$Tor_1^\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z})$

Question

How do I find $Tor_1^\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z})$? and is it free or at least projective?

I tried using the obvious short exact sequence then tensoring with $\mathbb{Z}/2\mathbb{Z}$ from either side, but I still need help.

Thanks

Answer 1

The long exact sequence associated to $0 \to I \to R \to R/I \to 0$ is

$0 \to Tor_1(R/I,M) \to I \otimes M \to M \to M/I \to 0.$

Hence, $Tor_1(R/I,M)$ is the kernel of $I \otimes M \to M$. If $M=R/J$, this map identifies with $I/IJ \to R/J$, so that $Tor_1(R/I,R/J) = (I \cap J)/(IJ)$.

If $R$ is a PID, we can write $I=(a)$, $J=(b)$, this shows easily $Tor_1(R/(a),R/(b)) \cong R/(\mathrm{ggT}(a,b))$.