Is $h$ forced to be lie in $F[X]$?

Question

I'm stuck at a step proving something.

Let $E/F$ be a field extension and $f,g\in F[X]$ and $h\in E[X]$ such that $f=gh$.

If $f$ is irreducible in $F[X]$, is $h$ forced to lie in $F[X]$?

Answer 1

Hint $\ $ It is an immediate consequence of the uniqueness of the quotient (and remainder) in the Division Algorithm (which is the same in $\rm\:E[x]\:$ and $\rm\: F[x]).$ Since dividing $\rm\:f\:$ by $\rm\:g\:$ in $\rm\:E[x]\:$ leaves remainder $\,0$, by uniqueness, the remainder is $\,0\,$ in $\rm\:F[x],\:$ i.e. $\rm\:g\mid f\ $ in $\rm\:E[x]\,$ $\, \Rightarrow\, $ $\rm\, g\mid f\ $ in $\rm\:F[x]$.

This is but one of many examples of the power of uniqueness theorems for proving equalities.