Am I missing something? According to mathworld, a free abelian group is an abelian group with no torsion, seemingly meaning that the only element with finite order is the identity (in this case the zero function). Clearly here, no function other than zero has finite order. Now if you ask for a set of generators, that could be a bit tricky. It's hard to list uncountably many things. --ms Michael Kleber wrote:
Dan asked last week:
Let P denote the direct product of countably many copies of the integers Z.
E.g., the set {f: Z+ --> Z | f is a function} under addition of functions.
Puzzle: Is P a free abelian group? Prove your answer.
And Gene just re-expressed interest in the subject. I started writing something about this yesterday, but realized I was on shaky ground.
I thought I had a proof that P was not free abelian, but then realized I was making some sort of unwarranted assumptiong about continuous functions on P which depended on some notion of topology of P, which made me uncomfortable. Given a function f : P->Z, just because you know all the f(ei), where ei=(0,0,...,0,1,0,0,...), doesn't mean you know f(a1,a2,a3,a4,...).
(The rest of yesterday's argument said that there were only countably many homomorphisms P->Z, since all but finitely many of the ei had to go to zero; but a free abelian group on uncountably many generators -- which P would have to be, it being uncountable itself -- should have uncountably many such homomorphisms. But as I said, this seems hollow to me now.)
I second the motion for an answer.
--Michael