Re: New paper, algebraic integers, Galois Theory

From: Marcel Martin (mm_at_ellipsa.no.sp.am.net)
Date: 10/12/04 

Date: Tue, 12 Oct 2004 17:29:34 +0200

James Harris a écrit :
>
> "*** T. Winter" <***.Winter@cwi.nl> wrote in message news:<I5G8yE.G0n@cwi.nl>...
> > In article <3c65f87.0410111713.4decc512@posting.google.com> jstevh@msn.com (James Harris) writes:
> > > "*** T. Winter" <***.Winter@cwi.nl> wrote in message news:<I5EF41.GtG@cwi.nl>...
> > ...
> > > > > would, supposedly, each have to share non-unit factors with 2 and 3.
> > > >
> > > > Define:
> > > > r1 = (-1 + sqrt(-23))/2 and r2 = (-1 - sqrt(-23))/2 (the roots),
> > > > p1 = (3 + sqrt(-23))/2 and p2 = (3 - sqrt(-23))/2,
> > > > q1 = (-2 - sqrt(-23)) and q2 = (-2 + sqrt(-23)).
> > > >
> > > > You can verify the following:
> > > > p1 * q1 = r1^3
> > > > p2 * q2 = r2^3
> > > > p1 * p2 = 8
> > > > q1 * q2 = 27
> > > > So the common factor between r1 and 2 is p1^(1/3), and between r1 and 3
> > > > it is q1^(1/3).
> > >
> > > Yeah, but how do you know that q2 doesn't have 27 itself as a factor?
> >
> > Perhaps because q2 is a root of x^2 + 4x + 27, and so q2^2 + 4.q2 + 27 = 0?
>
> So? It has *two* roots, so how do you know that one doesn't have have
> 27 as a factor?
>
> It's not a complicated question. I hope you'll at least try to answer
> fully.
> >
> > > In fact, let that be a test for you *** Winter.
> >
> > Passed.
> >
>
> Nope. I'm pointing out that you have a hidden assumption, which you
> use repeatedly though it can't be mathematically proven.

"it can't be mathematically proven" !?

If you ultimate goal is to convince people that you are totally
moronic, overall change nothing, your strategy is perfect. 

-- 
mm
http://www.ellipsa.net/
mm@ellipsa.no.sp.am.net  ( suppress no.sp.am. )
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