How do I compute the kernel of this map?

Question

How do I compute $\ker{(\mathbb{Z} \otimes A \longrightarrow \mathbb{Q} \otimes A)}$ where this map comes from the short exact sequence:

$0 \rightarrow Tor(\mathbb{Q}/\mathbb{Z}, A) \rightarrow \mathbb{Z} \otimes A \rightarrow \mathbb{Q} \otimes A \rightarrow \mathbb{Q}/\mathbb{Z} \otimes A \rightarrow 0$. Where $A$ is an abelian group.

Answer 1

The map from $\mathbb{Z} \otimes A \to \mathbb{Q} \otimes A$ is induced by the canonical inclusion, so is defined on generators by $n\otimes a \mapsto \frac{n}{1}\otimes a$ (for any $n \in \mathbb{Z}, a \in A$).

First, the torsion subgroup of $A$ is contained in the kernel. Recall that there is a canonical identification $A \simeq \mathbb{Z} \otimes A$, via the map $a \mapsto 1 \otimes a$. Let $a \in A$ be torsion, so that there is an integer $n \neq 0$ with $na =0$. Identify $a$ with $1 \otimes a$ and then note that $1 \otimes a \mapsto \frac{1}{1}\otimes a = \frac{n}{n}\otimes a = \frac{1}{n}\otimes na = \frac{1}{n}\otimes 0 = 0$, so $1 \otimes a$ is in the kernel.

For the opposite inclusion, consider why this doesn't work if $a$ isn't torsion.

Answer 2

The easiest way is to recall that $\operatorname{Tor}(\mathbf{Q}/\mathbf{Z}, A)$ is the torsion subgroup of $A$, and find its image in $\mathbf{Z} \otimes A$.