Corollary of Gauss's Lemma (polynomials)

Question

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 $f_0, g_0 \in R[X]$ primitive.

Write $\lambda=a/b$ and $\mu=c/d$ for some $a,b,c,d \in R$.

Clearing denominators yields $(bd)\cdot f = (ac)\cdot g_0f_0$ where both sides belong to $R[X]$.

By Gauss's lemma $g_0f_0$ 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?

Answer 1

The leading term of $f$ is the product of the leading terms of $g_0$ and $h_0$, hence these are units. Hence the factor $\lambda$ by which $g_0$ differs from $g$ is a unit.

Answer 2

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 $h$.