# Infinite direct product of the integers not a free module over the integers [duplicate]

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"?

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

If you know german: maddin.110mb.com/pdf/specker.pdf – Martin Brandenburg Nov 18 '10 at 16:54
This is basically a duplicate of (more precisely, a consequence of) mathoverflow.net/questions/10239/… – 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.

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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.

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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.

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