Calculating the kernel I'm trying to calculate the kernel of the map $$i\otimes_\mathbb{Z}id_A:\mathbb{Z}\otimes_\mathbb{Z}A\to \mathbb{Q}\otimes_\mathbb{Z}A$$ for any abelian group $A$ that way I know $Tor(\mathbb{Q}/\mathbb{Z},A)$ for any abelian group $A$. I know this will lead to this because I chose the flat resolution $$\cdots\to 0\to\mathbb{Z}\xrightarrow{i}\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to 0$$  which then leads to the tensoring of the deleted flat resolution $$\cdots\to 0\to \mathbb{Z}\otimes_\mathbb{Z}A\xrightarrow{i\otimes_\mathbb{Z}id_A} \mathbb{Q}\otimes_\mathbb{Z}A\to 0.$$ This means that since the image of the map before $i\otimes_\mathbb{Z}id_A$ is 0, that the first homology group of this sequence is equal to the kernel of $i\otimes_\mathbb{Z}id_A$. However, I am struggling to calculate the kernel of the map because of the generality that $A$ has. Can someone please assist me?
 A: For my answer I'm taking all tensor products over $\mathbb{Z}$, and dropping the subscript. I claim that $$\ker(i\otimes id_A) = 1\otimes T \subset \mathbb{Z} \otimes A$$
where $T\subset A$ is the subgroup of torsion elements.
First I claim the map $$\iota_\mathbb{Q}\colon A \to \mathbb{Q}\otimes A$$ sending $a$ to $1\otimes a$ has kernel $T$. Suppose $1\otimes a = 0 \in \mathbb{Q}\otimes A$. Since $\mathbb{Q}$ has no torsion this means there must be a $n\in \mathbb{Z}$ such that $\frac{1}{n}\otimes na = 0$, or in other words $na = 0$ so $a\in T$. Therefore $\ker(\iota_\mathbb{Q})\subset T$. Similarly if $a\in T$ so that $na = 0$ for some $n$ then $1\otimes a = \frac{1}{n}\otimes na =0$, therefore $a\in \ker(\iota_\mathbb{Q})$ and so $T\subset\ker(\iota_\mathbb{Q})$.
Now you can answer your original question by appealing to the fact that $\mathbb{Z}\otimes A \cong A$. If $\iota_{\mathbb{Z}} \colon A \to \mathbb{Z}\otimes A$ is defined by $\iota_{\mathbb{Z}}(a) = 1\otimes a$ then $\iota_\mathbb{Z}$ is an isomorphism of abelian groups taking $T$ isomorphically onto $1\otimes T$, and
$$\iota_\mathbb{Q} = (i\otimes id_A)\circ \iota_\mathbb{Z}.$$

This makes sense from the perspective of the $\mathrm{Tor}$ functor, which roughly-speaking computes the common torsion between the two inputs, since $\mathbb{Q}/\mathbb{Z}$ has $n$-torsion for every $n>1$.
