According to this lecture notes (in Lemma2.1) the statement $Ae\simeq Ae^{\prime}\to A(1-e)\simeq A(1-e^{\prime})$ is true for finite dimensional algebras by using Krull-Schmidt theorem. Can anyone please give me hint about that.
2026-03-25 09:22:34.1774430554
Krull-Schmidt theorem and internally cancellable modules?
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The Krull-Schmidt theorem literally says:
So the idea is that you take the two decompositions $Ae\oplus A(1-e)=A=Ae'\oplus A(1-e')$ and refine the summands into indecomposable modules. Since the pieces in $Ae$ and $Ae'$ pair up, the pieces in $A(1-e)$ and $A(1-e')$ pair up into isomorphic pairs. Thus their sums are isomorphic.