Divisibility of polynomials in a subfield of a field. 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.
 A: 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).
A: Let $\varphi:K[x]\to L[x]$ be the inclusion. Denote by $\langle f\rangle_L$ the ideal generated by $f$ in $L[x]$ and by $\langle f\rangle_K$ the ideal generated by $f$ in $K[x]$. I claim that $\varphi^{-1}(\langle f\rangle_L)=\langle f\rangle_K$. This will prove your statement.
Indeed, $\varphi^{-1}(\langle f\rangle_L)$ is an ideal of $K[x]$ and $K[x]$ is a principal ideal domain, therefore $\varphi^{-1}(\langle f\rangle_L)$ is generated by some polynomial $f'\in K[x]$. Since $f\in\varphi^{-1}(\langle f\rangle_L)=\langle f'\rangle_K$, this means that $f=u\cdot f'$ for $u\in K[x]$. Conversely, $\varphi(f')\in\varphi(\varphi^{-1}(\langle f\rangle_L))\subseteq\langle f\rangle_L$, so (remember that $\varphi$ is just an inclusion) basically $f'\in\langle f\rangle_L$, i.e. $f'=v\cdot f$ for some $v\in L[x]$. 
Sticking the two equations together, this means $f=uvf$ and canceling $f$ implies $u=v^{-1}$, so $u,v\in K$. This means that $\varphi^{-1}(\langle f\rangle_L)=\langle f\rangle_K$ as claimed.
Finally, $g\in\langle f\rangle_L$ is just the same as saying $f\mid g$ in $L[x]$ and equivalently over $K$.
A: Let $f\mid g$ in $L[x]$. Polynomial long division yields a polynomial $q\in L[x]$ with $f = gq$. A closer look at the polynomial long division algorithm shows that the coefficients of $q$ are computed by repeatedly applying field operations to the coefficients of $f$ and $g$. So if the coefficients of $f$ and $g$ are in $K$, then the computed coefficients of $q$ must be in $K$, too.
This shows that $q \in K[x]$ and therefore $f\mid g$ in $K[x]$.
A: Write $g=fq+r$, with $q,r \in K[x]$ and $r=0$ or $\deg(r)<\deg(f)$. Write $g=fh$, with $h \in L[x]$.
Then $r = f(h-q)$. If $r\ne0$, then $h-q\ne0$, and so $\deg(f(h-q))\ge \deg(f) > \deg(r)$. Thus, $r=0$.
The only detail is that the degree of a polynomial in $K[x]$ remains the same when it is considered as a polynomial in $L[x]$.
