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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.

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What do you mean by resolving the field of fractions? In what category do you want to resolve it? – Sasha Dec 5 '12 at 15:25
As a resolution of $Q(A) \otimes Q(A)^{op}$-modules (equivalently $Q(A)$−bimodules), I guess? The end goal is the Hochschild cohomology of $k(x,y)$. – eithil Dec 5 '12 at 15:46

1 Answer 1

up vote 4 down vote accepted

$Q(A)\otimes Q(A)$ 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 $Q(A)$.

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Lovely! Thank you! – eithil Dec 7 '12 at 10:06

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