homological algebra - Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1?

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 ($M mapsto M/langle rm-mr|rin Arangle$).

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

This is of the form $k[t] otimes k[t] to k[t] otimes k[t]$ with the map given by $1 otimes t - t otimes 1$, 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 $k[t] otimes k[t]$ 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.