0

I have a question, and it is if a Euclidean evaluation can be defined in a general way in any unique factorization domain. I've thought that Euclidean evaluation can be defined by factoring irreducibles, but I don't know how true that is.

EDIT Maybe I asked the question wrong, but what I am trying to ask is that if it is possible to define a Euclidean evulation such that a domain of unique factorization becomes a Euclidean domain. For an arbitrary UFD.

James A.
  • 824
  • 1
    The term “Euclidean evaluation” is not standard. I think you are asking if every UFD is a Euclidean domain. No. For example, $F[x_1,\ldots,x_n]$ where $F$ is a field and $n \geq 2$ is a UFD but is not a Euclidean domain. This is also true for $\mathbf Z[x]$. – KCd Jul 17 '22 at 04:48
  • Maybe I asked the question wrong, but what I am trying to ask is that if it is possible to define a Euclidean evulation such that a domain of unique factorization becomes a Euclidean domain. @KCd – James A. Jul 17 '22 at 04:57
  • If that was true, then every UFD was a PID. KCd gives an example above that shows that this is not the case. – Severin Schraven Jul 17 '22 at 05:12
  • Every Euclidean domain is a PID by remainder descent, but UFDs are PIDs if & only if if they have dimension $1,$ (i.e. every non0 prime ideal is maximal), for example $,\Bbb Z[x],$ and $,\Bbb Q[x,y],$ are non-Euclidean UFDs. – Bill Dubuque Jul 17 '22 at 09:56
  • If your notion of "Euclidean evaluation" means something different than the standard denotation (i.e. division with smaller remainder) then you need to precisely define it. – Bill Dubuque Jul 17 '22 at 10:02

0 Answers0