From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Prime and Irreducible elements of a Domain Date: 8 Dec 1999 04:56:56 GMT Newsgroups: sci.math In article <82bpq3$a52$1@nnrp1.deja.com>, Rogerio Brito wrote: >What I thought was that in a general domain D, if p \in D is a prime >element, then we know that p is an irreducible element. Let P be the >set of primes of D and I be the set of irreducible elements of D. The >above implication means that P is a subset of I. [...] >OK. Now, what happens if D is a general domain, not necessarily >Euclidean, not necessarily an Unique Factorization Domain? More >specifically, if D is a general domain, what are the sets > > Prod_1 = U * P^* > Prod_2 = U * I^*, > >were A^* is the set of all finite (possibly empty) products of >elements of A? The second is not too hard to understand. If an element x in your domain D is not irreducible, then it can be written as a product, say x = y * z. If y and z are irreducible, then x lies in Prod_2. If not, then factor each of them, getting x = a*b*c*d, say. If each is irreducible, then again x lies in Prod_2. If not, then some of them can be factored; do so and repeat... So you see that every element of D lies in Prod_2 IF this process is guaranteed to terminate. Well, that's NOT true in general rings, but it is true among Noetherian rings, an important and broad family of rings. The other question is a bit harder. Part of the problem is that there are very few "prime elements" in some rings. What you need to do is to redirect your focus from _elements_ in the ring to _ideals_ in the ring; it's much more common to have prime ideals than it is to have prime elements. (An element is prime iff the principal ideal it generates is a prime ideal.) If you're willing to restrict your attention to number fields (i.e. finite extensions of the rationals, or more precisely their rings of integers) then you have the happy situation that every ideal is uniquely factorized as a product of prime ideals. It's not hard to see then that the elements in your Prod_1 are precisely those which generate a principal ideal whose factorization into primes involves only principal prime ideals. If for example your ring is a PID (principal ideal domain), that's all elements of the domain, as you pointed out, but for number fields non-PID's always have prime ideals which are not principal, so there will be elements in the domain not in Prod_1. If you're new to ideals you might view what I've just said as a technical cop-out to avoid answering questions about the elements of a ring, but trust me, the ideals are central to the study of rings and in time you will see it is the elements of the ring which are "accidental" and the ideals which are of (ahem) primary importance. dave ============================================================================== From: kramsay@aol.commangled (Keith Ramsay) Subject: Re: Prime and Irreducible elements of a Domain Date: 08 Dec 1999 05:24:34 GMT Newsgroups: sci.math In article <82bpq3$a52$1@nnrp1.deja.com>, Rogerio Brito writes: |OK. Now, what happens if D is a general domain, not necessarily |Euclidean, not necessarily an Unique Factorization Domain? More |specifically, if D is a general domain, what are the sets | | Prod_1 = U * P^* | Prod_2 = U * I^*, | |were A^* is the set of all finite (possibly empty) products of |elements of A? I believe a general treatment of this question is somewhat complicated. In many commonly studied cases, every element is a product of irreducibles. A commonly assumed property is that the ring is "Noetherian", which means that in every chain of ideals I1, I2, I3,... where for each n In is a subset of I_{n+1}, there's some N for which I_N=I_{N+1}. In a Noetherian ring every element has a factorization into irreducibles. Then for any x_i which has no factorization into irreducibles, we can factor x_i (because it isn't irreducible itself) into non-units y*z. The factors y and z cannot both have factorizations into irreducibles, so let x_{i+1} be a factor of x_i which also doesn't have a factorization into irreducibles, and which isn't an associate of x_i. It would follow that given an element x not factorizable into irreducibles, there would exist a sequence x_1=x, x2, x3, x4,... of elements produced in such a way, and the sequence of principal ideals (x1),(x2),(x3),... generated by them cannot exist by the Noetherian property. When the ring is the ring of integers of a number field, the set of products of primes can be understood by considering what's called the class group of the number field. The class group for the case of the ring Q(sqrt(-5)) which has been mentioned by others, is a group of order 2, for example. There isn't unique factorization of elements, but there is unique factorization of nonzero ideals. Each ideal corresponds to an element of the class group. The nonzero principal ideals (which are generated by nonzero elements, with two elements generating the same ideal if and only if they differ by units) are associated with the identity element of the class group. The product of two ideals corresponds to the product of the elements of the class group associated with the two ideals. So in the simple case of Q(sqrt(-5)), each ideal is either principal, or corresponds to the other element of the class group. This means the product of two nonprincipal ideals in this ring is principal. From unique factorization of ideals and the class group, one can figure out facts about how factorization of elements works. It's only in special cases like these that it works out as nicely, though. Books in algebraic number theory and commutative algebra would tell more about this kind of stuff. Keith Ramsay