Having spent some significant time finding UFDs (and PIDs) within the family of quadratic rings of integers, we now explore how the property of being a UFD fares under the polynomial ring construction. The answer here is a simple and happy one: if \(R\) is a UFD, then so is \(R[x]\text{!}\) In particular, we see that our friend \(\Z[x]\) is a UFD, despite its not being a PID.
As you will see below, not only is the answer a simple one, but the proof approach is essentially the first thing that might cross your mind. In more detail, let \(F=\Frac R\) be the field of fractions of \(R\text{.}\) Identifying \(R\) with its image under the localization map \(\lambda\colon R\rightarrow F\) (which is injective in this case), we think of \(R[x]\) as a subring of \(F[x]\text{.}\) The latter is a PID, and hence a UFD. As such, we might naively hope to proceed as follows:
given a polynomial \(f(x)\in R[x]\text{,}\) consider it as an element of \(F[x]\text{;}\)
since \(F[x]\) is a UFD, we can factor \(f\) into powers of irreducibles as \(f=\prod_{i=1}^r f_i^{n_i}\text{,}\) and this factorization is unique in some sense;
In fact yes, it does stand to reason that \(f\) factors similarly over \(R[x]\text{,}\) but the argument is more delicate than you might expect. There are two key issues that the sketch above willfully steamrolls over: (a) the irreducible elements \(f_i\) are elements of \(F[x]\text{,}\) but not necessarily \(R[x]\text{;}\) (b) even if we have \(f_i\in R[x]\) for all \(i\text{,}\) being irreducible in \(F[x]\) does not a priori imply irreducibility in \(R[x]\text{.}\) Indeed, that implication is patently false: \(f(x)=2x\) is irreducible in \(\Q[x]\text{,}\) but not irreducible in \(\Z[x]\text{,}\) since \(f(x)=2\cdot x\) is a factorization into two nonunits of \(\Z[x]\text{.}\)
Accordingly, we will need to proceed with some caution in our proof, keeping straight where polynomials live (in \(R[x]\) or \(F[x]\)), and for a polynomial \(f\in R[x]\) we need to distinguish between being reducible/irreducible in \(R[x]\) and being reducible/irreducible in \(F[x]\text{.}\) (As you will see from Gauss’s lemma, the two notions are very close to being equivalent, but not quite.)
Let \(R\) be a unique factorization domain. A nonconstant polynomial \(f(x)=\sum_{i=0}^na_ix^i\in R[x]\) is primitive if the the coefficients \(a_i\) are relatively prime (i.e., if \(1\) is a greatest common divisor of \(a_0,a_1,\dots, a_n\)).
Let \(R\) be a unique factorization domain, let \(F=\Frac R\) be its field of fractions, and identify \(R\) with its image in \(F\) under the localization map.
If \(f\) and \(g\) are primitive elements of \(R[x]\text{,}\) then \(fg\) is primitive.
Given a nonconstant polynomial \(f\in F[x]\text{,}\) there exists a primitive polynomial \(f_0\in R[x]\) and constant \(c\in F\) such that \(f=c f_0\text{.}\) Furthermore, this representation satisfies the following uniqueness property: if \(f=cf_0=dg_0\) for primitive polynomials \(f_0,g_0\in R[x]\) and constants \(c,d\in F\text{,}\) then \(g_0=uf_0\) for some unit \(u\in R^*\) and \(c=du\text{.}\)
We will call a representation \(f=cf_0\) where \(c\in F\) and \(f_0\in R[x]\) is primitive a primitive decomposition of \(f\text{.}\) Additionally, we will say that \(f\) has content \(c\) in this case, and associated primitive \(f_0\text{.}\) Note that \(c\) and \(f_0\) are well-defined only up to multiplication by a unit in \(R\text{.}\) By abuse of notation, we call \(c\) the content of \(f\text{,}\) and we call \(f_0\) the primitive polynomial associated to \(f\text{.}\)
Let \(f\) and \(g\) be nonconstant polynomials in \(F[x]\text{.}\) If \(f\) has content \(c\) and \(g\) has content \(d\text{,}\) then \(fg\) has content \(cd\text{.}\)
Let \(R\) be a unique factorization domain, let \(F=\Frac R\) be its field of fractions, and identify \(R\) with its image in \(F\) under the localization map.
Let \(f\) and \(g\) be nonconstant polynomials in \(F[x]\) with associated primitive polynomials \(f_0,g_0\in R[x]\text{.}\) We have \(f\mid g\) if and only if \(f_0\mid g_0\text{.}\)
If \(f\) is a nonconstant polynomial in \(R[x]\) satisfying \(f(x)=g(x)h(x)\) for nonconstant polynomials \(g,h\in F[x]\text{,}\) then \(f(x)=cg_0h_0\text{,}\) where \(g_0\) and \(h_0\) are the associated primitive polynomials of \(g\) and \(h\) and \(c\in R\text{.}\)
Let \(R\) be a unique factorization domain, let \(F=\Frac R\) be its field of fractions, and identify \(R\) with its image in \(F\) under the localization map. Let \(f(x)=a_nx^n+\dots +a_0\in R[x]\text{,}\) with \(a_n\ne 0\text{.}\) If \(f\) has a root \(q\) in \(F\) (i.e., \(f(q)=0\)), then writing \(q=a/b\) with \(a\) and \(b\) relatively prime, we have \(a\mid a_0\) in \(R\) and \(b\mid a_n\) in \(R\text{.}\)
Suppose \(q=a/b\) is a root of \(f\) in \(F[x]\text{,}\) where \(a\) and \(b\) are relatively prime. It follows that \(f(x)=(x-a/b)h(x)\text{,}\) for some \(h\in F[x]\text{,}\) or equivalently, \(f(x)=(bx-a)g(x)\text{,}\) where \(g(x)=(1/b)h(x)\text{.}\) Note that \((bx-a)\) is primitive, since \(a\) and \(b\) are relatively prime. By Gauss’s lemma, we have \(f(x)=c(bx-a)g_0(x)\text{,}\) where \(c\in R\) and \(g_0\) is an associated primitive polynomial of \(g\text{.}\) Comparing leading terms and constant terms of the the left and right sides of this equation, we see that
where \(d_m\) and \(d_0\) are the leading and constant terms of \(g_0\text{,}\) respectively. Since \(c\in R\text{,}\) we conclude that \(b\mid a_n\) and \(a\mid d_0\text{,}\) as claimed.
Let \(R\) be a unique factorization domain, let \(F=\Frac R\) be its field of fractions, and identify \(R\) with its image in \(F\) under the localization map.
A constant polynomial \(f\in R[x]\) is irreducible if and only if \(f(x)=p\text{,}\) where \(p\) is irreducible in \(R\text{.}\)
