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This is probably something very simple but I got stucked with it. Consider the polynomial ring $k[x_1,\dots,x_n]$. How can one construct a free resolution of it in the category of $k$-algebras(not necessarily commutative)? Namely, I ask for an acyclic complex $$\dots\to R_2\to R_1\to k[x_1,\dots,x_n]=R_0\to 0$$ where all $R_i$ are free algebras of some number of variables.

It is clear how one should try to construct such resolution: take $R_1=k\langle x_1,\dots, x_n\rangle$ with evident map to $R_0$. The kernel of this map is clearly generated by $x_ix_j-x_jx_i$ so we take $R_2=k\langle x_{ij}\rangle$ with differential taking $x_{ij}$ to $x_ix_j-x_jx_i$. And here is a problem I can not resolve - how to determine the kernel of this map and proceed futrther?

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  • $\begingroup$ I am also interested in a similar question. I have asked many people, and done lots of Googling, but I haven't gotten anywhere. Have you made any progress on this? I am wondering if the term "noncommutative resolution" is only something that makes sense if you come from the commutative resolution side of things. It may just be that people who think about these things call them something else. $\endgroup$ – Trevor Jul 20 '16 at 12:41

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