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, for certain solutions.

(Updating October 26, 2017: Original argument was in error as assumes c

_{1}is a non-unit integer.)

**Proof:**

Let P(x) = (g

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

_{2}(x) + 2)

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

_{1}(0) = g

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

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

Introduce new functions f

_{1}(x), and f

_{2}(x), and non-zero integers c

_{1}and c

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

g

_{2}(x) = f

_{2}(x) + c

_{2}

where c

_{1}= c

_{2}+ 2

Now make the substitution:

P(x) = (g

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

_{2}(x) + c

_{2}+ 2) = (g

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

_{2}(x) + c

_{1})

Now multiply both sides by c

_{1},

c

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

_{1}g

_{1}(x) + c

_{1})(f

_{2}(x) + c

_{1})

And now, let g

_{1}(x) = f

_{1}(x)/c

_{1}, and make that substitution to get:

c

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

_{1}(x) + c

_{1})(f

_{2}(x) + c

_{1})

Indistinguishable, the f's must be roots of the same polynomial function, as symmetry has been forced.

But now the requirement that g

_{1}(x) = f

_{1}(x)/c

_{1}, means it is blocked in the ring of algebraic integers if c

_{1}is not 1 or -1, unless f

_{1}(x) is itself a polynomial function. But if P(x) is irreducible over Q, then that is not possible; therefore, g

_{1}(x) cannot be an algebraic integer function.

*Proof complete*.

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

Let c

_{2}= 5, 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.

Notice you can see the consequences if you let x = 1, then a

^{2}- 6a + 35 = 0, which would imply that just one of the roots has 7 as a factor, which is NOT possible in the ring of algebraic integers, as in that ring neither can have 7 as a factor.

So it is proven that algebra that is allowed for polynomial factorizations is blocked within the ring of algebraic integers for non-polynomial factorizations.

James Harris

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