I am thinking about the following statement, and I wonder if this is true: Every cyclic field is Galois. (we are in characteristics $0$).
I have started with a cubic case and tried to make use of the following proposition:
Prop. The splitting field of an irreducible polynomial $f(x)=x^3-px+q\in\mathbb{Q}[x]$ is cyclic iff the discriminant $\Delta(f)=4p^3-27q^2$ is a rational square.
Every cubic polynomial has one real root. So if we have a cubic polynomial whose discriminant is a rational square, then the splitting field $K$ (i.e Galois field) is cyclic and the real root generates the whole field. So all the remaining roots must also be real and belong to that field right? (As $K$ contains all the roots to the cubic polynomial)
Does it make any sense?
Does the statement hold in general for extensions of higher degrees ?
Many thanks