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Let $\Lambda$ be an artin algebra. We say $\Lambda$ is of finite representation type if the number of the isomorphism classes of indecomposable $\Lambda$-modules is finite.

Now suppose $\Lambda$ is not of finite type, how to get an simple module $S$ such that there is an infinite number of nonisomorphic indecomposable modules $X$ with $Hom_{\Lambda}(X,S) \not =0$?

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  • $\begingroup$ Artin algebras have a finite number of simples, don't they? $\endgroup$ – Tobias Kildetoft Jan 2 '17 at 9:48
  • $\begingroup$ I was to some extend asking you, since I don't recall how many of the properties of finite-dimensional algebras carry over to Artin algebras (but if it was true, it seems like the sort of thing you would have seen way before getting to studying algebras of finite representation type). $\endgroup$ – Tobias Kildetoft Jan 2 '17 at 10:03
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Since it is an artin algebra there are finitely many simple modules $S_1$,...,$S_n$ up to isomorphism (namely the summands of the module A/J, when J is the jacobson radical). Let $M=S_1 \oplus ... \oplus S_n$. Then for every indecomposable module $N: Hom(N,M) \neq 0$. Now $ Hom(N,M)=Hom(N,S_1) \oplus ... \oplus Hom(N,S_n)$ is always nonzero which would not be possible if for every simple module $S$ Hom(N,S)=0 for all but finitely many indecomposable modules $N$.

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  • $\begingroup$ I know fintie dimensional algebras have finite simple modules, but I don't know why it is hold for artin algebras. Also why $Hom(N,M) \not =0$ for every indecomposable module $N$? $\endgroup$ – Xiaosong Peng Jan 2 '17 at 13:16
  • $\begingroup$ look for example in the first two chapters of the book of auslander reiten smalo for the thing about simple modules. a concrete nonzero map in Hom(N,M) is easy to produce when you look at the top of N which has a summand of M. $\endgroup$ – Mare Jan 2 '17 at 13:18

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