Re: JSH: Weird but fascinating

On Jul 30, 9:48 pm, JSH <jst...@xxxxxxxxx> wrote:
Whew! Quite a bit of work from yesterday to today when I finally
realized I was brainstorming out a proof that I now call the wrapper
theorem. Turns out that the key to all that freaking arguing over the
years was noting that if you have a polynomial with integer
coefficients in an integral domain then the factorization

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

is BLOCKED when with non-zero integer x, the g's are non-rational,
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 being non-rational with
non-zero rational x.

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 get the symmetry that the ring of algebraic integers is requiring
if the functions are non-rational with rational non-zero x.

As then the h's can be algebraic integer functions, being roots of a
monic polynomial expression with integer coefficients as I've often
demonstrated with what I call non-polynomial factorization!

What a story! What a wacky ring! It is so cool. So oddly quirky.

No wonder mathematicians like the ring of algebraic integers so much,
it's almost human in its peculiarities!!!

James Harris


Somwhat less coherent than your usual, especially as no
one but (perhaps) you knows what you mean by BLOCKED.

However,


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 being non-rational with non-zero rational x.

So let Q(x) = p_1*p_2(x)*P(x)
g_1(x) = f_1(x) + (p_1 - 1)
g_2(x) = f_2(x) + (p_2 - 1)
so Q(x) is a polynomial with integer coefficients
and the g's are non-rational for non-zero rational x.

Then
Q(x) = (g_1(x) + 1) * (g_2(x) +1)

so it might be BLOCKED but it is also possible.

- William Hughes


.

Relevant Pages

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