Let \(R\) be a UFD. Let \(S\) be any multiplicative system of nonzero elements in \(R\), and let \(S^{-1}R\) be the localization of \(R\) at \(S\).
Let \(x\) be an irreducible element of \(R\). Show that the element \(\frac{x}{1}\) is either irreducible in \(S^{-1}R\) or a unit. Conversely, show that every irreducible in \(S^{-1}R\) is an associate of \(\frac{x}{1}\) for some \(x\) which is irreducible in \(R\).
Show that (i) every element of \(S^{-1}R\) is a finite product of irreducibles; and (ii) every irreducible element of \(S^{-1}R\) is prime. (It is well-known that these conditions imply that \(S^{-1}R\) is a UFD, so you have proved that \(S^{-1}R\) is also a UFD.)
Solution
1. Part (a): Irreducibles in \(S^{-1}R\):
1.1. Let \(x \in R\) be irreducible.In the UFD \(R\), every irreducible element is prime, so \((x)\) is a prime ideal.
prime and irreducible elements coincide in UFDs.
1.2. Case 1: \((x) \cap S \neq \emptyset\). There exists \(s \in S\) such that \(s = rx\) for some \(r \in R\).Then \(\frac{x}{1} \cdot \frac{r}{s} = \frac{rx}{s} = \frac{s}{s} = 1\), so \(\frac{x}{1}\) is a unit in \(S^{-1}R\).
definition of units in localization.
1.3. Case 2: \((x) \cap S = \emptyset\). Suppose \(\frac{x}{1} = \frac{a}{s_1} \frac{b}{s_2}\) for some \(\frac{a}{s_1}, \frac{b}{s_2} \in S^{-1}R\).Then \(s_1 s_2 x = ab\) in \(R\) (since \(R\) is an integral domain).Because \(x\) is prime in \(R\), \(x \mid ab \implies x \mid a\) or \(x \mid b\).Without loss of generality, assume \(x \mid a\), so \(a = x a'\) for some \(a' \in R\).Substituting gives \(s_1 s_2 x = x a' b \implies s_1 s_2 = a' b\).Thus \(b\) divides \(s_1 s_2 \in S\).Setting \(t = s_1 s_2 \in S\), we have \(\frac{b}{s_2} \cdot \frac{a'}{s_1} = \frac{a' b}{s_1 s_2} = 1\), so \(\frac{b}{s_2}\) is a unit in \(S^{-1}R\).Therefore \(\frac{x}{1}\) is irreducible in \(S^{-1}R\).
definition of irreducible element.
1.4. Converse: Let \(\frac{y}{s} \in S^{-1}R\) be irreducible.Since \(\frac{1}{s}\) is a unit, \(\frac{y}{s}\) is associate to \(\frac{y}{1}\).In the UFD \(R\), factor \(y = u p_1^{e_1} \cdots p_k^{e_k}\) into irreducibles.In \(S^{-1}R\), \(\frac{y}{1} = \frac{u}{1} \left(\frac{p_1}{1}\right)^{e_1} \cdots \left(\frac{p_k}{1}\right)^{e_k}\).Since \(\frac{y}{1}\) is irreducible (hence a non-unit), at least one \(p_i\) is a non-unit in \(S^{-1}R\).Because \(\frac{y}{1}\) is irreducible, there can be exactly one such non-unit factor and its exponent must be 1. All other factors are units in \(S^{-1}R\).Thus \(\frac{y}{s}\) is an associate of \(\frac{p_i}{1}\) for some irreducible \(p_i \in R\).
factorization in \(R\) maps to factorization in \(S^{-1}R\).
2. Part (b): \(S^{-1}R\) is a UFD:
2.1. (i) Existence of factorization into irreducibles: Let \(\frac{a}{s} \in S^{-1}R\) be a non-zero non-unit.Factor \(a = u p_1 \cdots p_m\) into irreducibles in \(R\).Then \(\frac{a}{s} = \frac{u}{s} \frac{p_1}{1} \cdots \frac{p_m}{1}\).By Part (a), each \(\frac{p_j}{1}\) is either a unit or an irreducible in \(S^{-1}R\).Absorbing all unit factors into \(\frac{u}{s}\), \(\frac{a}{s}\) is expressed as a unit times a product of irreducibles in \(S^{-1}R\).
factoring the numerator in \(R\).
2.2. (ii) Every irreducible in \(S^{-1}R\) is prime: Let \(q\) be irreducible in \(S^{-1}R\).By Part (a), \(q\) is associate to \(\frac{x}{1}\) for some irreducible \(x\in R\) such that \(\frac{x}{1}\) is a non-unit.Hence \((x)\cap S=\emptyset\).
It is enough to prove that \(\frac{x}{1}\) is prime, because associates of prime elements are prime. Suppose \[ \frac{x}{1} \mid \frac{a}{s_1}\frac{b}{s_2}. \] Then there exists \(\frac{c}{s_3}\in S^{-1}R\) such that \[ \frac{ab}{s_1s_2}=\frac{xc}{s_3}. \] Since \(R\) is a domain and every element of \(S\) is nonzero, equality in the localization gives \[ s_3ab=s_1s_2cx \] in \(R\). Since \(x\) is irreducible in the UFD \(R\), it is prime. Thus \[ x\mid s_3ab. \] Because \((x)\cap S=\emptyset\), we have \(x\nmid s_3\). Therefore \(x\mid a\) or \(x\mid b\). In the first case \(\frac{x}{1}\mid\frac{a}{s_1}\), and in the second \(\frac{x}{1}\mid\frac{b}{s_2}\). Hence \(\frac{x}{1}\) is prime, and therefore so is its associate \(q\).
This is the prime-element criterion, together with the associate classification from Part (a).
2.3. Since every non-zero non-unit factors into irreducibles and every irreducible is prime, \(S^{-1}R\) is a UFD.
standard criterion for unique factorization domains.
3. Conclusion: \(S^{-1}R\) satisfies all unique factorization domain axioms.Q.E.D.
1 and 2.