Suppose $p(x) = a(x)b(x)$ where $p,a,b$ are monic polynomials over the fraction field of $R$. Prove that if $a(x) \notin R[x]$ then $R$ is not a UFD.

Question

Question from Dummit and Foote:

Let $R$ be an integral domain. Let $F$ be fraction field of $R$. Let $p(x)$ be a monic polynomial in $R[x]$. Suppose $p(x) = a(x)b(x)$ where a$(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x)$. Prove that if $a(x) \notin R[x]$ then $R$ is not a UFD.

Attempt: Suppose $R$ is a UFD. Then $R[x]$ is a UFD. $F[x]$ is a UFD because $F$ is a field.

By hypothesis, $p(x)$ is monic and reducible in $F[x]$, hence is reducible in $R[x]$. By clearing denominators, we can write $d_1 d_2 p(x) = a'(x) b'(x)$ where $a'(x)$ and $b'(x)$ are primitive polynomials in $R[x]$. I think I need to use the fact they are monic to proceed from here but I am stuck.

Answer 1

In accordance with the statement of Gauss' lemma in Dummit and Foote, If $R$ is UFD and $p(x)=a(x)b(x)$ in $F[x]$ then $p(x)=a'(x)b'(x)$ in $R[x]$ where $a'(x)=ra(x)$ and $b'(x)=sb(x)$ and $r,s\in F^\times$. We have the following :

• Since $p(x)$ is monic, $rs=1$
• Since $a(x)$ is given to be monic and $ra(x)\in R[x]\implies r\in R$. Similarly $s\in R$.

$\implies\;r,s\in R^\times$. Hence $a(x)=r^{-1}a'(x)\in R[x]$.