2

The condition that $R$ is a GCD domain is sufficient for every irreducible element to be prime. However, is there a necessary condition?

It does seem to be the case that in an integral domain $R$ for which every element can be factored into irreducibles, $R$ is a UFD iff every irreducible is prime. But without this, is there a nice necessary condition?

rschwieb
  • 153,510
green frog
  • 3,404
  • 3
    See this answer for this and much more. – Bill Dubuque Jan 06 '20 at 05:43
  • Ok so it seems that without the ring being an atomic domain, AP doesn't necessarily have a nice implication? – green frog Jan 06 '20 at 21:30
  • 1
    An extreme case is when there are no atoms (antimatter domains), so AP holds vacuously, e.g. the ring of all algebraic integers where $,\alpha = \sqrt\alpha \sqrt\alpha.\ $ But when AP holds and there are atoms then we deduce as usual the uniqueness of products of atoms (but not every element need be such a product). – Bill Dubuque Jan 06 '20 at 22:01
  • "But when AP holds and there are atoms then we deduce as usual the uniqueness of products of atoms." - This follows from AP = PP for atomic a, right? Also this doesn't imply that the ring is UFD? All that it implies is that if an element has a factorization into atoms then the factorization is unique? – green frog Jan 06 '20 at 22:15
  • Generally in a domain if an element has a prime factorization then that is the unique atomic factorization, as the usual proof shows. – Bill Dubuque Jan 06 '20 at 22:24

0 Answers0