Surjectivity of a map between a module and its double dual

Solution 1:

As stated by Matt E in the comments, the question has been asked (and answered) on MathOverflow: Is it true that, as Z-modules, the polynomial ring and the power series ring over integers are dual to each other?. Hailong Dao's answer links to a text by Grzegorz Bobinski, based on a talk by Lutz Hille, proving the result for any non-local principal ideal domain. Here is the pdf file, and here is an html page it can be accessed from.

Here is a mild simplification of the proof.

Let $A$ be a principal ideal domain, let $(p)$ and $(q)$ be two distinct maximal ideals, let $$A^{\mathbb N}$$ be the $A$-module formed by all the sequences in $A$, and let $$A^{(\mathbb N)}$$ be the submodule consisting of the finitely supported elements of $A^{\mathbb N}$. We claim:

The canonical map from $A^{(\mathbb N)}$ into its double dual is an isomorphism.

Let $f:A^{\mathbb N}\to A$ be a nonzero $A$-linear map. Define $a\in A^{\mathbb N}$ by the condition $$ f(x)=\sum_i\ a_i\,x_i\quad\forall\ x\in A^{(\mathbb N)}. $$

It suffices to show:

(1) $a\not=0$,

(2) $a\in A^{\mathbb N}$.

Proof of (1). Let $x$ be an element of $A^{\mathbb N}$ satisfying $f(x)\neq0$. For each $i$ in $\mathbb N$ there are $a_i,b_i$ in $A$ such that $$p^i\,a_i+q^i\,b_i=x_i.$$ Then $a=0$ would imply $$ f((p^i)_{i\in\mathbb N})=0=f((q^i)_{i\in\mathbb N}), $$ and thus $f(x)=0$.

Proof of (2). Suppose by contradiction that $a$ is not in $A^{(\mathbb N)}$. We can assume $a_i\neq0$ for all $i$. Write $$a_i=p^{r(i)}b_i$$ with $b_i$ prime to $p$. We can also assume $r(i)\le r(i+1)$ for all $i$, and $r(0)=0$. Put $$x_0:=p^{1+r(1)}.$$ Let $y$ be in $A^{\mathbb N}$, and set $$x_i:=p^i\,q\,y_i\quad\forall\ i > 0.$$ It is easy to see that $y$ can be chosen so that $$ p^{n+r(n)}\ |\ a_0\,x_0+\cdots+a_{n-1}\,x_{n-1}\quad\forall\ n > 0. $$ Then $q$ doesn't divide $f(x)$, but $p ^n$ does for all $n$, contradiction.

Solution 2:

I'm almost sure what you are trying to prove is false. It is false if you replace $\mathbb{Z}$ by $\mathbb{Q}$. To see this, one would want to to define a functional $F:P^* \mapsto \mathbb{Q}$ which does not come from $P$. Obvious thing to try is to take one that is $0$ on all finite sequences and $1$ on the sequence $x=(1,1,1,....)$ (note that this is consistent, as $x$ is linearly independent of all finite sequences). Then one would want to extend it to an actual map $F$. For vector spaces this is Hahn-Banach theorem (in that context the definition of a functional requires continuity, which we can forget for our purposes). Unfortunately the proof does not carry over to $\mathbb{Z}$ in an obvious way (things are a bit tricky; for example one needs to be careful with the fact that $P^*$ is not free). Maybe some other axiom of choice/Zorn lemma/ultrafilter trick will work...