Osborne's text has the below basic example.
1. Example 9
$R = \mathbb Z_4$, $A = B = \mathbb Z_2$ A projective resolution of $\mathbb Z_2$ is
$$ \cdots \rightarrow \mathbb Z_4 \xrightarrow{\times 2} \mathbb Z_4 \xrightarrow{\times 2} \mathbb Z_4 \xrightarrow{\pi} \mathbb Z_2 \rightarrow 0 $$
Tensoring with $\mathbb Z_2$ and deleting gives $$ \cdots \rightarrow \mathbb Z_2 \xrightarrow 0 \mathbb Z_2 \xrightarrow 0 \mathbb Z_2 \rightarrow \cdots $$
So that $Tor_n^R \cong \mathbb Z_2$ for all $n$.
2. Not a projective resolution of $\mathbb Z_2$?
But now consider the projective resolution $$ \cdots \xrightarrow{0} 0 \xrightarrow{0} 0 \xrightarrow{0} \mathbb Z \xrightarrow{\times 2} \mathbb Z \xrightarrow{\pi} \mathbb Z_2 \rightarrow 0 $$
Tensoring with $\mathbb Z_2$ and deleting $$\cdots \rightarrow 0 \rightarrow 0 \rightarrow \mathbb Z_2 \xrightarrow{0} \mathbb Z_2 \rightarrow 0$$
So now we have $Tor_n^R(A,B) = Z_2$ for $n = 0, 1$ but $Tor_n^R(A,B) = 0$ for $n \ge 2$. What did I get wrong here?
3. Similar issue in the case where $R = \mathbb Z_8$, $A = B = \mathbb Z_4$
Proceeding in the same way as above, I get $Tor_n^R(A,B) = Z_4$ for $n = 0, 1$ but $Tor_n^R(A,B) = 0$ for $n \ge 2$
4. Osborne's definition of $Tor$.
I believe this definition is standard, but just in case provide it below.
