In the ring of algebraic integers the expression:
P(x) = (g1(x) + 1)(g2(x) + 2)
is blocked if P(x) is a primitive quadratic irreducible over Q, and g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.
Let P(x) = (g1(x) + 1)(g2(x) + 2)
where P(x) is a quadratic irreducible over Q, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.
Introduce new functions f1(x), and f2(x), and non-zero integers c1 and c2, where I'll use the second one first, as let:
g2(x) = f2(x) + c2
where c1 = c2 + 2
Now make the substitution:
P(x) = (g1(x) + 1)(f2(x) + c2 + 2) = (g1(x) + 1)(f2(x) + c1)
Now multiply both sides by c1,
c1*P(x) = (c1g1(x) + c1)(f2(x) + c1)
And now, let g1(x) = f1(x)/c1, and make that substitution to get:
c1*P(x) = (f1(x) + c1)(f2(x) + c1)
Indistinguishable, the f's must be roots of the same polynomial function, as symmetry has been forced.
But now the requirement that g1(x) = f1(x)/c1, means it is blocked in the ring of algebraic integers unless f1(x) is itself a polynomial function. But if P(x) is irreducible over Q, then that is not possible; therefore, g1(x) cannot be an algebraic integer function.
Here's an example of the blocking with actual equations that fulfill all of the restrictions.
Let c2 = 5, f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of
a2 - (7x-1)a + (49x2 - 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) = 175x2 - 15x + 2
which is not reducible over Q.
Notice you can see the consequences if you let x = 1, then a2 - 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.