# A potential resolution of $R/r$

### 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.

• 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.) 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.)