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Let $A= \mathbb{Q}[x]/(x^2)$ and define the module $M$ by $M=A/(x)$. I am supposed to compute $\text{Ext}_{A}^n(M,A)$. First we start by finding a free resolution for $M$ by $$\mathbb{Q}[x]/(x^2) \to A/(x) \to 0$$ where the first arrow is just the natural surjection. Now I am really confused in computing $\text{Ext}_{A}^n(M,A)$, because we obtain the complex $$0 \to \text{Hom}_A(\mathbb{Q}[x]/(x^2),\mathbb{Q}[x]/(x^2)) \to ...$$

The definition of $\text{Ext}_{A}^n(M,A)$ doesn't even allow me to compute anything here! There is only one arrow in the complex, what have I done wrong? Isn't my free resolution correct?

  • You have computed only the first term of a free resolution. You need to find the other terms (up to $n+1$ term at least). – Krish Sep 24 '17 at 04:24
  • @Krish in the definition there is no restriction on the length of a free resolution, would $(x)/(x^2) \to \mathbb{Q}[x]/(x^2) \to A/(x) \to 0$ serve as a free resolution? –  Sep 24 '17 at 04:26
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    How do you prove the existence of free resolution of a module? Use the same technique here. – Krish Sep 24 '17 at 04:29
  • @Krish Sorry to say that I haven't come to that part of the book, but I will take a look at it. However, is the free resolution I gave above correct? It lets me compute $\text{Ext}_{A}^n(M,A)$, but its only one computation. –  Sep 24 '17 at 04:31
  • This is not a free resolution. This only a part of a free resolution. Free resolution can be of infinite length. See here – Krish Sep 24 '17 at 04:38
  • @Krish i totally see my mistake now, thank you! –  Sep 24 '17 at 04:52
  • @Krish I have read through the way you construct free resolutions, if a module $M$ is finitely generated with generating set $X$, we may consider the free module $F_0$ on the set $X$, from here we keep constructing the components of the resolution by considering the kernel of $F_0 \to M$ and so on. I get the resolution $0 \to A/(x) \to A/(x) \to 0$ where the first $A/(x)$ is the free module. Although I feel that I have done something wrong here...would you help me out to understand? –  Sep 24 '17 at 05:32
  • This, this will help you to understand the idea. Also search for "free resolution" in this site, you will get more examples. – Krish Sep 24 '17 at 06:41

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