Re: root groups?


Arturo Magidin wrote:
Hm, that seems ridiculous to me. A divisible group can be given as an
abstract algebra with constants/operators: unity, inversion,
multiplication, n-th root for each natural n. With the equations 1x=x,
x^{-1}x=1, x(yz)=(xy)z, (r_n(x))^n = x. So the divisible groups form a
variety (because they are defined by a set of identities) and so there
must be a free divisible group for each cardinality of the set of
generators? Same with "root groups".

Note that the identities you propose do not suffice to define root
groups as originally given, since you are not including the relations
between roots (the (p/q)-th root times the (r/s)-th root being the
[(p/q) + (r/s)]-th root, etc). That is simple enough to fix, of
course.

Yes, yes. I was merely referring to divisible groups, because your
assertion already applied to divisible groups and they are simpler and
more standard. "same" was meant as "similar".

But you'll note that when I described the problem, I only talked about
a "unique group homomorphism", not requiring that it respect the 'root
operation.'

Yes, I see now.

That is not an additional property. That is the ->definition<- of
"free group generated by x1,...,xn". You must have a (set theoretic)
injection from {x1,...,xn} to the underlying set of the group F.

As far as I remember in universal algebra it is merely required that
the universal mapping property is satisfied. I.e. that each mapping of
the generators can be extended to a homomorphism. That one can for
example consider the variety of trivial groups and the free trivial
group. But maybe the framework is a bit different in category theory.

note that you will be restricting the allowable
morphisms if we proceed that way. Not every group homomorphism between
root groups will be a group-root-homomorphism.

Yes, but thats ok. If we introduce the additional operation p(x,m/n)
then the (root group) homomorphisms should respect this operation.

So then we are left with 3 questions:
Can every divisible group be made into a root group?
Is the free group in generators x1,...,xn embeddable into the free root
group in generators x1,...,xn?
Can equality be decided in the free root group (and what is the normal
form of a free root group or just of the free divisible group as given
above)?

.

Relevant Pages

  • Re: root groups?
    ... must be a free divisible group for each cardinality of the set of ... Same with "root groups". ... Note that the identities you propose do not suffice to define root ... a "unique group homomorphism", not requiring that it respect the 'root ...
    (sci.math.research)
  • Re: root groups?
    ... abstract algebra with constants/operators: unity, inversion, ... must be a free divisible group for each cardinality of the set of ... Same with "root groups". ... which is a "root group", contains distinguished elements x1,...,xn, ...
    (sci.math.research)
  • Re: Questions about field of Laurent series
    ... Since C is already algebraically closed, we don't have to worry these generators. ... By degree comparision, the equation ... to get algebraic closure we need to adjoint add n-th root of t to C). ... this should already imply there are infinitely many subfields between C) and its algebraic closure. ...
    (sci.math)