What is a PID? Give an example of a UFD that is not a PID. Is \(k[x]\) a PID when \(k\) is a field?
Solution
A principal ideal domain is an integral domain in which every ideal is generated by one element.
If \(k\) is a field, then \(k[x]\) is a Euclidean domain with Euclidean function \[ \deg:k[x]\setminus\{0\}\to\ZZ_{\ge0}. \] Indeed, polynomial division gives \[ f=qg+r, \qquad r=0\text{ or }\deg r<\deg g. \] Every Euclidean domain is a PID, so \(k[x]\) is a PID.
A standard UFD that is not a PID is \[ k[x,y]. \] Since \(k[x]\) is a UFD, Gauss’s lemma implies that \[ k[x,y]=k[x][y] \] is again a UFD.
However, the ideal \[ (x,y) \] is not principal. If \((x,y)=(f)\), then \(f\) divides both \(x\) and \(y\). Since \(x\) and \(y\) are relatively prime irreducibles, their only common divisors are units. Thus \(f\) would be a unit, forcing \((x,y)=k[x,y]\), contradiction because \(1\notin(x,y)\).
Therefore \(k[x,y]\) is a UFD but not a PID.