If $M$ and $N$ are divisible, abelian groups, show that their tensor product $M \otimes_\mathbb{Z} N$ is uniquely divisible

Question

Recall that an abelian group $M$ is divisible if for each $m \in M$ and $r \in \mathbb{Z}$, there is an $m' \in M$ such that $rm' = m$. It is uniquely divisible if that $m'$ is unique. If $M$ and $N$ are divisible, abelian groups, show that their tensor product $M \otimes_\mathbb{Z} N$ is uniquely divisible. Conclude $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$.

I am new to the subjects of tensor products, modules, and exact sequences. Here are my thoughts so far:

$M$ is divisible, so the map $\phi_1: M \longrightarrow M : m' \longmapsto rm'$ is surjective for any $r \in \mathbb{Z}$. Similarly, $N$ is divisible, so the map $\phi_2 : N \longrightarrow N : n' \longmapsto sn'$ is surjective for any $s \in \mathbb{Z}$.

To show that $M \otimes_\mathbb{Z} N$ is uniquely divisible, my idea was to construct the map $\phi: M \otimes_\mathbb{Z} N \longrightarrow M \otimes_\mathbb{Z} N : m' \otimes n' \longmapsto rm' \otimes sn'$ and show that it is bijective for any $r, s \in \mathbb{Z}$. I believe that surjectivity would follow easily from the fact that $\phi_1$ and $\phi_2$ above are both surjective. But, how can I show that this map is injective ? Am I on the right track ? I haven't yet used that both $M$ and $N$ are abelian.

As far as $\mathbb{Q}/\mathbb{Z}$, I'm aware of the purely algebraic way to see that $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$. However, how can I use the claim in this problem to show this ? $\mathbb{Q}/\mathbb{Z}$ is a divisible abelian group, so the claim is relevant. Upon proving it, it tells us that $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z}$ is uniquely divisible. But why does this tell us that $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$ ? Can we possibly involve an exact sequence to do this ? A relevant exact sequence might be $0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Q} \longrightarrow \mathbb{Q}/\mathbb{Z} \longrightarrow 0$. Tensoring is right exact, so we can get another exact sequence by tensoring each object in this sequence (over $\mathbb{Z}$) with $\mathbb{Q}/\mathbb{Z}$. Is that the right idea ?

Thanks for all of your help.

Answer 1

You are one the right track, here is a full answer (using exact sequence so you can compare).

First using prime decomposition we need only check that $M \otimes N \stackrel{\times p}\rightarrow M \otimes N$ is an isomorphism. We show the following, if $M, N$ are $p$-divisible then $M \otimes N$ is uniquely $p$-divisible (with the obvious definitions).

Denote by $M[p] = \ker(M \stackrel{\times p}\rightarrow M)$, then we have the following short exact sequence.

$$ 0 \rightarrow M[P] \rightarrow M \stackrel{\times p}\rightarrow M \rightarrow 0 $$

Tensoring (over $\mathbb{Z}$ which we don't write) with $N$ yields the following exact sequence :

$$ M[P] \otimes N \rightarrow M \otimes N \stackrel{\times p}\rightarrow M \otimes N \rightarrow 0 $$

(Note the we lost exactness on the left), to see that $M \otimes N \stackrel{\times p}\rightarrow M \otimes N$ is an isomorphism we only need to see that $M[P] \otimes N = 0$.

Let $m \otimes n \in M[P] \otimes N$, then as $N$ is $p$-divisible there exists $n'$ such that $n = pn'$, and thus $m \otimes n = p(m \otimes n') = pm \otimes n' = 0$.

Now to see that $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$ we can use the following :

This sequence is exact,

$$ 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0$$

We want to tensor with $\mathbb{Q}/\mathbb{Z}$, first it is easy to see that $\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$ and second we want to see that $\mathbb{Q} \otimes \mathbb{Q}/\mathbb{Z} = 0$. This implies that $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$ using the exact sequence :

$$\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{Q} \otimes \mathbb{Q}/\mathbb{Z} \rightarrow \mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} \rightarrow 0$$

To see that $\mathbb{Q} \otimes \mathbb{Q}/\mathbb{Z} = 0$, suppose it is non-zero, let $x = \frac{p}{q} \otimes \alpha \neq 0 \in \mathbb{Q} \otimes \mathbb{Q}/\mathbb{Z}$. The element $\alpha$ is torsion, let $r$ such that $r\alpha = 0$, then $rx = 0$ but by the lemma $\mathbb{Q} \otimes \mathbb{Q}/\mathbb{Z}$ is uniquely divisible, thus $x = 0$. We have a contradiction, this prove the claim.