# 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.

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As modules these each have only one generator, so determining Hom(M,N) should be the same as finding out where 1 can go, right? If that's true, then if $1\mapsto a$, then $x\mapsto ax$ and $y\mapsto ay$, so we need $ax=ay=0\in N$. –  Aaron Mazel-Gee Dec 17 '10 at 11:20
And I agree that your projective resolution of $N=\mathbb{C}[x,y]/(x-1)$ shows that the higher Ext's are 0. BTW, I think that as it's a cohomology-type thing Ext usually has its index as a superscript (i.e., $Ext^i$ instead of $Ext_i$). The subscript usually indicates the category, unless of course it's omitted. –  Aaron Mazel-Gee Dec 17 '10 at 11:25
To compute $\mathrm{Ext}(M,N)$ using a projective resolution, you need a projective resolution of $M$, not of $N$! –  Mariano Suárez-Alvarez Dec 17 '10 at 15:06
Oops! I always get this mixed up. Or you need an injective resolution of N. Incidentally, is there a good way to remember this that isn't just a mnemonic (i.e., a "moral" reason)? –  Aaron Mazel-Gee Dec 17 '10 at 22:51
Here are my two cents on the moral reason: $Ext(M,\bullet)$ is the derived functor of a left-exact covariant functor. So to compute it we use injective resolutions of $N$. Where as $Ext(\bullet , N)$ is covariant on $C^{op}$ so to compute it you need an injective resolution of M in $C^{op}$ that is a projective resolution in $C$. –  DBS Apr 29 at 10:15

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.