Let $F \subseteq E$ be a splitting field over $F$ both for $f \in F[x]$ and for $g \in F[x]$. I claim that $f$ is separable over $F$ iff $g$ is separable over $F$.
Suppose that $f$ is separable over $F$, that is each irreducible factor of $f$ in $F[x]$ has distinct roots. Also, $f(x)=a\prod_{i=1}^n(x-\alpha_i)$ and $g(x)=b\prod_{i=1}^m(x-\beta_i)$ in $E[x]$. I need to show that each irreducible factor of $g$ in $F[x]$ has distinct roots in order to $g$ is separable over $F$.
I need some help. How to prove this assertion ?
Since E is a splitting field of $f$ and $g$, assuming $f$ is separable, you have that $E|F$ is a Galois extension, thus in particular separable, so $g$ is separable.
You can prove that the following statemant holds:
Given a finite extension $E|F$ the following are equivalent: 1. $E|F$ is Galois 2. $E$ is a splitting field for a separable polynomial $0\ne f \in F[x]$
The interesting implication is 2 -> 1. To prove it you have to prove, or assume, these two:
If $E$ is a splitting field of a polynomial $f \ne 0$ in $F[x]$ and all roots are distinct, then $|Gal(E|F)| = [E:F]$
For a finite extension $E|F$ they are equivalent, being $G=Gal(E|F)$: