MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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"?

share|cite|improve this question

marked as duplicate by Kevin Buzzard, Andreas Thom, S. Carnahan Nov 19 '10 at 2:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If you know german: – Martin Brandenburg Nov 18 '10 at 16:54
This is basically a duplicate of (more precisely, a consequence of)… – Kevin Buzzard Nov 18 '10 at 20:41

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.

share|cite|improve this answer

See Example 3.5 at for an argument.

share|cite|improve this answer

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.

share|cite|improve this answer
Someone voted this answer down. May I ask why? – Todd Trimble Nov 20 '10 at 20:28

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.