An example of computing Ext I've been looking for less trivial examples of computing Ext than finitely generated abelian groups, which tends to be the standard example (and often the only example). Here's an interesting exercise I found in some notes:
Let $M = \mathbb{C}[x,y] / (x,y), N = \mathbb{C}[x,y] / (x-1)$. My question is how to compute $\text{Ext}_v(M,N)$ in the category of $\mathbb{C}[x,y]$-modules.
Well, first of all $\text{Ext}_0(M,N) = \text{Hom}(M,N)$. However, I'm not sure how to identify what this $\text{Hom}$ is! More generally, we have the short exact sequence
$0\rightarrow K \rightarrow \mathbb{C}[x,y] \rightarrow M \rightarrow 0$
where the second map is the inclusion, the third map is the quotient projection, and $K$ is the kernel of the projection. This sequence gives the exact sequence (a piece of the long exact sequence)
$0\rightarrow \text{Ext}_1(M,N) \rightarrow \text{Hom}(K,N) \rightarrow \text{Hom}(\mathbb{C}[x,y],N) \rightarrow \text{Hom}(M,N) \rightarrow 0$,
which means that $\text{Ext}_1(M,N)$ is the kernel of the map $\text{Hom}(K,N) \rightarrow \text{Hom}(\mathbb{C}[x,y], N)$. But again I'm having trouble determining this kernel.
Finally I think the projective resolution $0 \rightarrow \mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y]/(x-1)$ shows that the higher Ext's are zero.
Any help would be greatly appreciated.
 A: First, as Aaron mentioned, a homomorphism from $M$ to $N$ is uniquely defined by its image on the generator 1 of $M$, and it must commute with the action $x\cdot 1=y\cdot 1=0$. In particular, if $f$ is such a homomorphism and $\bar{f(1)}$ is a coset representative of $f(1)$, then you must have $x\cdot \bar{f(1)} \in (x-1)$. Since $\mathbb{C}[x,y]$ is a UFD and $x$ and $x-1$ are both irreducible, this implies that $\bar{f(1)}\in (x-1)$, i.e. that $f(1) = 0\in N$. So, that hom-space is 0, and that's good news for the next computation (see below).
As for your computation of $\text{Ext}^1$, you have forgotten that $N\mapsto \text{Hom}(,N)$ is a
contraviariant functor and you have to turn your long exact sequence around accordingly (indeed,
there is no obvious way of defining the map $\text{Hom}(K,N)\rightarrow \text{Hom}(\mathbb{C}[x,y],N)$,
while it's perfectly clear how to define the map the other way round - namely by restriction; similarly $\text{Hom}(M,N)\rightarrow \text{Hom}(\mathbb{C}[x,y],N)$ should be defined by composing homs with projection). See if that gets you anywhere
and feel free to report back.
