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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!

Ayman Hourieh
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    What is your definition of the Tor functor? – Bernard Sep 20 '19 at 15:42
  • The way I see it so far, is if we have a short exact sequence and tensor product it with some group, the Torsion group is what we need on the far left of our new sequence in order that the new sequence is also exact. Hatcher said that it measures the common torsion of the two groups in it's arguement. –  Sep 20 '19 at 15:44
  • Also, if we take the free resolution of a group and tensor that free resolution with a group $A$, then the Torsion group appears as a homology group of the tensored sequence. –  Sep 20 '19 at 15:45

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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 $$ A \xrightarrow{\cdot n} A. $$

Ayman Hourieh
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