Is there a way to turn a free resolution of a $k$-algebra $A$ into a resolution of the field of fractions $Q(A)$?


Specifically, I'm interested in the ring of polynomials in two variables: $A = k[x,y]$. It's Koszul, so I can write down a nice resolution of $A$ in terms of $A \otimes A^{op}$ modules. (Technically, I found some of my old notes where I worked it out for the quantum plane $k\_q[x,y]$ and set $q=1$; I'm going to happily assume this works, as it didn't require $q$ not to be a root of unity.)


How do I turn this into a resolution of $Q(A) = k(x,y)$?


I'm fairly sure the theory must exist, eg there's a notion of "evenly localizable" in some of Zhang and Yekutieli's papers for noncommutative dualizing complexes, but I haven't managed to trace it back to the commutative theory.


I wondered if I could just tensor the complex with $Q(A) \otimes\_A - $ and $- \otimes\_A Q(A)$, but maybe there's an easier way? Suggestions or references both welcome; essentially I only need the resolution for this one algebra, but I'd love to understand more generally how it works as well.



Turning a resolution of an algebra into a resolution of its field of fractions
$Q(A)\otimes Q(A)$ is a flat $A\otimes A^{op}$-module. Moreover, 
$$
A\otimes\_{A\otimes A^{op}}(Q(A)\otimes Q(A)) = Q(A)\otimes\_A Q(A).
$$
Note that the multiplication $Q(A)\otimes Q(A) \to Q(A)$ induces an isomorphism $Q(A)\otimes\_A Q(A) \cong Q(A)$. This shows that tensoring the free $A$-bimodule resolution of $A$ with $Q(A)\otimes Q(A)$ we obtain a free $Q(A)$-bimodule resolution of $Q(A)$.


