Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f$?.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.