Re: tomic polynomial: CONJECTURE 3
amy666
> of tommy polynomials.
>
> regards
>
> tommy1729
as you can see , i am a man of my word , coming back to this concept of tomic polynomials as mentioned in the OP.
despite its been a while , i havent forgotten ^_^
regards
tommy1729
Gerry Myerson
<4411947.12228717463...@nitrogen.mathforum.org>,
amy666 <tomm...@hotmail.com> wrote:
I have. What is a tomic polynomial? What is a tommy polynomial?
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
quasi
<ge...@maths.mq.edi.ai.i2u4email> wrote:
>In article
><4411947.12228717463...@nitrogen.mathforum.org>,
> amy666 <tomm...@hotmail.com> wrote:
>
>> > Every algebraic number is a ( finite ) sum of zero's
>> > of tommy polynomials.
>> >
>> > regards
>> >
>> > tommy1729
>>
>> as you can see , i am a man of my word , coming back to this concept of tomic
>> polynomials as mentioned in the OP.
>>
>> despite its been a while , i havent forgotten ^_^
>
>I have. What is a tomic polynomial? What is a tommy polynomial?
Presumably this:
<http://groups.google.com/group/sci.math/msg/088788eaae0ed110>
quasi
quasi
wrote:
>> Every algebraic number is a ( finite ) sum of zero's
>> of tommy polynomials.
I suspect your conjecture is true.
Moreover, I think it would still be true if the coefficient set
{1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any
nonzero a in Q.
As far as proving it (or disproving it), that may be quite hard.
While thinking about your conjecture, I came up with a few related
questions, but since they're not directly about "tomic" polynomials,
I'll pose those questions in a separate thread.
quasi
galathaea
> On Wed, 01 Oct 2008 10:35:15 EDT, amy666 <tommy1...@hotmail.com>
well
since all abelian extensions are cyclotomic
the "only" algebraic numbers one needs to study
are those that are members of nonabelian extensions
do the {0,1}- or {-1,0,1}-polynomials exhaust them?
i don't know
i'm just stating some obvious things...
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
Denis Feldmann
> On Wed, 01 Oct 2008 10:35:15 EDT, amy666 <tomm...@hotmail.com>
> wrote:
>
>>> Every algebraic number is a ( finite ) sum of zero's
>>> of tommy polynomials.
>
> I suspect your conjecture is true.
For which reasons?
>
> Moreover, I think it would still be true if the coefficient set
> {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any
> nonzero a in Q.
Again, for which reasons
>
> As far as proving it (or disproving it), that may be quite hard.
>
Which is why I ask again : why makes you believe it is true?
quasi
The counterexample of 1/3, noted by galathaea in my thread about the
field generated by the set of roots of unity works just as well as a
counterexample here. The roots of {-1,0,1}-polynomials are algebraic
integers, hence so is any finite sum.
I guess you could revise the conjecture to claim that you can get all
algebraic integers.
But even with that restriction, I've changed my opinion as to the
truth of your conjecture -- I now strongly doubt it.
quasi
quasi
<denis.feldm...@neuf.fr> wrote:
>quasi a écrit :
>> On Wed, 01 Oct 2008 10:35:15 EDT, amy666 <tomm...@hotmail.com>
>> wrote:
>>
>>>> Every algebraic number is a ( finite ) sum of zero's
>>>> of tommy polynomials.
>>
>> I suspect your conjecture is true.
>
>For which reasons?
Intuition.
>> Moreover, I think it would still be true if the coefficient set
>> {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any
>> nonzero a in Q.
>
> Again, for which reasons.
The same intuition.
>> As far as proving it (or disproving it), that may be quite hard.
>
> Which is why I ask again : why makes you believe it is true?
Well, that was then.
As of now, I think tommy's conjecture is false (even with the
restriction to algebraic integers).
But it may not be so easy to disprove.
As a simple test case, try this ...
Is sqrt(2) a finite sum of roots of {-1,0,1}-polynomials?
quasi
Robert Israel
> On Wed, 01 Oct 2008 10:35:15 EDT, amy666 <tomm...@hotmail.com>
> wrote:
>
> >> Every algebraic number is a ( finite ) sum of zero's
> >> of tommy polynomials.
>
> I suspect your conjecture is true.
>
> Moreover, I think it would still be true if the coefficient set
> {1,0,-1} was replaced by {0,1}, or for that matter {0,a} for any
> nonzero a in Q.
So a "tommy polynomial" is a polynomial (of degree >= 1) whose coefficients
are in {1,0,-1}? Then any root of such a thing is an algebraic integer.
A sum of algebraic integers is an algebraic integer. But not every algebraic
number is an algebraic integer, e.g. the only rationals that are algebraic
integers are integers.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Denis Feldmann
reliable tool 2) that tommy's conjectures are similar to JSH ones : not
very interesting, often very easy to refutate, not thought upon at all,
and worst of all, formulated as ego-booster and not out of ny genious
interest...
quasi
<denis.feldm...@neuf.fr> wrote:
Sure, my intuition is sometimes wrong, and it's subject to change as
new information is obtained -- that's just the nature of it. But it's
right enough of the time to have my trust, at least as a guidance as
to what to expect. Besides, right or wrong, it typically leads to more
questions, further insights, and hence it evolves, but always leading.
>2) that tommy's conjectures are similar to JSH ones
Not by a long shot.
> : not very interesting,
I find some of them very interesting,
>often very easy to refutate,
Sure, so what.
As long as the underlying concept has some validity, the conjectures
can usually be revised to reach the boundary where trivially true or
trivially false suddenly becomes non-trivial.
>not thought upon at all,
From my perspective, that doesn't really matter. If the question has
value, that's enough.
>and worst of all, formulated as ego-booster and not out of ny genious
>interest...
In my opinion, egoless posts are often very boring.
Sure, tommy's ego is sometimes way out of control, but then again, we
all have flaws. In tommy's case, as I see it, there are many redeeming
virtues.
tommy1729 loves math -- there's no question about that, and in my
opinion, his questions are driven not just by ego, but also by a
powerful curiosity, the lure of possible discovery, and the challenge
and joy of the hunt. I can easily relate to that.
quasi
alainv...@gmail.com
> On Thu, 02 Oct 2008 08:50:12 +0200, Denis Feldmann
>
>
>
>
>
> <denis.feldmann.sanss...@neuf.fr> wrote:
> >quasi a écrit :
> >> On Thu, 02 Oct 2008 08:14:00 +0200, Denis Feldmann
> >> <denis.feldmann.sanss...@neuf.fr> wrote:
>
> >>> quasi a écrit :
>
> - Afficher le texte des messages précédents -
Bonjour,
You wrote "tommy1729 loves math"
Like most of us on this site , but :
Does math love tommy1729 ...and any of us?
Alain
quasi
wrote:
>Like most of us on this site , but :
>Does math love tommy1729 ...and any of us?
Haha.
quasi
amy666
yes the OP of this very thread.
amy666
who is this math ?
is math male or female ?
if math is denis feldmann then math doesnt love tommy1729.
i think math is a male stubborn bodyguard, bouncer and ninja ; hard to deal with.
however it can be seductive like a beautiful lady in red.
considering the existance of undecidability ill go for the male ninja.
altough ramanujan prefers a female god who whispers things in his ears.
( JSH hears voices too but thats a different story ^_^ )
regards
tommy1729
Dan Hoey
> On Thu, 02 Oct 2008 03:01:32 GMT, Gerry Myerson wrote:
>> I have. What is a tomic polynomial? What is a tommy polynomial?
>
> Presumably this:
>
> <http://groups.google.com/group/sci.math/msg/088788eaae0ed110>
Calling polynomials "tomic" based on their coefficients being
among {-1,0,1} is presumably based on the misapprehension that
cyclotomic polynomials possess this property.
Can we say anything true and interesting about the possible
coefficients of cyclotomic polynomials? Perhaps an upper bound
U(n,m) on the number of coefficients of magnitude m in the nth
cyclotomic polynomial?
Dan Hoey
haoyuep at aol.com
amy666
for all clarity [0,1] could be a tommy polynomial too !
i.e we have the choice of 1 0 and -1 , not neccesarily all of them.
regards
tommy1729
galathaea
i know
it's just that the {0,1}s already generate
all cyclotomic extensions
so we have the abelian extensions covered with just those
does this collection also get all nonabelian extensions?
do we need to go to {-1,0,1}?
or maybe the coefficients still don't exhaust?
it seems possible the {0,1} case
is easier to get some properties from
since your just adding basis terms
but it might not make a difference in the end
it's at least nice to separate into cases
quasi
<gala...@gmail.com> wrote:
>
>On Oct 5, 9:05 am, amy666 <tommy1...@hotmail.com> wrote:
>>
>> for all clarity [0,1] could be a tommy polynomial too !
>>
>> i.e we have the choice of 1 0 and -1 , not neccesarily all of them.
>
>i know
>
>it's just that the {0,1}s already generate
> all cyclotomic extensions
>so we have the abelian extensions covered with just those
>
>does this collection also get all nonabelian extensions?
I doubt it.
>do we need to go to {-1,0,1}?
I doubt that as well.
>or maybe the coefficients still don't exhaust?
I don't believe any finite set of integer coefficients will suffice.
quasi
quasi
<gala...@gmail.com> wrote:
>well
> since all abelian extensions are cyclotomic
Wait! Isn't it the other way around?
All cyclotomic extensions (and their subfields) are abelian.
>the "only" algebraic numbers one needs to study
>are those that are members of nonabelian extensions
Based on my objection above, I don't accept this.
As a test case, try this question:
Is sqrt(2) an element of a cyclotomic field?
quasi
Robert Israel
Yes: if w is a root of the 4'th cyclotomic polynomial X^4 + 1, then
w + 1/w is a square root of 2.
galathaea
> On Wed, 1 Oct 2008 22:05:23 -0700 (PDT), galathaea
>
> >well
> > since all abelian extensions are cyclotomic
>
> Wait! Isn't it the other way around?
>
> All cyclotomic extensions (and their subfields) are abelian.
that direction is much easier
the galois groups of cyclotomic extensions
are cyclic
(the units of
and so abelian
but the other direction
was a famous conjecture at the end of the 19th century
which kronecker
weber
and eventually hilbert
helped crank through a proof of it
the simplest proof i know of it
is through ramification theory
basically
the proof reduces to
that if all abelian extensions of Q
of degree p^r (primepower extensions)
are cyclotomic
then all abelian extensions are cyclotomic
(you use the abelian structure theorem
but i won't explain it because i'll mention a paper
that gives everything)
this then is further reduced to the specific case
where those primepower extensions
have p as the only prime that ramifies
all other extensions
with multiple primes ramifying
can be shown to again inherit the property
if this special case has it
then this special case gets spit out
through a brute force calculation on differents
separated into even and odd cases
another way to prove it
uses the artin map and class field theory
http://www.math.tifr.res.in/~eghate/kw.pdf
http://www.math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Culler.pdf
> >the "only" algebraic numbers one needs to study
> >are those that are members of nonabelian extensions
>
> Based on my objection above, I don't accept this.
>
> As a test case, try this question:
>
> Is sqrt(2) an element of a cyclotomic field?
all quadratic extensions are abelian
and so cyclotomic
there is a tower of field extensions
Q(w_p)
|
|
Q('\/(+/- q))
|
|
Q
where + is used when q-1 = 0 or 1 (mod 4)
and - otherwise
so
just to underscore the proof
_all_ quadratics extension
are members of some extension given by
Q[x] / (p)
for some p a {0,1}-coefficiented polynomial
(in fact, one of the form x^m + 1)
robert israel has given the calculation for '\/2
quasi
<isr...@math.MyUniversitysInitials.ca> wrote:
>quasi <qu...@null.set> writes:
>
>> On Wed, 1 Oct 2008 22:05:23 -0700 (PDT), galathaea
>> <gala...@gmail.com> wrote:
>>
>> >well
>> > since all abelian extensions are cyclotomic
>>
>> Wait! Isn't it the other way around?
>>
>> All cyclotomic extensions (and their subfields) are abelian.
>>
>> >the "only" algebraic numbers one needs to study
>> >are those that are members of nonabelian extensions
>>
>> Based on my objection above, I don't accept this.
>>
>> As a test case, try this question:
>>
>> Is sqrt(2) an element of a cyclotomic field?
>
>Yes: if w is a root of the 4'th cyclotomic polynomial X^4 + 1, then
>w + 1/w is a square root of 2.
Cool.
I didn't expect that, but I wasn't really sure -- that's why I
regarded it as a test case.
Thanks.
quasi
quasi
Yes, I saw.
Thanks very much galathaea for the exposition.
A few questions:
For a given squarefree n in N, let m be the least positive integer
such that sqrt(n) is an element of Q(w), where w is a root of x^m + 1
(1) Is m an increasing function of n?
(2) Is there a simple formula for m as a function of n? If not, how
about an asymptotic formula?
quasi
galathaea
this number is known as the conductor of a field
for a quadratic extension
it is the absolute value of the discriminant
Hagen
Tomic polynomials by definition do not have roots of unity
as zeros - except -1,+1 if I read the definition correctly.
Gerry Myerson
> Can we say anything true and interesting about the possible
> coefficients of cyclotomic polynomials? Perhaps an upper bound
> U(n,m) on the number of coefficients of magnitude m in the nth
> cyclotomic polynomial?
There is a body of research on the magnitude of coefficients
of cyclotomic polynomials. Gennady Bachman has a book
about the problem in the AMS Memoirs series, and several
papers besides.
Which is my way of saying that I don't know the answers
offhand, but I know where I'd start looking if I wanted them.
Gerry Myerson
<68bdd644-5d2f-43af...@40g2000prx.googlegroups.com>,
galathaea <gala...@gmail.com> wrote:
Gauss sums do the trick for sqrt(n). If w is exp(2 pi i / p)
and you add up the numbers w^(r^2),
r = 0, 1, ..., p - 1,
you get sqrt(p) if p is 1 mod 4,
and i sqrt(p) if p is 3 mod 4.
That's if p is prime, but there are similar simple formulas
for composite values.
Then if i sqrt(p) isn't good enough for you,
and you really want sqrt(p),
just go up to Q(w, i),
which is still cyclotomic and only of degree 2 over Q(w).
quasi
That makes it very concrete -- thanks.
quasi
quasi
Problem (1):
Let w be a root of the polynomial x^3 - 3x +1. Find a cyclotomic field
which contains w.
Problem (2);
Let w be a root of the polynomial x^3 - 2. Can w be expressed as a
finite sum of roots of irreducible polynomials all of whose
coefficients are in {-1,0,1} ?
quasi
Gerry Myerson
quasi <qu...@null.set> wrote:
> Ok, how about 2 more test cases -- this time cubic irrationals.
>
> Problem (1):
>
> Let w be a root of the polynomial x^3 - 3x +1. Find a cyclotomic field
> which contains w.
Is this the one whose zeros are things like 2 cos (2 pi / 7)? If so,
easy-peasy, as 2 cos x = exp(i x) + exp(- i x).
On the other hand, if the Galois group is S_3, thus nonabelian,
then it can't be done.
> Problem (2);
>
> Let w be a root of the polynomial x^3 - 2. Can w be expressed as a
> finite sum of roots of irreducible polynomials all of whose
> coefficients are in {-1,0,1} ?
Is there some reason to be interested in such polynomials?
Robert Israel
> Ok, how about 2 more test cases -- this time cubic irrationals.
>
> Problem (1):
>
> Let w be a root of the polynomial x^3 - 3x +1. Find a cyclotomic field
> which contains w.
If r is a root of the 9th cyclotomic polynomial Z^6 + Z^3 + 1, then
x^3 - 3x + 1 = (x-r^4-r^5)(x+r-r^2+r^4)(x-r+r^2+r^5).
This can be found using the following Maple code:
for nn from 3 do
F:= factor(x^3-3*x+1,RootOf(numtheory[cyclotomic](nn,_Z)));
if type(F, `*`) then break end if
end do:
nn, numtheory[cyclotomic](nn,r),
subs(RootOf(numtheory[cyclotomic](nn,_Z))=r, F);
quasi
<ge...@maths.mq.edi.ai.i2u4email> wrote:
>In article <hcple4h6gql37ni2v...@4ax.com>,
> quasi <qu...@null.set> wrote:
>
>> Ok, how about 2 more test cases -- this time cubic irrationals.
>>
>> Problem (1):
>>
>> Let w be a root of the polynomial x^3 - 3x +1. Find a cyclotomic field
>> which contains w.
>
>Is this the one whose zeros are things like 2 cos (2 pi / 7)? If so,
>easy-peasy, as 2 cos x = exp(i x) + exp(- i x).
Ah, so there's the connection to trigonometry. Perhaps it's a simple
case of the phenomenon that galathaea keeps alluding to?
>On the other hand, if the Galois group is S_3, thus nonabelian,
>then it can't be done.
Which is why, for (2^(1/3), we need a bigger field, not just the field
generated by all roots of unity. Thus, problem (2).
>> Problem (2);
>>
>> Let w be a root of the polynomial x^3 - 2. Can w be expressed as a
>> finite sum of roots of irreducible polynomials all of whose
>> coefficients are in {-1,0,1} ?
>
>Is there some reason to be interested in such polynomials?
Only temporary interest.
It was suggested that perhaps every algebraic integer can be
represented as a finite sum of roots of such polynomials.
If true, I think that would be very interesting.
On the other hand, my (current) gut feeling is that it's false. If so,
the interest lies in the disproof.
I'll back the following counter-conjecture.
Conjecture:
There does not exist a finite subset A of Z such that the field (over
Q) generated by the roots of all polynomials with coefficients in A is
the field of all algebraic numbers.
Remarks:
(1) The above conjecture seems interesting to me, not because the
claim is surprising, but rather because it's _not_ surprising. Here we
have a simply stated result, which I think is probably true, but for
which I don't see an immediate attack.
(2) Problem (2) deals with a special case. I believe that 2^(1/3) is
_not_ an element of the field generated by the roots of all
polynomials whose coefficients are in {-1,0,1}. But how does one prove
such a claim? That's the challenge.
quasi
Hagen
>
> There does not exist a finite subset A of Z such that
> the field (over
> Q) generated by the roots of all polynomials with
> coefficients in A is
> the field of all algebraic numbers.
>
> Remarks:
>
> (1) The above conjecture seems interesting to me,
> not because the
> claim is surprising, but rather because it's _not_
> surprising. Here we
> have a simply stated result, which I think is
> probably true, but for
> which I don't see an immediate attack.
Comment: the maximal abelian extension of Q is the
splitting field of the polynomials X^n-1, n=1,2,3,4,... .
And the splitting field of the set of cyclotomic polynomials.
It is known that the set of all integer numbers appearing as
a coefficient of a cyclotomic polynomial equals Z.
Surprising?
quasi
>> Conjecture:
>>
>> There does not exist a finite subset A of Z such that
>> the field (over Q) generated by the roots of all polynomials with
>> coefficients in A is the field of all algebraic numbers.
>>
>> Remarks:
>>
>> (1) The above conjecture seems interesting to me,
>> not because the
>> claim is surprising, but rather because it's _not_
>> surprising. Here we
>> have a simply stated result, which I think is
>> probably true, but for
>> which I don't see an immediate attack.
>
>Comment: the maximal abelian extension of Q is the
>splitting field of the polynomials X^n-1, n=1,2,3,4,... .
>And the splitting field of the set of cyclotomic polynomials.
Ok.
>It is known that the set of all integer numbers appearing as
>a coefficient of a cyclotomic polynomial equals Z.
>Surprising?
Yes, somewhat surprising.
But I don't get your point.
Does provide any insight with regard to either tommy's conjecture or
my counter-conjecture?
quasi
amy666
> In article
> <68bdd644-5d2f-43af...@40g2000prx.googl
Ramanujan sums can be used as well ...
regards
tommy1729
amy666
for clarity , tomic polynomials do not have roots of unity , but they dont have to be irreducible.
amy666
correct.
but sum of there zero's might give roots of unity ...
>
> > does this collection also get all nonabelian
> > extensions?
> > do we need to go to {-1,0,1}?
> > or maybe the coefficients still don't exhaust?
> >
> > it seems possible the {0,1} case
> > is easier to get some properties from
> > since your just adding basis terms
> > but it might not make a difference in the end
> >
> > it's at least nice to separate into cases
> >
> > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
> > galathaea: prankster, fablist, magician, liar
regards
tommy1729
quasi
wrote:
Yes, my mistake -- I inadvertantly specified irreducibility.
quasi
amy666
very intresting.
amy666
i forgive you :)
however its still a related question.
btw my conjecture is probably false for algebraic numbers that are not algebraic integers ?
though i did not see gala's proof that 1/3 is not constructable ...
here i will present two concepts that might inspire you or others :
1) since tomic polynomials do not have roots of unity apart from [1,-1] , its a logical question -> does conjecture 3 hold for roots of unity ?
2) we can reduce conjecture 3 by replacing 'algebraic number' by ' real algebraic number between 0 and 1 '
i call this concept "V-theory" , but it might already have a name or be considered a basis , generator , vector origin , ... or a trivial ( and named ) theorem of them. ( perhaps you point me to the correct terminology if such )
3) what would galois or riemann do ? :p
***********
i think 1) holds , but requires some factorable tomic polynomials
***********
regards
tommy1729
Hagen
> however its still a related question.
>
> btw my conjecture is probably false for algebraic
> numbers that are not algebraic integers ?
Roots of tomic polynomials are algebraic integers, so your
conjecture can at most be true for algebraic integers.
> though i did not see gala's proof that 1/3 is not
> constructable ...
>
> here i will present two concepts that might inspire
> you or others :
>
> 1) since tomic polynomials do not have roots of unity
> apart from [1,-1] , its a logical question -> does
> conjecture 3 hold for roots of unity ?
No: the roots of unity generate a proper
subfield of the algebraic closure of Q - namely the maximal
abelian extension of Q.