TITLE: Divisibility of polynomials in a subfield of a field.
QUESTION [6 upvotes]: I am trying to prove the following assertion:

Let $K\subset L$ be fields, let $f,g\in K[x]$ be such that $f\mid g $ in $L[x]$, then $f\mid g$ in $K[x]$.

We clearly have that $fh=g$ for some $h\in L[x]$. We want to show that $h\in K[x]$. Suppose that $h\notin K[x]$. I am not sure how to argue towards a contradiction from here.

REPLY [4 votes]: Let $A$ and $B$ be domains with $A\subseteq B$. We say that the ring extension $B/A$ is good if the following implication holds:
$$\style{font-family:inherit;}{\text{For every}}\ f\in A\setminus\{0\}\ \style{font-family:inherit;}{\text{and}}\ h\in B,\ \style{font-family:inherit;}{\text{if}}\ fh\in A\ \style{font-family:inherit;}{\text{then}}\ h\in A\,.$$
We claim that if $B/A$ is good then $B[x]/A[x]$ is good as well. In fact, let $f\in A[x]\setminus\{0\}$ and $h\in B[x]\setminus\{0\}$ of degrees $m$ and $n$ respectively be such that $fh\in A[x]$. If $f=ax^m+f_1$ and $h=bx^n+h_1$, with $a\in A\setminus\{0\}$ and $b\in B\setminus\{0\}$, then the leading coefficient of $fh$, namely $ab$, lies in $A$ by hypothesis, so by goodness we obtain $b\in A$. This implies $fh_1=fh-bx^nf\in A[x]$, and now we can repeat the argument with $h_1$ instead of $h$. Continuing in this way we prove that all the coefficients of $h$ lie in $A$ as desired.
As a corollary, we obtain (as mentioned by user26857) that if $L/K$ is a field extension then the ring extension $L[x_1,\dots,x_n]/K[x_1,\dots,x_n]$ is good (because it is easily verified that $L/K$ is good).