Re: Some math, algebraic integers
From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 09/19/04
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Date: Sun, 19 Sep 2004 00:31:34 GMT
James Harris wrote:
> "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.
>
Note that there are plenty of reasons for the roots of irreducible
polynomials to be related to one another, in terms of divisibility
properties, just as there are plenty of reasons for those roots to
be related to the coefficients of their irreducible polynomials.
Your claims of corruption of the world of mathematicians (which is
what you mean by your code phrase "social reasons") are noted. I'll
note for the record that people who use code phrases are usually trying
to espouse something they're ashamed to admit. In your case, I can't
decide whether the shameful thing is your deeply-held bigotry, or
your profound ignorance.
> 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.
>
Proof? I didn't find a common factor between q_1 and 2; I have to admit
that I didn't spend a lot of time looking at that quadratic extension,
and it was well past time I would ordinarily head for bed. If one did
exist there, I'm sure I would have found it. Furthermore, I didn't
look at all for a common factor between all three of q_1, q_2, and 2.
Perhaps you could display such a factor. It's only fair, right?
>
>>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.
>
OK, so prove it. It could be the case; I haven't seen it just yet,
but I haven't spent lots of time on it. Note that you said that it
is an *assumption*. In order to claim that there must be a non-
unit factor as you say, it needs to be proven. You've said it's an
assumption. Get it? Proven? Assumption? Do these two words look
the same to you?
> Remember, in the ring of algebraic integers is VERY important to keep
> up with, as it turns out that ring has special characteristics.
>
No content here.
> 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.
>
No content here.
>
>>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.
>
How can you even write that phrase "overinterpretations of Galois
Theory", when you don't understand a word of it in the first place?
Isn't that like me pontificating on African-American culture because
I once saw an episode of "Good Times"? You say "overinterpretations
of Galois theory" to sound like you know what you're saying, while
I say "Dy No MITE!!!!" to give my peeps their props.
Guess what?
Neither of us is fooling anyone!
> 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.
>
There we have what? I've shown that the two numbers don't have
a common non-unit factor in the ring Z[z].
Since you apparently failed to understand the preceding argument,
I'll use baby words.
I showed that in the case N=4, the ring Q(z) does not have a non-unit
integer that divides both q and 2. At this point, I had not broached
the full ring of algebraic integers.
You had said that I *assumed* that the two numbers share a non-unit
algebraic integer factor. I denied that claim, and proceeded to prove
as much as I could to determine the issue. At this point in my
article, I had shown that the two numbers do not share a common
factor in the smallest ring containing both.
> 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.
>
What is the phrase "make some basic assumptions" supposed to mean?
What I assume are the definitions of the entities I'm working with,
and properties that either follow directly from the definitions or
that have been proven to hold. To assume any less is stupid; to
assume any more, without making such assumptions explicit, is improper.
Do you attempt to do math without *any* assumptions? Of course not,
but you just remain ignorant of the assumptions you're making.
> 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.
>
Well, you suggest that you were in fact *following* my line of argument,
when the evidence is that you dropped it immediately. Had you been
paying attention, you would have noticed that I wasn't arguing in that
direction at all.
> 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.
>
You aren't really saying anything here. Your oft-repeated examples are
irrelevant, and a waste of time.
> Notice the complexity of W. Dale Hall's work above, as that complexity
> is the hallmark of flawed thinking.
>
I see. You can't be bothered to understand it, so it's "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!!!
>
Kind of like Fourier, wasn't he? Fourier *decided* that periodic
functions (kind of like planetary orbits, right?) had to be made
up of sines and cosines (those are the x and y components of circles,
get it?). Well, Fourier did what mathematicians do, which is to prove
his claims. Later generations refined his results, and extended his
proofs, but that's what he did.
Note that the work of Fourier bears a striking similarity to the
Ptolemaic system of epicycles. JSH must now think that all of Fourier
analysis, with its complexity, is "the hallmark of flawed thinking".
> Human beings have an eery ability to come up with overly complicated
> and wrong systems and then fight for them, against all evidence.
>
That's "eerie". The dictionary is your friend.
What have I said in my article that is wrong? You've just suggested that
I have been defending an overly complicated (read: JSH can't get it)
and wrong system. Show me what non-factual statement I've made.
>
>>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!!!
>
No, you've made irrelevant statements and claimed they were "simple"
ways of understanding things. Virtually every one of your amazing facts
about the algebraic integers has been proven to be false. How can anyone
take what you say as having any form of validity?
> 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!
>
Isn't it interesting how your "mathematical" statements are vague
and essentially content-free, while the main content of your article
amounts to this sort of know-nothing commentary.
>
>>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.
>>
>
And the response?
>
> Notice my comment that follows is still dead on:
>
>
>>>Well, yeah, but the implication is wrong.
>>>
>
>
> Yup.
>
No response at all.
Note that I have not made the assumption you've claimed, a
and yet you claim I made some assumption that the numbers share
a non-unit integral factor. You stand by that claim, yet despite
my direct statement, you fail to back up your words with a single
shred of evidence.
> 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.
>
I see. You claim that the definition of "the ring of algebraic integers"
is a mistake. In what way?
Does it fail to be a ring?
Does it not have a property that has been proven?
???
You claim a mistake, make some wild assertions about corruption, and
weird assumptions, and produce not a word of evidence.
>
> And making a mistake is not the problem.
>
> The problem comes in if math society refuses to acknowledge the
> mistake, and keeps teaching it.
>
Please, for once, specify *what* the alleged mistake is.
> The question is, are math people big enough, like me, to own up to
> their mistakes?
>
Oh, you're finally admitting that the conclusion to your "Advanced
Polynomial Factorization" is in error? I must have missed it.
> Or are you too weak to accept mathematics that you don't like because
> it doesn't make you feel all that good?
>
Now, you bring in the bogus claim of emotion ruling all.
>
>>>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.
>
You probably mean "been seen by", or "been read by", but let's continue.
> Just so you know just how top level, one was Barry Mazur.
>
Has he written to agree with it? You are suggesting as much, or at least
that he believes it's correct. I claim that he does *not* agree with it.
Without his actual words, neither of us can say what the truth is.
However, you and I both know that I'm right. It's just that your
foolish pride won't let you say it.
> 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 have no fear of Barry Mazur. I make no claim to be in his league,
but I don't have any doubt that he and I would be on the same side
with regard to the errors of your paper.
> 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.
>
They are simply sitting quietly. You seem to read that as concurrence
with your personal opinion about the correctness of your paper.
Perhaps another interpretation should be suggested.
Have you ever been on the bus, only to have a mumbler or a raving
lunatic sit nearby?
How do you behave?
I'll tell you what I do: I don't do a thing. I don't acknowledge
their existence, by glance, or by body language, or by engaging
them in conversation. I should probably learn how to do the same
thing here.
I'll tell you what the big names do: they act as though they don't
hear or see. No eye contact, no gestures, no rolling of the eyes
or hunkering down behind their newspapers, nothing.
You may have received a form letter in response to your missal.
This is the "public communication" analogue of saying "OK" when
someone bumps past you in the aisle of the bus, and says "pardon
me": it holds no content as a communication, and it's a mistake
to take that brief token and wave it around as though it were a
serious conversation.
I assure you, that if Barry Mazur or Andrew Granville thought
your paper had *any* mathematical merit, you would have received
a real letter, discussing it in detail. If you had received such
a letter, I assure you that you would have mentioned it by now.
Instead, they are sitting quietly. No correspondence, no reaction.
Instead, they are your entree into the world of the mathematics
glitterati, the star-studded world where people discuss high-falutin'
stuff.... in your dreams, kid.
> 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.
>
Who in this newsgroup has discussed budgets, or prestige? I think it's
just you. Except for the people who get caught up in trading insults
with you, the point of view of the people here has almost entirely
to do with what's correct mathematically.
That fact is evidently lost on you.
> Making mistakes is human. Owning up to them, is more human still.
>
So, what about that errored conclusion in your paper? Time's a-wastin'.
>
> James Harris
Dale
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