From: "Charles H. Giffen" <chg4k@virginia.edu>
Subject: Re: 2 algebra questions
Date: Thu, 03 Feb 2000 12:35:07 -0500
Newsgroups: sci.math
Summary: [missing]

Andrei Prasolov wrote:
> 
> Hi everybody,
> 
> Let G be a product of countably many copies of Z,
> and H \subset G be the direct sum of countably many copies of Z.
> 
> 1. Is G free abelian?
> 2. Is G/H purely injective?
> 
> Purely injective = injective with respect for
> pure exact sequences ( = filtering direct limit of split exact
> sequences).
> 
> What is known: G is the set of sequences (n_i), i \in N, n_i \in Z.
> If we restrict ourselves with BOUNDED such sequences then
> the corresponding group G' is free abelian.
> 
> Yours,
> 
> Andrei Prasolov
> Tromso, Norway

In Kurosh (Theory of Groups, Vol. 1, Engl. transl. Chelsea Pub.,
1956/1960, section 32), it is proved that  Hom(G,Z) = H,  so  G 
cannot be free.  Of course,  Hom(H,Z) = G  is trivial.
==============================================================================

From: wcw@math.psu.edu (William C Waterhouse)
Subject: Re: algebra questions
Date: 3 Feb 2000 21:00:22 GMT
Newsgroups: sci.math

In article <38980E2E.F2F39FB3@math.uit.no>, 
Andrei Prasolov <prasolov@math.uit.no> writes:
>...
> Let G be a product of countably many copies of Z,
> and H \subset G be the direct sum of countably many copies of Z.
> 
> 1. Is G free abelian?
> 2. Is G/H purely injective?

The answer to 1) is NO, and there's a fairly standard proof as follows.

a) Let  P  be the subgroup of G consisting of sequences where the
power of 2 dividing  a_n goes to infinity.  This contains the
subgroup Z x 2Z x 4Z x..., so it has the cardinality of the continuum.

b) Evvery element in P can be changed in finitely many places to
become 2 times some other element in P.  Thus the elements in P with
only finitely many nonzero entries form a subgroup mapping onto
P/2P.

c) Now if G were free, its subgroup P would also be free. By a), P would 
have to have continuum-many elements in a basis, and so P/2P would be
a Z/2Z space of continuum dimension.  But by b), that's not true.


William C. Waterhouse
Penn State