Splitting idempotents

Question

Let $C$ be an additive category. An idempotent $e=e^2\in\mathrm{Hom}_C(X,X)$ is split if there are morphisms $\mu:Y\rightarrow X$, $\rho:X\rightarrow Y$ such that $\mu\rho=id_Y$ and $\rho\mu=e$. Suppose that in the category $C$ every idempotent splits. How can I prove that $\mathrm{End}(X)$ is local for every indecomposable object $X$?

Answer 1

EDIT: To see it more elementary let me take the category of free $\mathbb{Z}$-modules. Then $\operatorname{Hom}_{\mathbb{Z}}(X,X)$ is isomorphic to a full matrix ring and an idempotent is a projection matrix . Thus we have $X=eX\oplus (1-e)X$ and the inclusion $\mu:eX\to X$ and $\rho=e:X\to eX$ provide a splitting of $e$. However we have $\operatorname{End}(\mathbb{Z})=\mathbb{Z}$ is not local for the indecomposable $\mathbb{Z}$-module $\mathbb{Z}$.

Now for the case of modules of finite length the essential step is the Fitting lemma stating that if $M$ is a finite length $n$ module and $f$ is an endomorphism, then $M=\operatorname{Im}(f^n)\oplus \ker(f^n)$. Thus if $M$ is indecomposable either $\ker(f^n)=0$ implying $\ker(f)=0$, thus $f$ is invertible or $\operatorname{Im}(f^n)=0$ implying $f^n=0$. Thus $f$ is nilpotent. But this means that $(1-f)$ has an inverse namely $(1+f+f^2+\ldots+f^{n-1})$. Thus $\operatorname{End}(M)$ is a local ring.

For the proof of the Fitting lemma look at the increasing/decreasing sequences $\operatorname{Im}(f^i)$ and $\ker(f^j)$. Since $M$ is of finite length $n$ stay stabilize somewhere, at least from $n$ on. So to prove the direct sum decomposition let $f^n(x)\in \ker(f^n)$ Then $f^{2n}(x)=0$. But $\ker(f^{2n})=\ker(f^n)$. Thus $f^n(x)=0$. Therefore the intersection is zero. To prove that their sum equals $M$, let $m\in M$. Then $f^n(m)=f^{2n}(v)$ for some $v$ since $\operatorname{Im}(f^n)=\operatorname{Im}(f^{2n})$. Now $m=f^n(v)+(m-f^n(v))$, where the first summand is in $\operatorname{Im}(f^n)$ while the second term is in $\ker(f^n)$ by definition.