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 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