Is minimal polynomial self-reciprocal?

Question

I have finite extension $F/ \Bbb Q$ . I don't know if it is important that degree of extension is four (4) , and I have integral element $w \in L$ with norm equal to one, and this means N$^F_{\Bbb Q} (w) = 1$. Then $w$ is a unit in the integral ring $O_F$ and they tell me $w$ is not root of unity. If this helps, also $N^F_K(w)=1$ for quadratic extension $K/ \Bbb Q$, $K \subset F$.

Does above mean minimal polynomial of $w$ is self-reciprocal? This means the coefficients of polynomial are $a_k = a_{n-k}$ , if $n =$ degree of polynomial : if $p(x)$ is minimal polynomial of $w$, then reciprocal of $p$ is polynomial $p^*(x)=x^np(1/x)$

I am required to show $1/w$ is conjugate of $w$, but I know that if $p(x)$ is minimal polynomial of $w$ in $\Bbb Q$ then sure complex conjugate $\overline w$ is also a root of $p(x)$, but if also $1/w$ is conjugate and root of $p(x)$ then $p(x)=p^*(x)$ and I can not prove of this.

I wrote minimal $p(x)=x^4+ax^3+bx^2+cx+1$, because free coefficient = norm of $w$. I can not prove $a=c$ and then $p(x)$ is self-reciprocal, but the question requires this. Thank you.

Answer 1

You really need the fact that $w$ has norm $1$ w.r.t. some quadratic subextension $K$: its conjugate in $F/K$ is then its reciprocal $1/w$. Next, use that the minimal polynomial of $w$ in $F/K$ divides its minimal polynomial $p$ in $F/\mathbb Q$, so that $1/w$ is a root of $p$.

Without the information that $w$ has norm $1$ w.r.t. $F/K$, this is not true. It would imply that every irreducible monic polynomial of degree $4$ over $\mathbb Q$ with constant term $1$ is self-reciprocal.