For an abelian group $A$, put $DA=\text{Hom}(A,\mathbb{Z})$ and $D\_{(p)}A=\text{Hom}(A,\mathbb{Z}\_{(p)})$. It is a theorem of Specker that when $A$ is free abelian of countable rank, the natural map $A\to D^2(A)$ is an isomorphism. The proof uses the ordering of $\mathbb{Z}$ in an essential way: at a key step we have integers $n,m$ with $|n|<m$ and $n=0\pmod{m}$ and we conclude that $n=0$.


Is it also true that when $A$ is a free module of countable rank over $\mathbb{Z}\_{(p)}$, the natural map $A\to D\_{(p)}^2(A)$ is an isomorphism? As the order-theoretic properties of $\mathbb{Z}\_{(p)}$ are worse than those of $\mathbb{Z}$, it does not seem possible to adapt Specker's argument in any obvious way.



Double dual of free $\mathbb{Z}_{(p)}$-modules
There is at least one proof of Specker's theorem that can be adapted in an obvious way. I believe that the first half of this proof is due to Sąsiada, and the second half to Łoś.


Let $A$ be a free $\mathbb{Z}\_{(p)}$ module of countable rank, and $B=D\_{(p)}A$
its dual, whose elements I will think of as sequences of elements of
$\mathbb{Z}\_{(p)}$. Let $e\_{n}\in B$ be the sequence with $n$th term equal to
$1$ and all other terms zero.


It is easy to see that to prove that the natural map $A\to D^{2}\_{(p)}A$ is an
isomorphism, it is enough to prove the following two facts about elements
$\varphi:B\to\mathbb{Z}\_{(p)}$ of $D\_{(p)}B$.


**Proposition 1.** $\varphi(e\_{n})=0$ for all but finitely many $n$.


*Proof.* Suppose not. By ignoring those $n$ for which $\varphi(e\_{n})=0$, we can
assume, for simplicity, that $\varphi(e\_{n})\neq0$ for all $n$.


Construct a sequence $(x\_{n})\_{n\in\mathbb{N}}\in B$ by inductively choosing
$x\_{n}$ to be nonzero but divisible by a higher power of $p$ than
$\varphi(x\_{n-1}e\_{n-1})$.


Now consider all sequences whose $n$th term is either $x\_{n}$ or $0$. There are
uncountably many of these, but $\mathbb{Z}\_{(p)}$ is countable, so two of these
sequences must be sent to the same place by $\varphi$. Taking the difference, we
get a nonzero sequence $(y\_{n})\_{n\in\mathbb{N}}\in\ker(\varphi)$ with $y\_{n}$
equal to $0$ or to $\pm x\_{n}$. Choosing $m$ minimal such that $y\_{m}\neq0$ we
have
$$0=\varphi((y\_{n})\_{n\in\mathbb{N}})=
\varphi(y\_{m}e\_{m})+\varphi((y\_{n})\_{n\in\mathbb{N}}-y\_{m}e\_{m}),$$ but
$(y\_{n})\_{n\in\mathbb{N}}-y\_{m}e\_{m}$, and hence $\varphi((y\_{n})\_{n\in\mathbb{N}}-y\_{m}e\_{m})$, is divisible by a higher power of
$p$ than $\varphi(y\_{m}e\_{m})$. $\;\blacksquare$


**Proposition 2.** $\varphi$ is determined by the values of $\varphi(e\_{n})$.


*Proof.* If not, there is some nonzero $\varphi\in D\_{(p)}B$ with
$\varphi(e\_{n})=0$ for all $n$.


Choose a sequence $(a\_{n})\_{n\in\mathbb{N}}\not\in\ker(\varphi)$, and define a
homomorphism $\theta:B\to B$
by
$$\theta((b\_{n})\_{n\in\mathbb{N}})=((b\_{0}+b\_{1}+\cdots+b\_{n})a\_{n})\_{n\in\mathbb{N}}.$$


For each $m$, $\theta(e\_{m})$ differs from $(a\_{n})\_{n\in\mathbb{N}}$ in only
finitely many terms, so
$\varphi\theta(e\_{m})=\varphi((a\_{n})\_{n\in\mathbb{N}})\neq0$ for all $n$,
contradicting Proposition 1 for $\varphi\theta$. $\;\blacksquare$


