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