Let $K$ be a field, and consider a monic irreducible polynomial $f(x) \in K[x]$. Denote $d = \deg(f)$, and let $g(x) = f(x^2)$. Furthermore, let $\alpha \in \Omega^g_K$ (the splitting field of $g$ over $K$). Show that if $g(\alpha)=0$, we have that $[K[\alpha] : K] \in \{d,2d\}$.
This is what I tried so far: Since $f$ is irreducible over $K$, it has no roots in $K$. Also, if $g(\alpha)=0$, then $f$ has a factor $(x - \alpha^2)$ because $f(\alpha^2) =0$. Now, over $K[\alpha]$, $f$ thus has a root $\alpha$.
Furthermore, I guess that the assumption that $f$ is irreducible, is to prevent the possibility that $[K[\alpha]:K] = 1$, which could happen in the case that $\alpha$ is already in $K$.
How can I continue from here? Thanks!
EDIT: So I have made some progress. $f(\alpha^2) = 0$ implies that $[K[\alpha]:K]$ is at most $2d$ (the degree of $f(x^2)$). This follows as $\alpha^{2d+1}$ can be written as a combination of lower powers of $\alpha$ with coefficients in $K$. It could also be that $K[\alpha] = d$ in the case that $f$ has a factor $(x-\alpha)$. In that case, we know that $\alpha$ is not in $K$ as $f$ is irreducible over $K$. How do I show that these are the only two possibilities though?
Think about the tower, $K\subseteq K(\alpha^2)\subseteq K(\alpha)$. Clearly $\bigr[K(\alpha):K(\alpha^2)\bigl]=1$ or $2$, thus $\bigr[K(\alpha):K\bigl]=\bigr[K(\alpha):K(\alpha^2)\bigl]\bigr[K(\alpha^2):K\bigl]=d$ or $2d$, as required.