Tuesday, July 31, 2007
JSH: Blocking and ring of algebraic integers
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)
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 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!!!
Notice that the requirements on the functions is to prevent them from being polynomial or close enough to polynomial functions that the result is obscured.
Without them, for instance
P(x) = -x+1 = (-sqrt(x) + 1)*(sqrt(x) + 1)
is a simple counterexample.
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)
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 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!!!
Notice that the requirements on the functions is to prevent them from being polynomial or close enough to polynomial functions that the result is obscured.
Without them, for instance
P(x) = -x+1 = (-sqrt(x) + 1)*(sqrt(x) + 1)
is a simple counterexample.
# posted by James Harris @ 20:27 
