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Let $R=\mathbb{Z}/ 8 \mathbb{Z}$ and let $M=\mathbb{Z} / 4 \mathbb{Z}$ be an $R$-module. How can I compute $\text{Tor}^R_n(M,M)$?

I was just introduced to the theory of Tor, and I am having difficulties to compute it. I know that $R$ is a principal ring and that if $0 \to N' \to N \to N'' \to 0$ is exact, then so is $$N' \otimes M \to N \otimes M \to N'' \otimes M \to 0$$ (where $N',N,N''$ are right $R$-modules and $M$ is a left $R$-module). So, $L_n(- \otimes_R M) = 0$ for all $n \geq 2$. But I dont know how to proceed with this really. Can someone explain how I can compute Tor$(M,M)$?

Greg
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Careful, $R$ is a principal ideal ring, but it is not a domain! Can you find a resolution $$\cdots \longrightarrow R \longrightarrow R \longrightarrow M\longrightarrow 0?$$

The image and kernel of multipication by two in $R$ are $4R$ and $2R$ respectively, so you have a periodic resolution

$$\cdots \stackrel{\times 2}\longrightarrow R \stackrel{\times 4}\longrightarrow R \stackrel{\times 2}\longrightarrow R\stackrel{\times 4}\longrightarrow R \longrightarrow 0$$

of $M$.

Pedro
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  • I have already noted that $M=R/4R$. Is my obervation, that $Tor^R_n(M,M)=0$ for all $n \geq 2$ correct? – Greg Oct 15 '15 at 02:11
  • @Greg No, that's incorrect. – Pedro Oct 15 '15 at 02:12
  • Why? I dont see this, in my notes there is no requirment for $R$ to be a domain, only a principal ideal ring and $F=(- \otimes_R M)$ being right exact. – Greg Oct 15 '15 at 02:14
  • @Greg There should be a requierement that $R$ be a domain. – Pedro Oct 15 '15 at 02:15
  • Thanks for your answer. I will look closer to the hint. – Greg Oct 15 '15 at 02:19
  • How does the maps $... \to R \otimes M \to R \otimes M \to 0$ look like? Are they induced by our multiplication maps? I.e $r \otimes m \mapsto $$2r \otimes m$ and $r \otimes m \mapsto 4r \otimes m$? – Greg Oct 15 '15 at 03:25