Let us consider the polynomial ring $M=\mathbb{Z}[x,y]$ as a $\mathbb{Z}[C_4]$-module via the $C_4$ action given by $x\mapsto y \mapsto -x.$ Define a grading on $M$ by setting $|x|=|y|=2$. We denote by $M_{2k}$ as the $2k$-graded piece of $M$; in our case, it should be the set of all polynomials with degree $2k$. I am interested in computing the group cohomology $H^s(C_4; M_{2k})$ for all $s \ge 0.$
We can start with the projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[C_4]$-module
$$\cdots \mathbb{Z}[C_4] \to^N \mathbb{Z}[C_4] \to^{(1-g)} \mathbb{Z}[C_4] \to^{\epsilon} \mathbb{Z}.$$
Here $C_4= \langle g\colon g^4=1\rangle$ and $N= 1+g+g^2+g^3.$ Now if we apply $Hom_{\mathbb{Z}[C_4]}(-, M_{2k})$ we get $$0 \to M_{2k} \to^{(1-g)^\ast} M_{2k} \to^{N^\ast} M_{2k} \to^{(1-g)^\ast} \cdots $$
One observes $N^\ast(x)= 1.x+g.x+g^2.x+g^3.x= x+y-x-y=0$ and similarly $N^\ast(y)=0$. Also, since $(1-g)^\ast(x)=x-y$ and $(1-g)^\ast(y)=y+x.$ Therefore, $\ker((1-g)^\ast)=\{\mbox{constant polynomials}\} \cong \mathbb{Z}$ and $im((1-g)^\ast)\cong \{p(x,y) \in M_{2k}\colon p(0,0)=0\} = \ker(N^\ast).$
Am I missing something?
Any help will be appreciated—many thanks in advance.