**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 (\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 $\ast=1$?