Characterizations of Galois extensions

Remarks

The implication from 1 to 2 takes the product of the distinct minimal polynomials of a basis. From 2 to 3, one counts the extensions of the identity of $K$ to automorphisms of $L$ step by step, using that the roots are distinct. In general $\abs{\Aut(L/K)} \le [L:K]$, and Artin's theorem on fixed fields gives $[L:L^{G}] = \abs{G}$ for $G = \Aut(L/K)$, which links 3, 4 and 5. From 5 to 1, every element of $L$ is a root of the separable split polynomial $\prod (x - \beta)$ over its orbit.

Example: $\Q(\sqrt[3]{2})/\Q$ is not Galois. Its only automorphism is the identity, since the other roots of $x^3 - 2$ are not real; so condition 3 fails ($1 \ne 3$) and condition 4 fails (the fixed field is $\Q(\sqrt[3]{2})$).

Finiteness is needed for condition 3 to make sense; for infinite algebraic extensions, conditions 1 and 4 remain equivalent.

Going deeper

For a student

In practice one shows that an extension is Galois by exhibiting it as a splitting field of a separable polynomial, condition 2. One uses condition 3 or 4 to compute with the Galois group, for example to show that $\abs{\Gal(L/K)} = [L:K]$. Condition 5 is how Galois extensions arise from group actions. These equivalences are the input for the fundamental theorem of Galois theory.