I am taking undergrad level field theory and I am doing a problem:
Suppose $K[x]$ is a polynomial ring over the field $K$ and $F$ is a subfield of $K$. If $F$ is a perfect field, and $f \in F[x]$ has no repeated irreducible factors in $F[x]$, prove that $f(x)$ has no repeated irreducible factors in $K[x]$.
Here, the definition of a perfect field is either
(i) Char $F = 0$
or
(ii) Char $F = p$, $p$ is prime, and every element of $F$ is a $p$-th power.
My attempt is to use a theorem: $F$ is perfect iff every irreducible polynomial in $F[x]$ is separable. (Here a polynomial in $F[x]$ is separable over $F$ iff its roots have no multiplicity.) But I am not sure how to apply it.
Can someone give me some hint on this problem? Thanks!
Hint. the key point is to notice that if $f\in F[X]$, the set of roots in an algebraic closure $F_{alg}$ of $F$ is the same that the set of roots in an algebraic closure $K_{alg}$ of $K$.
Now if $f$ has no repeated irreducible factors in $F[X]$, then $f$ is separable (why?). So, what can you say about roots in $K_{alg}$ of an irreducible factor of $f$ in $K[X]$ ? Can you conclude ?