endomorphism rings of indecomposable objects Suppose $\mathcal C$ is a preadditive, Karoubi category with a zero object.  What further assumptions on $\mathcal C$ are required to ensure that the endomorphism ring of an indecomposable object is a local?
By an object $X$ being indecomposable, I mean that in any biproduct decomposition $X \cong X_1 \oplus X_2$, one has $X_i \cong 0$ for some $i=1,2$.
Thanks for your help.
 A: Hi benjamin.
You're asking if $\mathcal{C}$ is a so called Krull-Schmidt-category. There are several sufficient conditions know from ring theory if $\mathcal{C}$ is a module category. For example the argument George mentioned gives an affirmative answer if $\mathcal{C}$ is the category of finitely generated (left)modules over an artinian ring.
You get another characterisation by considering semiperfect rings. A ring is semiperfect iff the category of finitely generated projective (left)modules is Krull-Schmidt. Examples of semiperfect rings include all left or right artinian ring as well as all finitely generated $R$-algebras where $R$ is a complete local ring or a discrete valuation ring.
This connection between semiperfect rings and Krull-Schmidt-categories is in fact somewhat stronger: One can show that $\mathcal{C}$ is Krull-Schmidt iff $End_\mathcal{C}(X)$ is semiperfect for all objects $X\in\mathcal{C}$.
A: Suppose your additive category is an abelian one (for example a category of modules), and let $X$ be an object. Suppose $X$ is of finite length, that is there is a filtration 
$0=X_0\leq X_1\leq\cdots\leq X_n=X$, such that $X_i/X_{i-1}$ is simple for all $i=1,\cdots,n$.
Then it is easy to show that every object is both artinian and notherian, and for every endomorphism $f:X\to X$ the conditions to be monomorphism, epimorphism and automorphism are equivalent. Therefore, the endomorphism ring of $X$ is local.
Thus in a slightly more particular case, a sufficient condition is every object is of finite length. 
