From: hammond@csc.albany.edu (William F. Hammond)
Newsgroups: sci.math
Subject: Re: factorization in commutative rings
Date: 3 Jan 93 18:41:11 GMT
Distribution: sci.math
Organization: Dept of Math & Stat, SUNYA, Albany, NY

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 
> 
> A) D= factorial domain doesn't imply D= princ. idl. domain
> 
>    Standard example   Z[x] or F[x,y] where F=field

"PID" ==> every non-zero prime ideal is maximal ==> Krull dim = 1,
which is very special.  Note that  Z[x]  and  F[x,y]  both have
Krull dim = 2.   For example, if  A  is a domain that is finitely-
generated as a  C-algebra  of Krull dimension at least two, then the
maximal ideals of  A  correspond to points of an affine algebraic
variety.  These points are "smooth" points of that variety except
for those lying in a subset, possibly empty, of postive codimension.
The localizations of  A  at the maximal ideals corresponding to smooth
points are all factorial local rings that are not PID's.  (This is not
so elementary.)  To get more explicit look at the localizations at
"general" maximal ideals of  A = C[X_1, . . ., X_N]/F, where  N >= 3
and  F  is irreducible.  ["Usually" if  F(0,...,0) = 0, then the maximal
ideal corresponding to (0,...,0), i.e., that generated by X_1,...,X_N,
is "general".]

> 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.

If we identify  w  with one of its two complex conjugates, then  R becomes
a lattice in  C and  K  a subset of  C  that is dense in the Euclidean
topology on  C.  "Euclidean" is thus equivalent to the assertion that
every complex number in a dense subset is within distance 1 of a lattice
point, i.e., that the disks of radius 1 centered at lattice points cover
the plane.  If the discriminant of the imaginary quadratic field at hand
is large enough in absolute value, then the lattice points will be "too
spread out" for this to be the case.

Thus, it is easy to see that if the absolute value of the discriminant of
an imaginary quadratic field  K  is larger than a bound that can be made
explicit, then the ring of its integers  R  cannot be Euclidean.

> PS. I am preparing for my quals so that's why I asked.

Good luck with your qual.
----------------------------------------------------------------------
William F. Hammond                   Dept. of Mathematics & Statistics
518-442-4625                         SUNYA, Albany, NY 12222 (U.S.A.)
hammond@csc.albany.edu               FAX: 518-442-4731
----------------------------------------------------------------------