Re: Some math, algebraic integers

From: James Harris (jstevh_at_msn.com)
Date: 09/18/04 

Date: 18 Sep 2004 07:51:54 -0700

"W. Dale Hall" <mailtowd-hall@pacbell.net> wrote in message news:<1zU2d.17114$QJ3.13457@newssvr21.news.prodigy.com>...
> James Harris wrote:
> > Luckily for me the math that explains my work is actually quite
> > simple in many ways. All that is required is that you accept
> > algebra. That might sound weird, but pay attention and you'll
> > understand by the end.
> >
> > Now with rationals you can have a simple case like
> >
> > 2x^2 + 5x - 3 = 0
> >
> > where the roots are x=1/2 and x=-3, if I did my algebra correctly.
> >
> > Now notice you have an integer paired with a fraction, which, of
> > course, is NOT an integer.
> >
>
> So what? The polynomial is not irreducible, is it?
>
> 2x^2 + 5x - 3 = (2x - 1)(x + 3)
>
> Given any finite set of numbers, it is a triviality to produce
> a polynomial with those numbers as roots.
>

Like, yeah, the polynomial has rational roots, which is what I already
said and I gave those roots, as they are 1/2 and -3.

What mathematicians falsely assumed is that the picture changes if
both roots are irrational AND the polynomial they are roots of has
integer coefficients, as they made a false leap, which I explain
below.

> > Now consider some numbers q_1 and q_2, which I won't say a lot about
> > now, though in a bit I'll put them in a particular ring, where you
> > get a polynomial with
> >
> > x = q_1/2 and x=-3q_2 as the roots.
> >
> > That polynomial looks like
> >
> > 2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0
> >
> > and let q_1 q_2 = 1, and 6q_2 - q_1 be an integer.
> >
> > It turns out that q_2 can't be an algebraic integer.

OOPS! That should be that if

2x^2 + (6q_2 - q_1)x - 3q_1 q_2

is irreducible over rationals then q_2 can't be an algebraic integer.

Notice, I make mistakes but don't mind correcting them either.

The question is, are mathematicians as a group as big about their
mistakes as I am about mine?

> >
> > Mathematicians at this point make an assumption, which is that q_1
> > MUST have non-unit algebraic integer factors in common with 2.

That's the weird result that many people seem to have difficulty
grasping and posters who argue with me tend to step around such points
as if they don't readers might actually ask why mathematicians would
assume such a thing.

So, why then, does it matter that if q_1 q_2 = 1, and 6q_2 - q_1 is an
integer that q_1 and q_2 should, for some reason, be forced to have
non-unit algebraic integer factors in common with 2?

It turns out that there's no *mathematical* reason to believe as
mathematicians do, while the social reasons for why they believe as
they do are fascinating.

The weird point is that q_1 and q_2 DO in fact have non-unit algebraic
integer factors of 2 in common, but that's just because the ring of
algebraic integers is kind of screwed up.

> >
>
> At what point is such an assumption made? If your polynomial were
> irreducible over Z, then none of the roots could be algebraic
> integers. That isn't an assumption, it's readily proven. You can
> claim that we make whatever assumptions you like, but you fail to
> back up your claims. As far as q_1 having algebraic integer factors
> in common with 2, I have no assumption one way or the other.
>
> Let's look:
>
> 6q_2 - q_1 = N
> q_1 q_2 = 1, so q_1 = 1/q_2
>
> 6q_2 - 1/q_2 = N
> 6q_2^2 - 1 = N q_2
>
> 6 q_2^2 - N q_2 - 1 = 0
>
> and
>
> q_1^2 + N q_1 - 6 = 0
>
> OK, so q_1 is an algebraic integer; if N^2 + 24 is
> a square integer, then q_1 is an integer. For instance,
>
> N = 1, so N^2 + 24 = 5^2
>
> q_1^2 + q_1 - 6 = (q_1 + 3)(q_1 - 2)
>
> N = 5, so N^2 + 24 = 7^2
>
> q_1^2 + 5 q_1 - 6 = (q_1 + 6)(q_1 - 1)
>
> N = -1, N = -5 also yield integer values for q_1, presumably (I
> just checked the discriminant N^2 + 24; I didn't factor the
> polynomial).
>
> Otherwise, the polynomial is irreducible over Z. Then it's
> the case that q_1 divides 6 in the ring of algebraic integers.
> Does it have to share a non-unit factor with 2?

Actually, for readers, in the ring of algebraic integers, it does.

Remember, in the ring of algebraic integers is VERY important to keep
up with, as it turns out that ring has special characteristics.

An interesting bit of my work is about how mathematicians failed to
completely understand some of the special characteristics of the ring
of algebraic integers.
 
> Here's a case when it does:
>
> Let N = 3. The polynomial x^2 + 3x - 6 is irreducible
> over Z. If q is a root of this polynomial, then the
> number -4-q divides both 2 and q:
>
> Here's the product of -4-q with something to yield q:
>
> (-4 - q)(-3 + 2q) = 12 -5q - 2q^2
> = 12 - 5q - 2(6 - 3q)
> = 12 - 5q - 12 + 6q
> = q
>
> Here's the product of -4-q with something to yield 2:
>
> (-4 - q)(1 - q) = -4 + 3q + q^2
> = -4 + 3q + (6 - 3q)
> = -4 + 3q + 6 - 3q
> = 2
>
> Does it always happen? Well, Galois theory appears to say that, for any

For those who wonder about my talk of overinterpretations of Galois
Theory in my paper, well here you can see an example, as I've mostly
seen that on sci.math from various posters like W. Dale Hall and
Arturo Magidin.

Pay careful attention.

> algebraic integer a that divides q_1, there is a conjugate (under the
> action of the appropriate Galois group) of a that divides the other root
> of the polynomial x^2 + Nx - 6. The two roots multiply to yield 6, and
> so one might be tempted to infer that there would have to be divisors of
> 2 distributed among the two roots. If the ring of integers in the
> extension Q(q) (the extension of the rationals Q, generated by the root
> q ) is principal, this will always happen.
>
> However, if N=2, the situation is different (class number here is 2).
> The field Q(q) = Q(sqrt(40)) = Q(sqrt(10)). For convenience, let's
> write z = sqrt(10), and note that the ring of integers in Q(z) is Z[z],
> since 10 is congruent to 2 modulo 4. One of the roots is
>
> q = -2 + sqrt(10) = -2 + z,
>
> and we attempt to find a common factor for q and 2, as follows:
>
> 2 + z = (a + bz)(c + dz)
> 2 = (a + bz)(x + yz)
>
> Now, the norm of Q(z) is computed by taking the product of all
> Galois conjugates; since the conjugate of q = -2 + z is qbar = -2 - z,
> the norm of q is q qbar = 14. The norm of 2 is 4 (the Galois group
> of Q(z)/Q fixes the rationals). In general, the norm of the
> element a + bz is (a + bz)(a - bz) = a^2 + 10 b^2. In addition,
> the norm is a rational integer for all integral elements, and an
> integer of Q(z) is a unit in the ring of integers, if its norm is
> a unit (rational) integer, i.e., +/- 1.
>
> The norm is multiplicative, and so if there is such a common factor
> a + bz as written above, its norm a^2 + 10 b^2 would need to divide
> both 4 and 14, or more concisely, the norm would need to divide 2,
> which is gcd(4,14).
>
> So, what are the candidates?
>
> Norm = 1: UNIT.
>
> Norm = 2: a^2 + 10 b^2 = 2.
>
> Well, since 10 is congruent to 2 modulo 4, the ring of integers
> of Q(z) is just Z[z]: a and b must be (rational) integers.
> How can we get rational integers a and b so that a^2 + 10 b^2 = 2?
>
> Since the sum a^2 + 10 b^2 is at least 10 b^2, and b^2 is non-negative,
> it follows that b must be zero. At that point, we're led to conclude
> that a^2 = 2, an impossibility for any (rational) integer a.
>
> We must conclude that there is no common non-unit integral factor
> between q and 2, in the ring Z[z]. I will note, however, that
> q and 2 are not coprime, that is, they do not generate the
> ring Z[z] as an ideal.
>

And there you have it.

If you stay in the ring of algebraic integers, and make some basic
assumptions you come up with the conclusion that q_2 MUST have a
non-unit algebraic integer factor in common with 2.

But let's back up a bit as remember my first example was rational:
 
2x^2 + 5x - 3 = 0
 
Here the roots are x=1/2 and x=-3, and notice that -3 is coprime to 2
in the ring of algebraic integers, so for *rationals* you have the
possibility that one of the roots is not an algebraic integer, while
the other that is an algebraic integer can be coprime to 2.

But notice what happens with

2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0

where q_1 q_2 = 1, and 6q_2 - q_1 is an integer,

if the quadratic is irreducible over rationals as THEN suddenly,
supposedly a non-unit algebraic integer factor in common with 2 is
FORCED UPON q_2, as if in all of the infinity of possible numbers that
q_2 can be, it must always have a non-unit algebraic integer factor of
2.

Well, yes, IN THE RING OF ALGEBRAIC INTEGERS that is true, following
the line of argument that you see W. Dale Hall making here.

But it doesn't make sense, and the bigger picture is that the ring of
algebraic integers is itself misleading as consider this example:

Imagine that in evens, you have xy = 2, which I can write
algebraically, but how can you satisfy that with even numbers?

You can't because 1 is not even. So x=2, y=1, works, but because I've
*said* in the ring of evens, you have this weird block.

With algebraic integers, it's a little more complicated as if x and y
are irrational, and, get this, are roots of a non-monic quadratic with
integer coefficients irreducible over rationals, then *apparently*
both x and y must have factors in common with 2.

The real situation is like xy = 2 in evens, where because a rule
arbitrarily removes a possibility, to try and hold on to it, you have
to make mental gyrations.

Notice the complexity of W. Dale Hall's work above, as that complexity
is the hallmark of flawed thinking.

Kind of like Ptolemy's rings, and if you don't know about those you
should read up on the history as it's fascinating.

You see, thousands of years ago math people decided that circles were
perfect, and as God is perfection, God would only use circles in the
heavens.

So they *decided* that the Sun, Moon, stars and planets had circular
orbits, but that didn't work to actually figure out where they'd be!

So a guy named Ptolemy worked out a system where they used circles
within the circles called epicycles, to get it to work!!!

Human beings have an eery ability to come up with overly complicated
and wrong systems and then fight for them, against all evidence.

> Can I find an extension of Q(z), for which there *is* a common, non-unit
> integral factor of q and 2? Here's where I'm well out of my depth on the
> topic (and I would invite anyone who is better versed than I am to chime
> in at this point). I'll give it a blind stab:
>

Well, he goes on with his "epicycles" and I'd like to point out that
I've explained how things work with algebraic integers before...many
times before!!!

Certain posters have simply kept at it. They just keep repeating the
same things, and I think human nature is such that people WANT to
believe them, so they ignore the math, and accept what's wrong.

Sad, but also kind of fascinating to watch!

> Let's take the quadratic extension K of Q(z) that makes the ring of
> integers O_K into a PID (principal ideal domain: every ideal is
> generated by a single element . Then, there is a common factor of 2 and
> q in
> the
> extension ring. I suspect, however, that it may be a unit, from the
> following heuristic: the norm of q in the bigger field is the square
> of the norm of q in Q(z), since the Galois group Gal(K/Q(z)) fixes
> Q(z) (yielding an additional copy of each of the factors that
> contributed to the norm in Q(z)); similarly, the norm of 2 in
> the bigger field is squared; the norm of any common divisor will
> need to divide gcd(16, 196) = 4. The trick would be finding an
> element of small enough norm.
>
> Note that I've made no such "assumption" as you claim. I doubt that
> anyone who knows enough to correct whatever errors I've made above,
> or to complete the analysis of passing to the quadratic extension K/Q(z)
> will need to "assume" anything of the sort you're claiming. Either
> they'll prove something, or state known results (presumably with
> references), or they'll demur.
>

Notice my comment that follows is still dead on:

> > Well, yeah, but the implication is wrong.
> >

Yup.

Remember my example with xy = 2 in evens, where because 1 is not even,
you start having to do some mental gyrations, unless you get it.

> > That might sound complicated or weird, but think about the rational
> > example, where you have roots 1/2 and -3, and notice that -3 does not
> > have any non-unit factors in common with 2, but somehow, someway if
> > you have irrational factors, for some reason 3q_2 MUST have non-unit
> > factors in common with 2?
> >
> > It begs the imagination that despite the infinity of possibility that
> > 3q_2 is so constrained, and in fact, it's not.
> >
> > Mathematicians made a mistake.

And making a mistake is not the problem.

The problem comes in if math society refuses to acknowledge the
mistake, and keeps teaching it.

The question is, are math people big enough, like me, to own up to
their mistakes?

Or are you too weak to accept mathematics that you don't like because
it doesn't make you feel all that good?

> >
> > It turns out that q_1 and q_2 CAN be in a ring where 1 and -1 are the
> > only integers that are units, and to be more specific just in case
> > some are still confused, where 1 is the only NATURAL that is a unit.
> >
> > I figured out a rather basic way to prove that using some very simple
> > algebra.
> >
>
> Well, very simple algebra, and a heap of hand-waving that no algebra
> instructor would let pass.
>

Actually, my paper has been by quite a few top level mathematicians,
and to date, no one but sci.math posters have claimed it's in error.

Just so you know just how top level, one was Barry Mazur.

Now I don't know what it is with posters like W. Dale Hall, "Nora
Baron" and Arturo Magidin, but notice they don't pause.

I toss out a name like Barry Mazur, and they just keep on chugging as
if I didn't say anything!

Unfortunately they are people of limited real math ability, while top
mathematicians like Mazur and Granville, who also has looked over my
paper, and did not claim it had any errors in it, are simply sitting
quietly.

I'm hoping to get some movement eventually as I hope that
mathematicians like them learn to appreciate that it's not about their
budgets or their prestige, but about what's mathematically correct.

Making mistakes is human. Owning up to them, is more human still.

James Harris

Relevant Pages

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