###The DGA###

For $k$ some field, let $R$ be a $k$-algebra, and let $r\in R$.

Define a differential graded algebra $\mathbf{R}_r$ as follows. As a graded algebra, it is isomorphic to $R\langle t\rangle$, the free $k$-algebra over $R$ with non-central variable $t$ added. The algebra $R$ has degree zero, and $t$ has degree one. The differential $d$ is defined by $d(R)=0$ and $d(t)=r$, and extend by the Leibniz rule.

As a chain complex, the DGA $\mathbf{R}_r$ then looks like $$ R\leftarrow RtR\leftarrow RtRtR \leftarrow ...$$ Note that each $t$ may be replaced by $\otimes_k$ without changing the $R$-bimodule structure. So what are the homology groups? Clearly, $d(RtR)=RrR\subset R$, and so $H_0(\mathbf{R}_r)=R/r$ (the quotient by the two-sided ideal generated by $r$).

###The Question###

My question is, for what $R$ and $r$ do all higher homology groups of $\mathbf{R}_r$ vanish? Equivalently, when is $\mathbf{R}_r$ quasi-isomorphic to $R/r$?

###Examples and Counterexamples###

There are some immediate answers. When $r=1$, the complex $\mathbf{R}_1$ is the (augmented) bar resolution of $R$ as a bimodule over itself. As the name implies, the homology groups vanish. The standard argument for the exactness of the bar resolution (constructing a chain homotopy between the identity map and the zero map on $\mathbf{R}_1$) can be generalized to $\mathbf{R}_r$ for any $r$ which is a central unit.

If $R=k[x]$ and $r=x$, then a direct computation shows that the higher homology groups vanish, and so $\mathbf{R}_x$ is a resolution of $R/x\simeq k$.

As a counterexample, consider $R=k[x]$, and $r=x^2$. For $xt-tx\in (\mathbf{R}_{x^2})_1$, we have $$ d(xt-tx)=x(x^2)-(x^2)x=0$$ However, any 1-boundary $d(atbtc)$ must have $x$-degree at least 2, and so $xt-tx$ is not exact.

The DGA

For $k$ some field, let $R$ be a $k$-algebra, and let $r\in R$.

Define a differential graded algebra $\mathbf{R}_r$ as follows. As a graded algebra, it is isomorphic to $R\langle t\rangle$, the free $k$-algebra over $R$ with non-central variable $t$ added. The algebra $R$ has degree zero, and $t$ has degree one. The differential $d$ is defined by $d(R)=0$ and $d(t)=r$, and extend by the Leibniz rule.

As a chain complex, the DGA $\mathbf{R}_r$ then looks like $$ R\leftarrow RtR\leftarrow RtRtR \leftarrow ...$$ Note that each $t$ may be replaced by $\otimes_k$ without changing the $R$-bimodule structure. So what are the homology groups? Clearly, $d(RtR)=RrR\subset R$, and so $H_0(\mathbf{R}_r)=R/r$ (the quotient by the two-sided ideal generated by $r$).

The Question

My question is, for what $R$ and $r$ do all higher homology groups of $\mathbf{R}_r$ vanish? Equivalently, when is $\mathbf{R}_r$ quasi-isomorphic to $R/r$?

Examples and Counterexamples

There are some immediate answers. When $r=1$, the complex $\mathbf{R}_1$ is the (augmented) bar resolution of $R$ as a bimodule over itself. As the name implies, the homology groups vanish. The standard argument for the exactness of the bar resolution (constructing a chain homotopy between the identity map and the zero map on $\mathbf{R}_1$) can be generalized to $\mathbf{R}_r$ for any $r$ which is a central unit.

If $R=k[x]$ and $r=x$, then a direct computation shows that the higher homology groups vanish, and so $\mathbf{R}_x$ is a resolution of $R/x\simeq k$.

As a counterexample, consider $R=k[x]$, and $r=x^2$. For $xt-tx\in (\mathbf{R}_{x^2})_1$, we have $$ d(xt-tx)=x(x^2)-(x^2)x=0$$ However, any 1-boundary $d(atbtc)$ must have $x$-degree at least 2, and so $xt-tx$ is not exact.