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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
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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
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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

1 Answer 1

up vote 4 down vote accepted

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.

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