Re: Basic argument, algebraic integers

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

Date: 4 Oct 2004 16:58:32 -0700

So it's easy.

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).

There is no way to mathematically argue with the result.

Now take

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

and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside
by f^2(a_1 a_2 a_3) and you have

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

and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is

P(m)/f^2 =

        (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3).

So I have the ratios of algebraic integers:

uf/a_1, uf/a_2, and uf/a_3,

and now let

v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_2

where the v's and w's are algebraic integers in each case coprime to
each other.

Making the substitutions I have

P(m)/f^2 =

     (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3).

And I have from before that

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

so

(m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = u^3 f

as that is the last coefficient, which proves that

(m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let

(m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3

then I have

P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)

but I still have that

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

So finding algebraic integers w_1, w_2, w_3, and v_1, v_2 and v_3 are
irrelevant to that result, as they could always be found as proven
using some basic results where the only theorem used without naming or
giving proof covers the existence of coprime algebraic integers v and
w, given a ratio of algebraic integers.

James Harris

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