More clarification for problem 18.7 Isaacs' Algebra

Question

Consider the solved problem Problem 18.7 in I. Martin Isaacs' Algebra

Let $f(X),g(X) \in F[X]$ and suppose $E \supseteq F$ is the splitting field both for $f(X)$ and for $g(X)$ over $F$. Show that $f(X)$ is separable over $F$ if and only if $g(X)$ is separable over $F$

I can not see where we us that $E \supseteq F$ is the splitting field both for $f(X)$ and for $g(X)$ over $F$.

Any clarification please?

Thanks

Answer 1

In the solution you linked, $g_i(X)$ is an irreducible component of $g(X)$. Since $E$ is separable so if $g_i(X)$ has root $\alpha$ in $E$ then $g_i(X)$ will be minimal polynomial of $\alpha$ and hence $g_i(X)$ has no multiple root.

Since $E$ is splitting field of $g(X)$ so indeed, $\alpha\in E$, and we have the above result.