Harristotelian Logic

So I've been going on lately about what I call wrappers in the ring of algebraic integers, and clearly I'm excited about the explanation, but what does it actually mean? Researchers can be opaque at times in areas worked for years so I'm going to try and explain for those of you who haven't been pondering non-polynomial factorization for years.

It turns out that you can go beyond boring polynomial factorizations like

P(x) = x2 + 3x + 2 = (x+2)*(x+1)

into a wider world that I call non-polynomial factorization where you factor some polynomial into other than polynomial factors, excluding trivial extensions like

P(x) = x+1 = (sqrt(x) + 1)(sqrt(x) - 1)

which is still too close to a known polynomial factorization (you can see exclusions of such factorizations in my previous post on wrappers).

So is that even possible? How can you even have non-polynomial factorizations of polynomials? Just sounds odd, right?

So here's an example of one:

7*P(x) = 7*(175x2 − 15x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7)

where one of an infinity of possible solutions for the a's is as roots of

a2 − (7x + 4)*a + (49x2 − 14x − 7) = 0.

So now you can see how the a's are not themselves polynomials nor are they square roots of polynomials and the major issue for years about examples like the above when I argue with other people has been, what happens if you divide 7 off from both sides?

Well, the answer from my latest posts so that you can get a handle on the meaning is that you cannot divide the 7 off in general in the ring of algebraic integers.

So like you can't get to something like

P(x) = x2 + 3x + 2 = (x+2)*(x+1)

where you just have P(x) factored, and it turns out that you cannot show a factorization of P(x) in the ring of algebraic integers with these weird non-polynomial factorizations without multiplying it times some number like I did with 7.

The wrapper theorem I've been working on explains what the ring of algebraic integers does if you try.

One of the claims by people arguing with me is that the 7 is the product of two functions, forced by the quirkiness of the ring, which is an intriguing idea but it can't save you from being unable to show the factorization in general, and these people have dodged the issue by instead just saying that for any particular value of x, you can find numbers in the ring of algebraic integers to divide off.

ONE simple challenge, of course to those who claim the ring of algebraic integers is NOT blocking the factorization is to simply give the solution when the 7 is divided off.

Remember the non-polynomial factorization is

7*P(x) = 7*(175x2 − 15x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7)

where one of an infinity of possible solutions for the a's is as roots of

a2 − (7x + 4)*a + (49x2 − 14x − 7) = 0

and by the wrapper theorem you cannot in general divide the 7 off and give a solution for the result.

You may pick some value for x to appear to divide it off for a particular value but that is allowed by what I call wrappers, which have to shift from number to number so there is no general solution.

I call them wrappers because they multiply times the actual factors, wrapping them up in such a way that you can remain in the ring of algebraic integers, but they only work once.

You need new wrappers as you keep changing x.

I say wrapper theorem meaning that it is an absolute truth, so the clearest way to show I am wrong in that belief is to divide the 7 off with the solution I have for the a's in the ring of algebraic integers.

The wrapper theorem says it cannot be done in the ring of algebraic integers.

So it cannot be done in that ring. That is the absolute in front of you.