# 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 $f_0, g_0 \in R[X]$ 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?

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.
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 $$R[x]$$?
Similarly, because $$kg\in R[x]$$, then $$g=k^{-1}kg\in R[x]$$ too. The same for $$h$$.