A factorization over the fraction field descends to the UFD without changing factor degrees.Suggest a slogan
Let \(R\) be a unique factorization domain with field of fractions \(F\), and let \(p\in R[x]\). If \(p=gh\) with \(g,h\in F[x]\) and \(\deg g,\deg h\ge1\), then there are \(g',h'\in R[x]\) with \(\deg g'=\deg g\), \(\deg h'=\deg h\), and \(p=g'h'\). In particular, if \(p\) is reducible in \(F[x]\), then \(p\) is reducible in \(R[x]\).
If \(R\) is a unique factorization domain, then so is \(R[x]\); the proof uses Gauss’ lemma to reduce factorization in \(R[x]\) to factorization in \(R\) and in the principal ideal domain \(F[x]\). By induction, \(R[x_1,\ldots,x_n]\) is a unique factorization domain, and so is the polynomial ring over \(R\) in any set of variables, since each polynomial involves finitely many variables.
State Gauss' lemma.