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


$$ \ldots \longrightarrow A \otimes A \otimes A \longrightarrow A \otimes A \longrightarrow A$$


where the last two differentials are $$a \otimes b \otimes c \mapsto ab \otimes c-a \otimes bc + ca \otimes b$$ and $a \otimes b \mapsto ab-ba$, 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](http://www.ams.org/mathscinet-getitem?mr=1217970) (\S 3.2.2) and find a calculation, but I'd like a better explanation.


I know that the zero-th Hochschild homology $\operatorname{HH}\_0(k[t])$ 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 $\operatorname{HH}\_1(k[t])$.


**Edit**: Ben has a simple explanation below. Let me also rephrase the question, hoping for more. Here are a few things: If $A$ is semisimple, then $\operatorname{HH}\_{\ast}(A)$ is concentrated in degree 0. Is there something about $k[t]$ that ensures it's concentrated in degrees 0 and 1? Conversely, can I conclude anything about $A$ from the fact that $\operatorname{HH}\_{\ast}(A)$ is zero above $\ast=1$?



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 ($M \mapsto M/\langle rm-mr|r\in A\rangle$).


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.


