$ \operatorname{Ext}_{k[x]}^n(k,k)$ for a field $k$

Question

Consider the polynomial ring $k[x]$ for a field $k$ and the $k[x]$-module $k$, letting $x$ act trivially on $k$. What is $ \operatorname{Ext}_{k[x]}^n(k,k)$ for $n\geq0$?

Answer 1

A free resolution of $k$ is $k[x] \overset{x}\to k[x]$. Now apply $ \operatorname{Hom}(-,k)$. This gives the complex $k\overset{0}\leftarrow k$. So $ \operatorname{Ext}^0, \operatorname{Ext}^1 = k$ and higher exts are zero (which we already knew because $k[x]$ has homological dimension 1).