Skip to main content
3 of 4
edited tags
Ilya Nikokoshev
  • 15.1k
  • 12
  • 77
  • 129

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

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

... --> A (x) A (x) A --> A (x) A --> A

where the last two differentials are a(x)b(x)c \mapsto ab(x)c-a(x)bc+ca(x)b and a(x)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 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 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 HH_*(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 HH_*(A) zero above *=k?

Kim Morrison
  • 7.8k
  • 7
  • 48
  • 75