteristic-0 field is impossible for a nonconstant f, so in characteristic 0 every irreducible polynomial is separable and every algebraic extension is separable. The trouble lives only in characteristic p.
In characteristic p, f′ = 0 happens exactly when f is a polynomial in xᵖ. The standard cautionary tale is t in 𝔽_p(t): over the field K = 𝔽_p(tᵖ), the polynomial xᵖ − tᵖ = (x − t)ᵖ has t as a p-fold root, so K(t)/K is inseparable. Fields where this *cannot* happen — where every irreducible is separable — are called perfect fields; these include all of characteristic 0 and all finite fields.
The finite fields, completely classified
Here is one of the cleanest classification theorems in algebra. A finite field has exactly q = pⁿ elements for a prime p and integer n ≥ 1, and for each such q there is one and only one field of that size up to isomorphism, written 𝔽_q. It is the splitting field of x^q − x over 𝔽_p — precisely the q roots of that polynomial. And 𝔽_{p^m} ⊆ 𝔽_{p^n} exactly when m divides n. The whole lattice of subfields is just the divisor lattice of n.
Two structural gems. First, the multiplicative group 𝔽_q^× is cyclic of order q − 1 — a generator is called a primitive element of the finite field. Second, 𝔽_q/𝔽_p is a Galois extension whose Galois group is cyclic of order n, generated by Frobenius x ↦ xᵖ. Galois groups of finite fields are about as simple as Galois groups ever get: always cyclic.
Build F_8 = F_2[x]/(x^3 + x + 1).
x^3 + x + 1 is irreducible over F_2 (no root: f(0)=1, f(1)=1). Let a = class of x.
Elements: all c0 + c1*a + c2*a^2 with ci in {0,1} -> 2^3 = 8 elements.
Rule for reducing: a^3 = a + 1 (since a^3 + a + 1 = 0, and -1 = 1 in F_2).
Powers of a (check it generates F_8^*, a cyclic group of order 7):
a^1 = a
a^2 = a^2
a^3 = a + 1
a^4 = a*a^3 = a^2 + a
a^5 = a*a^4 = a^3 + a^2 = a^2 + a + 1
a^6 = a*a^5 = a^3 + a^2 + a = a^2 + 1
a^7 = a*a^6 = a^3 + a = (a+1) + a = 1 <-- order exactly 7, so a is primitive.
Galois group of F_8/F_2 = <Frobenius>, Frob(y) = y^2, cyclic of order 3.
The 3 roots of x^3+x+1 are a, a^2, a^4 (= Frob orbit of a).One element to rule them all
The payoff of separability is the primitive element theorem: every *finite separable* extension is simple, L = K(γ) for a single γ. So in characteristic 0, any finite extension — no matter how many generators you started with — is generated by one element. We saw this concretely in Guide 2: ℚ(√2, √3) = ℚ(√2 + √3). The primitive element theorem is what makes the minimal-polynomial machinery of Guide 2 apply to *all* finite separable extensions, not just the obviously simple ones.