### Saturday, August 30, 2008

## Algebraic integer result with quadratics

A rather remarkable bit of mathematics casts a darker view on the ring of algebraic integers where quadratics thankfully are useful for a somewhat subtle result, and the distributive property is almost bizarrely the linchpin of the proof.

Consider in an integral domain

P(x) = 175x^2 - 15x + 2

and

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

where the a's are roots of

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

So I have a non-monic quadratic with integer coefficients which is multiplied times 7 and then factored using functions given by the roots of a monic quadratic expression, so I've entangled one quadratic with a quadratic generator in such a way as to allow myself to probe at the ring of algebraic integers in a way never before done without such remarkable tools.

Of course with a regular polynomial factorization like

7*(x^2 + 3x + 2) = (7x + 7)*(x+2)

it's trivial to just divide the 7 off to get

(x^2 + 3x + 2) = (x + 1)*(x+2)

but by entangling the two expressions and making the functional values roots of monic quadratics with integer coefficients when x is an integer, I've managed to FORCE a division of the 7 across non-rational solutions, and in that way remove the ability to in general divide it off from

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

in the ring of algebraic integers.

That was done by a clever re-grouping of terms:

7*P(x) = 1225x^2 - 105x + 14

and with the re-grouping I have

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2

to get

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)

which clearly gives

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

But what's the point of the exercise?

Well by entangling the function into the roots of a quadratic generator I can put the distributive property to the test! As there isn't a lot of difference really between

7*(x^2 + 3x + 2) = (7x + 7)*(x+2)

and

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, because the distributive property doesn't care what is being multiplied, so in the first case you have (x+1)*(x+2), while in the second you have some weird unseen factorization where more complicated than linear functions are involved as yes, x+1 and x+2 are just linear functions.

So in EACH case you have FUNCTIONS involved but in the more complicated case I forced the functions to be more complicated by making them roots of a quadratic generator.

So the two examples are the same mathematically except in the second case the functions are not linear, which is why it's a non-polynomial factorization, which allows non-rational solutions with rational x.

So as they are mathematically the same in all key ways, not surprisingly, in each case you simply have 7 multiplied rather simply, which can be seen with the more complex example by letting x=0 as then

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

gives

a^2 + a = 0

so one of the a's is 0 and the other is -1, so I have from

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

that

7*P(0) = (7)*(2) or (2)*(7).

So one of (5a_1(x) + 7) or (5a_2(x)+ 7) is coprime to 7 at that value just like with

7*(x^2 + 3x + 2) = (7x + 7)*(x+2)

one of them is coprime to 7 at x=0, and since I've mentioned coprimeness I'll go ahead and say that we'll try to be in the ring of algebraic integers.

I say try because you may already know that now there is a HUGE problem as if integer x generates a quadratic that is irreducible over Q, then you already know that NEITHER of the a's can have 7 as a factor in that ring!!!

For example with x=1, the result would need that for

a^2 - 6a + 35 = 0

one of the roots has 7 as a factor, but provably NEITHER of them do in the ring of algebraic integers.

And that is the quick demonstration of the major issue over which so many arguments have raged in the roughly six years or so since I discovered the problem.

After pondering things for a while I concluded that the problem was not with my understanding of the distributive property—which is after all rather basic—but with the ring of algebraic integers itself, which lead me to wondering what ring might be ok, and years ago I came up with what I call the ring of objects:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:

P(x) = (5b_1(x) + 1)(5a_2(x)+ 7) or (5a_1(x) + 7)(5b_2(x)+ 1)

where a_1(x) = 7*b_1(x) or a_2(x) = 7*b_2(x),

where you have an ambiguity as to which one unless you have rational values for the functions.

AND I went ahead and figured out how the ring of algebraic integers can allow dividing off of the 7 as well, using what I call wrappers, and you can find my research on that subject by searching in Google on "wrapper theorem".

The problem though with this issue is that it shows that the ring of algebraic integers has additional properties beyond those previously known, and you can do some odd things with it.

Unfortunately some of those things include coming up with mathematical arguments that are not actually proofs as they rely on the quirks of the ring to imply something false.

Consider in an integral domain

P(x) = 175x^2 - 15x + 2

and

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

where the a's are roots of

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

So I have a non-monic quadratic with integer coefficients which is multiplied times 7 and then factored using functions given by the roots of a monic quadratic expression, so I've entangled one quadratic with a quadratic generator in such a way as to allow myself to probe at the ring of algebraic integers in a way never before done without such remarkable tools.

Of course with a regular polynomial factorization like

7*(x^2 + 3x + 2) = (7x + 7)*(x+2)

it's trivial to just divide the 7 off to get

(x^2 + 3x + 2) = (x + 1)*(x+2)

but by entangling the two expressions and making the functional values roots of monic quadratics with integer coefficients when x is an integer, I've managed to FORCE a division of the 7 across non-rational solutions, and in that way remove the ability to in general divide it off from

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

in the ring of algebraic integers.

That was done by a clever re-grouping of terms:

7*P(x) = 1225x^2 - 105x + 14

and with the re-grouping I have

7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2

to get

7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)

which clearly gives

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

But what's the point of the exercise?

Well by entangling the function into the roots of a quadratic generator I can put the distributive property to the test! As there isn't a lot of difference really between

7*(x^2 + 3x + 2) = (7x + 7)*(x+2)

and

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, because the distributive property doesn't care what is being multiplied, so in the first case you have (x+1)*(x+2), while in the second you have some weird unseen factorization where more complicated than linear functions are involved as yes, x+1 and x+2 are just linear functions.

So in EACH case you have FUNCTIONS involved but in the more complicated case I forced the functions to be more complicated by making them roots of a quadratic generator.

So the two examples are the same mathematically except in the second case the functions are not linear, which is why it's a non-polynomial factorization, which allows non-rational solutions with rational x.

So as they are mathematically the same in all key ways, not surprisingly, in each case you simply have 7 multiplied rather simply, which can be seen with the more complex example by letting x=0 as then

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

gives

a^2 + a = 0

so one of the a's is 0 and the other is -1, so I have from

7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)

that

7*P(0) = (7)*(2) or (2)*(7).

So one of (5a_1(x) + 7) or (5a_2(x)+ 7) is coprime to 7 at that value just like with

7*(x^2 + 3x + 2) = (7x + 7)*(x+2)

one of them is coprime to 7 at x=0, and since I've mentioned coprimeness I'll go ahead and say that we'll try to be in the ring of algebraic integers.

I say try because you may already know that now there is a HUGE problem as if integer x generates a quadratic that is irreducible over Q, then you already know that NEITHER of the a's can have 7 as a factor in that ring!!!

For example with x=1, the result would need that for

a^2 - 6a + 35 = 0

one of the roots has 7 as a factor, but provably NEITHER of them do in the ring of algebraic integers.

And that is the quick demonstration of the major issue over which so many arguments have raged in the roughly six years or so since I discovered the problem.

After pondering things for a while I concluded that the problem was not with my understanding of the distributive property—which is after all rather basic—but with the ring of algebraic integers itself, which lead me to wondering what ring might be ok, and years ago I came up with what I call the ring of objects:

The object ring is defined by two conditions, and includes all numbers such that these conditions are true:

- 1 and -1 are the only rationals that are units in the ring.
- Given a member m of the ring there must exist a non-zero member n

such that mn is an integer, and if mn is not a factor of m, then n

cannot be a unit in the ring.

P(x) = (5b_1(x) + 1)(5a_2(x)+ 7) or (5a_1(x) + 7)(5b_2(x)+ 1)

where a_1(x) = 7*b_1(x) or a_2(x) = 7*b_2(x),

where you have an ambiguity as to which one unless you have rational values for the functions.

AND I went ahead and figured out how the ring of algebraic integers can allow dividing off of the 7 as well, using what I call wrappers, and you can find my research on that subject by searching in Google on "wrapper theorem".

The problem though with this issue is that it shows that the ring of algebraic integers has additional properties beyond those previously known, and you can do some odd things with it.

Unfortunately some of those things include coming up with mathematical arguments that are not actually proofs as they rely on the quirks of the ring to imply something false.