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Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr))$M \mapsto M/\langle rm-mr|r\in A\rangle$).

Your definition used the "bar-complex" resolution of the form --> A (x) $\to A (x) \otimes A --> \otimes A (x) \to A \otimes A$ but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> $k[t] \otimes k[t] (x) \to k[t] \otimes k[t]$ with the map given by $1 (x) \otimes t - t (x) 1\otimes 1$, so when you apply coinvariants, you get two copies of k[t] $k[t]$ with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of k[t] (x) k[t] $k[t] \otimes k[t]$ is 2 gets you halfway there.

EDIT AGAIN: Actually, any Koszul algebra has its Hochschild homology bounded above by its global dimension. This is clear from the existence of the diagonal Koszul resolution.

3 added 175 characters in body

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of k[t] (x) k[t] is 2 gets you halfway there.

2 added 231 characters in body

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants (M -> M/(rm-mr)).

Your definition used the "bar-complex" resolution of the form --> A (x) A (x) A --> A (x) A but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form k[t] (x) k[t] -> k[t] (x) k[t] with the map given by 1 (x) t - t (x) 1, so when you apply coinvariants, you get two copies of k[t] with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

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