# Infinite Integer Product Not Free

Posted by Jason Polak on 01. February 2012 · Write a comment · Categories: commutative-algebra, modules · Tags: ,

### Introduction

Assuming the axiom of choice, any vector space possesses the pleasant but prosaic property* that it is determined up to isomorphism by the cardinality of its basis.

For instance, consider $\prod_\omega \mathbb{Z}/2$ and $\oplus_{2^\omega} \mathbb{Z}/2$. Both are vector spaces over the finite field $\mathbb{Z}/2$ so to show that they are isomorphic, we need to show that their respective bases have the same cardinality. The vector space on the right is written as a direct summand and so we can see that its basis must have size $\mathfrak{c}$. On the other hand, the vector space $\prod_\omega \mathbb{Z}/2$ has cardinality $\mathfrak{c}$ over a finite field, so its basis must have the same cardinality as the space itself. Aren’t vector spaces a walk in the park; a piece of cake; easy as pie (ok, enough metaphors?!)?

### From $\mathbb{Z}/2$ to $\mathbb{Z}$ Modules

But what if we sent the above proof to a publisher who didn’t yet have the “2” character or the “/” installed on her printing press? Then all hell would break loose because $\prod_\omega \mathbb{Z}$ and $\oplus_{2^\omega} \mathbb{Z}$ aren’t vector spaces any more, and the previous paragraph would be rife with errors. But they certainly are abelian groups, and they have a bit more spice than those vector spaces. So are they isomorphic? They do have the same cardinality. Fortunately for us, Baer (“Abelian Groups without Elements of Finite Order”, Duke Math J. 3 (1937), pp. 88-122) answered this question in the negative (fortunately, because otherwise abelian groups would be less exciting). In fact, this question is particularly interesting to me because I had wondered about it a few months ago, and now I have the answer, thanks to Faith’s book for the references.
More »