Minimality of the Koszul resolution

Question

Let $R = \mathbb{C}[x,y]$ and $V = \mathbb{C}x\oplus\mathbb{C}y$. Then, the Koszul resolution of $R$ (as an $R$-bimodule) is given by \begin{align*} 0\to R\otimes_{\mathbb{C}}\wedge^2V\otimes_{\mathbb{C}} R& \xrightarrow{\partial_2} R\otimes_{\mathbb{C}}V\otimes_{\mathbb{C}} R \xrightarrow{\partial_1} R\otimes_{\mathbb{C}} R \xrightarrow{\mathrm{mult}} R \to 0\,,\\ \partial_1(1\otimes v\otimes 1) &= 1\otimes v - v\otimes 1\,,\\ \partial_2(1\otimes x\wedge y\otimes 1) &= 1\otimes x\otimes y + x\otimes y\otimes1 - 1\otimes y\otimes x - y\otimes x\otimes 1\,. \end{align*}

I want to know whether this resolution is minimal (i.e. each step is a projective cover). How can I (dis)prove this?

Since $R$ is not an Artinian ring, I'm not sure even a minimal resolution exists. Nonetheless, it seems somewhat plausible. Any comments or references on other general results are appreciated.

Answer 1

In the category of ungraded bimodules, the multiplication map $R\otimes_\mathbb{C}R\to R$ is not a projective cover. For example, the proper sub-bimodule of $R\otimes_\mathbb{C}R$ generated by $1\otimes1+x\otimes1-1\otimes x$ surjects onto $R$.