On indecomposable modules without selfextensions (Q1369267)

Denote by \(k\) an algebraically closed field and by \(A\) an associative finite-dimensional basic and connected \(k\)-algebra. The paper deals with the question when two indecomposable \(A\)-modules have the same composition factors. It was proved by \textit{I. Reiten}, \textit{A. Skowroński} and \textit{S. O. Smalø} [Proc. Am. Math. Soc. 117, No. 2, 343-354 (1993; Zbl 0794.16013)] that if \(M\) is an indecomposable \(A\)-module which does not lie on a short cycle \(M\to X\to M\) of nonzero nonisomorphisms, where \(X\) is indecomposable, too, then \(M\) is uniquely determined by its class \([M]\) in the Grothendieck group \(K_0(A)\). Moreover it was shown in the same paper that if \(M\) does not lie on a short cycle then \(\text{Ext}^1_A(M,M)=0\). The authors give an example showing that in general an indecomposable \(A\)-module \(M\) with \(\text{Ext}^1_A(M,M)=0\) is not uniquely determined by \([M]\). The main results of the paper are the following. Theorem 1. Let \(A\) be an algebra such that every component of the Auslander-Reiten quiver of \(A\) is generalized standard. Then the number of isomorphism classes of indecomposable \(A\)-modules \(M\) with \(\text{Ext}^1_A(M,M)=0\) and \(\text{End}_A(M)\neq k\) is finite. Theorem 2. Let \(A\) be a strongly simply connected algebra of polynomial growth. If \(M\) and \(N\) are indecomposable \(A\)-modules with \([M]=[N]\) and \(\text{Ext}^1_A(M,M)=0\), then \(M\simeq N\).