12
$\begingroup$

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"?

Improve this question
$\endgroup$
1

5 Answers 5

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
4
$\begingroup$

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

Improve this answer
$\endgroup$
3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Someone voted this answer down. May I ask why? $\endgroup$
    – Todd Trimble
    Commented Nov 20, 2010 at 20:28
2
$\begingroup$

See Example 3.5 at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/dualmod.pdf for an argument.

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .