an odd degree polynomial with cyclic Galois group has root all real I am thinking about following question:
Let $f\in\mathbb{Q}[x]$ be an odd degree polynomial with cyclic Galois group. Prove that all the roots of $f$ are real. 
I tried to prove it by contraction. 
Suppose $f$ has some complex roots. Let $K$ be its splitting field. Then $[K:\mathbb{Q}]=odd$. 
I tried to derive something from the cyclic Galois group, but I failed. Or maybe I am reviewing for the prelim so I am really terrified...
Any hints or explanations are really appreciated!!!
 A: Without the assumption that $ f $ is irreducible, the statement is false; for instance $ f(x) = x(x^2 + x + 1) $ is an odd degree polynomial with cyclic Galois group which has nonreal roots.
If we assume that $ f $ is irreducible then the statement is true: if $ n $ is the degree of $ f $ we know that the Galois group is a transitive subgroup of $ S_n $, and the only cyclic transitive subgroup of $ S_n $ is one generated by an $ n $-cycle, so has odd order $ n $. If $ f $ had nonreal roots, complex conjugation would give an automorphism of order $ 2 $ in the Galois group, which is impossible by Lagrange since the order of the Galois group is odd.
A: Starfall's answer, of course, settles the question. I can't resist adding the following alternative argument. Assuming $f(x)$ is irreducible.


*

*As an odd degree polynomial $f(x)$ has a real zero $\alpha$.

*The usual complex conjugation $\tau$ is an automorphism, possibly trivial, of the splitting field of $f(x)$.

*If $\beta$ is another zero of $f(x)$, there exists an automorphism $\sigma$ of the splitting field such that $\sigma(\alpha)=\beta$.

*Because the Galois group is abelian (don't need cyclicity!) $\tau\sigma=\sigma\tau$, and hence $$\tau(\beta)=\tau(\sigma(\alpha))=\sigma(\tau(\alpha))=\sigma(\alpha)=\beta.$$

