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- From: hammond@csc.albany.edu (William F. Hammond)
- Newsgroups: sci.math
- Subject: Re: factorization in commutative rings
- Message-ID: <HAMMOND.93Jan3134111@annemarie.albany.edu>
- Date: 3 Jan 93 18:41:11 GMT
- References: <Jan.3.02.05.44.1993.24643@spade.rutgers.edu>
- Sender: hammond@sarah.albany.edu (William F. Hammond)
- Distribution: sci.math
- Organization: Dept of Math & Stat, SUNYA, Albany, NY
- Lines: 71
- In-Reply-To: cadet@spade.rutgers.edu's message of 3 Jan 93 07:05:44 GMT
-
- 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
- ----------------------------------------------------------------------
-