Let $A$ be a ring (e.g. an $R$-algebra over a noetherian local ring $R$), $M$ an indecomposable $A$-module, and $f$ an endomorphism, which is not an isomorphism. Now, we assume that the endomorphism ring of $M$ is a local ring.
Then, does a power of $f$ factor through a projective $A$-module?
If $A$ is a finite-dimensional algebra over a field, then this is true by the Harada-Sai lemma.
Thank you.
In full generality, the answer is no. Take $A = K[[x,y]]$, the ring of power series in variables $x$ and $y$ with coefficient in a field $K$. Take $M = K[[x,y]]/\langle y \rangle$. Then the endomorphism ring of $M$ is isomorphic to a ring of formal power series $K[[\phi]]$, where an endomorphism $p(\phi) = \sum_{i=0}^{\infty} a_i \phi^i$ applied to a $z\in M$ is defined by $$ p(\phi)(z) := \sum_{i=0}^{\infty} a_i xz. $$ Finally, take $f=\phi$. Then $f$ is not an isomorphism, and no power of $f$ vanishes. However, there are no non-zero morphisms from $M$ to $A$ (and thus to any projective module). So $f$ does not factor through a projective module.
If, moreover, the endormosphism ring of $M$ is Artinian (which happens if $A$ is an algebra over a field and $M$ is of finite length over $A$, for instance), then any non-invertible endomorphism of $M$ is nilpotent. Thus a power of $f$ vanishes, and trivially factors through a projective module.