Let K/F be a finite extension, given a polynomial in K[x] find another so that their product is in F[x] Let $K$ be a finite extension of a field $F$, and let $f(x)$ be in $K[x]$. Prove that there is a nonzero polynomial $g(x)$ in $K[x]$ such that $f(x)g(x)$ is in $F[x]$. 
Should I do this by induction on the degree of $f(x)$? 
Obviously if $n=0$, then $g(x)=1/f(x)$
Let $f(x) = a_nx^n+...a_1x+a_0$       then I know that there exists a h(x) so that $(f(x)-a_nx^n)h(x)$      is in $F[x]$. I want now to find a $g(x)=h(x)+i(x)$ so that $f(x)g(x)$ is in $F[x]$. Thus I need to find an $i(x)$ so that $a_nx^nh(x)+i(x)f(x)$ is in $F[x]$. I feel like this is wrong because I have no control over the degrees of $h(x)$.
Any suggestions?  
 A: It becomes easy if you know about integrality over rings. (Here and in the following, "ring" always means "commutative ring with $1$".)
Since $K$ is a finite extension of $F$, we see that $k\in K$ is integral over $F$ for every $k\in K$. Thus, $k\in K\left[x\right]$ is integral over $F\left[x\right]$ for every $k\in K$. Hence, $kx^i\in K\left[x\right]$ is integral over $F\left[x\right]$ for every $k\in K$ and $i\in\mathbb N$ (since $k$ and $x^i$ are both integral over $F\left[x\right]$, and the product of two integral elements is integral). Hence, $f\left(x\right) \in K\left[x\right]$ is integral over $F\left[x\right]$ (since $f\left(x\right)$ is a sum of elements of the form $kx^i$ for $k\in K$ and $i\in\mathbb N$, and since the sum of integral elements is integral). Now, all we need to prove is the following fact:
(1) If $ A\subseteq B$ is a ring extension, and $ u\in B$ is integral over $ A$, then there exists a nonzero $ v\in B$ such that $ uv\in A$.
Proof of (1). Let $ n$ be the smallest positive integer such that there exists a monic polynomial $ P\in A\left[Y\right]$ of degree $ n$ satisfying $ P\left(u\right)  =  0$. (Such an $n$ exists since $u$ is integral over $A$.) Write the polynomial $P$ in the form $ P\left(Y\right)  =  \sum\limits_{i  =  0}^{n  -  1}a_iY^i  +  Y^n$ with all $a_i$ lying in $A$. Then, set $ v  =  \sum\limits_{i  =  1}^{n  -  1}a_iu^{i  -  1}  +  u^{n  -  1}$. Then, $uv = \sum\limits_{i  =  1}^{n  -  1}a_iu^{i}  +  u^{n} = \underbrace{\sum\limits_{i  =  0}^{n  -  1}a_iu^{i}  +  u^{n}}_{ = P\left(u\right) =0} - a_0 = - a_0 \in A$. Also, $ v\neq 0$ follows from the minimality of $ n$. Thus, (1) is proven.
[I have copied some of this from one of my AoPS posts.]
A: Assume that the extension is Galois. By assumption, there are only finitely many automorphisms of $K$ that fix $F$. The polynomial
$$
\omega(x) = \prod_{\sigma \in \mathrm{Aut}(K/F)} \sigma(f(x))
$$
is a polynomial such that $f(x) \, | \, \omega(x)$ and $\omega$ is fixed by every automorphism of $K/F$, because $\mathrm{Aut}(K/F)$ is a group, so that when one tries to apply an automorphism on $\omega(x)$, the conjugate factors get permuted... hence all the coefficients of $\omega$ lie in the fixed field $F$. You can choose $g(x) = \omega(x) / f(x)$.
Hope that helps,
A: A proof using field theory and assuming only that $K/F$ is an algebraic extension . 
Let $K^a$ be an algebraic closure of $K$ , $f(x)\in K[x]$ splits in $K^a[x]$ as $f(x)=(x-a_1)...(x-a_n)$ 
where $a_1,..,a_n \in K^a$ ($a_i$'s are not necessarily distinct) . Each $a_i$ is 
algebraic over $K$ and $K$ is algebraic over $F$ , so each $a_i$ is algebraic over $F$ ; let $g_{a_i}(x)=Irr(a_i , F)$ 
be the irreducible polynomial of $a_i$ over $F$ . Then $h(x)=\prod_{i=1}^n g_{a_i}(x) \in F[x]$ is a non-zero 
polynomial . Moreover , $x-a_i|g_{a_i}(x)$ in $K^a[x]$ , so $f(x)|h(x)$ in $K^a[x]$ . Now $f(x),h(x) \in K[x]$ ; 
by division algorithm , $\exists q(x) , r(x)\in K[x]$ such that $h(x)=f(x)q(x)+r(x)$ where either 
$r(x)=0$ or $\deg r(x)< \deg f(x)$ . But then , as $f(x),h(x),q(x),r(x) \in K^a[x]$ and $f(x)|h(x)$ in 
$K^a[x]$ , so $f(x)|r(x)$ in $K^a[x]$ , then $\deg r(x) <\deg f(x)$ is not possible , hence $r(x)=0$ . Hence 
$h(x)=f(x)q(x)$ , where $h(x) \in F[x]$ is a non-zero polynomial and $f(x) , q(x) \in K[x]$ , 
which is what we wanted to prove .
