Gauss lemma over a UFD

Let $\pi \in R$ be irreducible. By [L1], any factorization of $ab$ into irreducibles is obtained by concatenating factorizations of $a$ and $b$, so if $\pi \mid ab$ then $\pi$ is associate to one factor from $a$ or one factor from $b$. Therefore every irreducible element of $R$ is prime.

Let $f={\sum }_i{a}_i{x}^i$ and $g={\sum }_j{b}_j{x}^j$ be primitive. Suppose some irreducible $\pi \in R$ divides every coefficient of $fg$. Choose the least indices $i$ and $j$ with $\pi \nmid a_i$ and $\pi \nmid b_j$. Then the coefficient of $x^{i+j}$ in $fg$ is $a_i{b}_j+{\sum }_{u+v=i+j,(u,v)\mathrm{\ne }(i,j)}{a}_u{b}_v\mathrm{.}$ Every summand in the sum is divisible by $\pi$, while $a_i{b}_j$ is not by step 1.1. This contradicts the choice of $\pi$. So no irreducible divides all coefficients of $fg$, and hence $fg$ is primitive.

If $p$ is reducible in $R[x]$, then it is reducible in $K[x]$ because $R[x]\subseteq K[x]$. Conversely, suppose $p=qr$ in $K[x]$ with both factors of positive degree. Choose nonzero $a,b\in R$ with $aq,br\in R[x]$, and factor out the greatest common divisor of the coefficients to write $aq=c{q}_0,br=d{r}_0,$ where $q_0,r_0\in R[x]$ are primitive. Then $abp=cd{q}_0{r}_0\mathrm{.}$ By step 2.1 the product $q_0{r}_0$ is primitive, so the right-hand side has content associate to $cd$, while the left-hand side has content associate to $ab$ because $p$ is primitive. Thus $ab$ and $cd$ are associates. Absorbing the unit into one factor yields $p=q_1{r}_1$ with $q_1,r_1\in R[x]$ of positive degree, contradicting irreducibility in $R[x]$.

Therefore a primitive positive-degree polynomial is irreducible in $R[x]$ exactly when it is irreducible in $K[x]$.