# Example computation of $\operatorname{Tor_i}{(M,N)}$

Let $M = \mathbb Z / 284 \mathbb Z$ and $N = \mathbb Z / 2 \mathbb Z$.

Can you tell me if my computation of $\operatorname{Tor_i}{(M,N)}$ is correct:

(i) First we want a projective resolution of $M$: $$0 \to \mathbb Z \xrightarrow{\cdot 284 } \mathbb Z \xrightarrow{\pi} M \to 0$$ Then we chop off $M$ to get $$0 \to \mathbb Z \xrightarrow{d_1 = \cdot 284 } \mathbb Z \xrightarrow{d_0 = 0} 0$$ And apply $- \otimes N$ to get $$0 \to \mathbb Z \otimes N \xrightarrow{d_1^\ast } \mathbb Z \otimes N \xrightarrow{d_0^\ast} 0$$

(ii) Now we see that $\operatorname{Tor_i}{(M,N)} = 0$ for $i \geq 2$.

(iii) We know that $\operatorname{Tor_0}{(M,N)} = M \otimes N = \mathbb Z / 284 \mathbb Z \otimes \mathbb Z / 2 \mathbb Z$

(iv) $\operatorname{Tor_1}{(M,N)} = \operatorname{Ker}{d_0^\ast} / \operatorname{Im}{d_1^\ast} = 0 / 0 = 0$.

In your calculation of the group $\textrm{Tor}^1(M, N)$, it seems that the calculation of the kernel is incorrect. The kernel must be the whole $\mathbb{Z} \otimes N$ since an elementary tensor $k \otimes [n]$ is mapped to $284k \otimes [n] = 142k \otimes [2n] = 142k \otimes [0] = 0$.
By the way, $i$ is generally written as a subscript in $\textrm{Tor}_i(M, N)$ since $\textrm{Tor}_i(M, N)$ is a homology group.
Thank you so much! It's now clear to me that $\mathrm{ker}(d_0^\ast)= \mathrm{ker}(0) = \mathbb Z \otimes N$. But I don't remember what I was thinking when I wrote that $\mathrm{im}(d_1^\ast) = 0$. – Rudy the Reindeer Jul 28 '12 at 19:37