Re: New paper, algebraic integers, Galois Theory

From: Nora Baron (norabaron_at_hotmail.com)
Date: 10/06/04 

Date: 6 Oct 2004 07:37:39 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410051423.62102f68@posting.google.com>...
> Edited, added Section 3:
>
>
> ----------------------------------------------------------------------
>
> I. First section
>
> 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)?
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf)
>
> which shows that at least two of the a's have to equal 0 at m=0, while
> one equals 3.
>
> Since, at m=0, two of the a's have to equal 0, 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 DON'T have to "go from uf to u". You are assuming that the
only way to factor out f^2 is to factor f out of each of
a1*x + uf and a2*x + uf. That is true when m = 0 but
not true otherwise. See my argument in the "cheaters"
thread to see the correct factorization.

> 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.
>

  But you cannot conclude that a_1/f etc. are algebraic
integers. In fact you don't say what kind of number
a_1/f is. Conclusion: technically, Section 1 doesn't say
anything. If you intend it to assert that a_1/f is an
algebraic integer, it is wrong. The root of the error
is assigning too much significance to the "constant term".

> II. Second section
>
> 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).
>
>
> Now consider the case that m, f, and u are algebraic integers, then 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.
>

  Dedekind's theorem used here ...

> 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) = f
>
> as that is the last coefficient from the last term u^3 f, which proves
> that
>
> (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let
>

  No, right here your argument is incomplete. You are thinking that
w1*w2*w3 is coprime to v1*v2*v3. It is true that PAIRWISE, you
have coprimeness: that is w1 is coprime to v1, w2 is coprime to
v2, and w3 is coprime to v3. But you have NOT proved that
w1 is coprime to v2, etc.

  Therefore you cannot conclude from this that w1 divides

      (m^3 f^4 - 3m^2 f^2 + 3m).

> (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)
>

  Almost. *If* you had done some more work, you can arrive
at a factorization in this form in which all of w1, w2,
w3, v1, v2, v3 are algebraic integers.

  And this would disprove your central claim following APF:
that there is no such factorization in the algebraic integers.

  Thus, if you had done it right - and it CAN be done right -
you would yourself have undermined your own beloved paper.

  You are in the soup. In fact, your claimed major result
is in the toilet. Where it belongs.

> but I still have that
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
>

  You *have* that certainly, but it leads nowhere.
You have chosen the wrong factorization of f^2 even
though you almost got it. Of course you cannot
accept the implications of what you yourself are
saying. It would mean two years of "work" for naught.

> III. Third section
>
> So, even if a_1/f is not an algebraic integer, you can find w_1 an
> algebraic integer.
>
> But if a_1/f is an algebraic integer and w_1 is not, they cannot be
> equal.
>
> But I have
>
> P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)
>
> and
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf)
>
> so how do you reconcile a case where a_1 x/f is not an algebraic
> integer?
>
> There must exist algebraic integers z_1, z_2, and z_3 such that
>
> w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3
>
> and z_1 z_2 z_3 = 1,
>

  No - first, that "x" should not be in there. Second, there
is no reason to conclude that z_1 z_2 z_3 = 1. You are
trying to resolve your own favored factorization and the
correct factorization. The problem arises out of the fact
that your factorization does not work in the algebraic
integers. You have come fairly close to showing that
another factorization DOES work as I have shown in the
"cheaters" thread.

> so algebraically the z's are units, but they cannot be units in the
> ring of algebraic integers, since a_1 and f do not share factors in
> that ring, as if they did, then a_1/f would be an algebraic integer.
>
> So z_1, z_2, and z_3 are not roots of a monic polynomial with integer
> coefficients that has a last coefficient of 1 or -1, but they are
> roots of a monic polynomial with integer coefficients.
>
> Often objectors to my work have found the existence of algebraic
> integers like w_1 and claimed that existence proves that my work is
> wrong, but the basic arguemnt in Section 1 can't be wrong, or algebra
> is wrong.
>

  Since you do not specify what kind of number a_1/f is, Section
1 technically doesn't say anything. If you want to claim
that dividing by f^2 in the way you specify produces a factorization
with algebraic integer coefficients, then Section 1 IS wrong.
That however does not eliminate the possibility that there is
another way to divide out f^2 which DOES produce an algebraic
integer factorization. This is what you have come fairly close
to showing with your w_1, v_1 etc. factorization. But you dare
not accept the implications of that factorization, because it
denies the central claim in APF.

  Result: you are left wondering what the heck is going on. You
have an inconsistency and you cannot figure out the source.
It's simple. You think that the factorzation which works when
m = 0 must work for other m also. You have been misled by your
own tautologies about the "constant term".

> Then as algebra is valid, it simply follows that algebraic integers
> z_1, z_2 and z_3 must exist.
>

  They exist but they are not units. The correct argument here,
which shows exactly what the right factorization looks like,
is in the "cheaters" thread.

> Then it is simply a fallacy to claim counterexamples to Section 1
> based on algebraic integers found, as finding algebraic integers can
> be explained as shown in Section 2, and, most importantly, algebra is
> in fact valid, and each step in Section 1 is correct.
>
> Challenging Section 1 is challenging algebra itself.
>

  See above.

> That leaves Galois Theory to be handled next.
>

  Galois theory is in no danger.

  Nora B.

>
> James Harris

Relevant Pages

  • Re: JSH: Resolution now possible
    ... >>with the factorization ... and constant multiples of polynomials are not functions, ... What I know is that constant terms are not varying functions. ... That is, given two algebraic integers a and b, ...
    (sci.math)
  • Re: JSH: Resolution now possible
    ... We all concede that your factorization ... >reason to EVER believe that P/2 has constant terms that vary ... That is, given two algebraic integers a and b, ... divides both a and b, ...
    (sci.math)
  • JSH: Overinterpertations of Galois Theory Exposed
    ... ADVANCED POLYNOMIAL FACTORIZATION ... Determining the distribution of factors within irrational algebraic integers ... one of the a's is coprime to 5. ... that are themselves polynomials. ...
    (sci.math)
  • JSH: The "Published" paper he dosent what you to know about.
    ... ADVANCED POLYNOMIAL FACTORIZATION ... Determining the distribution of factors within irrational algebraic integers ... one of the a's is coprime to 5. ... that are themselves polynomials. ...
    (sci.math)
  • JSH: Secret paper that was published, then withdrawn
    ... ADVANCED POLYNOMIAL FACTORIZATION ... Determining the distribution of factors within irrational algebraic integers ... one of the a's is coprime to 5. ... that are themselves polynomials. ...
    (sci.math)
Quantcast