Consider:

P(x) = (g

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

_{2}(x) + 2)

where P(x) is a primitive quadratic irreducible over Q, g

_{1}(0) = g

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

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

If you think that is without problems, watch as I do some simple algebra. So at this point you just have a polynomial which is a product of two functions of x, which are normalized, how hard can that be?

Let's play.

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) + 5.

No objections there, right? If there are, go away, you don't know algebra.

Now make the substitution:

P(x) = (g

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

_{2}(x) + 5 + 2) = (g

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

_{2}(x) + 7)

Wow, boring. But let's multiply both sides by 7, why? Just follow along...

7*P(x) = (7g

_{1}(x) + 7)(f

_{2}(x) + 7)

If you're wondering why I multiplied by 7 the way I did, you're not thinking symmetry!

As now of course, let g

_{1}(x) = f

_{1}(x)/7, and make that substitution to get:

7*P(x) = (f

_{1}(x) + 7)(f

_{2}(x) + 7)

And we're done.

It turns out that if P(x) is an irreducible quadratic polynomial, then g

_{1}(x) does NOT exist in the ring of algebraic integers.

Neat!

Here's an example of the blocking with actual equations that fulfill all of the restrictions.

Let 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. That then requires that

P(x) = 175x

^{2}- 15x + 2

which is not reducible over Q.

What's wrong with that? Well it implies that one of the f's has 7 as a factor, which it cannot in the ring of algebraic integers, except for at x = 0.

Why do you have a problem? Goes back to the asymmetry in the beginning:

P(x) = (g

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

_{2}(x) + 2)

Turns out you can't have the 1 and 2 with normalized functions of x, and be in the ring of algebraic integers if P(x) is irreducible.

Of course if P(x) IS reducible, then you can, like have g

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

_{2}(x) = x, as simple examples that ARE possible. Then, of course P(x) = x

^{2}+ 3x + 2.

But if P(x) is

*irreducible*, then P(x) = (g

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

_{2}(x) + 2) cannot exist in the ring of algebraic integers. It's what I call a non-polynomial factorization.

So notice, algebraically we can write the expression, but regardless if P(x) is irreducible it's not allowed to exist within the ring of algebraic integers.

The ring of algebraic integers actually blocks algebra in that situation.

However, the expression

*does*exist in the field of complex numbers!

Oh yeah, notice I chose to multiply by 7 above, but could have used 5, or 11 or any other nonzero integer, so why exactly is the ring of algebraic integers blocking?

Inquiring minds would want to know!!!

James Harris