Re: JSH: Wrappers in ring of algebraic integers

James Harris wrote:

Whew!

We've heard that before.

[...]

The full remarkable result is that

P(x) = (g_1(x) + 1)*(g_2(x) + 1)

CANNOT exist in the ring of algebraic integers when the g's are non-
rational with non-zero rational x.

REMEMBER P(x) is a polynomial with integer coefficients!!!

Counterexample: let

P(x) = -2x^2 + 1

g_1(x) = Sqrt(2)*x

g_2(x) = -Sqrt(2)*x

Whew!!! Good bit of thinking was required to figure out the details
here.

Intriguingly, seeing everything the ring of algebraic integers has to
do to comply with its definition of only having roots of monic
polynomials with integer coefficients requires use of my object ring,
where you can watch it work in detail, and you can see that detail on
my math blog.

I quote from that blog below:

***Begin Quote***

In an integral domain let

d1d2P(x) = (f1(x) + d1)(f2(x) + d2)

where the d's are non-zero integers, P(x) is a polynomial with integer
coefficients where P(0) is coprime to d1 and d2, and where f1(0) =
f2(0) = 0.

If the ring is the ring of algebraic integers, with non-zero integer
x, and d1 = p1 and d2 = p2 where p1 and p2 are differing prime
numbers, and f1(x) and f2(x) are non-rational there cannot exist g1(x)
and g2(x) such that

d1d2P(x) = (d1g1(x) + d1)(d2g2(x) + d2)

where

d1g1(x) = f1(x) and d2g2(x) = f2(x).

***End Quote***

A counterexample to this theorem is easily constructed from the
example I gave above. More from your blog:

Proof:
Since P(x) is a polynomial with integer coefficients f1(x) and f2(x)
must have values that are roots of the same polynomial.

What? If x is a fixed algebraic integer then P(x) is a number, not a
polynomial. If it is a variable, then f1(x) and f2(x) are functions,
not numbers. Do you mean that, for any algebraic integer x, f1(x) and
f2(x) are roots of the polynomial P? If so then this quite simply
isn't true, why do you claim it is?

But if f1(x) has d1 as a factor and f2(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.

You can find a non-monic primitive polynomial with integer
coefficients irreducible over the rationals such that *what*? I'm sure
that there is no theorem of number theory that no non-monic
polynomials with integer coefficients irreducible over rationals
exist.

To summarise: what on Earth are you talking about?

.

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