Finite Galois extensions are separable and normal

Let $K\subseteq L$ be a finite field extension. Then:

Theorem. The following are equivalent:

$L/K$ is a Galois extension.
$L/K$ is both separable and normal.

Under these conditions, the extension degree equals the order of the Galois group:

[L:K] = |\mathrm{Gal}(L/K)|

(cf. degree equals group order).

Remarks

Over a perfect field, separability is automatic for finite extensions (see perfect implies separable), so “finite Galois” is equivalent to “finite normal.”

Examples

Quadratic extensions over $\mathbb{Q}$ . $L=\mathbb{Q}(\sqrt2)$ is the splitting field of $x^2-2$ , hence normal (see normality via splitting fields); characteristic $0$ gives separability. Thus $L/\mathbb{Q}$ is Galois with $|\mathrm{Gal}(L/\mathbb{Q})|=2$ .

A separable but non-normal extension. $L=\mathbb{Q}(\sqrt[3]{2})$ is separable (char $0$ ) but not normal, so it is not Galois. Its normal closure is the splitting field $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ .

A normal but inseparable extension (characteristic $p$ ). Let $K=\mathbb{F}_p(t)$ and $L=K(t^{1/p})$ . Then $L/K$ is inseparable (its defining polynomial has zero derivative), hence not Galois even though purely inseparable extensions satisfy a strong form of “no new embeddings.” This illustrates why the separability hypothesis is essential.