$f \in \mathbb Z[x]$ be irreducible and suppose $f(x)$ has two roots in $\mathbb C$ with product $1$ , then degree of $f$ is even ?

Question

Let $f \in \mathbb Z[x]$ be irreducible and suppose $f(x)$ has two roots in $\mathbb C$ with product $1$ . Then is it true that degree of $f$ is even ?

Answer 1

Another idea :

Put $d=$ degree of $f$; we suppose $d\geq 2$. Then the hypothesis say that $f(z)$ and $z^d f(1/z)$ have a commun root. As they are irreductibles, there exists $c\not=0$ in $\mathbb{Q}$ such that $z^df(1/z)=cf(z)$. Putting $z=1$ and using $f(1)\not =0$ we get $c=1$. Putting $z=-1$ and using $f(-1)\not =0$, we get $(-1)^d=1$.

Answer 2

Yes. The assumption is that there exists $z \in \mathbb C$ so that $f(z)=f(z^{-1}) = 0$, and $z \neq z^{-1}$.

It follows that, for any root $w$ of $f$, $w \notin \{0, 1\}$ and $w^{-1}$ is also a root of $f$. To prove this: let $K$ be the splitting field of $f$, and let $\phi: K \to K$ be an automorphism which carries $z$ to $w$.