If \(R\) is a UFD, is \(R[x]\) again a UFD?
Solution
Yes. This is Gauss’s lemma.
Let \(K=\operatorname{Frac}(R)\). For a nonzero polynomial \(f\in R[x]\), write \[ f=c(f)f_0, \] where \(c(f)\in R\) is the content (a gcd of the coefficients, defined up to a unit) and \(f_0\) is primitive.
Gauss’s lemma says:
- the product of primitive polynomials is primitive;
- if a primitive polynomial in \(R[x]\) factors in \(K[x]\), then it factors in \(R[x]\) after multiplying the factors by units of \(K\).
Since \(K[x]\) is a PID and therefore a UFD, every primitive polynomial factors uniquely into irreducibles in \(K[x]\). By Gauss’s lemma, these factors may be chosen primitive in \(R[x]\), and irreducibility over \(K\) agrees with irreducibility for primitive polynomials over \(R\).
The content \(c(f)\) factors uniquely in the UFD \(R\). Combining the factorization of the content with the primitive factorization gives unique factorization in \(R[x]\).
Therefore \[ R\text{ UFD}\quad\Longrightarrow\quad R[x]\text{ UFD}. \] By induction, the same holds for \(R[x_1,\ldots,x_n]\).