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Let $\Lambda$ be a finite-dimensional $k$-algebra with $k$ algebraicallyclosed. Bongartz has recently shown that the existence of an indecomposable$\Lambda$-module of length $n > 1$ implies that also indecomposable$\Lambda$-modules of length $n-1$ exist. Using a slight modification of hisarguments, we strengthen the assertion as follows: If there is anindecomposable module of length $n$, then there is also an accessible one.Here, the accessible modules are defined inductively, as follows: First, thesimple modules are accessible. Second, a module of length $n \ge 2$ isaccessible provided it is indecomposable and there is a submodule or a factormodule of length $n-1$ which is accessible.

Indecomposables live in all smaller lengths