Re: Basic argument, algebraic integers

From: James Harris (jstevh_at_msn.com)
Date: 10/05/04 

Date: 5 Oct 2004 14:53:08 -0700

rupertmccallum@yahoo.com (Rupert) wrote in message news:<d6af759.0410042349.552149bb@posting.google.com>...
> 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"?
>

Start with

P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)

with the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

and note that at

m=0, P(0) = u^2 f^2(3x + uf),

which gives you terms that do not vary as m varies.

So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)?

Well, at m=0, two of the a's have to equal 0, and it's convenient to
just arbitrarily select a_1 and a_2 as those two.

Then you have uf for the first, uf for the second and 3x + uf for the
third as terms that do not vary when m varies.

Now then, if m=1, what are the *constant* terms?

They are uf, for the first, uf for the second, and 3x + uf for the
third.

That's logical because they do not vary with m, so if m=1003909273,
what are the constant terms?

They are uf, for the first, uf for the second, and 3x + uf for the
third.

Now divide f^2 from both sides, which gives

P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f

P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2

and you note that P(0)/f^2 = u^2(3x + uf), which means that now your
constant terms are u for the first, u for the second and 3x + uf for
the third.

Now then, if m=1, what are the constant terms now?

They are u for the first, u for the second, and 3x + uf for the third.

If m = 2938479378, what are the constant terms now?

They are u for the first, u for the second, and 3x + uf for the third.

How can the constant terms of the first two go from uf to u?

They must be divided by f.

Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided
from P(m), it is u; therefore, a_1 x + uf is divided by f, and you
have

a_1 x/f + u

and the constant term of a_2 x + uf is uf, but when f^2 is divided
from P(m), it is u; therefore, a_2 x + uf is divided by f, and you
have

a_2 x/f + u

while the constant term of a_3 x + uf is 3x + uf, and after f^2 is
divided off, it is 3x + uf, so you have

a_3 x + uf

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).

And that's what it means for it to follow algebraically.

James Harris

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