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  <rbrito@my-deja.com> 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 <rbrito@my-deja.com> 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