Corollary of Gauss's Lemma (polynomials) I am trying to prove the following result. I have outlined my attempt at a proof but I get stuck.
Any help would be welcome!
Theorem:
Let $R$ be a UFD and let $K$ be its field of fractions.
Suppose that $f \in R[X]$ is a monic polynomial.
If $f=gh$ where $g,h \in K[X]$ and $g$ is monic, then $g \in R[X].$
Proof Attempt:
Clearly we have $gh=(\lambda \cdot g_0)(\mu\cdot h_0)$ for some $\lambda, \mu \in K$ and $g_0, h_0 \in R[X]$ primitive.
Write $\lambda=a/b$ and $\mu=c/d$ for some $a,b,c,d \in R$.
Clearing denominators yields $(bd) \cdot f = (ac) \cdot g_0h_0$ where both sides belong to $R[X]$.
By Gauss's lemma $g_0h_0$ is primitive and so taking contents yields $\lambda\mu=1$.
This proves that $f=g_0h_0$ is a factorization in $R[X]$ but not necessarily that $g \in R[X]$.
I can't seem to get any further than this - any help greatly appreciated?


why does $g$ and $h$ being monic imply that $k$ and $t$ are in $R$? 

Because $kg$ and $th$ are primitive. In particular, they belong to $R[x]$. Since the highest coefficient of $kg$ is $k$, and the highest coefficient of $th$ is $h$, both $t$ and $h$ belong to $R$.

why does $k$ and $t$ being invertible in $R$ imply that $g$ and $h$ are in $R[x]$?

If the proof is still not clear to you, feel free to ask more questions.
Similarly, because $kg\in R[x]$, then $g=k^{-1}kg\in R[x]$ too. The same for $h$.
