Re: New paper, algebraic integers, Galois Theory

From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 10/14/04 

Date: Thu, 14 Oct 2004 06:05:17 GMT

James Harris wrote:
> "W. Dale Hall" <mailtowd-hall@pacbell.net> wrote in message news:<5Mebd.3436$6q2.1455@newssvr14.news.prodigy.com>...
>
>>James Harris wrote:
>
> ...
>
>
>>>But I have
>>>
>>>P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)
>>>
>>>and
>>>
>>>P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf)
>>>
>>>so far simultaneously true without contradiction, so there must exist
>>>z_1, z_2, and z_3 such that
>>>
>>>w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3
>>>
>>>and z_1 z_2 z_3 = 1,
>>>
>>>where algebraically the z's are units, but z_1, z_2 and z_3 are not
>>>units in the ring of algebraic integers.
>>>
>>
>>If you do not specify where all the action takes place, that does not
>>make any sense. Clearly, the z's are not algebraic integers.
>>
>
>
> The ring is a commutative ring.

This is too vague. Clearly, you require more, and just as
clearly you are unaware of what you do require.

>
> When something more specific is required I then mention it.
>

You mean when a dozen people inform you of your gaffes?

>
>>>That raises a question about units as if z_1, z_2 and z_3 are units
>>>algebraically but are not algebraic integers, what does that mean?
>>>
>>
>>It has been said on countless occasions that any mention of "unit"
>>requires the specification (this can usually be done implicitly, but
>>in your case it seems necessary to include an explicit mention) of
>>a ring in which the element is invertible. To pretend otherwise is
>>simply nonsense.
>>
>
>
> Actually, no, as unit means factor of 1.
>

Actually, you have no business telling me what the word unit means
with regard to a ring; I have worked with rings since before you
were born, and I know exactly what the meaning is. If you state that
an element of a ring is a unit, the assumption is that the element
is a unit *in that ring*. Without that, the statement "z is a unit"
has no meaning whatsoever.

Now, is a unit a factor of 1? Is a unit an invertible element?
In fact, there is no difference between the two descriptions:

        r invertible <==> r^(-1) exists in the ring
        r a factor of 1 <==> for some s, rs = 1.

        Note that r r^(-1) = r^(-1) r = 1, so r^(-1),
        the (multiplicative) inverse of r, is the
        "other" factor of 1, in the product rs = 1.

> Now yes, that can mean a lot depending on the ring, like in fields,
> it's rather trivial, but what unit means is not open to debate.
>
> Unit means, factor of 1.
>

Righto.

>
>>>Consider properties that cover rings like the ring of algebraic
>>>integers, and the ring of integers itself, and it can be shown that
>>>two are required:
>>>
>>
>>You clearly don't know what you're talking about.
>>
>
>
> You're trying to cheat as usual by setting up your audience with
> sniping.
>

Ah, "sniping". You'd know about that.

> Now my position is that you don't read more into the math than
> follows, but so far you have taken the rather extreme position of not
> presenting any math at all!
>

I'm sorry, but what you've said so far doesn't really require
more of a response that what I've given. Your work here is
void of mathematical content.

> But you have managed to criticize quite a bit already.
>
>
>>>1. No rational in the ring except 1 and -1 is a unit.
>>>
>>
>>Therefore, the intersection of this ring with the rationals is
>>precisely the set of integers.
>>
>
>
> That is true.
>

But unstated by you. In fact, one may infer from this
that you believed that there are more rationals in the
ring than the integers.

>
>>>2. No non-unit number within the ring is a factor of any two
>>>rationals that are coprime to each other.
>>>
>>
>>What do you mean by a factor of a rational number? If you are
>>thinking of another rational number, you surely know that every
>
>
> Now you're just going off on a tangent as there is the first
> requirement to consider.
>

However, you did not state the fact that the only rationals in
this ring are integers. From the first item, one must assume
that you believed that there may be non-integral rationals in
this ring.

> Now even math students with only a little bit of training know that if
> there are *two* requirements, it's dissembling to jump on the second
> as if the first does not exist!
>

Sorry, "dissemble" is not a term of mathematics. I have attempted to
mislead no one, and my remark is not confusing. If you meant that
no non-unit is a common factor of any two coprime integers, you
could have said just that. You didn't.

>
>>nonzero rational number is a factor of every other nonzero
>>rational number. Further, given two rational numbers, what
>
>
> So now you're talking as if the first requirement wasn't mentioned!!!
>

I've stated why it made sense to make that statement.

> That's the basic kind of lying that can explain to some people what
> kind of person you are, and how you could send an email to the
> Southwest Journal of Pure and Applied Mathematics lying about my
> paper.
>

Oh, *lying*. Somehow, telling the truth is *lying*. Asking for a
definition to give a clear description of the thing being
defined is *lying*, requiring concepts to fit their definitions
amounts to *lying*.

You claim that I am *lying*.

How many times have we had to read your castigation of those who
point your errors out, as *liars*? How often has the epithet
*liar* been slung by you, only to have events show that the
alleged *liar* was telling the truth? How much longer will you
resort to the basest attacks you can muster, making *liar*
the first word that flies from your mouth, to be followed by
every threat that you can dream up? When will you grow up and
drop that pacifier?

> You lied to that journal the same way you're lying to these groups
> now: knowingly, and with malice aforethought.
>

You just don't get it, do you? You don't understand my argument, can't
believe that no one on earth buys yours, and the only rationale that
occurs to you is that you are beset by *liars*. I have to think that
you must lead a sort of miserable existence, if this is how you react
in ordinary life.

        At the bakery:
                ... "Take a number"
                        LIAR!!!!
        At the bank:
                ... "There will be a $2.50 charge for that, sir"
                        LIAR!!!!

        At the store:
                ... "Sorry, this line is for five items and under;
                     you have seven"
                        LIAR!!!

        At school:
                ... "This test is closed book"
                        LIAR!!!

        With your sweetie:
                ... "That Billy Bob Thornton is a DRREEEEEMBOOAATT!!!
                        LIAR!!!

        With your ma:
                ... I'm sorry, I just can't seem to open this jar.
                        LIAR!!!

>
>>do you mean by "coprime"? Since all nonzero rational numbers
>>are units ("in the rationals" is understood, since the numbers
>
>
>>were *only* specified as rational), they are all coprime to
>>one another.
>>
>
>
> Well 2 and 3 are rationals, but in the ring of integers they are
> coprime.
>
> But they are still rationals, are they not?
>

Yes. However, that's a bogus example, since 2 and 3 will be coprime
in any ring containing both of them. However, 6 and 8 will be coprime
whenever 2 is invertible, while they aren't coprime in the ring of
integers.

That's the point: whether numbers are coprime is a function of what
ring you're using.

> It's not complicated. Given the requirement 1. you also have
> requirement 2. and the two requirments give the ring.
>

Hmm. I don't think that specifies a ring, even allowing for
the possibility that it contains the integers.

> Now, so much arguing already and you STILL haven't given any math.
>

Hey, score's even so far, so what's your gripe?

>
>>>So there are two basic factor properties of rings like the ring of
>>>integers and the ring of algebraic integers.
>>>
>>
>>How about the absence of zero divisors?
>>
>
>
> Rings come with *two* operations: addition and multiplication.
>
> Fields have division, but rings do not necessarily have division
> unless they are fields.
>

Ha, hahaha HEHEEHEE haha HOHO HO haha HaHaha! HEHEEHEE haha HOHO HO haha
HOHO HO haha HaHaha!Ha, hahaha HEHEEE haha HOHO HO haha HaHaha!

Ah, me!

Too bad I forgot to mention how you don't know anything...No, wait, I
said that above, and you thought I was just sniping!

HOAHAHAHAHAHA!!!!

Oh, you weren't in on the joke, so I'll share it.

One of the standard terms in ring theory is ZERO DIVISOR.

It doesn't have to do with division, appearances notwithstanding.

A ZERO DIVISOR in a ring is an element s, for which there is a
nonzero element t, such that

                st = 0.

ZERO DIVISOR, get it?

Like the equivalence classes of 2,3, and 4 in the ring of
integers modulo 6.

Like the class of (2x-1) in the ring Q[x]/<2x^2 + x - 1>

You know, ZERO DIVISOR.

> More arguing and still no math from W. Dale Hall.
>

Ah, you're just sore you didn't get the joke.

>
>>>Therefore, it suffices for z_1, z_2, and z_3 to be in a ring where
>>>properties 1. and 2. hold, which includes the ring of algebraic
>>>integers, and I have named that ring, the ring of objects.
>>>
>>
>>*WHAT* suffices? I could take *any* ring in which the z's are invertible
>>and make the above statements:
>>
>> w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3
>>
>>and
>>
>> z_1 z_2 z_3 = 1,
>>
>>without your presumed "object ring".
>>
>>BTW, when are you going to exhibit a member of this "object ring"
>>that is not an algebraic integer? You haven't done it yet.
>>
>
>
> Now he's off into non sequiturs.
>

Hey, I asked what on earth you thought you were saying by using
the phrase "it suffices". You weren't proving anything, no matter
how valiantly you were flapping your wings.

And on the reference to "non sequiturs", you have made allusion to
pixies, er, "objects" in discussions like this, and my suspicion
is that you were going to claim that the z_i are "objects". I decided
to point out the obvious, that you've never even shown us an object.

Reason: if you show me an object that's not an algebraic integer,
I'll show you a non-object that can't be discerned from your
object using ordinary arithmetic.

>
>>>Some seem intent on challenging the results here with positions that
>>>necessarily contradict the algebra in Section 1.
>>>
>>
>>I'm sorry. I didn't see any result. You have asserted the existence of
>>some w's and v's and z's. I didn't see anything that proved they exist,
>>nor have I seen anything like a proof that there is a unique "object
>>ring". Others have pointed out that it is not difficult to extend the
>>ring of algebraic integers without acquiring any non-integral rationals.
>>There are infinitely many ways to do this, and you are not doing your
>>cause any service by your continued failure to exhibit a non-integral
>>object, let alone your refusal to give any sort of way to determine what
>>is and what is not an object.
>
>
>
> Actually in Section 1 constants terms are shown to require a
> particular factorization.
>

No, they are not. All you've done is to shuffle symbols around
under the delusion you're doing mathematics.

> In Section 2, I show that factorization does not conflict with a
> factorization into algebraic integers, which refutes your claims of
> counterxamples using such algebraic integers.
>

Again, no. For you to refute my claim, you would have to show
algebraic integers m and n, that make the following true:

        m a + n 5 = 1

where "a" represents the coefficient that you claim to be coprime to 5.

Can do, no can do?

Which will it be, chump?

> A mathematician would follow the math, but you're trying to obfuscate.
>

Oh! You mean obfuscate, as in "JSH can't follow any math that wasn't
in his high school algebra course"!

THAT obfuscate!!!

Sorry, but homey don't play that.

>
>
>>>At times these people have tried to invoke Galois Theory, claiming it
>>>invalidates the argument in Section 1, but that requires a belief that
>>>mathematics can be inconsistent.
>>>
>>
>>Please produce the contradiction. If you think you've proven that two
>>of the a's are coprime to f in the ring of algebraic integers, then
>>you'll have to find a numerical example where that holds. If your
>>argument were correct, then *each* example resulting from substituting
>>numbers in for m,u,f would give such an example. However, not a single
>>one works when the resulting cubic is irreducible. In each such case
>>examined, we have found non-unit algebraic integer factors in common
>>between each of the a's and f, and recently you have said how easy it
>>is to do so, but then claimed that this does not invalidate your
>>argument.
>
>
> That is covered in Section 2. where I show *why* algebraic integer
> factorizations can be found.
>

No, you haven't shown a blamed thing. You may think you've shown
something, but then you think I'm a *liar*. That's OK, you don't have
the wherewithall to hurt my feelings.

Care to venture a guess about how threatened I feel by dingleberries
such as yourself?

>
>>If you are maintaining that this coprimeness result holds in a larger
>>ring than the algebraic integers, it's not what the others are arguing
>>about, and it surely is not what your paper "Advanced Polynomial
>>Factorization" claims:
>>
>> This paper will show, using basic algebraic methods, that
>> given the factorization, in the ring of algebraic integers,
>>
>> 65x^3 + 12 x+ 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x+ 1)
>>
>> one of the a's is coprime to 5.
>>
>>If you had intended this coprimeness claim to be outside the ring
>>of algebraic integers, you would have said as much.
>
>
> The coprimeness claim *does* hold in that ring, and in fact you can't
> prove that it does not.
>

Hey, big dope! The fact that I have found non-unit factors in
the ring of algebraic integers, shared by each of the a's and 5,
says just that!

Here's the definition of coprime:

        r and s are coprime in the commutative ring R (a ring with
        identity) if and only if there are elements u and v in R
        with the following:

                ur + vs = 1.

Note the repetition of "in R"? That's because the property is very
much dependent on what the ring is.

Now, suppose I have a non-unit factor q in R, shared by r and s.
That is: r = qr', s = qs' , where q,r',s' are all elements of R.
Then, every combination of r and s will ALSO be a multiple of q:

                ur + vs = u(qr') + v(qs')
                        = q(ur' + vs').

I'll repeat, since you seem incapable of following this exact point:

        Every combination ur + vs is a multiple of q.

Now, the only way for r and s to be coprime is to get the
number 1 by means of such a combination. What's a legal
combination? I've already told you: to show r and s are
coprime in R, you need to produce

        ur + vs = 1,

with u and v elements of R.

Somehow, you think that the conclusion of this argument:

        q(ur' + vs') = 1, thus q is a unit in R, contradicting
        the assumption q is not a unit of R

        (where I've added the gratuitous "in R" to emphasize
         that *all* the action here is IN R)

does not lead to the result that I am claiming: r and s
are not coprime in R.

Instead, you seem to be claiming some flaw with the argument. However,
neither have you been specific about *what* the flaw is, nor have you
done what *would* show the elements coprime: produce the required u
and v, within the ring of algebraic integers, giving (in the case I
proved in my note to SWJPAM)

        u a + v 5 = 1.

Can do, no can do? You tell me, champ.

>
>>>Notice that the conclusions here follow logically and do not extend
>>>beyond the mathematics itself. Social issues have no relevance to
>>>mathematical truth.
>>>
>>
>>I have shown common factors in the ring of algebraic integers. You
>>maintain that this does not imply the numbers fail to be coprime.
>>The domain of discussion is the ring of algebraic integers, unless
>>otherwise specified.
>
>
> There you go again, falling back to the ring of algebraic integers.
>

Hey, the claim has to do with coprimeness *in the ring of
algebraic integers*. It's not any sort of flaw. Instead:

        BY DEFINITION, THAT'S WHERE THE PROOF NEEDS TO OCCUR!

Do you have writers to develop this stuff for you, or are you naturally
this ignorant? I mean, if you're actually *this* ignorant, it's a true
feat. I apologize for making fun of your pride in your own ignorance,
this is really World-Class Ignorance. The Stuff Of Legend:

        God's Own Ignorance.

It's well worth having pride in, being a veritable le Pétomane of
mathematical argument.

                                                        ### ### ###
##### ###### ###### ###### ##### ##### ##### ##### ### ### ###
# # # # # # # # # ### ### ###
# # ##### ##### ##### # # # # # # #
##### # # # # # # #
# # # # # # # # ### ### ###
# # # # # # # # ### ### ###

> In Section 1, I show why a particular factorization is necessary using
> rather basic algebra. To refute it, you need to refute algebra
> itself.
>

No, this is just your standard claim. You've retracted it precisely
the number of times you've claimed it, less one. With rare exceptions,
this will always be the case

> In Section 2, I address the issue of algebraic integers, and show why
> a factorization in algebraic integers exists.
>

So what? Your arguments never actually prove anything!

> In Section 3, I resolve the conclusion from Section 1, with the result
> from Section 2, in a way that is mathematically consistent.
>

Of course, you don't. If you could, you would produce the two
algebraic integers necessary to show that the relevant coefficient
a_i and 5 are coprime in the ring of algebraic integers. The fact
that you don't do this proves my case, although not to the extent
that my actual arithmetic does. The former is merely a compelling
plausibility argument, while the arithmetic is mathematically
tight.

> You, on the other hand, besides cheating the proper process by sending
> your email against my paper, also come here and TALK as if that can
> defeat a math proof.
>

Hey. Give the numbers.

Two algebraic integers u and v so that a_i and 5 are shown to be
coprime in the ring of algebraic integers. No one can deny the
truth then, bucko. Here it is:

        a_i u + 5 v = 1.

Just fill in the u and v, and

## ## ####### ## ## ## ## #### ## ## #### ####
  ## ## ## ## ## ## ## ## ## ## ### ## #### ####
   #### ## ## ## ## ## ## ## ## #### ## #### ####
    ## ## ## ## ## ## ## ## ## ## ## ## ## ##
    ## ## ## ## ## ## ## ## ## ## ####
    ## ## ## ## ## ## ## ## ## ## ### #### ####
    ## ####### ####### ### ### #### ## ## #### ####

        

>
>
>>Let's see who agrees. The following are in reverse chronological order.
>>Look quickly before these articles are yanked to preserve the image of
>>an honest debater:
>>
>
>
> Math is not a debate.
>

Ah, the curse of the small-minded.

> Maybe that's why you think there's still room to argue, and that you
> can talk your way out of a math proof.
>

Do you want anyone to believe you? Hand over the numbers.

That's all it takes.

Can do, No can do? It's up to you, meester poopy-pants.

> I've handled all of your objections, and done so with a rather basic
> exposition, which is math.
>

Here's the objection: you have not, and can not, produce two algebraic
integers that exhibit the co-primeness of the a_i and 5.

You have not handled that. You have not shown any error in the above
proof that if two elements of a commutative ring with identity share
a common non-unit factor, they cannot be coprime.

>
> James Harris

Don't imagine that you have settled anything.

Dale.
        

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