Let \(R\) be a commutative ring.
Say what it means for \(R\) to be a unique factorization domain (UFD).
Say what it means for \(R\) to be a principal ideal domain (PID)
Give an example of a UFD that is not a PID. Prove that it is not a PID.
Solution
A commutative ring \(R\) with \(1\neq0\) is a unique factorization domain if \(R\) is an integral domain, every nonzero nonunit \(r\in R\) is a finite product \(r=p_1\cdots p_k\) of irreducible elements, and whenever \(r=q_1\cdots q_m\) is another such product, \(m=k\) and there is \(\sigma\in S_k\) with \(q_i\) associate to \(p_{\sigma(i)}\) for every \(i\).
A commutative ring \(R\) with \(1\neq0\) is a principal ideal domain if \(R\) is an integral domain and every ideal of \(R\) has the form \(aR\) for some \(a\in R\).
The ring \(\ZZ[x]\) is a UFD that is not a PID.
1. \(\ZZ[x]\) is a UFD.
\(\ZZ\) is a UFD, and by Gauss’s lemma a polynomial ring over a UFD is a UFD.
2. The ideal \(I=(2,x)=\{h\in\ZZ[x]:h(0)\in2\ZZ\}\) is not principal.
Every \(2f+xg\) has constant term \(2f(0)\), and conversely \(h=h(0)+x\,\frac{h-h(0)}{x}\) with \(h(0)\) even lies in \(I\). Suppose \(I=(d)\). Since \(2\in(d)\), \(d\) divides \(2\), so \(\deg d=0\) and \(d=c\in\{\pm1,\pm2\}\). If \(c=\pm1\), then \(1\in I\), but \(1\) has odd constant term. If \(c=\pm2\), then \(x\in(2)\), but the coefficient \(1\) of \(x\) is odd.
3. Q.E.D.
By steps 1 and 2, \(\ZZ[x]\) is a UFD with a non-principal ideal.