field generated by the set of roots of unity
quasi
conjecture about "tomic" polynomials ...
Let K be the field of algebraic numbers, and let F = Q(S) where
S = {x in K | x^n = 1 for some n in N}.
In other words, F is the subfield of K generated by the set of roots
of unity.
Questions:
(1) Does F = K?
(2) Is every element of F a finite linear combination, over Q, of
elements of S?
(3) Assuming the answer to (2) is yes, is every element of F a finite
sum of elements of S?
quasi
Arturo Magidin
quasi <qu...@null.set> wrote:
>Here are some questions, related to and inspired by tommy1729's
>conjecture about "tomic" polynomials ...
>
>Let K be the field of algebraic numbers, and let F = Q(S) where
>
> S = {x in K | x^n = 1 for some n in N}.
>
>In other words, F is the subfield of K generated by the set of roots
>of unity.
>
>Questions:
>
>(1) Does F = K?
No. F is the inverse limit of the cyclotomic fields, under the usual
Galois correspondence in the infinite case.
>(2) Is every element of F a finite linear combination, over Q, of
>elements of S?
I believe so; look at Q(a), which will be contained in a finite
extension, given by a finite number of elements of S, which can always
be enriched to include powers of those elements since S is closed
under powers to get a spanning set.
--
======================================================================
"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
quasi
(Arturo Magidin) wrote:
>In article <qtk8e4tqa4hhlmsrf...@4ax.com>,
>quasi <qu...@null.set> wrote:
>>Here are some questions, related to and inspired by tommy1729's
>>conjecture about "tomic" polynomials ...
>>
>>Let K be the field of algebraic numbers, and let F = Q(S) where
>>
>> S = {x in K | x^n = 1 for some n in N}.
>>
>>In other words, F is the subfield of K generated by the set of roots
>>of unity.
>>
>>Questions:
>>
>>(1) Does F = K?
>
>No. F is the inverse limit of the cyclotomic fields, under the usual
>Galois correspondence in the infinite case.
Yes, I follow.
But I should have seen that F can't equal K, for the simple reason
that every element of F can be expressed as an arithmetic combination
of radicals.
>>(2) Is every element of F a finite linear combination, over Q, of
>>elements of S?
>
>I believe so; look at Q(a), which will be contained in a finite
>extension, given by a finite number of elements of S, which can always
>be enriched to include powers of those elements since S is closed
>under powers to get a spanning set.
Yes, I see.
So that takes care of questions (1) and (2).
So only question (3) remains.
quasi
galathaea
> Here are some questions, related to and inspired by tommy1729's
> conjecture about "tomic" polynomials ...
>
> Let K be the field of algebraic numbers, and let F = Q(S) where
>
> S = {x in K | x^n = 1 for some n in N}.
>
> In other words, F is the subfield of K generated by the set of roots
> of unity.
>
> Questions:
>
> (1) Does F = K?
no
examples can be generated from methods
that solve the embedding problem
in galois theory
examples of low degree are difficult
i don't think an extension exists
for instance
for the quaternion group of order 8
but i may be wrong...
> (2) Is every element of F a finite linear combination, over Q, of
> elements of S?
by definition of generated by?
maybe i'm not understanding...
> (3) Assuming the answer to (2) is yes, is every element of F a finite
> sum of elements of S?
integers?
or rationals?
if i understand
then the answer is no
no sum would give 1/3 (w_11)
(i think)
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
quasi
<gala...@gmail.com> wrote:
>On Oct 1, 10:10 pm, quasi <qu...@null.set> wrote:
>> Here are some questions, related to and inspired by tommy1729's
>> conjecture about "tomic" polynomials ...
>>
>> Let K be the field of algebraic numbers, and let F = Q(S) where
>>
>> S = {x in K | x^n = 1 for some n in N}.
>>
>> In other words, F is the subfield of K generated by the set of roots
>> of unity.
>>
>> Questions:
>>
>> (1) Does F = K?
>
>no
>
>examples can be generated from methods
> that solve the embedding problem
> in galois theory
>
>examples of low degree are difficult
>
>i don't think an extension exists
> for instance
>for the quaternion group of order 8
>but i may be wrong...
>
>> (2) Is every element of F a finite linear combination, over Q, of
>> elements of S?
>
>by definition of generated by?
Well, S generates F as a field, but question (2) asks whether S
generates S as a vector space. But as Arturo Magidin outlines, the
answer to question (2) is "yes".
>maybe i'm not understanding...
>
>> (3) Assuming the answer to (2) is yes, is every element of F a finite
>> sum of elements of S?
>
>integers?
>or rationals?
>
>if i understand
> then the answer is no
>
>no sum would give 1/3 (w_11)
> (i think)
Yes, of course -- you're absolutely right -- 1/3 can't be a finite sum
of elements of S.
Every element of S is an algebraic integer, hence so is any finite sum
of elements of S.
Then I guess the revised question should be ...
(3) [revised]:
Is every algebraic integer in F a finite sum of elements of S?
quasi
Hagen
> > Here are some questions, related to and inspired by
> tommy1729's
> > conjecture about "tomic" polynomials ...
> >
> > Let K be the field of algebraic numbers, and let F
> = Q(S) where
> >
> > S = {x in K | x^n = 1 for some n in N}.
> >
> > In other words, F is the subfield of K generated by
> the set of roots
> > of unity.
> >
> > Questions:
> >
> > (1) Does F = K?
>
> no
>
> examples can be generated from methods
> that solve the embedding problem
> in galois theory
>
> examples of low degree are difficult
>
> i don't think an extension exists
> for instance
> for the quaternion group of order 8
> but i may be wrong...
>
The Galois extension F|Q is abelian - it is in fact the maximal
abelian extension of Q. Hence F cannot contain a Galois
extension having a non-abelian Galois group.
Hagen
Yes. Every algebraic integer x is contained in a cyclotomic
extension Q(z), where z denotes a primitive nth root of
unity for some n. The ring of algebraic integers in Q(z) is
known to be Z[z], so that x can be written as a linear
combination of powers of z with integer coefficients.
H
Jannick Asmus
> (3) [revised]:
>
> Is every algebraic integer in F a finite sum of elements of S?
Yes: S is a finite group, hence cyclic - generated by a primitive root
of unity, say, w. Then the powers of w are an integral basis of the ring
of integers of Q(S)=Q(w).
--
Best wishes,
J.
Hagen
quasi considers ALL roots of unity.
Jannick Asmus
I loosely stated that w was primitive. If K is normal, hence Galois, w
is. If not that might give problems here ?!? At the moment I cannot see
an argument to assume wlog that K is Galois.
--
Best wishes,
J.
Jannick Asmus
Ok, I misread it. I thought K is a number field.
--
Best wishes,
J.
Anvita
This statement seems to contradict the inequality F =/= K, doesn't it?
Can we take x from K\F ? Then x is in Q(z) which, in turn, is a subset
of F ... What am I missing?
Anvita
Jannick Asmus
I stay to the idea by "descent on finite steps": Since S is the union of
finite subgroups S' of S, F is the union of such Q(S'). Wlog we can
assume that S' is generated by a primitive root of unity.
If a is algebraic in F, then a is already algebraic in some Q(S') with a
finite subgroup S' of S generated by a primitive root of unity, say w.
Algebraic elements in Q(w) can be represented as Z-linear combination of
powers of w. Hence the claim.
--
Best wishes,
J.
Jannick Asmus
This was already proven by Hagen. Sorry for the confusion, but my
newsserver seems to spit out the postings with some nasty delay today.
--
Best wishes,
J.
Hagen
> tommy1729's
> conjecture about "tomic" polynomials ...
>
> Let K be the field of algebraic numbers, and let F =
> Q(S) where
>
> S = {x in K | x^n = 1 for some n in N}.
>
> In other words, F is the subfield of K generated by
> the set of roots
> of unity.
>
> Questions:
>
> (1) Does F = K?
>
Just for insiders and with no intention to mock somebody:
This question reminds me of a >>brain storming<< of a
similar problem that took place at a certain math department
in Germany and that ended with the outstanding >>result<<
that the absolute Galois group of the rationals is trivial.
H
Hagen
It should read 'every algebraic integer in F'.
Sorry.
galathaea
rereading my post
i see it makes absolutely no sense
without my other post in the tomic thread
i thought after the no i had explained
that it F was abelian from kronecker-weber
which was the whole point of me
looking for a nonabelian example of low degree
and thus the look at quaternions
(or p^3 groups in general as a starting point)
but that was in the other thread
thanks for filling in the gap here
because it certainly makes no sense otherwise
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
but new question:
in what sense is this maximal?
is this maximal over algebraic extensions of Q?
or over all extensions of Q?
i've been pouring over kroenecker-weber
which really isn't that long of a proof really
and i still haven't figured out
if it excludes some of the transcendental extensions
built from the generalised polynomials
that come from the generalised trigonometry
i.e.
are there abelian subfields of
lim W[x] / (g)
<--
g
where W_n[x] is the ring of generalised polynomials
of the form
---
\ e_a
/ a x
---
a e A
A finite
A c C_n
e_a c C^_n
where C_n is a cyclotomic field
and C^_n it's ring of integers
(ie. as you've pointed out in the past
the semigroup ring built from cyclotomics)
?
or are all such limits' center always F?
i've been able to find a lot of algebraic properties
of these transcendental extensions
but i'm still very fuzzy
about this very foundational property...
Hagen
Yes, only algebraic extensions are considered.
Hagen
I (still) don't understand what precisely the generalised
polynomials in W_n[x] are.
To simplify notation a bit let K be a cyclotomic field and let
O be the ring of integers of K.
Sticking to the theory of semi-group rings (which my be the
wrong approach here) you seem to consider maps
f: O --> K
having finite support.
Such maps are added pointwise.
Multiplication however - if we follow the situation for polynomial
rings - should be defined using some sort of Cauchy product
rule. Can you explain how it works?
Given W[x] is defined, what is g in the projective limit appearing
in your post?
H
galathaea
products work just like polynomial products
as an example
look at the cyclotomic field of degree 3
example "polynomials" might be
w
3
x + x + 1
and
2
w w
3 3
x + x - 1
where w_3 is the cube root of unity
multiplying these together gives
term-by-term
2 2
1+w 1+w 2w w + w
3 3 3 3 3
x + x - x + x + x
2
w w w
3 3 3
- x + x + x - 1
which might be simplified to
2 2
2w -w -w w
3 3 3 -1 3
x + x + x + x - x + x - 1
> Given W[x] is defined, what is g in the projective limit appearing
> in your post?
g was meant to range over "irreducibles"
(with a few ambiguities handwaved away)
perhaps a better characterisation
of the generalised polynomial rings i am interested in
can be built on the definition starting with the form
K[x0, x1, ..., x_(n-1)]
and quotienting out terms like
x0 x1 ... x_(n-1) = 1
(i.e. modding over the ideal (x0 x1 ... x_(n-1) - 1))
it's much easier to see the formal variables
are all units
and similar properties from this form
then
taking this construction as the "polynomial" ring
taking quotient rings with
ideals generated from irreducible elements
should give something that looks like a number field extension
the point or motivation behind the construction
where the projective limits were used
was to build extensions of the algebraic numbers
that still have many usable properties
or at least
that's the hope
i'm still banging my head against them
to try to figure out which properties are attackable
and i've made some basic characterisations
but i'm still not practiced enough in the tools
these rings come from the generalised trigonometry
where they are the rings that are used
to invert the trigonometrics
(much as in the standard log forms
of the classical trigonometrics)
these rings have tchebyshef maps on important elements
(and also horizontal maps
and other tools from the trigonometry)
galathaea
let me try to give a little better motivation
these polynomials can actually be solved in C
(the complex numbers)
and these solutions behave a lot like polynomial solutions
there is a sense in which these generalised polynomials
have "zeroes" in number
(on a given branch)
corresponding to a "depth" or "level"
of the exponents
(by an application of cauchy's theorem
and some algebraic trickery)
these "zeroes" are kind of weird
when compared to polynomials
because they have a lot more structure
and most of these zeroes are not algebraic
so looking at these extensions
some W_n[x] / (g) extension over W_n
seems an accessible target of study
in that world beyond algebraic extensions of Q
it's a manageable transcendental field
and an interesting new structure
to extend galois analysis
and it's not just any transcendental extension
it corresponds to the exponential extensions
through the maps with the generalised trigonometry
so this could be a useful way to extract information
important to "big" conjectures
like that of schanuel
of course it might not give anything useful
(and i'm likely missing many stupid things)
but some study transcendence theory
and a couple of interesting leads
at least give a path for study
but ideas like trying to understand
even just it's abelian structure
seem like a task for many
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, topposter
Hagen
Ok, this really looks like the group ring Q[(R,+)],
where (R,+) denotes the additive group of the
integral closure of Z within a number field K (the
3rd cyclotomic field in your case).
(R,+) is a free abelian group of rank n=[K:Q].
The structure of Q[(R,+)] then is known: take n
variables x_1,...,x_n and consider the rational function
field F=Q(x_1,...,x_n) . Q[(R,+)] is isomorphic to the
Q-algebra A in F generated by the variables x_k and
their inverses x_k^(-1).
galathaea
i'm not sure i understand
in what way is (R,+) "free"?
2 n-1
1 + w + w + ... + w = 0
n n n
or
as a relationship on the exponents
> perhaps a better characterisation
> of the generalised polynomial rings i am interested
> ed in
> can be built on the definition starting with the form
> K[x0, x1, ..., x_(n-1)]
> and quotienting out terms like
> x0 x1 ... x_(n-1) = 1
am i misunderstanding your point?
Hagen
(R,+) as an abelian group is isomorphic to
Z^n for example through chosing an integral
basis of R over Z.
galathaea
i'm very sorry
something is flying over my head
and i have no clue what it is
maybe my point was not clear because of the ascii art
my point was
in the ring of integers of a cyclotomic field
the additive group doesn't seem to be free
there is the nontrivial linear relationship
1 + w_n + (w_n)^2 + ... + (w_n)^(n-1) = 0
for instance
i don't see how an isomorphism with Z^n could exist
because in Z^n
(1, 1, ..., 1) =/= (0, 0, ..., 0)
but in the cyclotomics they are actually the same element
in fact
in this additive group of cyclotomic integers
(as also found in the polysign formulation
and other combinatorial equivalents)
any vector (a0, a1, ..., a_(n-1))
is equivalent to some vector
with a zero in (at least) one of it's entries
by repeatedly using the above relation
also
relations like
sum of all pairs of products = 0
sum of all triplets of products = 0
...
are other relations unrelated to the additive group
that can be read off of the symmetric polynomial coefficients
of x^n - 1 = 0
i'm sorry if i'm completely misunderstanding your point
Gerry Myerson
<6eacce6f-43ea-40be...@l33g2000pri.googlegroups.com>,
galathaea <gala...@gmail.com> wrote:
> On Oct 13, 1:58 am, Hagen <k...@itwm.fhg.de> wrote:
> > (R,+) as an abelian group is isomorphic to
> > Z^n for example through chosing an integral
> > basis of R over Z.
>
> i'm very sorry
>
> something is flying over my head
> and i have no clue what it is
>
> maybe my point was not clear because of the ascii art
>
> my point was
> in the ring of integers of a cyclotomic field
> the additive group doesn't seem to be free
>
> there is the nontrivial linear relationship
>
> 1 + w n + (w n)^2 + ... + (w n)^(n-1) = 0
I may be in over my head here, but I think this is a little
like saying that R^2 can't be free because there is the
non-trivial linear relationship
(1, 0) + (0, 1) + (-1, -1) = 0.
It is true that, say, 1 + i + i^2 + i^3 = 0,
but the ring of integers, seen as the set of a + b i,
is isomorphic, as an abelian group, to Z^2.
Maybe the problem is that the n in Z^n
isn't the same n
as the n in n-th root of unity.
4th root of unity - Z^2
5th root of unity - Z^4
6th root of unity - Z^2 again
etc.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
galathaea
wrote:
lol
the extension is only phi(n)...
and any ring of integers is a free Z-module...
thank you for letting me see
what should have been pretty obvious
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: swoooosh!
galathaea
i will just repeat your claims
to help me understand and not miss things
what i have written in the past as W_n[x]
the generalised polynomial rings
associated to the generalised trigonometry
is the group ring Q[(O(C_n), +)]
O(C_n) is the ring of integers
of the n-th cyclotomic field
this ring of integers is actually a free group
of rank phi(n)
(all rings of integers are free Z-modules)
and this
Q[<x1, x2, ..., x_(phi(n))>]
is known to be to be the subfield
Q(x1, x2, ..., x_(phi(n)))
that is generated by x1, x2, ..., x_(phi(n))
and their inverses
okay
so for W_3[x]
phi(3) = 2
and one can take the integral basis {1, w_3}
here we are looking at the subfield of
Q(x1, x2) generated by x1, x1^(-1), x2, x2^(-1)
i will write x1 <-> x
w
3
x2 <-> x
and then
2
w
3
the element that i called x
is translated in this language to
-1 -1
x x
1 2
okay
this makes a lot of sense
now
does this mean the transcendence degree of W_n[x]
is phi(n)?
does it inherit the full transcendence from
Q(x1, x2, ..., x_(phi(n)))
or does the subfield lose any?
Hagen
The group ring Q[(O(C_n), +)] is not a field, but
an integral domain generated as a Q-algebra by
the xk and their multiplicative inverses.
> so for W_3[x]
> phi(3) = 2
> and one can take the integral basis {1, w_3}
>
> here we are looking at the subfield of
> Q(x1, x2) generated by x1, x1^(-1), x2, x2^(-1)
Ok.
> i will write x1 <-> x
>
> w
> 3
> x2 <-> x
>
> and then
> 2
> w
> 3
> the element that i called x
> is translated in this language to
>
> -1 -1
> x x
> 1 2
>
> okay
> this makes a lot of sense
>
> now
> does this mean the transcendence degree of W_n[x]
> is phi(n)?
Yes.
amy666
perhaps you are very well aware of it , but just in case ;
i guess its about time i said your concepts can probably be completely "translated" in the so-called polysigned combined with differential equations.
( and their associated matrices if you like matrices )
since you based it on roots of unity and multisections ...
regards mathbabe :)
tommy1729