Consider the quiver $Q\colon 1\xrightarrow{\alpha} 2\xrightarrow{\beta} 3\xrightarrow{\gamma} 4$ and the algebra $A=k[Q]/(\gamma\beta\alpha)$. Denote the simple $A$-modules by $L(-)$ and let $M$ be the direct sum of the simple submodules.

What is the algebra $\operatorname{Ext}^*(M,M)$, and, more importantly, how can it be computed? I'd guess that the simples have extensions

$$\begin{aligned}\operatorname{Ext}^1(L(2),L(1))\colon\quad & 0 \to L(1) \xrightarrow{\alpha} e_2A =\langle e_2, \alpha\rangle \to L(2)\to 0\\ \operatorname{Ext}^1(L(3),L(2))\colon\quad & 0\to L(2) \xrightarrow{\beta} e_3A =\langle e_3, \beta\rangle \to L(3)\to 0\end{aligned}$$

and so on, but I have the impression that with this approach, I don't end up with the correct algebra. For example, how do I know that $\operatorname{Ext}^1(L(3), L(1))=0$ and does not contain e.g.

$$0\to L(1)\to \langle e_3, \alpha\beta\rangle \to L(3)\to 0?$$

I guess it's because the middle term is no $A$-module… but still: even if I find some extensions, how do I know I find all, and then, their relations?

  • $\begingroup$ I have to admit that this was an extremely stupid question. I wonder why it got upvoted. $\endgroup$ – Bubaya May 14 '20 at 22:17
  • $\begingroup$ Is it really extremely stupid? You should type up an answer for anyone who disagrees. ;) Also, this question is so nicely typeset, if I were in a hurry I'd upvote it without reading it. :D $\endgroup$ – Mike Pierce May 15 '20 at 22:36
  • $\begingroup$ @MikePierce I guess, yes. But as you complimented me for my writing, I have written an answer. $\endgroup$ – Bubaya May 18 '20 at 12:56

Ext is an additive functor, so it suffices to know the$\operatorname{Ext}^*(L(i), L(j))$ and their compositions. We can compute Ext-algebras using projective resolutions. We choose

$$\begin{aligned} 0\to P(1) \to L(1)\to 0\\ 0 \to P(1) \to P(2) \to L(2) \to 0\\ 0 \to P(2) \to P(3) \to L(3) \to 0\\ 0 \to P(1) \to P(3) \to P(4) \to L(4) \to 0 \end{aligned}$$

as projective resolutions of the respecitve simple modules. Then, for instance

$$\operatorname{Hom}^2(P(4), P(1)) = \left\langle \begin{matrix} P(1) & \to & P(3) & \to & P(4)\\ \downarrow\\ P(1)\end{matrix}\right\rangle$$

and all other $\operatorname{Hom}^*(P(4), P(1))$ are zero. This yields that $\operatorname{Ext}^*(P(4), P(1))$ is concentrated in degree 2.

As another example, one sees that

$$\operatorname{Hom}^2(P(1), P(4)) = \left\langle\begin{matrix} & & & & P(1)\\ & & & \swarrow & \downarrow\\ P(1) & \to & P(3) & \to & P(4) \end{matrix}\right\rangle$$

is spanned by a single degree-0-map that factors through the indicated degree-1-map and thus is null-homotopic. Taking homology gives that $\operatorname{Ext}^*(P(1), P(4))=0$.

All other Ext-spaces are obtained similarly. Composition is given by composition of these diagrams.


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