In an integral domain let

d

_{1}d

_{2}P(x) = (f

_{1}(x) + d

_{1}c

_{1})(f

_{2}(x) + d

_{2}c

_{2})

where the d's and c's are non-zero integers, where the d's are coprime to the c's and c

_{1}is coprime to c

_{2}, P(x) is a non-monic primitive polynomial with integer coefficients, and where f

_{1}(0) = f

_{2}(0) = 0 so P(0) is coprime to d

_{1}and d

_{2}, as then P(0) = c

_{1}c

_{2}.

If the ring is the ring of algebraic integers, with non-zero integer x, and d

_{1}= p

_{1}and d

_{2}= p

_{2}where p

_{1}and p

_{2}are differing prime numbers, and f

_{1}(x) and f

_{2}(x) are non-rational there cannot exist g

_{1}(x) and g

_{2}(x) such that

d

_{1}d

_{2}P(x) = (d

_{1}g

_{1}(x) + d

_{1}c

_{1})(d

_{2}g

_{2}(x) + d

_{2}c

_{2})

where

d

_{1}g

_{1}(x) = f

_{1}(x) and d

_{2}g

_{2}(x) = f

_{2}(x).

Proof:

Since P(x) is a polynomial with integer coefficients f

_{1}(x) and f

_{2}(x) must have values that are roots of the same polynomial. But if f

_{1}(x) has d

_{1}as a factor and f

_{2}(x) does not, then you can find a non-monic primitive polynomial with integer coefficients irreducible over rationals, which contradicts a well established elementary theorem in number theory that none exists.

So then, how can you divide off d

_{1}and d

_{2}?

One way is for the ring to employ what I call wrappers.

Consider w

_{1}and w

_{2}, where

d

_{1}d

_{2}P(x) = ((w

_{1}d

_{1})(w

_{1}(g

_{1}(x)+c

_{1}-d

_{2}) + (w

_{1})

^{2}d

_{1}d

_{2})((w

_{2}d

_{2})(w

_{2}(g

_{2}(x)+c

_{2}-d

_{1}) + (w

_{2})

^{2}d

_{1}d

_{2})

where w

_{1}d

_{1}and w

_{2}d

_{2}are roots of a monic polynomial with integer coefficients, so those products are in the ring of algebraic integers while the factors are not.

In my ring of objects the wrappers are units and w

_{1}w

_{2}= 1 or -1, while they are NOT units in the ring of algebraic integers.

So the ring of algebraic integers can multiply the factors (f

_{1}(x) + d

_{1}) and (f

_{2}(x) + d

_{2}) by numbers that are units in the object ring while also shifting them internally so that it is possible to divide off the d's.

What is left is then

P(x) = ((w

_{1}(g

_{1}(x)+c

_{1}-d

_{2}) + w

_{1}d

_{2})((w

_{2}(g

_{2}(x)+c

_{2}-d

_{1}) + w

_{2}d

_{1})

where that detail is available in the object ring as the individual elements are not all available in the ring of algebraic integers.

James Harris

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