JSH: Object Ring Challenges
Frank J. Lhota
are deficient in some way, e.g. they lack "coverage". As an alternative, he
proposes the "Object Ring" as an alternative. The object ring is defined on
the "My Math" blog as follows:
<quote>
The object ring is defined by two conditions, and includes all numbers such
that these conditions are true:
1. 1 and -1 are the only rationals that are units in the ring.
2. Given a member m of the ring there must exist a non-zero member n such
that mn is an integer, and if mn is not a factor of m, then n cannot be a
unit in the ring.
</quote>
This highly unconventional definition has perplexed many a Math newsgroup
poster. Some have complained (quite rightly) that the definition is unclear.
Others have questioned whether the object ring exists at all.
To help resolve the issue, let me introduce some terminology. I define a
Harris ring to be a subring H of the complex numbers that satisfies these
two conditions:
1. The intersection of H and Q is Z; and
2. Given a member m of H, there is a non-zero member n of H such that mn
is in Z.
I *think* that James Harris defines the object ring to be the unique maximal
Harris ring. (If I'm wrong, I think we'd all really appreciate it if James
would clarify what he /does/ mean by the object ring!)
In the quest to determine if the object ring exists, I propose two
challenges:
1. By definition, any rational number that is not an integer cannot be in
a Harris ring. Outside of the rationals, is there any other complex number
that can be shown to be outside of every Harris ring?
2. Does there exist two complex numbers z1 and z2 such that there exists
a Harris ring containing z1, and there exists a Harris ring containing z2,
but there is no Harris ring that contains both z1 and z2? I think that is an
issue that is in the back of a lot of posters minds. If such a pair of
numbers exist, it poses the question of which (if any) of these numbers are
in the object ring.
--
"All things extant in this world,
Gods of Heaven, gods of Earth,
Let everything be as it should be;
Thus shall it be!"
- Magical chant from "Magical Shopping Arcade Abenobashi"
"Drizzle, Drazzle, Drozzle, Drome,
Time for this one to come home!"
- Mr. Wizard from "Tooter Turtle"
Frank J. Lhota
news:PY2dnZsRCM41OhvY...@rcn.net...
> ... As an alternative, he proposes the "Object Ring" as an alternative.
^^^^^^^^^^^^^^^^^^
My apologies, I meant to delete that before pressing "Send". Due to my
error, I inadvertantly came up with wording that sounds like it came from
the Department of Redundancy Department.
Arturo Magidin
Yes.
> I think that is an
>issue that is in the back of a lot of posters minds. If such a pair of
>numbers exist, it poses the question of which (if any) of these numbers are
>in the object ring.
It is easy to check that any subring of the algebraic numbers satisfies
the second condition in the definition.
Suppose d is a nonzero algebraic integer. let
f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0
be its monic irreducible polynomial. Then
d(d^{n-1} + a_{n-1}d^{n-2} + ... + a_1) = -a_0
Now, d^{n-1}+...+a_1 is an element of the ring; it is nonzero, since
f(x) is the monic irreducible of d; and a_0 is a nonzero
integer. Thus, there is a nonzero n in the ring such that nd is an
integer.
Now, suppose that a is a nonzero algebraic number in the ring. Then
there exists a nonzero integer d such that da is an algebraic
integer. And by the argument above, there is a nonzero n in the ring
such that (da)n is an integer in the ring. Thus, a(dn) is an integer,
with dn in the ring.
So any subring of the algebraic numbers satisfies the second
condition.
Now consider for example x^2 - x + (1/2).
Let a1 and a2 be the two roots of this polynomial. It is not hard to
check that Z[a1]/\ Q = Z (I'm pretty sure; if not, then one can find
polynomials that will work). Symmetrically, Z[a2]/\Q = Z. Since both
satisfy condition 2 of the definition, both Z[a1] and Z[a2] are
"Harris rings". But any ring that contains both a1 and a2 will
necessarily contain a1*a2 = 1/2, hence not be a "Harris ring".
If the example doesn't work, look up some old posts of Bill Dubuque,
where he produced some explicit examples.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Frank J. Lhota
news:em6aah$1ugb$1...@agate.berkeley.edu...
Come to think of it, there are some fairly simple examples of such numbers.
James, you've been looking at this issue longer than I have. Can you provide
us with an irrational that is outside of every Harris Ring?
Well done! This, of course, raises the question of whether the object ring
contains a1 or a2. James, do you have any thoughts on this?
> If the example doesn't work, look up some old posts of Bill Dubuque,
> where he produced some explicit examples.
I've checked the Bill Dubuque posts. Dubuque makes a convincing argument,
but his posts dealt with an earlier definition of the object ring.
hagman
Frank J. Lhota schrieb:
> "Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
> news:em6aah$1ugb$1...@agate.berkeley.edu...
> > In article <PY2dnZsRCM41OhvY...@rcn.net>,
> > Frank J. Lhota <FrankLh...@rcn.com> wrote:
> >>2. Does there exist two complex numbers z1 and z2 such that there
> >>exists
> >>a Harris ring containing z1, and there exists a Harris ring containing z2,
> >>but there is no Harris ring that contains both z1 and z2?
> > Now consider for example x^2 - x + (1/2).
> >
> > Let a1 and a2 be the two roots of this polynomial. It is not hard to
> > check that Z[a1]/\ Q = Z (I'm pretty sure; if not, then one can find
> > polynomials that will work). Symmetrically, Z[a2]/\Q = Z. Since both
> > satisfy condition 2 of the definition, both Z[a1] and Z[a2] are
> > "Harris rings". But any ring that contains both a1 and a2 will
> > necessarily contain a1*a2 = 1/2, hence not be a "Harris ring".
Z[a1] contains 1-a1 = a2 and vice versa (and of course a1^2-a1 = -1/2).
By the same argument x^2 + n*x + q with n in Z, q in Q\Z won't work.
Maybe x^2 - 1/2 x + 1 ? But then Z[a1] contains a1/2 =a1^2+1, so
Z[a1]=Z[b1] where b1=a1/2 is a root of x^2-x+4, i.e. an algebraic
integer.
It looks like it is not /that/ trivial after all to find two Harris
rings that are not
contained in a common Harris ring...
fishfry
JSH doesn't believe in fancy words like "intersection" and "maximal." If
you read his posts, it's clear he doesn't know what a ring is. He sure
is a good troll though. He gets everyone to pay attention.
Arturo Magidin
Fair enough. There were some explicit examples produced some time ago;
you just need to find an algebraic number r, not an algebraic integer,
such that Z[r] /\ Q = Z. Once you have that, it will work for any
conjugate of r, and then you get a countererxample. Such numbers are
definitely known to exist.
jst...@gmail.com
I am waiting on an apology as well as a withdrawal of your claims of
disproof of my paper.
But I suspect you care more about what sci.math readers think of you
than with what is mathematically correct and will risk your math career
to hold and wait, just in case I still can't make any progress versus
admitting the truth now.
Ok then, along with Ullrich you roll the dice.
And I will have no guilt when you are completely out of academia.
James Harris
James Burns
> Arturo Magidin wrote:
>
[...]
>>Fair enough. There were some explicit examples produced some time ago;
>>you just need to find an algebraic number r, not an algebraic integer,
>>such that Z[r] /\ Q = Z. Once you have that, it will work for any
>>conjugate of r, and then you get a countererxample. Such numbers are
>>definitely known to exist.
>
> I am waiting on an apology as well as a withdrawal of your claims of
> disproof of my paper.
>
Harris, you have this amazing ease with which you reshape
your "reality" in order to match your wishes. You just pretend
you never told Arturo Magidin never to respond to you, and,
voila!, you become the victor of whatever one-sided "debate"
you want to hold with him.
It is, of course, Arturo who holds his promise to you, so
neither you nor I control it, but if I were in his place,
I would declare your addressing him directly as falling
under the "may answer questions of interest" clause and
respond to you this once.
Jim Burns
jst...@gmail.com
> jst...@gmail.com wrote:
> > Arturo Magidin wrote:
> >
> [...]
> >>Fair enough. There were some explicit examples produced some time ago;
> >>you just need to find an algebraic number r, not an algebraic integer,
> >>such that Z[r] /\ Q = Z. Once you have that, it will work for any
> >>conjugate of r, and then you get a countererxample. Such numbers are
> >>definitely known to exist.
> [...]
> >
> > I am waiting on an apology as well as a withdrawal of your claims of
> > disproof of my paper.
> >
> [...]
>
> Harris, you have this amazing ease with which you reshape
> your "reality" in order to match your wishes. You just pretend
> you never told Arturo Magidin never to respond to you, and,
> voila!, you become the victor of whatever one-sided "debate"
> you want to hold with him.
That's irrelevant to the reality that he has made false claims against
my paper on non-polynomial factorization.
I have given simple proof that his claims against my paper were indeed
FALSE.
He is duty bound to apologize and retract those claims regardless of
any gentleman's agreement not to reply to me and he can do so in a new
thread without replying to me.
> It is, of course, Arturo who holds his promise to you, so
> neither you nor I control it, but if I were in his place,
> I would declare your addressing him directly as falling
> under the "may answer questions of interest" clause and
> respond to you this once.
>
> Jim Burns
He can start a new thread and not reply to me directly.
I have refuted the claims made against my paper with direct and easy
algebra.
I think Magidin believes that posters can just let this storm slide by
and still have the trust and faith of the newsgroup so he is waiting to
see if he can get away with false assertions rather than looking to act
in a civilized manner.
I think he believes that this is just another storm that will pass with
the newsgroup still trusting him and posters who deliberately ignore
valid mathematical proofs no matter how basic they are to push lies on
the newsgroups, even though he and his group destroyed a math journal
in the process.
James Harris
David Moran
<jst...@gmail.com> wrote in message
news:1166490719.5...@n67g2000cwd.googlegroups.com...
Prove to me Magidin is lying. I've read many of his posts and they make
perfect sense to me. Why would that be, James? Could it be you've got no
clue what you're doing? Try doing something productive with your life.
Dave
Frank J. Lhota
> Fair enough. There were some explicit examples produced some time ago;
> you just need to find an algebraic number r, not an algebraic integer,
> such that Z[r] /\ Q = Z. Once you have that, it will work for any
> conjugate of r, and then you get a countererxample. Such numbers are
> definitely known to exist.
In a post on sci.math, hagman presents such an example, using the roots of a
cubic. See
http://groups.google.com/group/sci.math/msg/a5be5d7bbd2b13a4
jst...@gmail.com
Um, reading over your posts I detect the possibility of a weird kind of
self-destructiveness in them. Especially at this point, as with the
simple proof I have which is an easy disproof of the assertion that a
non-monic with integer coefficients irreducible over Q cannot have an
algebraic integer root, there is little doubt that this news should
travel.
Now Ullrich and Magidin are playing the same games they've always
played as it has worked before, where when I come up with something new
they just hold their line hoping it all blows over and no one gets it
that I'm right.
But you seem to be running into the path of the train full tilt for no
good reason I can figure out.
So I guess I'll stay away from your posts from this point on, except to
down-rate them, but I'll try not to reply.
James Harris
Rupert
No.
You do not have a proof. Your argument is wrong. I have shown you the
mistake.
I have also recently posted for your convenience two proofs that a
non-monic primitive polynomial over Z that is irreducible over Q cannot
have an algebraic integer as a root, one self-contained and one using a
lemma from Galois theory and the fact that the algebraic integers form
a ring. For a bonus I also threw in a really, really easy, and
completely self-contained argument that neither (3+sqrt(-26))/7 nor
(3-sqrt(-26))/7 is an algebraic integer.
http://groups.google.com/group/alt.math.undergrad/msg/376ed9ea05ff3f71?dmode=source
> Now Ullrich and Magidin are playing the same games they've always
> played as it has worked before, where when I come up with something new
> they just hold their line hoping it all blows over and no one gets it
> that I'm right.
>
You're not right.
> But you seem to be running into the path of the train full tilt for no
> good reason I can figure out.
>
There is a reason why your picture of the world doesn't make sense.
> So I guess I'll stay away from your posts from this point on, except to
> down-rate them, but I'll try not to reply.
>
He politely asks you to back up your claims, you say you'll stay away
from his posts. That's going to convince people, all right.
>
> James Harris
Arturo Magidin
>news:em771d$2jcr$1...@agate.berkeley.edu...
>> Fair enough. There were some explicit examples produced some time ago;
>> you just need to find an algebraic number r, not an algebraic integer,
>> such that Z[r] /\ Q = Z. Once you have that, it will work for any
>> conjugate of r, and then you get a countererxample. Such numbers are
>> definitely known to exist.
>
>In a post on sci.math, hagman presents such an example, using the roots of a
>cubic. See
>
> http://groups.google.com/group/sci.math/msg/a5be5d7bbd2b13a4
That he did; I should have remembered the extension needs to be
non-normal. Thanks!
Arturo Magidin
>It is, of course, Arturo who holds his promise to you, so
>neither you nor I control it, but if I were in his place,
>I would declare your addressing him directly as falling
>under the "may answer questions of interest" clause and
>respond to you this once.
There is no question of interest there. The objections I raised at the
time were accurate, the argument in that paper was, and continues to
be, incorrect where intelligible, and mostly unintelligible. The
current regression to rejecting, yet again, a basic result that he had
already managed to digest the proof for, is perhaps a bit unexpected,
but hardly interesting and hardly a question.
Odysseus
I figured the phrases must be nested in order to obfuscate the argument
self-referentially. ;)
--
Odysseus