A positive-degree separable polynomial is irreducible exactly when its Galois group is transitive on the roots

For the forward direction, suppose $f$ is irreducible and let $\alpha ,\beta$ be roots. The rule sending $\alpha$ to $\beta$ gives an $F$-isomorphism $F(\alpha )\to F(\beta )$ because both have minimal polynomial associated to $f$; by [L1] it extends to an $F$-automorphism of $L$. Thus some Galois element sends any chosen root to any other, so the action is transitive. This includes degree one.

For the reverse direction, suppose the action is transitive and let $g$ be a monic irreducible factor of $f$ containing one root $\alpha$. For every $\sigma \in G_f$, the coefficients of $g$ are fixed, so $g(\sigma \alpha )=\sigma (g(\alpha ))=0$. Transitivity puts every root of $f$ among the roots of $g$; since $f$ is separable, $g$ has the full degree of $f$, so $f$ is a scalar multiple of $g$ and is irreducible. A root equal to zero causes no exception.