Question about the numbers of distinct roots of a polynomial

Question

Let $K/F$ be a field extension and $u \in K$ algebric over $F$ and $\beta \in F(u) $. Let $f=Irr(u , F)$, the irreducible monic polynomial of $u$ over $F$ , $g=Irr(\beta , F) $, $h=Irr(u, F(\beta)) $ and $L$ be a splitting field of $fg$ over $F$

If $n_f$, $n_g$ and $n_h$ are the number of distinct roots of $f$, $g$ and $h$ in $L$, respectively, how do I show that $n_f =n_g n_h$?

Answer 1

If $f$ is separable then $F(u)/F$ is a separable extension. $n_f=F(u):F$ , $n_g=F(\beta):F$, $n_h=F(u):F(\beta)$ Hence we have $n_f=n_g.n_h$. Suppose $f$ is not separable. Then $f(X)=P(X^{p^m})$ where $P(X)$ is separable. Then $n_f=[F(u):F]_{sep}$, $n_g=[F(\beta):F]_{sep}$, $n_h=[F(u):F(\beta)]_{sep}$ and hence we have $n_f=n_g.n_h$