Let $f(x)$ be irreducible so that $\operatorname{Gal}(f(x)) \simeq Z_{p}$ for p an odd prime. Show that $f(x)$ has degree p and all real zeros.

Question

I know that if $\operatorname{Gal}(f(x)) \simeq Z_{p}$, then $|\operatorname{Gal}(f(x))| = |Z_{p}| = p$. Does this imply that $deg(f(x)) = p$? How do I go about showing that $f(x)$ only has real roots? Thanks in advance.

Answer 1

If $f(x)$ is irreducible, then its degree must be $p$.

If $f$ is irreducible with some non-real zeros, and $L$ is its splitting field within $\Bbb C$ then complex conjugation induces an automorphism of $L$ with order $2$. Thus the Galois group of $L$ must have even order...