Re: Compromise, and you leave my papers?

From: Rick Decker (rdecker_at_hamilton.edu)
Date: 10/21/04 

Date: Thu, 21 Oct 2004 10:40:35 -0400

James Harris wrote:
> Rick Decker <rdecker@hamilton.edu> wrote in message news:<BJSdna7IQNKsl-rcRVn-iA@hamilton.edu>...
>
>>James Harris wrote:
>>
>><snip>
>>
>>>Like remember that 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)
>>>
>>>and
>>>
>>>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>>>
>>>the a's are 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 you can check at m=1, f=sqrt(2), that my argument is correct.
>>
>>And you can check that with m = 1 and f = 2 your argument is incorrect.
>>
>>
>>Regards,
>>
>>Rick
>
>
>
> Readers may think Decker is insinuating that m=1, f=sqrt(2) is a
> special case that might never occur again.

I'm not insinuating that. I'm saying that your claim--that
in your factorization exactly two of the a's are divisible
by f and the other is coprime to f--is not true in general.

> Well that's what it looks
> like, but in fact the only thing special about that case is that with
> it one of the a's is rational.
>
> With one of the a's rational, you can of course, look at it and just
> see that my work does give the right result.

Evidently you didn't try it. With m = 1 and f = 2 you also have
the case that one of the a's is rational. The factorization is

P(1) = 4(7 x^3 - 9 x + 2) = (a_1 x + 2)(a_2 x + 2)(a_3 x + 2)

where a_1 = -2, a_2 = (-7 + sqrt(105))/2, a_3 = (-7 - sqrt(105))/2

See, one of the a's is rational, just as in the case f = sqrt(2).
However, in this case one of the a's is divisible by 2, but neither of
the other two are. In addition, none of the a's are coprime to 2 (in the
algebraic integers).
>
> There are an *infinity* of such results, and they will all give the
> same information that my argument is correct.

Even if that were true, it would still leave open the possibility
that for infinitely many of your polynomials your argument is
wrong.
>
> So with an argument that works over infinity posters are sure to try
> and make you think that you're dealing with a single case as then they
> can't justify their claims that ALL of my work should be tossed.

You haven't shown that there are infinitely many cases in which
the conditions of your claim are true (though that is in fact true
as long as you allow non-integral values for m and f).

Even so, as long as you claim that your condition is *always* true
all it takes to make your argument incorrect is *one* counterexample
like the one above. You first need to restate your claim like this:

    If P(m, f, x) = f^2((m^3f^4 - 3m^2f^2 + 3m)x^3 - (-1 + mf^2)x + f)
    is factored as (a_1 x + f)(a_2 x + f)(a_3 x + f), with a_i algebraic
    integers, then as long as [some condition on m and f], exactly
    two of the a's will be divisible by f in the algebraic integers
    and the other a will be coprime to f.

and then you would have to prove the result above.
>
> From my prime counting function to my work with factorizations, notice
> that posters want it all to just go away. They want me to not post on
> it, and not to send it to journals.

Unlike you, I can't claim to know what the readers of sci.math want,
but I don't care whether you post here or not. However I wouldn't
advise you to send an incorrect result to a journal, if only because
doing so would (1) waste the referees' time and (2) result in yet
another rejection.

<snip>

Regards,

Rick

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