sher@rochester.UUCP (02/10/86)
From: David Sher <sher>
I thank you out there for the response to my simple group theory
problem. I of course meant to only consider finite abelian groups. I
did not realize that there was such a simple number theoretic proof.
Number theory and its brother field of combinatorics is the form of
mathematics that I always felt least comfortable with because of the
many counterintuitive results from those fields.
So I will now try to construct a generalization of the my problem
that gets away from number theory. Consider the ring of polynomials
over Z, Z[x]. I think an ideal in a ring is the set of elements that
contain all the ring elements divisible by a subset. If I got that
wrong then call what I call an ideal a pseudo ideal and continue with
the problem. So if I is an ideal over Z[x] then Z[x] mod I is the ring
over the equivalence classes formed by the relationship a - b is a
member of I. I think this is well defined. The minimal subset of the
ideal for which every element of the ideal is divided by at least one
of the subset I call the minimal generator of the ideal or the
generator of the ideal for short. I believe this is unique and well
defined for ideals over Z[x] anyway. So I will refer to an ideal by
its generator. Thus if I say Z[x] mod {x} then I am refering to Z[x]
mod the ideal generated by x. Thus I state that Z[x] mod {x} is
isomorphic to Z. The set of nonzero divisors in a ring form a group
under multiplication that I call a multiplicative group. So is every
finite abelian group a multiplicative group of Z[x] mod I for some
ideal?
As an example consider the cyclic group of order 3. It has been
shown in this news group several times that there is no multiplicative
group for Z mod I that is cyclic of order 3. Consider Z[x] mod { 2 ,
x**2 + x + 1 }. It has 4 elements: 0, 1, x, x + 1. 1,x,x+1 is an
abelian group under multiplication hence the cyclic group of order 3 is
a multiplicative group for Z[x] mod I where I is the ideal generated by
{ 2 , x**2 + x + 1 }.
--
-David Sher
sher@rochester
seismo!rochester!sher