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'm not sure what your point is here. Are you conceding that your argument is circular?
: 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.
Just for future reference, there is standard terminology for this. A ring in which AB = 0 implies A = 0 or B = 0 is called an integral domain.
There are simple examples of infinite rings which are not integral domains. Take ordered pairs of integers with coordinatewise addition and multiplication.
You wanted to use a "result" (Every infinite ring is an integral domain) to justify the next step of your argument. But you were unsure whether the result was true. That means you have a *gap*. There is no such thing as proof by I-can't-think-of-a-counterexample. In this case, the result you wanted to use is false.