I am trying to solve the following problem in I. Martin Isaacs' Algebra: A graduate course, p.290:
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 $F$.
To prove this, one only needs to show one direction since $f(X)$ and $g(X)$ are interchangeable. To be honest, I have no idea where to begin. By definition, $E$ would have to be the smallest field containing all the roots of both $f(X)$ and simultaneously $E$ would have to be the smallest field containing all the roots of both $g(X)$. I cannot, however, see how I can relate repeated roots in $f(X)$ to repeated roots in $g(X)$ with this information.
Here is my first attempt:
Let $\alpha_1,...,\alpha_n$ and $\beta_1,...,\beta_m$ be the roots of $f(X)$ and $g(X)$ in the algebraic closure of $F[X]$. Using Isaacs' definition of a splitting field we see that $E=F(\alpha_1,...,\alpha_n)$ and $E=F(\beta_1,...,\beta_m)$, and so $F(\alpha_1,...,\alpha_n)=F(\beta_1,...,\beta_m)$. Assume $f(X)$ is separable over $F$. Then every irreducible component of $f(X)$ has distinct roots. This implies that the minimal polynomial of the roots of $f(X)$ are separable. Let $g_i(X)$ be an irreducible component of $g(X)$ and assume $g_i(X)$ has a multiple root, say $\beta_k$. What can we say about $\beta_k$? Well, since $\beta_k \in F(\alpha_1,...,\alpha_n)$ we know
$\beta_k=a_1\alpha_1+...+a_n\alpha_n$
for some $a_1,...,a_n \in F$.
I'm thinking that there must be some way to obtain a contradiction about the non separability of $g_i(X)$ from the fact that the minimal polynomials of all the $\alpha_j$'s are separable.