Background: the Hochschild homology of an associative algebra is the homology of the complex
... --> A (x) A (x) A --> A (x) A --> A$$ \ldots \longrightarrow A \otimes A \otimes A \longrightarrow A \otimes A \longrightarrow A$$
where the last two differentials are a(x)b(x)c \mapsto ab(x)c-a(x)bc+ca(x)b$$a \otimes b \otimes c \mapsto ab \otimes c-a \otimes bc + ca \otimes b$$ and a(x)b \mapsto ab-ba$a \otimes b \mapsto ab-ba$, 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]$k[t]$, the Hochschild homology is concentrated in degrees 0 and 1, and in both of those degrees it's k[t]$k[t]$. 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])$\operatorname{HH}_0(k[t])$ must just be k[t]$k[t]$, because the zero-th Hochschild homology is just coinvariants, and k[t]$k[t]$ is commutative.
What I'd like is a "good" explanation for HH_1(k[t])$\operatorname{HH}_1(k[t])$.
Edit: Ben has a simple explanation below. Let me also rephrase the question, hoping for more. Here are a few things: If A$A$ is semisimple, then HH_*(A)$\operatorname{HH}_{\ast}(A)$ is concentrated in degree 0. Is there something about k[t]$k[t]$ that ensures it's concentrated in degrees 0 and 1.? Conversely, can I conclude anything about A$A$ from the fact that HH_*(A)$\operatorname{HH}_{\ast}(A)$ is zero above *=k$\ast=1$?
