# Does a free $R$-module of countable rank ever arise as a (double) dual of some $R$-module?

The usual counterexample to all vector spaces being canonically isomorphic to their double duals goes something like this: if $F$ is a field, take $F^{\omega}$. Since contravariant hom takes colimits to limits, its dual is an infinite product rather than an infinite direct sum, and so assuming the axiom of choice has uncountable dimension. This generalizes to show that dimension always jumps when you take duals of infinite-dimensional vector spaces.

It's clear that for vector spaces, $F^{\omega}$ never arises as a (double) dual. I'm curious about the case for modules over a commutative ring.

By a theorem of Specker, $$Hom(\bigoplus_n \mathbb{Z},\mathbb{Z})\cong \prod_n \mathbb{Z}$$ $$Hom(\prod_n \mathbb{Z},\mathbb{Z})\cong \bigoplus_n \mathbb{Z}$$ where $n$ ranges through the integers. Hence $\bigoplus_n \mathbb{Z}$ is the double dual of itself.
• To be precise: The second equation is the theorem of Specker; $Hom(\oplus_I -,-)\cong \prod_I Hom(-,-)$ is always true. Mar 17 '16 at 14:13