email@example.com wrote: > > In article <942neb$dd5$1@nntp.Stanford.EDU>, > Michael Hochster <michael@rgmiller.Stanford.EDU> wrote: > > > > > > : (Sort of like if AB = 0, A or B = 0. These people are saying that > must > > : be proven, and that it is a "gap" in my proof that I don't do so.) > > > > : If so, I'd like them to say that is their position here and we can > see > > : if we can't work that one out. > > > > Yes, that is my position. I would like an explanation of why > > it is true that if AB = 0, then A = 0 or B = 0. I grant that > > this statement is true when A and B are integers. However, > > I would like you to verify it when A and B are funny things > > like x + sqrt(-1)y and x - sqrt(-1)y (x, y integers). > > > > Hey, I've already seen the post where someone says you guys proved that > AB = 0, when A = 0, or B = 0 by using x^2 + y^2 = 0. > > I concede that one could debate the question of whether or not there > might exist some objects in an infinite ring that could be nonzero and > multiply times each other to give 0. After all, it's trivally done in > a finite ring. > > So, to me it's become a moot point. Maybe one of you will come back > with some numbers like quarternions or something where that's not true, > but hey, I don't mind that as I'd simply find that interesting, and > worth thinking about whether or not I could work around that.
Are you asking for an example of a ring on an infinite set in which fg = 0 doesn't imply f = 0 or g = 0? Consider the real functions on [0, 1] (e.g., many other examples can be used). Define (fg)(x) = f(x)g(x) and (f + g)(x) = f(x) + g(x). Define the ring zero as the constant function 0. Etc.