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- Path: sparky!uunet!mcsun!uknet!comlab.ox.ac.uk!oxuniv!wbs
- From: wbs@vax.oxford.ac.uk
- Newsgroups: sci.math
- Subject: Re: factorization in commutative rings
- Message-ID: <1993Jan6.125203.11070@vax.oxford.ac.uk>
- Date: 6 Jan 93 12:52:03 GMT
- References: <Jan.3.02.05.44.1993.24643@spade.rutgers.edu> <HAMMOND.93Jan3134111@annemarie.albany.edu>
- Distribution: sci.math
- Organization: Oxford University VAX 6620
- Lines: 44
-
- In article <HAMMOND.93Jan3134111@annemarie.albany.edu>, hammond@csc.albany.edu (William F. Hammond) writes:
- > In article <Jan.3.02.05.44.1993.24643@spade.rutgers.edu>
- > cadet@spade.rutgers.edu (Uniquely TiJean) writes:
- [....]
-
- >> Well, Given D an integral domain
- >>
- >> Then D = euclidean domain ==> D = principal ideal domain ==> D = factorial
- >> domain.
- >>
- >> I am looking for counterexamples
- >>
- >> B) D= princ. idl. domain doesn't imply D= euclidean domain.
- >>
- >> Well, I found in Hungerford ( Algebra )
- >>
- >> the following Z( (1+sqrt(-19)) / 2 )
- >>
- >> Question: How come? I have no cue as to why the above domain isn't euclidean.
- >
- > Is there a standard definition of "Euclidean", i.e., of what it means to
- > have a ring in which there is a division algorithm? Anyway, that is not
- > a problem here.
- >
- > Let w = (1+sqrt(-19))/2 , R = Z[w] = Z + Zw , and K = Q(w) = Frac(R).
- > In this context "Euclidean" means that given a, b in R with b not
- > 0, one can find q, r in R such that (i) a = qb + r and
- > (ii) Nm(r) < Nm(b), where Nm(x) is the product of x and its
- > conjugate. (Note: Nm(x) > 0 for x not 0.)
- >
- > "Euclidean" is equivalent to the statement that given an element x of
- > K there is an element a of R such that Nm(x-a) < 1.
-
- Am I being very silly in not seeing why there are not other
- candidates for the Euclidean function?
- (If so please tell me why.)
-
- [...]
-
- > Good luck with your qual.
- > ----------------------------------------------------------------------
- > William F. Hammond
-
- B.
-