$\DeclareMathOperator\Ext{Ext}$
Example 8.15. Let $a ≥ b ≥ c$ be integers. We will compute $\Ext_{ℤ / 2^a}^*(ℤ / 2^b, ℤ / 2^c)$.
Let $R=ℤ / 2^a ℤ$. Then $ℤ / 2^b / ℤ$ has projective resolution
\[
⋯ → R \xrightarrow{2^{a-b}} R \xrightarrow{2^b} R \xrightarrow{2^{a-b}} R \xrightarrow{2^b} R \xrightarrow{1} ℤ / 2^b ℤ .
\]
Therefore $\Ext_R^*(ℤ / 2^b, ℤ / 2^c)$ is the cohomology of
\[
⋯ ← ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{2^b}{←} ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{2^b}{←} ℤ / 2^c .
\]
Since $b>c$, the map $2^b: ℤ / 2^c → ℤ / 2^c$ is zero, so $\Ext_R^*(ℤ / 2^b, ℤ / 2^c)$ is the cohomology of
\[
… ← ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{0}{←} ℤ / 2^c \stackrel{2^{a-b}}{←} ℤ / 2^c \stackrel{0}{←} ℤ / 2^c .
\]
Therefore, we have
\[
\Ext_R^i(ℤ / 2^b, ℤ / 2^c)= \begin{cases}ℤ / 2^c & \text{ if } i=0 \\ \ker(f) & \text{ if } i ≥ 1 \text{ is odd } \\ \operatorname{coker}(f) & \text{ if } i ≥ 2 \text{ is even, }\end{cases}
\]where $f$ is the map $ℤ / 2^c \xrightarrow{2^{a-b}} ℤ / 2^c$. Suppose that $a-b ≥ c$. Then $f=0$, so
\[
\Ext_R^i(ℤ / 2^b, ℤ / 2^c)=ℤ / 2^c, \text{ for all } i
\]
Suppose that $a-b