Finite Galois extensions are separable and normal

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

Theorem. The following are equivalent:

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

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

(cf. degree equals group order ).

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

Examples

1. 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 $|\mathrm{Gal}(L/\mathbb{Q})|=2$.

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 $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$.

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.