Separability controls multiplicity of roots. It is invisible over characteristic zero and finite fields because all algebraic extensions there are separable, but it becomes essential in characteristic \(p\).
Separable elements and extensions
Definition. An algebraic element \(\alpha\) over \(K\) is separable if its minimal polynomial over \(K\) has no repeated root in an algebraic closure. An algebraic extension \(L/K\) is separable if every element of \(L\) is separable over \(K\).
Example. Every algebraic extension of a field of characteristic \(0\) is separable. In characteristic zero, irreducible polynomials have nonzero derivative, so they cannot have repeated roots.
Derivative criterion
Theorem. A polynomial \(f\in K[x]\) has a repeated root in an algebraic closure iff \(\gcd(f,f')\ne1\). An irreducible polynomial is inseparable iff \(f'=0\).
Proof. A repeated root is a common root of \(f\) and \(f'\), so the gcd is nontrivial. If \(f\) is irreducible, any common divisor is either \(1\) or an associate of \(f\), hence inseparability forces \(f\mid f'\). Since \(\deg f'<\deg f\) unless \(f'=0\), this is equivalent to \(f'=0\).
Perfect fields
Definition. A field \(K\) is perfect if every algebraic extension of \(K\) is separable.
Theorem. Every field of characteristic \(0\) is perfect. A field of characteristic \(p>0\) is perfect iff the Frobenius map \(x\mapsto x^p\) is surjective.
Corollary. Every finite field is perfect.
Purely inseparable examples
Example. Let \(K=\mathbb F_p(t)\), and let \(L=K(u)\) with \(u^p=t\). Then \(u\) has minimal polynomial \(x^p-t\), whose derivative is \(0\). Thus \(L/K\) is inseparable.
Remark. In a purely inseparable extension, elements acquire powers in the base field. This behavior has no analogue over \(\mathbb Q\), \(\mathbb R\), or finite fields.
Separable degree
Definition. For finite \(L/K\), the separable degree \([L:K]_s\) is the number of \(K\)-embeddings \(L\hookrightarrow\overline K\). The inseparable degree is \([L:K]_i=[L:K]/[L:K]_s\).
Proposition. A finite extension \(L/K\) is separable iff \([L:K]_s=[L:K]\).
Separable plus normal equals Galois
Theorem. A finite extension \(L/K\) is Galois iff it is normal and separable. In that case \(|\operatorname{Gal}(L/K)|=[L:K]\).
Embeddings and the primitive element theorem
Primitive element theorem. Every finite separable extension \(L/K\) is simple: \[ L=K(\theta) \] for some \(\theta\in L\).
Proof. It suffices to combine two separable generators, \(L=K(\alpha,\beta)\). If \(K\) is finite, then \(L^\times\) is cyclic and any generator of \(L^\times\) generates \(L\) as a field. Suppose \(K\) is infinite. Let \(\sigma_1,\ldots,\sigma_n\) be the distinct \(K\)-embeddings of \(L\) into an algebraic closure. For all but finitely many \(c\in K\), the values \[ \sigma_i(\alpha+c\beta) \] are pairwise distinct: equality for \(i\ne j\) excludes at most one value of \(c\), unless both images already coincide. Choose such \(c\) and put \(\theta=\alpha+c\beta\). Then \(\theta\) has at least \([L:K]\) distinct conjugates, so \([K(\theta):K]\ge[L:K]\), forcing equality.
Trace pairing
Theorem. A finite extension \(L/K\) is separable if and only if the bilinear trace pairing \[ (x,y)\longmapsto\operatorname{Tr}_{L/K}(xy) \] is nondegenerate.
For a separable simple extension, the embedding matrix has nonzero Vandermonde determinant, making the trace Gram matrix nonsingular. Inseparable extensions have fewer embeddings; in the purely inseparable case every trace is zero. This criterion explains why discriminants detect separability.