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Kim Morrison
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Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1?

Background: the Hochschild homology of an associative algebra is the homology of the complex

... --> A (x) A (x) A --> A (x) A --> A

where the last two differentials are a(x)b(x)c \mapsto ab(x)c-a(x)bc+ca(x)b and a(x)b \mapsto ab-ba, and you can guess the rest. More generally, it's "derived coinvariants": take a projective resolution of your algebra, then coinvariants of that.

For k[t], the Hochschild homology is concentrated in degrees 0 and 1, and in both of those degrees it's k[t]. I know that I can go look up Loday (\S 3.2.2) and find a calculation, but I'd like a better explanation.

I know that the zero-th Hochschild homology HH_0(k[t]) must just be k[t], because the zero-th Hochschild homology is just coinvariants, and k[t] is commutative.

What I'd like is a "good" explanation for HH_1(k[t]).

Kim Morrison
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