prove that if $p(x)\in R[x]$ is reducible over $F[x]$ then $p(x)$ is reducible over $R[x]$.

Question

let $R$ be a unique factorization domain and let $F$ be its field of fractions.
Prove that if $p(x)\in R[x]$ is reducible over $F[x]$ then $p(x)$ is reducible over $R[x]$.

Answer 1

Note that any polynomial in $R[x]$ is uniquely (up to associates) the product of an element of $R$ and a primitive polynomial in $R[x]$ (that is, one whose coefficients have gcd 1). If $p$ was not primitive it would be reducible already, so assume that it is.

Suppose $p(x) = G(x) H(x)$ is a non-trivial factorization in $F[x].$ By clearing denominators, and then extracting common factors, we get $G(x) = \dfrac{g_1}{g_2} g(x)$ for some coprime $g_1,g_2\in R$ and a primitive $g(x)\in R[x].$ Similarly we write $H(x) = \dfrac{h_1}{h_2}h(x).$ Write $\dfrac{g_1h_1}{g_2h_2} = \dfrac{\alpha}{\beta}$ with $\alpha,\beta\in R$ coprime. Then we have $$\beta p(x) = \alpha g(x) h(x).$$

By Gauss' lemma, the product of primitive polynomials is primitive, so $g(x)h(x)$ is primitive here. Then by the uniqueness of such forms, we have that $\alpha$ and $\beta$ are associates, so $\dfrac{\alpha}{\beta}$ is a unit in $R.$ Now the equation $p(x) = \dfrac{\alpha}{\beta} g(x) h(x)$ gives a non-trivial factorization of $p$ in $R[x].$

Answer 2

Hint $ $ If $\, f = g/c\ h/d,\,\ g,h\in R[x],\,\ c,d \in R\,$ then $\, cd\,f = gh.\,$ If $\,cd\,$ is a nonunit then some prime $\,p\mid cd.\,$ But $\,\color{#c00}{p\ \, \text{remains prime in}\,R[x]}\,$ by $\,R[x]/p \cong (R/p)[x]\,$ a domain, so $\,p\mid fg\,\Rightarrow\,p\mid f\,$ or $\,p\mid g.\,$ Hence$\,p\,$ may be canceled from $\, cd\,f = gh\,$. In this way, by induction, we can cancel all of the prime factors of $\,cd,\ $ leaving $\ u f = g'h',\, $ so $\, f = u^{-1}g'h'\,$ for unit $\,u\in R,\,$ $\,g',h'\in R[x].\ \ $ QED