### Thursday, October 30, 2008

## Proof of problem with ring of algebraic integers

Some simple algebra with a basic polynomial reveals a serious problem with the naive use of the ring of algebraic integers, bringing into question over a hundred years of algebraic number theory. This post will step through the extremely, short, simple, but overwhelming proof of the problem.

In an integral domain, consider a simple polynomial

P(x) = 175x^2 - 15x + 2. Multiply it times 7, to get

7*P(x) = 1225x^2 - 105x + 14. Cleverly re-group terms:

1225x^2 - 105x + 14 = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2

and now factor into non-polynomials:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where you'll note that the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

But now consider our polynomial again with a factorization before any multiplying by 7:

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Now multiply by 7, to get

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5(7*b_2(x))+ 7)

and use the substitutions b_1(x) = c_1(x) + 1, and 7*b_2(x) = c_2(x), and you have

7*(175x^2 - 15x + 2) = (5c_1(x) + 7)(5c_2(x)+ 7)

and of course if c_1(x) = a_1(x) and c_2(x) = a_2(x), I have my original factorization, but in so doing I'm PICKING that 7 multiplies times just one of the factors of 175x^2 - 15x + 2, but what if I picked wrong?

For instance, consider again

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

and again multiply times 7, but split it up so that each factor is multiplied times sqrt(7):

7*(175x^2 - 15x + 2) = (5*sqrt(7)*b_1(x) + 2*sqrt(7))(5*sqrt(7)b_2(x) + sqrt(7))

but there's an immediate problem!

If you let x=0, then you have the factorization:

7*(2) = (5*sqrt(7)*b_1(0) + 2*sqrt(7))(5*sqrt(7)b_2(0)+ sqrt(7))

which contradicts at x= 0 with

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

unless b_2(0) divides off sqrt(7), or b_1(0) divides off sqrt(7), as

7*(2) = (5a_1(0) + 7)(5a_2(0)+ 7) = (0 + 7)(-5+ 7)

because then the a's are roots of

a^2 + a = 0.

Therefore, there is no other way to multiply

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

by 7, and get the factorization

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

as ANY other way other than multiplying (5b_2(x)+ 1) by 7, will contradict with

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

at x=0, as demonstrated above with sqrt(7).

Therefore, one of the a's must have 7 as a factor for all x, but it is trivial to show that NEITHER of them can have 7 as a factor for any integer x, for which the a's are not rational, in the ring of algebraic integers, so there is proven a problem with that ring.

QED.

Note that attacking this proof requires that you show how given any factorization

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

you can get to the factorization

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

without multiplying (5b_2(x)+ 1), by 7.

I emphasize that as mathematicians have resisted this result for years now, and in fact, destroyed an entire mathematical journal to avoid it, killing SWJPAM, and smeared me for years rather than accept it.

Since the result cannot be refuted mathematically their only choice will be to delete out the mathematics in reply, and address non-issues, because the mathematical proof is so easy.

When they do that I want you to understand they are deliberately doing so, avoiding a simple proof in order to hold on to an error, and to keep teaching it to their students which is as basic a betrayal as a professor in any field can make: to deliberately train your students in error.

In an integral domain, consider a simple polynomial

P(x) = 175x^2 - 15x + 2. Multiply it times 7, to get

7*P(x) = 1225x^2 - 105x + 14. Cleverly re-group terms:

1225x^2 - 105x + 14 = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2

and now factor into non-polynomials:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where you'll note that the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

But now consider our polynomial again with a factorization before any multiplying by 7:

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Now multiply by 7, to get

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5(7*b_2(x))+ 7)

and use the substitutions b_1(x) = c_1(x) + 1, and 7*b_2(x) = c_2(x), and you have

7*(175x^2 - 15x + 2) = (5c_1(x) + 7)(5c_2(x)+ 7)

and of course if c_1(x) = a_1(x) and c_2(x) = a_2(x), I have my original factorization, but in so doing I'm PICKING that 7 multiplies times just one of the factors of 175x^2 - 15x + 2, but what if I picked wrong?

For instance, consider again

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

and again multiply times 7, but split it up so that each factor is multiplied times sqrt(7):

7*(175x^2 - 15x + 2) = (5*sqrt(7)*b_1(x) + 2*sqrt(7))(5*sqrt(7)b_2(x) + sqrt(7))

but there's an immediate problem!

If you let x=0, then you have the factorization:

7*(2) = (5*sqrt(7)*b_1(0) + 2*sqrt(7))(5*sqrt(7)b_2(0)+ sqrt(7))

which contradicts at x= 0 with

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

unless b_2(0) divides off sqrt(7), or b_1(0) divides off sqrt(7), as

7*(2) = (5a_1(0) + 7)(5a_2(0)+ 7) = (0 + 7)(-5+ 7)

because then the a's are roots of

a^2 + a = 0.

Therefore, there is no other way to multiply

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

by 7, and get the factorization

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

as ANY other way other than multiplying (5b_2(x)+ 1) by 7, will contradict with

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

at x=0, as demonstrated above with sqrt(7).

Therefore, one of the a's must have 7 as a factor for all x, but it is trivial to show that NEITHER of them can have 7 as a factor for any integer x, for which the a's are not rational, in the ring of algebraic integers, so there is proven a problem with that ring.

QED.

Note that attacking this proof requires that you show how given any factorization

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

you can get to the factorization

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are functions of x that are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

without multiplying (5b_2(x)+ 1), by 7.

I emphasize that as mathematicians have resisted this result for years now, and in fact, destroyed an entire mathematical journal to avoid it, killing SWJPAM, and smeared me for years rather than accept it.

Since the result cannot be refuted mathematically their only choice will be to delete out the mathematics in reply, and address non-issues, because the mathematical proof is so easy.

When they do that I want you to understand they are deliberately doing so, avoiding a simple proof in order to hold on to an error, and to keep teaching it to their students which is as basic a betrayal as a professor in any field can make: to deliberately train your students in error.