Infinite direct product of the integers not a free module over the integers 
Possible Duplicate:
Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other? 

Is there an easy proof? I only found citations but have no access. By the way: If we cross over to the rationals every vector space is free (using Zorn's lemma). But can one construct a basis of "Countable infinite product of the rationals"?
 A: It is known (this is Specker's theorem) that the natural map
$$\iota :\bigoplus_{n \in \mathbb N} \ \mathbb Z \to Hom_{\mathbb Z} \left( \prod_{n \in \mathbb N} \mathbb Z,\mathbb Z \right)$$
is an isomorphism of abelian groups.
In particular, $\prod_{n \in \mathbb N} \mathbb Z$ cannot be a free abelian group. The crucial part of the proof appeared here. Also interesting: Nöbeling showed that the abelian group of bounded sequences in $\mathbb Z$ is free as an abelian group.
A: This paper addresses the question
http://www.math.uni-duesseldorf.de/~schroeer/publications_pdf/infinite_product-1.pdf
Also there is a wiki link 
http://en.wikipedia.org/wiki/Baer%E2%80%93Specker_group
A: I don't know if it counts as "easy", but a proof of this result appears as some notes in the American Mathematical Monthly, here. 
A: See Example 3.5 at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/dualmod.pdf for an argument.
A: Another reference to a proof of Specker's theorem is Zagier's St Andrews problems.
Added
Also rings such as $\mathbb{Z}$ with this property are called
slender rings.
