One of my favorite demonstrations uses a seemingly simple construction:

P(x) = (g

_{1}(x) + 1)(g

_{2}(x) + 2)

where P(x) is a quadratic with integer coefficients, g

_{1}(0) = g

_{2}(0) = 0, but g

_{1}(x) does not equal 0 for all x.

Turns out that expression is in the ring of algebraic integers, only if the g's are linear functions, which is equivalent to a trivial factorization.

For example, like have g

_{1}(x) = x, and g

_{2}(x) = x, then, of course P(x) = x

^{2} + 3x + 2.

But I pushed the envelope by imagining the quadratic is what's called primitive and irreducible over Q, which means that it does NOT factor into polynomials.

And I

showed stepping through some simple algebraic manipulations where I multiplied by 7, but this time I will multiply by 11:

Introduce new functions f

_{1}(x), and f

_{2}(x), where I'll use the second one first, as let

g

_{2}(x) = f

_{2}(x) + 9.

Now make the substitution:

P(x) = (g

_{1}(x) + 1)(f

_{2}(x) + 9 + 2) = (g

_{1}(x) + 1)(f

_{2}(x) + 11)

Multiply both sides by 11:

11*P(x) = (11g

_{1}(x) + 11)(f

_{2}(x) + 11)

Now, let g

_{1}(x) = f

_{1}(x)/11, and make that substitution to get:

11*P(x) = (f

_{1}(x) + 11)(f

_{2}(x) + 11)

Now multiply it out:

11*P(x) = f

_{1}(x)*f

_{2}(x) + 11(f

_{1}(x) + f

_{2}(x)) + 121

And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:

f

_{1}(x) + f

_{2}(x) = H(x), so: f

_{2}(x) = -f

_{1}(x) + H(x),

and make that substitution, to get:

11*P(x) = f

_{1}(x)*(-f

_{1}(x) + H(x)) + 11H(x) + 121

So you have:

11*P(x) = -f

_{1}^{2}(x) + H(x)f

_{1}(x) + 11H(x) + 121

Which means:

f

_{1}^{2}(x) - H(x)f

_{1}(x) - 11H(x) - 121 + 11*P(x) = 0

And you can solve for f

_{1}(x) using the quadratic formula.

Notice you can see the ambiguity between the f's right there, which prevents you from knowing which one has 11 as a factor, as of course looks the same with f

_{2}.

And if H(x) is in the ring of algebraic integers, since the coefficients of P(x) are integers, f

_{1}(x) must be in the ring of algebraic integers, for algebraic integer x.

Notice you can start with:

P(x) = (g

_{1}(x) + 1)(g

_{2}(x) + 2)

and multiply by 13 if you wish, as of course it is bare. What you multiply with is

*human choice*. I've chosen 7 and 11, for my own reasons. And you may now wonder, so how do you pick H(x)?

Does it matter?

One way of looking at it, going in this direction, you can choose H(x) and in that way force the g's, which means it is a

*handle* on the original factorization.

Your choice of H(x) tells the math how the original factorization must go.

Are there any circumstances under which the f's that result from your choice could NOT be in the ring of algebraic integers?

Well the only way is if you can deliberately make H(x) not be in the ring of algebraic integers, for instance if you have it be a polynomial and can pick coefficients outside of the ring, but even if you try I think it's kind of hard to do and have all the equations work. For instance since f

_{1}(0) = 0, and f

_{2}(0) = -9, you know it must be that H(0) = -9.

Actually I don't know if you can force H(x) to not be an algebraic integer for algebraic integer x and have all the equations work. But even if you can, you'd have to be an ornery (and clever) person who just made that happen.

With my own research H(x) has been linear with integer coefficients.

Those original g's are weird then, eh? Solutions for algebraic integer x cannot all be seen, as they are not both expressible as algebraic integers, so can't be seen directly, but you can multiply the one that is not by an integer and see the result!

And they're not too picky about the integer either. It just needs to be nonzero and not 1 or -1, and they will appear, as the result is an algebraic integer.

I came up with the

object ring to categorize these peculiar numbers, and call them and integers, objects. So you can say mathematically that given an object o

_{1}, which is a solution to one of the g's from my original factorization which is not an algebraic integer, it is visible with 7o

_{1} or 11o

_{1} with the examples I've used. It doesn't care.

Oh yeah, so what about o

_{2}? Well it is an algebraic integer, but which is which? No way to tell. But, yup, o

_{1}o

_{2} is an algebraic integer.

Non algebraic integer objects are fascinating critters!

Their refusal to be seen directly is NOT intuitive.

These non-rational objects which behave integer-like but are not algebraic integers are an undiscovered country of number theory. If you don't believe they exist and are being befuddled by the idea of fractions then how can I multiply the same number by 7 or 11 to get an algebraic integer? Would it help if I did it again with 9? Or 16?

James Harris