Graded Fitting decomposition for degree-zero endomorphisms

The degree-zero endomorphisms of $M$ are closed under pointwise addition, additive inverses, and composition, and contain $1_M$, because each such map preserves every $M_d$. Thus $R:={\mathrm{End}}_{A,0}(M)$ is a unital subring of ${\mathrm{End}}_A(M)$, whose ring operations are those of [L3] and [L4].

The kernels $\ker (f^m)$ form an increasing sequence of subspaces and the images $\mathrm{im}(f^m)$ form a decreasing sequence. Finite-dimensionality makes both sequences stabilize; choose $n\ge 1$ after stabilization, so $\ker (f^n)=\ker (f^{2n})$ and $\mathrm{im}(f^n)=\mathrm{im}(f^{n+1})$. Since each $f^n$ is degree-zero, [L1] makes these stabilized subspaces graded $A$-submodules.

If $x\in \ker (f^n)\cap \mathrm{im}(f^n)$, write $x=f^n(y)$. Then $f^{2n}(y)=f^n(x)=0$, so stabilization gives $y\in \ker (f^{2n})=\ker (f^n)$ and hence $x=0$. By [L2] applied to $f^n$, the two submodules have dimensions summing to ${\dim }_{k}M$; their zero intersection therefore gives $M=\ker (f^n)\oplus \mathrm{im}(f^n)$. For $M=0$ this reads $0=0\oplus 0$; for $f=0$ it reads $M=M\oplus 0$, and for invertible $f$ it reads $M=0\oplus M$.

Now suppose $M\mathrm{\ne }0$ and let $N$ be the set of nonunits of $R$. It contains $0$, and $-a\in N$ whenever $a\in N$. If $ra$ were invertible, $a$ would be injective; if $ar$ were invertible, $a$ would be surjective. Since $M$ is finite-dimensional, either property makes $a$ bijective, with degree-zero $A$-linear inverse. Thus $N$ absorbs multiplication on both sides by every $r\in R$.

Suppose in addition that $M$ is graded-indecomposable. The decomposition in step 2.1 forces $\ker (f^n)=0$ or $\mathrm{im}(f^n)=0$. In the first case $f$ is injective, hence bijective by finite-dimensionality; its inverse is again degree-zero and $A$-linear. In the second case $f^n=0$. Thus $f$ is invertible or nilpotent, and every nonunit is nilpotent. This includes the zero endomorphism in the nilpotent case. If ${\dim }_{k}M=1$, nonzero indecomposability is automatic because two nonzero graded direct summands would have total dimension at least two.

If $f$ is a nonunit and $f^m=0$, then $1-f$ has two-sided inverse $1+f+\cdots +f^{m-1}$. Hence for each $f\in R$, at least one of $f$ and $1-f$ is invertible.

If $a,b\in N$ but $u=a+b$ were invertible, then $x=u^{-1}a$ and $1-x=u^{-1}b$ would both be nonunits: otherwise $a=ux$ or $b=u(1-x)$ would be invertible. This contradicts step 4.1. Therefore $N$ is closed under addition; together with step 2.2 and [L5], it is a two-sided ideal. It is proper because $1_M$ is a unit. Every proper left or right ideal contains no unit and is therefore contained in $N$, so $N$ is the unique maximal left ideal and the unique maximal right ideal.