Re: Object Ring Challenges
Frank J. Lhota
it should also have been posted on the other newsgroups that has "object
ring" threads.
"Frank J. Lhota" <FrankLh...@rcn.com> wrote in message
news:PY2dnZsRCM41OhvY...@rcn.net...
> The James Harris posts often complain that the ring of algebraic integers
> are deficient in some way, e.g. they lack "coverage". As an alternative,
> he proposes the "Object Ring". 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"
>
>
hagman
Frank J. Lhota schrieb:
or every principle ideal of H intersects Q in a non-zero principal
ideal of 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?
Trivially, for example z=(i+1)/2 is not in any Harris ring because z^4
is in Q\Z.
Frank J. Lhota
news:em6aah$1ugb$1...@agate.berkeley.edu...
> In article <PY2dnZsRCM41OhvY...@rcn.net>,
> Frank J. Lhota <FrankLh...@rcn.com> wrote:
>>The James Harris posts often complain that the ring of algebraic integers
>>are deficient in some way, e.g. they lack "coverage". As an alternative,
>>he
>>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.
>>
>>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?
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?
>>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?
>
> 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".
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.
> --
> ======================================================================
> "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
>
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...
jst...@gmail.com
Frank J. Lhota wrote:
> I started this thread on alt.math.recreational, but it occurred to me that
> it should also have been posted on the other newsgroups that has "object
> ring" threads.
>
Now that things are cleared up about the ring of algebraic integers,
you can just say that an algebraic number is an object if it is an
algebraic integer.
Outside of that ring things get more interesting, as the object ring
includes transcendentals.
For instance it includes 2^{sqrt(2)}.
But it excludes pi and e.
That shouldn't be hard to prove by showing that pi and e are NOT
constructible in any ring without the use of members not in the ring of
objects.
I have looked over posts in this thread and think I should focus on
other issues right now, so I'll leave things with the comments above.
James Harris
hagman
Quadratic extensions are not good enough to try to find such numbers
because they are normal. One needs a non-normal extension.
Consider the polynomial P(X)=X^3+2X+4, which is irreducible in Z[X]
(because -4,-2,-1,1,2,4 are no roots).
The discriminant is D=-4b^2-27c^3 < 0 and hence not a square in Q.
Therefore the splitting field is of degree 6 and not just 3 over Q (in
other words we have a not normal extension).
Let a[1],a[2],a[3] be the roots of P in C.
Then b[1]=a[1]/2, b[2]=a[2]/2,b[3]=a[3]/2 are roots of
R(X)=P(2X)=X^3+1/2 X + 1/2, which is of course also irreducible.
The ring Z[b[1]] contains neither b[2] nor b[3], because otherwise we
would conclude from b[1]+b[2]+b[3]=0 that Q(b[1]) splits P.
The only rational elements of Z[b[1]] are the integers:
All elements have the form p*b[1]^2+q*b[1]+r with p,q,r in Q with the
additional property that p+q+r is an integer(!).
But if p*b[1]^2+q*b[1]+r=s is rational then either
1) p=q=0 and hence s is an integer because p+q+r is; or
2) (p,q)!=(0,0), i.e. p*X^2+q*X+r-s is a polynomial in Q[X] of degree
1 or 2 that has b[1] as root, in contradiction to the irreducibility of
R(X).
(Actually, I had to adjust a few signs in my argument and I think it
has become buggy here;
maybe someone else can pin that down / correct it / decide that one has
to take a different polynomial; I'm too tired to do so right now)
That an element x=p*b[1]^2+q*b[1]+r can be multiplied by a ring element
to yield an integer is clear since x has a multiplicative x^{-1}
inverse in Q[b[1]] and a suitable integer multiple n*x^{-1} of this is
in Z[b[1]].
Conclusion: Z[b[1]] is a Harris ring.
By symmetry, Z[b[2]] and Z[b[3]] are Harris rings.
Z[b[1],b[2]] contains b[1]*b[2]*(b[1]+b[2]) = -b[1]*b[2]*b[3]=1/2 and
is therefore not Harris.
-hagman
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
Jesse F. Hughes
> Arturo Magidin wrote:
[snip]
>
> I am waiting on an apology as well as a withdrawal of your claims of
> disproof of my paper.
But you told him never to post in response to you. Have you
forgotten?
http://mathforum.org/kb/plaintext.jspa?messageID=523004
How can you expect an apology?
(Not that I'm suggesting that Arturo has any reason to apologize. But
even if he did, JSH has said he shouldn't.)
--
Jesse F. Hughes
"A factor is simply something that multiplies against another factor
to produce a 'product'." -- James Harris offers a definition.
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.
Arturo Magidin
Jesse F. Hughes <je...@phiwumbda.org> wrote:
>> I am waiting on an apology as well as a withdrawal of your claims of
>> disproof of my paper.
>
>But you told him never to post in response to you. Have you
>forgotten?
>
>http://mathforum.org/kb/plaintext.jspa?messageID=523004
Clearly he's forgotten at least the facts. Otherwise, how could anyone
call my offer a "gentlemen's agreement"? I never claimed to be "a
gentleman" in any case, and James demonstrated in his reply (as well
as much sooner) that he wasn't one either. (-;
>(Not that I'm suggesting that Arturo has any reason to apologize. But
>even if he did, JSH has said he shouldn't.)
For someone with James's track record to say that an apology is called
for would have fried all nearby irony meters if I didn't have my
computer well isolated against such an occurrence. (-:
hagman
hagman schrieb:
> Consider the polynomial P(X)=X^3+2X+4, which is irreducible in Z[X]
> (because -4,-2,-1,1,2,4 are no roots).
> The discriminant is D=-4b^2-27c^3 < 0 and hence not a square in Q.
> Therefore the splitting field is of degree 6 and not just 3 over Q (in
> other words we have a not normal extension).
> Let a[1],a[2],a[3] be the roots of P in C.
> Then b[1]=a[1]/2, b[2]=a[2]/2,b[3]=a[3]/2 are roots of
> R(X)=P(2X)=X^3+1/2 X + 1/2, which is of course also irreducible.
> The ring Z[b[1]] contains neither b[2] nor b[3], because otherwise we
> would conclude from b[1]+b[2]+b[3]=0 that Q(b[1]) splits P.
>
> The only rational elements of Z[b[1]] are the integers:
> All elements have the form p*b[1]^2+q*b[1]+r with p,q,r in Q with the
> additional property that p+q+r is an integer(!).
> But if p*b[1]^2+q*b[1]+r=s is rational then either
> 1) p=q=0 and hence s is an integer because p+q+r is; or
> 2) (p,q)!=(0,0), i.e. p*X^2+q*X+r-s is a polynomial in Q[X] of degree
> 1 or 2 that has b[1] as root, in contradiction to the irreducibility of
> R(X).
> (Actually, I had to adjust a few signs in my argument and I think it
> has become buggy here;
> maybe someone else can pin that down / correct it / decide that one has
> to take a different polynomial; I'm too tired to do so right now)
I was really tired and in the form given, my proof is just a piece of
extreme mathematics. :)
Let I(X)=2X^3+X+1, so my b:=b[1] above is a root of I(X).
To show that Z[b]/\Q=Z proceed like this:
An arbitrary element a of Z[b] is of the form a=P(b) where P in Z[X].
If n=deg P < 3, a cannot be in Q\Z because it either is irrational
(n>0) or is an integer (if n<1).
Therefore, assume n>=3.
Consider the polynomials P'(X):=X^n * P(1/X) and
I'(X):=X^3*I(X)=X^3+X^2+2.
Then P'(X)=Q(X)*I'(X)+ r*X^2+s*X+t with Q(X) in Z[X], r,s,t in Z.
Hence P(X)=Q'(X)*I(X)+X^(n-2)*(r+s*X+t*X^2) and
a=P(b)=b^(n-2)*(r+s*b+t*b^2).
Among all (n,r,s,t) such that a is in Q\Z select one with minimal n.
Since n>2, we can rewrite a as
a = b^(n-3)*(r*b+s*b^2+t*b^3)
= b^(n-3)*(-t/2 + (r-t/2)*b + s*b^2)
If t is even, this contradict the minimality of n.
Hence t is odd.
Let |x| denote 2adic value, i.e. for x!=0, |x| is a power of 2 such
that x*|x| is an odd integer (it is clear that all possible fractions
occurring have at most a power of 2 in the denominator), and |0|:=0.
Hence a is of the form
(1) u+v*b+w*b^2 with u,v,w in Q, |u|>|v|, |u|>|w|.
(Namely v=w=0)
Multiplying u+v*b+w*b^2 by b^{-1}=-(2b^2+1), we obtain
u'+v'*b+w'*b^2 with u'=v-u,v'=w,w'=-2u, which is again of the form (1)
because |u'|=|u|, |v'|=|w|<|u'|, |w'|=|u|/2 < |u'|.
Repeat this n-2 times to see that for
a/b^(n-2) = r+s*b+t*b^2
we must have
|r|>|s|, |r|>|t|.
But since t is odd, this implies that r is not an integer --
contradiction.