Re: Basic argument, algebraic integers
From: Rupert (rupertmccallum_at_yahoo.com)
Date: 10/05/04
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Date: 5 Oct 2004 00:49:47 -0700
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410040233.f2c35b1@posting.google.com>...
> rupertmccallum@yahoo.com (Rupert) wrote in message news:<d6af759.0410031504.4c8c171a@posting.google.com>...
> > jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410030904.402a133f@posting.google.com>...
> > <snip>
> >
> > > so, dividing P(m) by f^2 gives
> > >
> > > P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
> > >
> >
> > But there's no reason why a_1/f and a_2/f should be algebraic integers.
> >
> > [rest deleted]
>
> But at times they *are* algebraic integers.
>
> At times they are, other times they are not.
>
> So there's a factorization that follows algebraically that's not
> always true in the ring of algebraic integers.
>
Why does the factorization "follow algebraically"?
> Or do any of you wish to deny that?
>
> Mathematicians will not deny what is mathematically true, now will
> they?
>
> If you dispute and I'm right then you cannot be a mathematician.
>
>
> James Harris
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