From: djb@koobera.math.uic.edu (D. J. Bernstein) Subject: Re: Matrices: If AB = I, then BA = I. Date: 19 Mar 1999 01:45:18 GMT Newsgroups: sci.math Keywords: Does rank of a module always make sense? > Surely it doesn't have to be that nasty a ring? Nasty but elementary: Endomorphisms of the group of positive rationals. Define four endomorphisms of that group G by their effect on the primes: r takes 2 to 2, 3 to 5, 5 to 11, etc.; s takes 2 to 3, 3 to 7, 5 to 13, etc.; u takes 2 to 2, 3 to 1, 5 to 3, 7 to 1, 11 to 5, 13 to 1, etc.; v takes 2 to 1, 3 to 2, 5 to 1, 7 to 3, 11 to 1, 13 to 5, etc. Then (u,v) is an isomorphism from G to GxG, with inverse (r s). In other words, over End G, (r s)(u,v) = (1) and (u,v)(r s) = (1 0,0 1). Several books credit Eilenberg with the observation that rank doesn't make sense over End G if G=GxG, but I don't know the original reference. ---Dan ============================================================================== From: "Charles H. Giffen" Subject: Re: Matrices: If AB = I, then BA = I. Date: Fri, 19 Mar 1999 14:50:18 -0500 Newsgroups: sci.math To: ullrich@math.okstate.edu ullrich@math.okstate.edu wrote: > > In article <7cjrhn$dss$1@panix3.panix.com>, > lrudolph@panix.com (Lee Rudolph) wrote: > > ullrich@math.okstate.edu writes: > > > > > Assuming that there is such a thing as B^(-1), meaning > > >a two-sided inverse, yes. How do you know there is? We were > > >given only that A and B are nxn matrices with AB = I. Why > > >does it follow that B is invertible? > > > That's more or less assuming what we're to prove. > > >A proof using nothing but fiddling at this level can't > > >be right, or it would prove the same thing for operators > > >on vector spaces in general. > > > > Note that P.M. Cohn gives an example of matrices (over, > > I admit, a *really* *nasty* ring; can't happen with a > > field), one 2-by-3, the other 3-by-2, such that both > > their products are identity matrices of the appropriate > > size. > > Huh, I didn't know that. (For anyone who doesn't see > why this doesn't contradict what everyone else has been saying, > we've been taking "matrix" to mean "real or maybe complex matrix", > possibly "matrix over a field".) > > Surely it doesn't have to be that nasty a ring? Surely > some rings even I know would do this? (Now I have today's > excuse for not getting any work, um, I mean today's exercise.) > > > Gotta use *some* properties of *something* in the proof. > > I do and you do. But that's just us - some people don't > even need definitions of things to prove things about them. > > > Lee Rudolph > > > > -----------== Posted via Deja News, The Discussion Network ==---------- > http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own Here are a couple of "nasty" rings for your consideration (actually they are pretty interesting to an algebraic K-theorist): (1) Let L be the free ring Z{s,t} on two indeterminates s and t modulo the ideal generated by (1 - st). This ring has the interesting property that there is an invertible upper triangular matrix M over L whose inverse is a lower triangular matrix, viz. | t 1-ts | | s 0 | M = | | M^{-1} = | | | 0 s | | 1-ts t | (2) Let B be the free ring Z{t,t^{-1},x,y} on an invertible generator t and two more free generators x and y modulo the ideal generated by 1 - xy, ytx, yxt + tyx - t, xt^2 - tx, t^2 y - yt Let | x | X = | | Y = | y ty | | xt^{-1} | Then XY = I_2 in M_2(B) and YX = I_1 in M_1(B), where I_n denotes the identity in the ring M_n(B) of n x n matrices over B. Thus a 2 x 1 matrix can have a 1 x 2 matrix two-sided inverse. The ring B also has the magical property of shifting algebraic K-theory by one dimension, ie. K_i(R) \iso K_{i+1}( B(R) ), where B(R) = B \tensor R. --Chuck Giffen