For $A=\mathbb{Z}/n\mathbb{Z}$ and $d$ a divisor of $n$, let $M=A/(d)\simeq\mathbb{Z}/d\mathbb{Z}$.
I'm trying to compute $\text{Tor}_i^A(M,N)$ for a general $A$-module $N$.
I'm using the following projective resolution of $M$:
$$\cdots\xrightarrow{\times e} A\xrightarrow{\times d} A\xrightarrow{\times e} A\xrightarrow{\times d} A\xrightarrow{\pi} M\to 0$$
(where $e:=n/d$ and $\pi$ is the natural projection)
Applying the tensor $-\otimes_A N$, we have $A\otimes_A N\simeq N$, so we get the complex:
$$\cdots\xrightarrow{\times e} N\xrightarrow{\times d} N\xrightarrow{\times e} N\xrightarrow{\times d} N\to M\otimes_A N$$
From what I've understood of the definition, $\text{Tor}_{2i-1}^A(M,N)=\ker(\times d)/\text{im}(\times e)$ and $\text{Tor}_{2i}^A(M,N)=\ker(\times e)/\text{im}(\times d)$ for all $i\geq 1$.
Now, in many different places, I've seen that $\text{Tor}_1^A(A/(d),N)=\{n\in N\mid dn=0\}$, and even $\text{Tor}_0^A(A/(d),N)=N/dN$. I'm not being able to verify this 2 facts in this example (actually, I don't really know what $\text{Tor}_0^A$ means), and I have no idea how to calculate $\text{Tor}_{2i}^A$ for a general $N$.
I'm clearly missing something here. Any hints?
It appears that you are thinking of $\text{Tor}_1^{\Bbb Z}(\Bbb Z/(d),N) =N_d$ where $N_d$ is the $d$-torsion of the Abelian group $N$. But $\Bbb Z$ and $\Bbb Z/(n)$ are different rings, so it should be no surprise that they have different Tor-functors. So if $N$ is a $A=\Bbb Z/(n)$ module (an Abelian group with $nN=0$) then $\text{Tor}_k^A(A/(d),N)\cong N_d/eN$ for odd $k$, and $\text{Tor}_k^A(A/(d),N)\cong N_e/dN$ for even $k>0$.