{group theory} positive rationals under multiplication = polynomials under addition
Sniz Pilbor
Been reading "The Theory of Groups: An Introduction (2nd
Edition)", a huge and advanced tome by Rotman. In one of the
exercises, he casually mentions that Z[x], the polynomials with
integer coefficients, considered as a group under addition, is
isomorphic to the positive rationals considered as a group under
multiplication. The exercise was to prove this by exhibiting such an
isomorphism explicitly, and I solved it thus:
a
- = (p1^k1) * (p2^k2) * (p3^k3) * ...
b
where p1,p2,... are the prime numbers and k1,k2,... are integers
which, by the fundamental theorem of arithmetic, exist and are unique.
(They are not necessarily nonnegative, as for example 1/2 = 2^(-1))
Therefore, given a positive rational q, we are also given k1,k2,k3,...
and we use these to construct
i(q) = k1*(x^0) + k2*(x^1) + ... + kn(x^(n-1)) + ...,
a function from the positive rationals to Z[x].
It's not hard to show that i is an isomorphism.
As some examples:
i(0) = the zero polynomial
i(1/2) = -1 (the polynomial)
i(3/5) = x - x^2
i(3.14159265) = -8 + 2x - 7x^2 + x^3 + x^30 + x^991
This all seems incredibly interesting and profound. It means, for
example, that we can sensibly refer to the "zeros", the "degree", the
"maxima", the "derivative", etc. of a positive rational, or of "the
value at x=5 of" the same. Given a rational, we can figure it's
isomorphic image among Z[x], differentiate this and go back to Q to
obtain a new rational.. like so:
d/dx 3.14159265 = i^-1(2 - 14x + 3x^2 + 30x^29 + 991x^990)
= (2^2)*(5^3)*(113^30)*(7841^991)/(5^14)
It seems there is no limit to the amount of new operations we can
invent using this isomorphism, the above are just a few examples. We
could speak of "irreducible" rationals, "the minimum rational" of an
algebraic (ie, the rational image of its minimal polynomial), etc.
etc. etc.
Unfortunately, the book only briefly mentioned this fascinating
isomorphism in the exercise, and that's the end. Now I'm horribly
filled with unsated curiosity. Has anyone studied this isomorphism in
more detail? Are there any papers on it?
Julien Santini
845b431.03120...@posting.google.com...
> Hello!
>
Hello
> Been reading "The Theory of Groups: An Introduction (2nd
> Edition)", a huge and advanced tome by Rotman. In one of the
> exercises, he casually mentions that Z[x], the polynomials with
> integer coefficients, considered as a group under addition, is
> isomorphic to the positive rationals considered as a group under
> multiplication. The exercise was to prove this by exhibiting such an
> isomorphism explicitly, and I solved it thus:
>
> a
> - = (p1^k1) * (p2^k2) * (p3^k3) * ...
> b
>
> where p1,p2,... are the prime numbers and k1,k2,... are integers
> which, by the fundamental theorem of arithmetic, exist and are unique.
> (They are not necessarily nonnegative, as for example 1/2 = 2^(-1))
>
First thing: the decomposition is unique, but a/b (the representant of your
rational q, is not unique; I suppose you implicitely assumed that
gcd(a,b)=1).
Second thing: what's happening when q = 1 ?
> Therefore, given a positive rational q, we are also given k1,k2,k3,...
> and we use these to construct
> i(q) = k1*(x^0) + k2*(x^1) + ... + kn(x^(n-1)) + ...,
> a function from the positive rationals to Z[x].
> It's not hard to show that i is an isomorphism.
>
And what about i(1) = ?
> As some examples:
> i(0) = the zero polynomial
> i(1/2) = -1 (the polynomial)
> i(3/5) = x - x^2
> i(3.14159265) = -8 + 2x - 7x^2 + x^3 + x^30 + x^991
>
> This all seems incredibly interesting and profound. It means, for
> example, that we can sensibly refer to the "zeros", the "degree", the
> "maxima", the "derivative", etc. of a positive rational, or of "the
> value at x=5 of" the same. Given a rational, we can figure it's
> isomorphic image among Z[x], differentiate this and go back to Q to
> obtain a new rational.. like so:
> d/dx 3.14159265 = i^-1(2 - 14x + 3x^2 + 30x^29 + 991x^990)
> = (2^2)*(5^3)*(113^30)*(7841^991)/(5^14)
>
> It seems there is no limit to the amount of new operations we can
> invent using this isomorphism, the above are just a few examples. We
> could speak of "irreducible" rationals, "the minimum rational" of an
> algebraic (ie, the rational image of its minimal polynomial), etc.
> etc. etc.
>
> Unfortunately, the book only briefly mentioned this fascinating
> isomorphism in the exercise, and that's the end. Now I'm horribly
> filled with unsated curiosity. Has anyone studied this isomorphism in
> more detail? Are there any papers on it?
Eheh... this is merely useless, by definition: you've found (maybe, but
what's sure is that you can find) an isomorphism between (Z[X],+) and the
(non-negative rationals, X) which means -very simply) that you identified
those two groups, ie those are the SAME structure.
As a result, you're not finding out anything new, just shifting operations
from one group to one other (which is the "same").
William Elliot
>"Sniz Pilbor" <snizp...@yahoo.com> a crit dans le message
>> Z[x], the polynomials with integer coefficients, considered
>> as a group under addition, is isomorphic to the positive rationals
>> considered as a group under multiplication. The exercise was to
>> prove this by exhibiting such an isomorphism explicitly,
>
>> a
>> - = (p1^k1) * (p2^k2) * (p3^k3) * ...
>> b
a_m x^m + a_m-1 x^{m-1} + ... + a_1 x + a_0
--> p_m ^ a_m * p_m-1 ^ a_m-1 * ... * 3^a_1 * 2^a_0
p_m = m-th odd prime
>> where p1,p2,... are the prime numbers and k1,k2,... are integers
>> which, by fundamental theorem of arithmetic, exist and are unique.
>I suppose you implicitely assumed that gcd(a,b)=1).
>Second thing: what's happening when q = 1 ?
It gets mapped to 0.
>> Therefore, given a positive rational q, we are also given
>> k1,k2,k3,... and we use these to construct
>> i(q) = k1*(x^0) + k2*(x^1) + ... + kn(x^(n-1)) + ...,
>> a function from the positive rationals to Z[x].
>> It's not hard to show that i is an isomorphism.
>
>And what about i(1) = ?
i(1) = 0
>> As some examples:
>> i(0) = the zero polynomial
Doesn't compute as 0 isn't a positive rational.
>> i(1/2) = -1 (the polynomial)
>> i(3/5) = x - x^2
>> i(3.14159265) = -8 + 2x - 7x^2 + x^3 + x^30 + x^991
>
So for i:(0,oo) -> Z[[x]], gives i(pi) = some infinite series ???
If rational sequence q_j -> pi, then i(q_j) -> some thing in Z[[x]].
If rational r_j -> pi, then does i(r_j) -> some thing in Z[[x]]?
> Given a rational, we can figure it's isomorphic image among Z[x],
>differentiate this and go back to Q to obtain a new rational..
>
Let f:positive Q -> positive Q, f(q) = i^-1(Di(q))
Is f continuous?
If q_j -> q, does f(q_j) -> f(q)?
Can f be (uniquely) extended to a real function?
Is the extension f:[0,oo) -> [0,oo) differentiable, integrable?
For all q in Q, some n in N with f^n(q) = 0
f^n(q) = i^-1(D^n i(q))
--
Let g:Q -> Q, g(q) = i^-1(integral(0,x) i(q) dx)
Is g continuous?
If q_j -> q, does g(q_j) -> g(q)?
Can g be (uniquely) extended to a real function?
Is the extension g:R -> R differentiable, integrable?
lim(n->oo) g^n(q) -> ??
g^n(q) = i^-1(integral(0,x) integral(0,x) ... integral(0,x) i(q) dx^n)
--
The answer to these and many more tantalizers like
For p in Z[x], let r:Q -> Q, r(q) = i^-1(p(i(q)))
r^n(q) = i^-1(p^n(i(q))
can be found in a diligent search of mathematical research papers,
and if perchance by some unforeseen improbability, there's nothing there,
then forthwith immediately you'll find wondrous answers and even more
unseemly phantasies in your thesis proposal. ;-)
----
matt grime
but none of these is meaningful, as you can define as many different
isomorphisms as you may wish, and the information you want to be encoded
depends on the isomorphism. for instance, just swap over any two
coefficients in you polynomials. this alters the 'degree' of some of the
rationals, the 'maxima' and the roots too.
for something like this, you would need the information you want to study
to be independent of any choices made, or something akin to that anyway.
as a further example, you can artificially make the degree independent of
the choice of isomorphism by claiming two degrees are equivalent if there
is an isomorphism of Z[x] which sends polys of the first degree to the
second. however you can map swap any two coeffs, so that identifies all
degrees.
you do things like that all the time. for instance, the trace of a linear
operator (on a finite dimensional vector space). pick a basis and work it
out. then show that for any two bases the answer is the same.