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.

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    $\begingroup$ I don't have an answer for your question. If R is a free algebra with r mapping to one of the generators, then the higher homology groups all vanish (this is because R is the free DGA on a chain complex weakly equivalent to a complex in degree zero obtained by forgetting the generators). I'm pretty sure that this means you're computing the derived pushout of $k\leftarrow k[x]\to R$ in associative DGAs, where $x\mapsto r$. This, to me, suggests that the higher homology groups may be simply "what they are". (Also, if r is any unit, $H_n$ is an $H_0=0$-module and so is zero.) $\endgroup$ Jun 14 '11 at 1:55

This complex (in simultaneously a more general setting, where you have several elements $t_1,\ldots,t_p$, and a more special setting, because only the case of $R$ being a free algebra was studied then) was introduced by Shafarevich [E. S. Golod, I. R. Shafarevich, “On the class field tower”, Izv. Akad. Nauk SSSR Ser. Mat., 28:2 (1964), 261–272], and was studied somewhat thoroughly recently, see, e.g. [E. S. Golod, “Homology of the Shafarevich complex and noncommutative complete intersections”, Fundam. Prikl. Mat., 5:1 (1999), 85–95] - this paper again dealing with the case of free $R$. As the latter paper suggests, this complex is a noncommutative generalisation of the Koszul complex in the commutative case, detecting the "noncommutative complete intersections" usually known as "strongly free sets" from a paper of Anick [D. J. Anick, Non-commutative graded algebras and their Hilbert series, J. Algebra, Volume 78, Issue 1, September 1982, Pages 120-140].

Finally, in the case of arbitrary (graded) $R$ this complex is discussed in a paper of Piontkovski [http://arxiv.org/abs/math/0606279] for the purpose of studying "relative noncommutative complete intersections". You would be especially interested in "Theorem-Definition 2.3" from that paper.

(For those who cannot be bothered to check the references, one condition of Piontkovski's paper is completely analogous to the complete intersection result stating that $A=B[f_1,\ldots,f_k]$ for a regular sequence $f_1,\ldots,f_k$: the complex in question is acyclic iff $R=(R/r)\langle r\rangle$. Here one has to make some assumptions, e.g. assume $R$ graded and $r$ being homogeneous of positive degree, so this does not cover central units mentioned in the question, whereas Tyler Lawson's example is the simplest instance of this result.)

  • $\begingroup$ Thank you Vladimir, this was the kind of survey answer I was looking for. $\endgroup$ Jun 14 '11 at 16:33

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