TITLE: Computing an example of Ext
QUESTION [9 upvotes]: Let $k$ be a field. I want to compute $\operatorname{Ext}_{k[x] / \langle x^2 \rangle}(k,k)$. 

However I have no idea how to do this? I cannot even think how to construct a projective resolution that would give me a useful answer. Any help would be appreciated.

REPLY [7 votes]: Full credit goes to Mariano for this answer.
Define $R:= k[x] / \langle x^2 \rangle$. We think of $k$ as an $R$-algebra via the homomorphism that sends a poly $f \mapsto f(0)$. It is very clear that $k$ is generated by one element, namely the element $1 \in R$. So there is a surjection $R \twoheadrightarrow k$. Clearly $\ker (R \twoheadrightarrow k)$ is the ideal generated by the polynomial $x$ in $R$, again one element. So using the standard construction I now have an exact sequence $R \to R \twoheadrightarrow k$, where the left map is the homomorphism generated by $1 \mapsto x$. Again, after some thought it's clear that $\ker(R \to R)$ is again the ideal generated by the poly $x$ in $R$. We can keep repeating this construction to get a projective (free) resolution $$\cdots \to R \to R \to R \to k \to 0$$
which gives the chain complex (chopping off $k$) $$ \cdots \to R \to R \to R \to 0$$
which yields the complex of hom-sets
$$0 \to \operatorname{Hom}_R(R,k) \to \operatorname{Hom}_R(R,k) \to \operatorname{Hom}_R(R,k) \to \cdots $$
You can stop here by noticing each of these maps are zero, which tells us the homology at each point is $\operatorname{Hom}_R(R,k) \cong k$. We have this isomorphism because of the fact that $\operatorname{Hom}_R\left(\coprod_{\alpha \in \mathcal{A}}R, M\right) \cong \prod_{\alpha \in \mathcal{A}}M$ in general.
Hence for all $i \ge 0$, $$\operatorname{Ext}_R^i (k,k) \cong k.$$