Showing $\operatorname{Tor}(\mathbb{Z}_n,A)\cong \ker (\phi$) where $\phi: A \xrightarrow{\;n\;} A$ (multiplication by $n$).
Okay, so according to page 265 of Hatcher, this is easy to see, and all we have to do is is tensor $0 \rightarrow \mathbb{Z} \xrightarrow{\;n\;} \mathbb{Z} \rightarrow \mathbb{Z}_n \rightarrow 0$ with $A$. This leads us to the exact sequence: $$0 \rightarrow \ker(n \otimes 1_A) \rightarrow \mathbb{Z} \otimes A \xrightarrow{n \otimes 1_A} \mathbb{Z} \otimes A \rightarrow \mathbb{Z}_n \otimes A\rightarrow 0$$
Right? Where do I go from here? I'd appreciate helping me finish what I'm sure are trivial details, along with any general insight into the Tor functor! Thanks!
We start with a free resolution of $\mathbb Z/(n)$: $$ 0 \to \mathbb Z \xrightarrow{\cdot n} \mathbb Z \to \mathbb Z/(n) \to 0. $$
We trim the resolution at $\mathbb Z/(n)$ and tensor by $A$ to get the following sequence: $$ 0 \to \underbrace{\mathbb Z \otimes_{\mathbb Z} A}_{*} \xrightarrow{(\cdot n) \otimes 1} \mathbb Z \otimes_{\mathbb Z} A \to 0. $$
Now $\operatorname{Tor}(\mathbb Z/(n), A)$ is the homology of the resulting complex at the term marked by $*$. Using the natural isomorphism $\mathbb Z \otimes_{\mathbb Z} A \cong A$, we see that this is nothing but the kernel of $$ A \xrightarrow{\cdot n} A. $$