Regarding \(\mathbb{Z}_2\) as a module over the ring \(\mathbb{Z}_4\), construct a resolution of \(\mathbb{Z}_2\) by free modules over \(\mathbb{Z}_4\) and use this to show that \(\operatorname{Ext}_{\mathbb{Z}_4}^n(\mathbb{Z}_2, \mathbb{Z}_2)\) is nonzero for all \(n\).
Solution
Regard \(\mathbb Z_2\) as the quotient \(\mathbb Z_4/(2)\). Consider the periodic free resolution \[ \cdots\xrightarrow{\,2\,}\mathbb Z_4 \xrightarrow{\,2\,}\mathbb Z_4 \xrightarrow{\,2\,}\mathbb Z_4 \longrightarrow\mathbb Z_2\longrightarrow0. \]
1. This sequence is exact.
For multiplication by \(2\) on \(\mathbb Z_4\), \[ \ker(\times2)=\{0,2\}=\operatorname{im}(\times2). \] At the right end, the quotient map \(\mathbb Z_4\to\mathbb Z_2\) has kernel \(\{0,2\}\), again the image of multiplication by \(2\). Every term before \(\mathbb Z_2\) is free over \(\mathbb Z_4\), so this is a free resolution.
2. Applying \(\operatorname{Hom}_{\mathbb Z_4}(-,\mathbb Z_2)\) gives the cochain complex\[ 0\longrightarrow\mathbb Z_2\xrightarrow{0}\mathbb Z_2 \xrightarrow{0}\mathbb Z_2\xrightarrow{0}\cdots . \]
For every degree, \[ \operatorname{Hom}_{\mathbb Z_4}(\mathbb Z_4,\mathbb Z_2)\cong\mathbb Z_2 \] by evaluation at \(1\). The coboundary is precomposition with multiplication by \(2\). If \(\varphi:\mathbb Z_4\to\mathbb Z_2\) is \(\mathbb Z_4\)-linear, then \[ (\varphi\circ\times2)(1)=\varphi(2)=2\varphi(1)=0 \] in the \(\mathbb Z_4\)-module \(\mathbb Z_2\). Hence every coboundary is zero.
3. Therefore\[ \boxed{\operatorname{Ext}_{\mathbb Z_4}^{\,n}(\mathbb Z_2,\mathbb Z_2)\cong\mathbb Z_2} \]for every \(n\ge0\).
The cohomology of the zero-differential cochain complex in 2 is one copy of \(\mathbb Z_2\) in each degree.
In particular these Ext groups are nonzero in every degree.