Suppose $f$ is an irreducible polynomial in $\mathbb{F}_q[x]$ of degree $d$, where $\mathbb{F}_q$ denote the finite field with $q$ elements. We can assume that $q$ is odd. Now, let $\eta$ be a root of $f$ in $\mathbb{F}_{q^d}$. Now, $\eta^2 \in \mathbb{F}_{q^d}$. We further assume that minimal polynomial of $\eta^2$, say $g$ is an irreducible polynomial of degree $d$. My question is that can we express $g$ solely in terms of $f$? It is clear that $g(x^2)$ has $f$ as one of its irreducible factors.
Thanks for any kind of help.
Posting something to get this started (getting late here).
A recurring trick is that the polynomial $ f(x)f(-x) $ has only even degree terms, and thus there exists a polynomial $F[x]\in\Bbb{F}_q[x]$ such that $$F(x^2)=f(x)f(-x).$$ Therefore $F(\eta^2)=0$.
Furthermore, $\deg F=\deg f$, so if you know that $\eta$ and $\eta^2$ generate the same extension you are done – we know that $g(x)=\pm F(x)$. Observe that we haven't used the fact that the fields are finite. At least not yet.