Hochschild homology - motivation and examples

Set $R = k[x]/(x^{n+1}),\,u=x\otimes 1-1\otimes x,\,v=\sum_{i=0}^n x^i\otimes x^{n-i} \in R^e := R \otimes_k R$.

First, let's recall from Weibel (Ex. 9.1.4) that in the ungraded case a projective resolution of $R$ over $R^e$ is given by the periodic complex $$\cdots \xrightarrow[]{v} R^e \xrightarrow[]{u} R^e \xrightarrow[]{v} R^e \xrightarrow[]{u} R^e \xrightarrow[]{\mu} R \to 0$$

Now suppose $R$ is a DGA with $\deg(x)=d$ and zero differentials. The latter implies that the notions of the Hochschild homology of $R$ as DGA and as graded algebra agree. Hence we can compute the Hochschild homology of $R$ by a projective resolution of $R$ over $R^e$ in the category of graded $R^e$-modules.

For a graded $R^e$-module $M$ let $\Sigma^kM$ be the shifted graded $R^e$-module given by $(\Sigma^kM)_i := M_{i-k}$. Set $e_k := (0,\ldots,1\otimes 1,\ldots 0) \in (\Sigma^kR^e)_k$. Then $\Sigma^kR^e=R^e\cdot e_k$ is a free graded $R^e$-module (in particular it's a projective object in the category of graded $R^e$-modules).

Taking into account $\deg u = d, \,\deg v=nd$, we can adjust the projective resolution from Weibel above and find the following projective resolution of $R$ over $R^e$ (taken in the category of graded $R^e$-modules): $$\cdots \to \Sigma^{(n+1)d}R^e \xrightarrow[]{d_2} \Sigma^dR^e \xrightarrow[]{d_1} R^e \to R \to 0$$ $$\cdots \to \Sigma^{(n+1)di}R^e\xrightarrow[]{d_{2i}}\Sigma^{(n+1)di-nd}R^e \xrightarrow[]{d_{2i-1}}\Sigma^{(n+1)d(i-1)}R^e\to\cdots $$ where $d_{2i}: e_{(n+1)di} \mapsto v\cdot e_{(n+1)di-nd},\,d_{2i-1}: e_{(n+1)di-nd} \mapsto u \cdot e_{(n+1)d(i-1)}$.

Now $HH_\ast(R,M)$ can be computed by tensoring this complex with $M$ (over $R^e$) and taking the homology. Using the relation $M \otimes_{R^e}\Sigma^kR^e=\Sigma^k M$ we obtain, for example, for $M=R$ the complex $$\displaystyle\cdots \to \Sigma^{(n+1)di}R\xrightarrow[]{d_{2i}}\Sigma^{(n+1)di-nd}R \xrightarrow[]{0}\Sigma^{(n+1)d(i-1)}R\to\cdots $$ where $d_{2i}: e_{(n+1)di} \mapsto (n+1)x^n\cdot e_{(n+1)di-nd}$. Hence

If $n+1$ is invertible in $k$ then (as graded $R$-module) $$HH_{2i}(R,R)=\Sigma^{(n+1)di}Rx,\quad HH_{2i-1}(R,R)=\Sigma^{(n+1)di-nd}R/(x^n), \quad H_0(R,R)=R.$$