Difference between revisions of "UFD not implies PID"
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Latest revision as of 16:35, 12 May 2008
More general results
There are many ways of seeing this -- one approach is to observe that a polynomial ring over a UFD continues to remain a UFD; however, the polynomial ring over a PID is not necessarily a PID (in fact, a polynomial ring over a ring is a PID iff the base ring is a field). This gives examples of UFDs which are not PIDs:
- Polynomial ring in two or more variables over a field
- Polynomial ring over the integers
It is true that under certain circumstances, being a UFD is the same as being a PID. An example of such circumstances is a Dedekind domain. In particular, all rings of integers in number fields are Dedekind domains, and hence any such ring is a PID iff it is a UFD.