When talking about field extensions of degree two I understand the automorphisms in the Galois group intuitively as analogous to complex conjugation. I lose my understanding when going to field extensions generated by cubics.
Suppose we have roots $a_1, a_2, a_3$ so that $K=F(a_1, a_2, a_3)$. It's possible for $G(K/F)$ to have order 3 even though there are six possible permutations of these roots. So where did the other 3 go? I see two possibilities:
- Some permutations of the roots are not automorphisms. This seems unlikely because I think $F(x,y)\cong F(y,x)$ always, correct?
- Some permutations are equivalent. I don't understand how this could be.
Is there a third option I didn't think of? Am I missing the point entirely?