Re: Resolving the a's

From: Nora Baron (norabaron_at_hotmail.com)
Date: 10/08/04 

Date: 8 Oct 2004 11:42:18 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410080249.2727d873@posting.google.com>...
> Given
>
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
>
> and
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> it's fairly easy to solve for the a's, as it follows that
>
> a_1 a_2 a_3 = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)
>
> a_1(a_2 + a_3) + a_2 a_3 = 0
>
> and
>
> a_1 + a_2 + a_3 = - 3(-1 + mf^2)
>
> so the a's are the negatives of the three roots of the cubic
>
> a^3 - 3(-1 + mf^2)a^2 + f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) = 0
>
> and I've shown that algebraically *two* of the roots have f as a
> factor.
>

  It has also been shown RIGOROUSLY that (1) none of the roots
have f as a factor (for most values of m), and (2) all three
roots are non-coprime to f. Your "proof" is incorrect as has
been demonstrated elsewhere.

> That can actually be seen with m=1, f=sqrt(2), which gives
>
> a^3 - 3a^2 + 2 = 0, and it's true that
>
> a^3 - 3a^2 + 2 = (a-1)(a^2 - 2a - 2)
>
> so obviously two of the a's have f as a factor, and one is coprime to
> it.
>

  This is because in this exceptional case, the resulting
polynomial is REDUCIBLE. See also below -

  
> So the theory actually works.
>

  Yes. Here is what theory (not your theory) says. For
reducible monic polynomials, one of the roots can be coprime
to a factor of the constant term. For IRREDUCIBLE monic polynomials,
this is not true; each of the roots must have factors in common
with all algebraic integer factors of the constant term.

> In fact, in any case where you can get a rational root you will always
> find that it is either coprime to f, or it has f as a factor, as my
> work proves.
>

  What you say is essentially true (not because your work
shows it) but it is somewhat incomplete. Specifically, you
should include a statement to the effect that if the monic
polynomial of which the a's are roots is irreducible, then
EACH of the roots is non-coprime to each nonunit factor of the
constant term. Right?

  A problem with what you have tried to do is that your polynomial
is extremely specialized and it is cubic. If your main conclusions
are correct, they would apply in more generality. Consider the
following discussion of a quadratic:

==============================================================

  Assume f is a prime greater than 3 and m is nonnegative integer,
and u is an integer coprime to f.

  Consider the function

    P(m) = f (m x^2 + (2*m + 3) x u + u^2 f)

  Suppose P(m) is factored in the form

    P(m) = (a1 x + uf)*(a2 x + uf)

where multiplying out shows that

    a1 a2 = m f.

  Therefore, at least one of the a's cannot be
coprime to m, and at least one of the a's must
equal 0 when m = 0.

  (Note: the a's are roots of a monic polynomial
with algebraic integer coefficients so they are
algebraic integers.)

 Notice that the constant term P(0) is given by
P(0) = f*(3*u*x + u*f) and also that
P(0)/f = 3*u*x + u*f, which is coprime to f.

  Then I have the factor of P(m), g1, where
g1 = a1 x + uf, where here I also have that a1 is
not coprime to m.

  It is clear that when m = 0,

     g1 = uf,

meaning f is a factor of the constant term.

  Therefore, exactly one of the a's equal 0, when m = 0,
to get the factor f in the constant term P(0), while
the other one must not equal 0, or f^2 would be the
factor.

  Now as noted before in general P(m) has a factor that
is f, and separating that factor off, gives a constant term
coprime to f; therefore, given g1 = a1 x + uf where with
m = 0, g1 gives a factor of f it must have that same
factor in general, proving that one of the a's has a
factor that is f.

==============================================================

  Does this seem to be consistent with what you have done in
"Advanced Polynomial Factorization" ? [The wording should
look familiar.] Do you find any errors in it ? Please point
them out explicitly if you do. Is the conclusion correct?

  Nora B.

> Posters disagreeing with me tend to stay away from this area, which is
> further proof that they *knowingly* lie to you.
>
>
> James Harris

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