Re: JSH: Critique means slow, and thorough



jstevh@xxxxxxx wrote: 
W. Dale Hall wrote:

jstevh@xxxxxxx wrote: 


	... stuff deleted ...

Then coprimeness in the ring of algebraic integers does not mean
coprimeness in the more inclusive ring.


This is incorrect. If R and R' are commutative rings with identity,
R a subring of R', then whenever elements r,s of R are coprime in R,
then they are coprime in R'.


But if you don't realize that possibility, and worse, assume that
you've included all rings where the two key properties hold, you
can have an odd thing, where you can prove "coprimeness" by relying on coprimeness in the ring of algebraic integers, and algebraically
find that two numbers are not coprime, and thus have the appearance of proving two different and opposite things.


Can't happen.  Here's why.

Let r,s be coprime in R. That means there are u and v in R such that

		ur + vs = 1.

Now, r,s and u,v are in R. R is a subring of R'. Thus, r,s and u,v
are elements of R'. The operations of R extend to operations of R', so the equation

		ur + vs = 1

also holds in R'. Thus, r and s are coprime in R'.



That's correct, but can you now prove that for every case

a/b

where a is an algebraic integer and b is an algebraic integer, and a is
coprime to b that you can find a construction

ax + by = 1

where x and y are algebraic integers?


Yes, and I've included a proof below. It relies on a difficult (for me)
result that I believe is due to Dedekind: the finiteness of the class
number for any number field (= finite extension of Q). I also use the
existence and properties of the Hilbert Class Field (conjectured by
David Hilbert, proved by Furtwängler)

If so I'd think that'd be a powerful argument against my claims.

Why don't you try?

So that there's less confusion, assume you start with a/b being the
root of a non-monic polynomial irreducible over Q, as if a/b is
rational that's trivial to handle.


Here's a proof:

Suppose r is an algebraic number, with irreducible polynomial p(x).
Then p(x) is in Q[x], the ring of polynomials with rational
coefficients, and is monic. Suppose p(x) is given by:

	p(x) = x^N + p_1 x^(N-1) + ... + p_(N-1) x + p_N

where each p_j = n_j/d_j, with (n_j,d_j) = 1 [all coefficients
of p are given in lowest terms].

If m = lcm(denom(d_j: j = 0 ... N)), then the number s = mr is
an algebraic integer. For instance, the polynomial q(x) given by

	q(x) = x^N + q_1 x^(N-1) + ... + q_(N-1) x + q_N

with

	q_j = m^j p_j

has the property that

	q(s) = q(mr) = m^N r^N + m p_1 m^(N-1) r^(N-1) + ...

	               + m^j p_j m^(N-j) r^(N-j) + ...

	               + m^(N-1) p_(N-1) m r + m^N p_N

	      = m^N (r^N + p_1 r^(N-1) + ... + p_(N-1) r + p_N)

	      = m^N p(r) = 0.

Now, this exhibits r as a fraction with an algebraic integer as its
numerator and the (rational) integer m as its denominator:

	r = s/m

Let F denote the field Q(r) = Q(s), obtained from the rationals by
adjoining the element r (note that since m is in Q, it makes no
difference whether you adjoin r or s to Q). Let O_F denote the
integers of F (i.e., those elements of F that are algebraic integers).

Define I to be the ideal of O_F generated by s and m:

	I = <s,m> = {x s + y m | x,y in O_F}

LEMMA: If I is principal (meaning that it is generated as the
set of all multiples g*f, for some (fixed) g, and all f in O_F),
then there is an element g of O_F, with

	s = s'g, m = m'g,

and there are u,v, elements O_F, with

	us' + vm' = 1.

PROOF:  Suppose I = <g> = { x g | x in O_F }.

	I = <s,m> = <g>,

so there are u and v in O_F for which

	g = us + vm.

Since g generates I over O_F, we also have elements s', m' in O_F for
which

	s = s' g, m = m' g.

Putting these together, we have

	g = u (s'g) + v (m'g) = (us') g + (vm') g

so

	g = (us' + vm') g,

and since O_F is an integral domain,

	us' + vm' = 1.

QED

The desired goal is then met: the algebraic number

	r = s/m

is also

	r = (s'g)/(m'g) = s'/m'

We see from the Lemma that there are u and v in O_F for which

	us' + vm' = 1.


THE GENERAL CASE:

Note that I needed to assume I is principal in O_F to obtain this
element g, which might be thought of as a gcd (greatest common
divisor) of s and m.

Suppose I is not principal. This is where the finiteness of the
class number of F = Q(m) enters into the picture, as well as a
particular field extension called the Hilbert class field of F.

The idea then is that there is an extension field of F, with
the property that the ideal I becomes principal.

The class number of F (or of O_F, its ring of integers) is the
order of (i.e., number of elements in) a particular group, called
the ideal class group of O_F. For F a number field (i.e., a finite
extension of Q), this group is always a finite abelian group. If I
recall correctly, Dedekind proved this fact; I personally don't know
how to prove it, but I would be willing to bet real money that either
Arturo Magidin or Bill Dubuque would know this.

Suppose the class number of O_F is k. Then the Hilbert class field
of F is F', and extension of order k; for every ideal I in O_F, the
extension O_F' contains an ideal I', lying over I, and I' is principal.
(I have to admit that I am no expert. One source appears to give this
result only for prime ideals; several state the result I'm using here)

Thus, we can take I = <s,m> in O_F. Construct the Hilbert class field
F', and know that I' = <s,m> in O_F' is principal:

Then, from the above argument, we are finished, since the ideal of O_F'
generated by s and m is principal.

Now, given an algebraic number a/b that is the root of a non-monic
polynomial irreducible over Q, can you prove that you can find x and y
such that

ax + by = 1?


A simple analogy that I've given before is to consider 6 and 2 in 

the

ring of evens.


Bogus example: no identity element.



Not really, as in fact, that's essentially what happens with the ring
of algebraic integers where people *assume* that given a and b coprime
algebraic integers that you can always find

ax + by = 1

where x and y are algebraic integers, but that is never proven.

It's a BFC, where by assuming that coprimeness in algebraic integers
proves global coprimeness, people assume that you can find x and y
above.


No. Coprimeness is expressible in terms of an equation that has
a specific form. Once that equation can be satisfied in some ring,
a larger ring cannot change its validity.

It's a circular position, or can you prove otherwise?


Not circular at all. It's analogous to the question of divisibility:
suppose you have three elements r, s and t of some commutative ring R,
and also you know that

	r = s t

as elements of R. Now, let S be a larger ring, containing R as a
subring. We still have

	r = s t

in S. Why?

	It's because any equation that holds in a smaller
	ring does not cease to hold when the ring is enlarged.

I think that you may *believe* you can prove that without ever having
seeing a proof and now with the challenge you can see that, or you can
show I'm wrong.


You're wrong.

If I'm wrong here, then I really have to reconsider quite a few things.


Start your reconsidering at your earliest convenience. 

James Harris

.

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