
Re: FLT Discussion: Simplifying
Posted:
Jan 21, 2001 8:39 AM


In article <94cvf0$4p6$1@nnrp1.deja.com>, jstevh@mydeja.com wrote: > In article <3a683db5.481647214@news.newsguy.com>, > randyp@visionplace.com (Randy Poe) wrote: > > On Fri, 19 Jan 2001 11:24:12 GMT, "Dik T. Winter" <Dik.Winter@cwi.nl> > > wrote: > > > > > > I'm looking for something that pushes me outside of integers > because > > > > that's what I want to prove must happen! > > > > > >Tsk. Your initial use of sqrt(1) pushes you out of the integers > already. > > > > You misunderstood this statement. He believes that the fact that > > sqrt(1) pushes him out of the integers is the contradiction and that > > the proof can end at that point, just with the factorization. That > > writing down a statement not in integers, though his first equation > > was in the integers, is a contradiction. > > > > Here's where there's the issue between what I've recently called > patterns and regular rings. > > My understanding is that mathematicians have avoided this through > coupling. > > That is, polynomials have counting number exponents *and* counting > number coefficients.
Polynomials are not limited to just counting number coefficients. I am sure that you are aware of this, so I don't know why you made this slip.
> However, I have argued that this coupling is simply one way of doing > things and there's no mathematical or logical requirement for it.
Polynomials in X can have coefficients in any ring R (which is then fixed). The polynomials themselves then form a ring, called the ring of polynomials in X and Y over R. This polynomial ring is denoted by R[X, Y]. Of course, you can have more or less variables than just X and Y and you need not even call them X, Y etc.
> The only remaining issue then has been an insistence that it must be > proven that (x+sqrt(1)y)(xsqrt(1)y) = 0 means that x+sqrt(1)y = 0 > or x  sqrt(1)y = 0, as some have insisted I must move to complex > numbers to get this result.
No one has suggested that you *must* move to the complex numbers to get this result. I suggested it as one way to get the ring that you are working in to be a known one, so that you would not have to prove that it exists and you could use many of the theorems already proved about the complex numbers: for example, that it is a commutive field with an isomorphic copy of the integers embedded in it.
> But, I've repeatedly brought up the fact that it's true for other > rings, so why if you guys are acting like using rings is so fundamental > and important, do you wish to make a result that's a hack depending on > what ring you're using? > Don't understand what I mean?
I don't understand what you mean. First, the *fact* is not true for all rings. But, more importantly, as I have been trying to emphasize, you need to state what ring you are working in so that 1) your statements in your proof have meaning; 2) the words you use that have been defined can be "unwind" to their more basic meaning; and 3) the properties known to be true about that ring can be quoted (since you seldom need to start from scratch with a ring that nobody has ever thought of).
An example of 2 is the following. Suppose I say that 2 is prime. If I want to "unwind" prime and eliminate that word, then I cannot do it unless you tell me what ring you are working in. For, 2 is prime in the integers, but 2 is a unit in the rationals, and 2 = (1+i)(1i) is composite in the gaussian integers. I need to know what ring the statement "2 is prime" is referring to, in order to know whether or not 2 has that property of primeness.
 Bill Hale
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