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I'm working on an assignment, in which in the end I'm trying to show that the countable direct product $\prod_{i\in \mathbb Z}\mathbb Z$ is not reflexive. I've already made some progress on the subject, and reduced the problem to showing that: $$\oplus_{i\in \mathbb Z}\mathbb Z\cong\hom_{\mathbb Z}(\hom_{\mathbb Z}(\oplus_{i\in \mathbb Z}\mathbb Z,\mathbb Z)\mathbb Z)$$ In order to make notation somewhat easier, I use $M=\oplus_{i\in\mathbb Z}\mathbb Z$ and also $P^*$ to denote the $R$ module $\hom_R(P,R)$ where $R$ is a ring and $P$ is an $R$ module. Being so what I'm trying to prove is that $$M\cong (M^*)^*$$ The proper way to show this seems to be via the evaluation homomorphism: any $x=\langle x_i\rangle\in M$ is mapped to $\epsilon_x\in (M^*)^*$ which maps a morphism $\varphi\in M^*$ to the integer $\varphi(x)\in\mathbb{Z}$.

It isn't hard to show that this is an injective, well defined homomorphism, but the surjectivity part is a little tricky..

It isn't hard to find the pre-image of any element $F\in (M^*)^*$: it holds that if $F$ has a pre-image in $\mathbb Z$, then for any $n\in\mathbb Z$, (we denote the $n$-th coordinate projection by $p_n$) then $$F(p_n)=\epsilon_x(p_n)=p_n(x)=x_n$$ then the pre-image of $F$ is simply the sequence $\langle F(p_n)\rangle_{n\in\mathbb Z}$.

the problem here is showing that $\langle F(p_n)\rangle$ is indeed an element of $M$, that is that the set $S=\lbrace n\in\mathbb Z\mid F(p_n)\neq 0\rbrace$ is finite. I've so far been working on two possible ways of solving this:

  1. Using an argument based on cardinality: Since $M^*$ is isomorphic to the direct product $\prod_{i\in\mathbb Z}\mathbb{Z}^*\cong\prod_{i\in\mathbb Z}\mathbb{Z}$ (I've shown both isomorphisms), we have that $F$ maps a set of cardinality $\aleph$ into $\mathbb Z$ and therefore must have an uncountable kernel. I've tried assuming by contradiction that the set $S$ is infinite, and trying to deduce that $F$ induces an injection of an uncountable set intothe integers, but with no luck so far.
  2. Trying to show boundedness: I tried considering the set $S^+=\lbrace n\in\mathbb Z\mid F(p_n)>0\rbrace$, and showing that the formal sum $\sum_{n\in S^+}p_n$ is an element of $M^*$, and thus $F(\sum_{n\in S^+}p_n)\in\mathbb Z$, and tried showing that $F(\sum_{n\in S^+}p_n)=\sum_{n\in S^+}F(p_n)$, and thus the sum is finite. But I got stuck on showing why $F$ should decomposes so nicely on infinite sums. Additionally I tried arguing that the partial sums $\lbrace F(p_{n_1})+\ldots+F(p_{n_k})\mid n_1,\ldots,n_k\in S^+\rbrace$ are uniformly bounded by $F(\sum_{n\in S^+}p_n)$ and thus there is at most finitely many of them (being that the sums are of strictly positive integers)

Once this has been shown I'm pretty sure I know how to sure I know how to deduce that $\epsilon_x=F$, but I'm a little at a lost here.. If anyone has any insight to give on one of the ideas I've proposed, or alternatively has another possible way of showing this, I would be most grateful.

Thank you very much :)

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I might be a bit turned around by your notation, but you are on the right track. You don't actually have to worry about whether "$F$ decomposes so nicely on infinite sums." Instead of thinking of $\sum_{n\in \mathbb{Z}^+} p_n$ as a formal sum, just think of it as the map which sends $\langle x_n\rangle$ to $\sum_{n\in \mathbb{Z}^+} x_n$. It's clearly well-defined and linear on $M$, so it's a member of $M^*$. – Adam Saltz Dec 26 '11 at 14:50
Dear kneidell: You may want to take a look at this answer. - I wonder if your question is not a duplicate of the one linked to above. – Pierre-Yves Gaillard Dec 26 '11 at 15:00
@Pierre-YvesGaillard, I definitely see the connection, but still am failing at trying to deduce the answer. I'll try working from it, thanks. – kneidell Dec 26 '11 at 16:20
@IAmBrianDawkins, the problem isn't to show that $\sum x_n$ is a member of $M^*$, but rather to show that $F(\sum x_n)=\sum F(x_n)$. I think this is true, but I'm not sure if its obvious... – kneidell Dec 26 '11 at 16:21
Here is a complete proof. – Paolo Capriotti Dec 26 '11 at 21:05
up vote 3 down vote accepted

After working hard on understanding the $\mathbb Z$-module $((\oplus_{i\in\mathbb Z}\mathbb Z)^\vee)^\vee$ I'm happy to say that with the help of good people and their useful comments (that is- @Pierre, @Brian and @Paolo) and of some more friends, I've manage to find a way to find an aswer to my question, using mostly basic algebraic tools. I post it here so that it might help anybody else. Though I do not expect anyone to grade me here, if anyone has any comments I would very much appreciate posting them.

Let's get proving!

First, let review what we know here- we denote $M=\oplus_{i\in\mathbb Z}\mathbb Z$, and we write $A^\vee$ to denote the $R$-module $\hom_R(A,R)$ whenever $R$ is a ring, and $A$ is an $R$-module (since all modules in this context are $\mathbb Z$-modules, we anticipate no confusion on that count).

By previous knowledge have that $M^\vee\cong \prod_{i\in\mathbb Z}\mathbb Z$. We use the notation $\pi_n$ to denote the projection of the $n$'th coordinate of $\prod_{i\in\mathbb Z}\mathbb Z$ onto $\mathbb Z$. The above isomorphism asserts us that $\sum_{j\in S}\pi_j$ is a well-defined element of $M^\vee$, for any $S\subseteq\mathbb Z$ (finite or infinite).

Our main aim here is to show that the canonical map $M\to (M^\vee)^\vee$, which sends any element $x\in M$ to the map $\epsilon_x:M^\vee\to\mathbb Z$, defined by $$M^\vee\ni \varphi\mapsto \varphi(x)\in \mathbb Z$$

The hard part of showing this is proving the surjectivity of this map. One of the reasons for this is the fact that it is unclear how a $\mathbb Z$-linear map "reacts" to infinite sums, but we'll get to dealing with this later.

Let $F\in (M^\vee)^\vee$ be arbitrary. Finding a candidate $\langle x_n\rangle=x\in M$, such that $\epsilon_x=F$ is easy, since for any $n$, both maps should agree on the projection map, that is $$x_n=\pi_n(x)=\epsilon_x(\pi_n)=F(\pi_n)$$ So we have a sequence $\langle x_n\rangle\in\prod_{i\in\mathbb Z}\mathbb Z$. There are two facts that need to be shown here:

  1. That $x$ is indeed an element of the direct sum $M$, that is- that the set $\lbrace n\in\mathbb Z\mid F(\pi_n)\neq 0\rbrace$ is finite.
  2. That $\epsilon_x(\varphi)=F(\varphi)$ for any $\varphi\in M$

We prove (1):

Assume toward contradiction that there are infinitely many indices $n$ such that $F(\pi_n)\neq 0$. We make a few useful ovservations:

  • Up to change of indexation, we may assume that for any $n\in\mathbb N$, $F(\pi_n)\neq 0$.
  • Better yet, since there are infinitely many indices where $F(\pi_n)>0$ or infinitely many such that $F(\pi_n)$ (if both are finite there is nothing to prove), we may assume WLOG that for any $n\in\mathbb N$, $F(\pi_n)>0$
  • Lastly- we take notice of images of infinite sums $F(\sum_{j=N}^\infty\pi_j)$, where $j\in\mathbb N$: Since for any $N\in\mathbb N$, it holds that $$F(\sum_{j=1}^\infty\pi_j)=\sum_{j=1}^N F(\pi_j)+F(\sum_{j=1}^\infty\pi_j)$$ As mentioned before, we cannot say what the most-right element is equal to, but since the finite sums $\sum_{j=1}^N F(\pi_j)$ incerase as $N$ increases, and the LHS is a constant integer, it holds that $\lbrace F(\sum_{j=N}^\infty \pi_j)\rbrace_{N=1}^\infty$ is a decreasing sequence of integers. hence WLOG, upto discarding finitely many indices, we may assume that $F(\sum_{j=1}^\infty\pi_j)<0$

To come to a contradiction, we wish to find a sequence of "weights" on $\sum_{j=1}^\infty\pi_j$, to produce our contradiction. We define this sequence by induction:

  1. Let $D_1=1$
  2. Assume by induction that $D_1,\ldots, D_n$ have been defined for some $n$, and find $D_{n+1}$ to satisfy the following conditions: (a) $D_n\mid D_{n+1}$, (b) $D_n< D_{n+1}$, and (c) $D_1 F(\pi_1)+\ldots+D_nF(\pi_n)<D_{n+1}$ (obviously condition (b) is superfluous, but we mention it anyway).

Condiser the series $\sum_{j=1}^\infty D_j\pi_j$. We know that it is an element of $M$, and hence it is in the domain of $F$, so what is its image under $F$?

Let $n$ be an arbitrary natural- we have the following: $$F(\sum_{j=1}^\infty D_j\pi_j)=\sum_{j=1}^nD_j F(\pi_j)+D_{n+1}F(\sum_{j=n+1}^\infty\frac{D_j}{D_{n+1}}\pi_j)=(*)$$ where $\frac{D_j}{D_{n+1}}\in\mathbb Z$ by condition (a), and by condition (c) we have that $$(*)<D_{n+1}(1+F(\sum_{j=n+1}^\infty\frac{D_j}{D_n}\pi_j))$$ Note that $F(\sum_{j=n}^\infty\frac{D_j}{D_n}\pi_j)$ decreases as $n$ increases, since: $$F(\sum_{j=n}^\infty\frac{D_j}{D_n}\pi_j)=\frac{D_n}{D_n}F(\pi_n)+\frac{D_{n+1}}{D_n}F(\sum_{j=n+1}^\infty\frac{D_j}{D_{n+1}}\pi_j)>F(\sum_{j=1}^\infty \frac{D_j}{D_{n+1}}\pi_j$$ by condition (b) and the fact that $F(\pi_n)>0$. Given that $1+F(\sum_{j=n+1}^\infty\frac{D_j}{D_n}\pi_j)$ decreases as a sequence of integers, it becomes strictly negative at some point, from which on we get, since $\lbrace D_j\rbrace$ is strictly decreasing, that $F(\sum_{j=1}^\infty D_j\pi_j$ is smaller the any element of a strictly decreasing sequence of integers. A contradiction.

We thus have that the sequence defined by $x_n=F(\pi_n)$ is indeed an element of the direct sum $M$.

Now we turn to prove (2):

To do this, we have to prove another useful lemma:

Suppose that for $\lbrace \pi_{n_j}\rbrace_{j\in\mathbb N}$ it holds that $F(\pi_{n_j})=0$ for any $j$. Then $F(\sum_{j=1}^\infty\pi_{n_j})=0$:

To show this, for any $j\in\mathbb N$ let $\alpha_j,\beta_j$ be such that $$2^j\alpha_j+3^j\beta_j=1$$ and observe the series $\sum_{j=1}^\infty2^j\alpha_j\pi_{n_j}$: $$\forall n\in\mathbb N: \: F(\sum_{j=1}^\infty 2^j\alpha_j\pi_{n_j})=\sum_{j=1}^n\alpha_j2^jF(\pi_{n_j})+2^{n+1}F(\sum_{j=n+1}^\infty 2^{j-n-1}\alpha_j\pi_{n_j})\equiv 0(\mod 2^n)$$ since $n$ is arbitrary, this implies that $F(\sum_j\alpha_j\pi_{n_j})=0$.

Likewise, one shows that $F(\sum_j\beta_j3^j\pi_{n_j})=0$, which implies that $$F(\sum_j\pi_{n_j})=F(\sum_j2^j\alpha_j\pi_{n_j})+F(\sum_j3^j\beta_j\pi_{n_j})=0$$

Now, let $\varphi\in M^\vee$ be arbitrary. We use the notation $\iota_n:\mathbb Z\to M$ to denote the $n$th coordinate injection $$a\mapsto \langle a_j\rangle=\begin{cases}a&\text{if }j=n\\0&\text{otherwise}\end{cases}$$ For any $x\in M$, it holds that $$\varphi(x)=\sum_{j\in\mathbb N}\varphi(\iota_j(1))\pi_j(x)$$ and so $\varphi$ is the same as the formal sum $\sum_j\varphi(\iota_j(1))\pi_j$

We conclude by noting that for $x=\langle F(\pi_n)\rangle_{n\in\mathbb Z}$ it holds that $$\epsilon_x(\varphi)=\varphi(x)=\sum_j\varphi(\iota_j(1))F(\pi_j)\stackrel{*}{=}F(\sum_j\varphi(\iota_j)\pi_j)=F(\varphi)$$

Where the equality (*) is justified by breaking the sum into a finite part on which $F$ decomposes linearily, and an infinite sum on which it is $0$, using the above lemma.

Added To finish the proof off, I added how this implies that the countable product $\prod_{i\in\mathbb Z}\mathbb Z$ is not a free $\mathbb Z$ module.

Assume towards contradiction that $\prod_{i\in\mathbb Z}\mathbb Z$ is free, and let $\lbrace e_i\rbrace_{i\in I}$ be a $\mathbb Z$- basis for it. That is $$\prod_{i\in\mathbb Z}\mathbb Z\cong \oplus_{i\in I}\mathbb Ze_i\cong\oplus_{i\in I}\mathbb Z$$

Two more useful lemmas to which we omit the proof, give the following:

  1. $\mathbb Z\cong\hom_{\mathbb Z}(\mathbb Z,\mathbb Z)\
  2. For any ring $R$, and $\lbrace A_j\rbrace_{j\in J}$ a family of $R$-modules, it holds that $$\hom_R(\oplus_{j\in J}A_j,R)\cong\prod_{j\in J}\hom_R(A_j,R)$$

Consider the following: Let $\oplus_{i\in\mathbb Z}\mathbb Z$ be the direct sum. By the above proof we have that $$\oplus_{i\in\mathbb Z}\mathbb Z\cong\hom_{\mathbb Z}(\hom_{\mathbb Z}(\oplus_{i\in\mathbb Z}\mathbb Z,\mathbb Z),\mathbb Z)=(*)$$ which by the two above lemmas is $$(*)\cong \hom_{\mathbb Z}(\prod_{i\in\mathbb Z}\hom_{\mathbb Z}(\mathbb Z,\mathbb Z),\mathbb Z)\cong \hom_{\mathbb Z}(\prod_{i\in\mathbb Z}\mathbb Z,\mathbb Z)=(**)$$ By our assumption, this is the same is the same as $$(**)\cong \hom_{\mathbb Z}(\oplus_{i\in I}\mathbb Z,\mathbb Z)$$ By applying the two lemmas again this gives that $$\oplus_{i\in\mathbb Z}\mathbb Z\cong\prod_{i\in I}\mathbb Z$$ But this is a contradiction, since if $I$ is infinite then the LHS is countable, whereas the RHS is uncountable (in particular, both sides are not bijectible), and if $I$ is finite, then the RHS is a finitely generated $\mathbb Z$ module, and the LHS is not.

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