Re: Some math, algebraic integers

From: *** T. Winter (***.Winter_at_cwi.nl)
Date: 09/20/04 

Date: Mon, 20 Sep 2004 14:13:24 GMT

In article <3c65f87.0409200232.4da4f358@posting.google.com> jstevh@msn.com (James Harris) writes:
...
> The result that is being claimed is rather basic and odd, and while I
> gave one simple quadratic--quite deliberately--it's easy to give
> another.
>
> The assertion by posters "Nora Baron", W. Dale Hall, and Arturo
> Magidin is that given
>
> x^2 + Ax + B
>
> where A and B are integers, iff the quadratic is reducible, then roots
> of it CAN be coprime to a prime factor of B, but if the quadratic is
> irreducible over Q, that is, if it has irrational roots, then their
> claim is that NONE of its roots can be coprime to any of the prime
> factors of B.

Yes, that is a very basic result of Galois theory. If you had looked
at the answers you receive you would have seen a few proofs of this
claim. Moreover, in *all* cases that you came up with that you thought
would contradict it, it has been shown that they indeed did show that
none of its root can be coprime to any of the prime factors of B.

> The rational case is easy to show as consider A = 5, B=6, which gives
> x^2 + 5x + 6 = (x+2)(x+3)
> and note that the root -2 is coprime to the root -3, so a root of that
> quadratic CAN be coprime to a prime factor of B.

Yup. That is a reducible quadratic. So it is possible indeed. Now
come up with a monic irreducible quadratic.

> If they were right, irrationals would have a special property, which
> would basically come out of the blue with no apparent mathematical
> reasoning for it.

Nope. It comes from the specific properties of irreducible polynomials
for a very good mathematical reason.

> It'd have to be a VERY DEEP property of irrationals.

It is a property of the polynomials.

> Like imagine s_1 and s_2 such that
> x^2 + (2s_1 + 3s_2)x + 6s_1 s_2 = (x + 2s_1)(x + 3s_2)
> where s_1 s_2 = 1, and 2s_1 + 3s_2 is an integer.

Again possibly a reducible quadratic, namely when 2s_1 and 3s_2 are
both integer.

> Now one might assume that s_1 and s_2 can be units

Because s_1 s_2 = 1, s_1 and s_2 *are* units in the ring they are in.

> such that you can
> maintain a situation where the roots are coprime to each other AND to
> 2 or 3, but the position of "Nora Baron", W. Dale Hall, and Arturo
> Magidin is that EACH root MUST have non-unit factors of BOTH 2 and 3.

Let's have s_1 = 3/2 and s_2 = 2/3, we have 2s_1 + 3s_2 = 5 is integer,
s_1 s_2 = 1, and the polynomial is x^2 + 5x + 6 = (x + 3)(x + 2), reducible.
It is only true when the polynomial is irreducible. But actually your
use of s_1 and s_2 is just obfuscation.

> The actuality is that ***in the ring of algebraic integers***it is
> impossible for both s_1 and s_2 to be in that ring, if they are
> irrational and because of that s_1 can't be a unit in the ring of
> algebraic integers!

Eh? So what? It is 2s_1 and 3s_2 that are algebraic integer. Why
must s_1 and s_2 be algebraic integers? 

-- 
*** t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~***/
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