From: rusin@teton.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Dim of an operator space on C(infinity)
Date: 15 Nov 1997 06:41:44 GMT

In article <34673a97.0@newnews.inch.com>,
John McGowan  <jmcgowan@shell.inch.com> wrote:

>Is it true than any module over a ring must have a basis?

If by that you mean a set of module elements  {m_i} such that  M  is
the internal direct sum  M = R*m_1 + R*m_2 + ..., then the answer is
certainly "no". For example,  the matrix ring  R = Mat_n(F)  over any
field obviously has the set of column vectors  M = F^n  as an  R-module,
but this can't possibly have a "basis" -- simply compare dimensions
of these vector spaces over  F.  For a commutative example, consider
the ideal  M = (x,y)  in the polynomial ring  R = F[x,y]  over a field.
Here  {x,y} is a set of generators, but we have the relation
r1.x + r2.y = 0 for (duh!) r1=y, r2=-x  in  R  Finally, consider any
2-sided ideal  I  in any ring  R  and let  M = R/I ; then  M
certainly has no "basis" since for any  r  in  I  and any  m  in  M
we have  r.m = 0.

"Modules with bases" are the "free modules". The only rings for which 
all modules are free are division rings.

Maybe you meant something else by "basis"?

dave