Re: JSH: Blocking and ring of algebraic integers

On Jul 31, 4:27 pm, JSH <jst...@xxxxxxxxx> wrote:
Brainstorming out a new research approach can difficult in that
corrections are needed to cover errors, omissions or missed logical
steps.

So I'm doing a new post which has the full result with correction
necessitated by counterexamples to previous ones and omissions that I
noticed.

Turns out that the key to all that freaking arguing over the years was
noting that if you have a polynomial with integer coefficients and a
positive leading coefficient in an integral domain then the
factorization

P(x) = (g_1(x) + 1)*(g_2(x) + 1)

I was wrong. Turns out there is a simple fix though.

Oddly enough, with the factorization that IS blocked you also get
another crucial result.


where g_1(0) = g_2(0) = 0, is BLOCKED if the g's are not polynomials
nor are they square roots of polynomials, which is just this
remarkable, odd little result that NO ONE would notice unless they
went looking for it, as you cannot see it with polynomial
factorizations, of course.

So now you get this odd thing with the distributive property as then

p_1*p_2*P(x) = (p_1*g_1(x) + p_1)*(p_2*g_2(x) + p_2)

where the p's are differing prime numbers is blocked from the ring of
algebraic integers as well, when the g's are non-rational with non-
zero rational x.

But it gets weirder!!!

Because you CAN have the factorization

p_1*p_2*P(x) = (f_1(x) + p_1)*(f_2(x) + p_2)

in the ring of algebraic integers with the f's not polynomials nor
square roots of polynomials as long as they are not both equal to 0
when x =0.

And you can because you can keep substituting to get to

p_1*p_2*P(x) = (h_1(x) + p_1*p_2)*(h_2(x) + p_1*p_2)

And now we're at the nitty-gritty, as assume now that p_1*p_2 can be
divided off in the ring of algebraic integers as a function!

Then assuming that w_1(x)*w_2(x) = p_1*p_2

you'd have

P(x) = (h_1(x)/w_1(x) + w_2(x))*(h_2(x)/w_2(x) + w_1(x))

but now you can so something simple as let

h_1(x)/w_1(x) + w_2(x) = j_1(x) + 2

and

h_2(x)/w_2(x) + w_1(x) = j_2(x) + 1

and you have

P(x) = (j_1(x) + 2)*(j_2(x) + 1)

and the blocked factorization.

Yeah I cheated a bit as I gave the proper form for a blocked
factorization when the functions are 0 when x=0 with functions that
are not, correcting the prior incorrect form.

Notice that p_1*p_2 dividing off as functions is specifically blocked
by the ring of algebraic integers casting the objections of posters
like *** Winter, Arturo Magidin, Rick Decker and William Hughes into
the abyss of failure.

Note then as I now give the correct form that

P(x) = (g_1(x) + 2)*(g_2(x) + 1)

cannot exist in the ring of algebraic integers when P(x) is a
polynomial with integer coefficients, g_1(0) = g_2(0) = 0, and the g's
are not rational with rational x.

THAT is the ultimate set of conditions which not only proves my case
but directly blocks prior claims.

Now I think that much of the arguing with me has been about weird
American class warfare, which traces back to lower class Brits looking
at their upper class in a peculiar and distorted way which they
transmitted to their descendants, many of whom are now US citizens.

If so, then posters arguing with me, now that it is over, may see this
as life or death and KEEP FIGHTING as if their lives depended on it.

My hope is that instead they will now concede to what is
mathematically correct and end the Math Wars.

Remember, what is mathematically correct was ALWAYS so, but we just
had to discover the truth.


James Harris


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