Let $R$ be an integral domain and $F$ be its field of fraction. We know that if $R$ is a UFD then for $f(X), g(X) \in R[X],$ $f(X)g(X)$ is primitive iff $f(X)$ and $g(X)$ are primitive. BTW by $f(X)$ primitive we mean $C(f)$=$\text{gcd}$ of all non zero coefficiens of $f(X)$ in $R$ is $1.$
Now my question is the following: Does it hold if we are given that $R$ is integrally closed.
Actually I want to show that if a monic polynomial $f(X) \in R[X]$ is irreducible in $F[X]$ then it is also irreducible in $R[X],$ where $R$ is integrally closed. Can I show that even if the above doesn't hold. I need some help. Thanks.
Best Answer
No, GL = Gauss's Lemma is not true for any integrally closed domain because it implies that atoms (irreducibles) are prime. Indeed, if an irreducible $\,p\,$ is not prime then there exists $a,b\,$ such that$\,p\mid ab,\, p\nmid a,p\nmid b\,$ so $\,f= px+a,\,g = px+b\,$ are primitive but $\,p\mid fg\,$ so $fg$ is not, so GL fails. So a counterexample lies in any integrally closed domain having a non-prime irreducible (e.g. any non-UFD quadratic number ring).
To learn more anout generalizations of Gauss's Lemma I highly recommend the following extensive survey GCD Domains, Gauss’ Lemma, and Contents of Polynomials by D. D Anderson (free link).
In particular see the results around superprimitive polynomials (dating back to Hwa Tsang Tang's thesis under Kaplansky in the early 70's). Tang showed that $R$ being integrally closed is equivalent to this: $ $ if $\,f\in R[x],\,g\in F[x]\,$ then $\,fg\in R[x]\iff fg_i\in R[x]\,$ for all coefs $\,g_i$ of $g.\,$ This naturally leads to the notion of superprimitive polynomials, and the study of so-called PSP domains where primitive $\Rightarrow$ superprimitive and related notions, all discussed at length in said survey.