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Suppose that $A = k[x_1, ..., x_n]$. I want to compute $HH_*(A) = \mathrm{Tor}_*^{A \otimes_k A}(A, A)$. To do this we have to take a resolution of $A$ as a module over $A \otimes_k A$, which is nice because $A \otimes_k A = k[x_1, ..., x_n, y_1, ..., y_n]$. So now we should be able to take the Koszul resolution:

$$0 \to \wedge^n(A \otimes_k A)^n \to .... \to \wedge^2(A \otimes_k A)^n \to (A \otimes A)^n \to A \otimes_k A \to A,$$

and tensor with $A$ then compute homology. This is where I'm confused; is there some cleverness that one must do with $\wedge^n(A \otimes A)^n \otimes_k A$ to make it clear that the homology should be $\Omega^n_A$? Maybe i'm not familiar enough with the differential on the Koszul complex, but it seems really hard to describe the image or kernel of these maps otherwise.

ufabao
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1 Answers1

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There is a useful change of coordinates trick to clear the way here. As you said, $A^e$ is just a polynomial algebra on $2n$-variables, which you called $x$ and $y$. Consider now $x'= x-y$, so that $A^e$ is also a polynomial algebra on the $2n$-variables $x'$ and $x$, with $x'$ acting by zero on $A$.

Now the Koszul resolution above works like the 'usual one', where $K = k[x_1,\ldots,x_n]$ is a trivial module over $K[x_1',\ldots,x_n']$. In this case, you already what that $\text{Tor}$ is, right? You can now imitate what happens in this case to deduce what happens in this one.

Pedro
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