Galois group acts freely?

Question

If $L/K$ is a Galois extension of fields such that $L$ is the splitting field of an irreducible polynomial $f$ with coefficients in $K$, then does $Gal(L/K)$ act freely on the roots of $f$?.

Answer 1

Not necessarily. For example, if $K=\bf Q$ and $f=x^3-2$, then the Galois group does not act freely on roots of $f$: one of the automorphisms of $L$ is complex conjugation, which fixes $\sqrt[3] 2$. Any splitting field of an irreducible polynomial which has at least one real root and at least one nonreal root will have the same property.