Double dual of free $\mathbb{Z}_{(p)}$-modules

Question

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.

Answer 1

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 $\varphi\theta$. $\;\blacksquare$

Answer 2

E. E. Enochs, “A note on reflexive modules”, Pacific J. Math. 14 (1964), 879–881 shows that, over a discrete valuation ring $R$, the free module with denumerable base is reflexive (meaning that the canonical linear map to its bidual is an isomorphism) iff the base ring $R$ is not complete. In particular, this answers your question affirmatively by applying to $R = \mathbb{Z}_{(p)}$ (which is not complete).