P(x) = x

^{2}+ 3x + 2 = (x+1)(x+2)

That is such an obvious thing to do it's hard to think of factoring any other way, but of course can have quadratics irreducible over Q, which just means they do not so simply factor, like:

P(x) = x

^{2}+ 5x + 2

Though if P(x) = 0, you can solve for x, but what about just not factoring into a polynomial at all?

What I discovered was that nonpolynomial factorizations were a wide open territory, which I know because they indicated weird things, like consider the following result.

In the ring of algebraic integers the expression:

P(x) = (g

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

_{2}(x) + 2)

is blocked if P(x) is a primitive quadratic irreducible over Q, and g

_{1}(0) = g

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

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

That is something where I recently posted a proof, and had to make a bit of an adjustment to what I had originally as that last bit, is necessary to force a factorization versus having:

P(x) = (g

_{2}(x) + 2) if g

_{1}(x) does equal 0 for all x.

Where of course then there's no factorization! And I was thinking factorization but had to write that down so that it was clear. And the result is weird because it shows algebra being blocked for a particular ring, which is the ring of algebraic integers.

That result has important consequences as it means you cannot display an example of:

P(x) = (g

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

_{2}(x) + 2)

if P(x) is irreducible, the g's equal 0 at 0, and g

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

Looking at what you can show:

P(x) = x

^{2}+ 3x + 2 = (x+1)(x+2)

It's clear that polynomial factorizations work just fine. And notice here all the other requirements are met, as you have: g

_{1}(x) = x, g

_{2}(x) = x

So you can't show that factorization when P(x) is irreducible, but guess what? If you multiply by some nonzero integer other than 1 or -1, and do some simple algebraic manipulations to force symmetry, then you

*can*show the result of that!!!

c

_{1}*P(x) = (f

_{1}(x) + c

_{1})(f

_{2}(x) + c

_{1})

Here c

_{1}is some nonzero integer not 1 or -1, and you can force that factorization, which is not polynomial when P(x) is irreducible over Q, and actually see the results!

And it's worth emphasizing that forcing symmetry is important as that is what makes it possible to get close as you can visually see the f's expressed.

So what does such a nonpolynomial factorization look like? I've used one basic example for years, where I have c

_{1}= 7:

7*P(x) = 7(175x

^{2}- 15x + 2) = (5a

_{1}(x) + 7)(5a

_{2}(x) + 7)

So f

_{1}(x) = 5a

_{1}(x), and f

_{2}(x) = 5a

_{2}(x), where the a's are roots of

a

^{2}- (7x-1)a + (49x

^{2}- 14x) = 0

so the f's are algebraic integer functions, since the polynomial expression is monic, as long as x is an algebraic integer.

And if you want you can solve for the a's using the quadratic formula. So yes, you can see them, though I think it looks a bit messy or I'd show it here.

In contrast g

_{1}(x) cannot be seen directly, but 7*g

_{1}(x) = 5a

_{1}(x), which can be seen, as multiplying by 7 makes it visible. And I matched indices but there's no way really to pick which root is which! So it's more of something that works for human eyes.

The f's are roots of the same quadratic expression, which is forced on them as they're indistinguishable.

So you have this weird thing: the base expression cannot be shown, with actual values though you can show it algebraically in abstract, but you can show with actual equations if you multiply by an integer constant.

So we can always see:

P(x) = (g

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

_{2}(x) + 2)

as an abstraction, but we can't fill in the g's, if P(x) has integer coefficients but doesn't factor into two polynomial factors with integer coefficients, if we don't allow g

_{1}(x) equal 0 for all x.

But we can multiply P(x) by some integer and display the f's, when:

c

_{1}*P(x) = (f

_{1}(x) + c

_{1})(f

_{2}(x) + c

_{1})

As long as c

_{1}is not zero, 1 or -1.

However, what we can see displayed in a way familiar to our human eyes does not change what exists! And of course in the field of complex numbers it exists, but we still can't see it displayed.

So the ring of algebraic integers doesn't tell us what doesn't exist, only what we can

*see*in a familiar way or not.

So you may wonder: what happens if we try to force the issue and see it anyway?

Well the multiplying integer is entangled. It's not visually separable.

For example with:

a

^{2}- (7x-1)a + (49x

^{2}- 14x) = 0

If you try x = 1, that is:

a

^{2}- 6a + 35 = 0, which solves as a = 3 +/- sqrt(-26)

The 7 is entangled in there and NOT separable to our human eyes. But mathematically it is possible to remove it. So it's something we can't see that the math can do anyway.

That the math can remove that 7 as easily as we can from polynomials is fascinating to me, as what our minds can visualize is not a limitation mathematically, as the math is not human.

This result shifted my thinking about mathematics as it reveals a fundamental limitation of creatures like us to visualize mathematics. It also shows how our biases for the familiar can lead us away from surprising answers!

For most people doing polynomial factorizations is so obvious it might never occur to them to do a nonpolynomial one, whereas for the math, one thing is hardly different from the other.

James Harris