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From: sher@rochester.UUCP
Newsgroups: net.math
Subject: A generalization of the simple group theory problem
Message-ID: <15198@rochester.UUCP>
Date: Mon, 10-Feb-86 08:30:15 EST
Article-I.D.: rocheste.15198
Posted: Mon Feb 10 08:30:15 1986
Date-Received: Wed, 12-Feb-86 20:50:06 EST
Sender: sher@rochester.UUCP
Organization: U. of Rochester, CS Dept.
Lines: 37

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