I was trying to do the following exercise. Consider the category of graded $\mathbb{C}[x]$-Modules, it is clear that we can regard $\mathbb{C}[x]$ as a graded module setting $\operatorname{deg}(x)=1$. Given the module $M=\mathbb{C}[x]/(x)$ I want to show that $\operatorname{RHom}(M,\mathbb{C}[x](-1)[1])\cong M$ where $\mathbb{C}[x](-1)$ is the shift of the grading of $\mathbb{C}[x]$ by $-1$.
First of all $\operatorname{RHom}: \mathcal{D}(\mathcal{A}) \to \mathcal{D}(\mathcal{Ab})$ where $\mathcal{A}$ is the category of graded $\mathbb{C}[x]$-Modules. I think I should show that $RHom(M,\mathbb{C}[x](-1)[1])$ is a complex $B^{\bullet}$ with the property that $H^{n}(B^{\bullet})=0$ for every $n$ except $n=0$.
To do this I could replace $\mathbb{C}[x](-1)$ by a graded injective resolution $I^{\bullet}$ (I have no idea of how to find this) and then take $\operatorname{Hom}(M,I^{\bullet}[1])$. Now some questions:
- How do I find $I^{\bullet}$?
- Should I try to find a short exact sequence to compute this instead of the resolution?
- What's the point of shifting $I^{\bullet}$?
- What's the point of shifting $\mathbb{C}[x]$?
First it is easy to see that $\operatorname{Hom}(M,\mathbb{C}[x](-1))\cong 0$ since $M$ has only elements in degree $0$ and must be sent to $C[x]_{-1}=0$.
There is an easy projective resolution of $M$ given by: $$0 \to \mathbb{C}[x](-1) \xrightarrow{d_1} \mathbb{C}[x] \xrightarrow{d_0} M \to 0$$
$d_1(p(x))=xp(x)$ and $d_0$ is the quotient map. This resolution is also a short exact sequence so we can get a long exact sequence:
$$ 0 \to \operatorname{Hom}(M,\mathbb{C}[x](-1)) \to \operatorname{Hom}(\mathbb{C}[x],\mathbb{C}[x](-1)) \to \operatorname{Hom}(\mathbb{C}[x](-1),\mathbb{C}[x](-1))\to\\ \to\operatorname{Ext}^{1}(M,\mathbb{C}[x](-1)) \to \operatorname{Ext}^{1}(\mathbb{C}[x],\mathbb{C}[x](-1)) \to \operatorname{Ext}^{1}(\mathbb{C}[x](-1),\mathbb{C}[x](-1)) \to 0$$
Clearly since $\mathbb{C}[x]$ is free the last two term must vanish. Also ·$\operatorname{Hom}(\mathbb{C}[x],\mathbb{C}[x](-1))$ must be zero by degree reasons s finally $$\operatorname{Hom}(\mathbb{C}[x](-1),\mathbb{C}[x](-1))\cong \operatorname{Ext}^{1}(M,\mathbb{C}[x](-1)) \cong \mathbb{C} \cong M$$