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Re: FLT Discussion: Simplifying
Posted:
Jan 18, 2001 7:57 PM
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jstevh@my-deja.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.
>
> So, bring it on.
>
> James Harris
>
> Sent via Deja.com
> http://www.deja.com/
--
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peter dot percival at cwcom dot net
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