Fitting decomposition in a finite-length abelian category

A proper inclusion $A⊊B$ of subobjects of a finite-length object has $\mathrm{\ell }(A)<\mathrm{\ell }(B)$, because $B\mathrm{/}A\mathrm{\ne }0$ contributes at least one composition factor; consequently any ascending chain of subobjects stabilizes, and a nonzero object has a proper subobject of maximal length, hence a maximal proper subobject.

Every object $X$ is a finite direct sum of indecomposable objects, by induction on $\mathrm{\ell }(X)$: for $X=0$ take the empty sum; if $X\mathrm{\ne }0$ is indecomposable there is nothing to prove; otherwise $X\cong Y\oplus Z$ with $Y,Z\mathrm{\ne }0$, and $\mathrm{\ell }(Y),\mathrm{\ell }(Z)<\mathrm{\ell }(X)$, so the induction hypothesis applies to $Y$ and $Z$.

For an endomorphism $f:X\to X$, the image and kernel chains stabilize by [F1]. Choose $N$ with $\mathrm{im}{f}^N=\mathrm{im}{f}^{2N}$ and $\ker f^N=\ker f^{2N}$, and put $I=\mathrm{im}{f}^N$, $K=\ker f^N$. The restriction $f^N{\mathrm{\mid }}_I:I\to I$ is epic by image stabilization, hence is an isomorphism: length additivity makes its kernel zero. If $p:X\twoheadrightarrow I$ is the image factorization of $f^N$, then $(f^N{\mathrm{\mid }}_I{)}^{-1}p:X\to I$ retracts the inclusion $I\hookrightarrow X$ and has kernel $K$. The split exact sequence therefore gives $X\cong I\oplus K$.

If $X$ is indecomposable, step 2.1 forces $I=0$ or $K=0$. The first case gives $f^N=0$. In the second case $f^N$ is monic; length additivity makes its cokernel zero, so $f^N$ is an isomorphism. Since $f$ commutes with $f^N$ and its inverse, $f^{N-1}(f^N{)}^{-1}$ is an inverse of $f$. Thus every endomorphism of $X$ is either invertible or nilpotent.

Let $X$ be indecomposable and $f,g\in \mathrm{End}(X)$ with $f+g=u$ an isomorphism. If neither $f$ nor $g$ is an isomorphism, both are nilpotent by step 3.1; then $a:=u^{-1}f$ and $b:=u^{-1}g$ satisfy $a+b={\mathrm{id}}_X$ and are non-units, hence nilpotent, and $a={\mathrm{id}}_X-b$ gives $ab=b-b^2=ba$, so the commuting nilpotents $a,b$ have nilpotent sum: ${\mathrm{id}}_X=a+b$ is nilpotent, forcing ${\mathrm{id}}_X=0$ and $X=0$ by [F2], contrary to indecomposability. Hence $f$ or $g$ is an isomorphism.

More generally, if $f_1+\cdots +f_k={\mathrm{id}}_X$ with $X$ indecomposable, then some $f_i$ is an isomorphism: induct on $k$, the cases $k=1$ and $k=2$ being trivial and step 4.1; for $k\ge 3$ put $g=f_2+\cdots +f_k$, so that $f_1+g={\mathrm{id}}_X$, and if $f_1$ is not an isomorphism then $g$ is an isomorphism by step 4.1, and $g^{-1}{f}_2+\cdots +g^{-1}{f}_k={\mathrm{id}}_X$ has $k-1$ terms, so some $g^{-1}{f}_i$ is an isomorphism by induction and then $f_i=g(g^{-1}{f}_i)$ is an isomorphism. Consequently $\mathrm{End}(X)$ is local: in a ring $R$, locality is equivalent to the criterion that for every $a\in R$ either $a$ or $1-a$ is a unit, and here $a+(1-a)={\mathrm{id}}_X$ is the identity, so the two-term case applies. Also, a nonzero idempotent in a local ring is the identity.

Let $X=X_1\oplus A=Y_1\oplus \cdots \oplus Y_m$ with $X_1,Y_j$ indecomposable, and let $a_j:Y_j\to X_1$, $b_j:X_1\to Y_j$ be the composites of the inclusions and projections. Then ${\sum }_j{a}_j{b}_j={\mathrm{id}}_{X_1}$; by step 5.1 some $a_{j_0}{b}_{j_0}$ is an isomorphism, and after relabelling $j_0=1$ and setting $\beta :=a_1{b}_1$ we obtain $\epsilon :=b_1{\beta }^{-1}{a}_1\in \mathrm{End}(Y_1)$ with ${\epsilon }^2=b_1{\beta }^{-1}(a_1{b}_1){\beta }^{-1}{a}_1=\epsilon$, so $\epsilon$ is an idempotent; it is nonzero because ${\beta }^{-1}{a}_1$ is a left inverse of $b_1$ and $X_1\mathrm{\ne }0$, and its image is $b_1(X_1)$. By the last sentence of step 5.1, $\epsilon ={\mathrm{id}}_{Y_1}$; hence ${\beta }^{-1}{a}_1$ and $b_1$ are mutually inverse isomorphisms $X_1\cong Y_1$.

Let $P$ be an indecomposable projective object and let $Q,Q^{\mathrm{\prime }}\subseteq P$ be proper subobjects with $Q+Q^{\mathrm{\prime }}=P$. The addition morphism $Q\oplus Q^{\mathrm{\prime }}\to P$ is an epimorphism, so by [F3] the identity of $P$ lifts to $h:P\to Q\oplus Q^{\mathrm{\prime }}$; writing $h=(h_1,h_2)$ and $\phi :=i_Q{h}_1$, $\psi :=i_{Q^{\mathrm{\prime }}}{h}_2$ in $\mathrm{End}(P)$, one has $\phi +\psi ={\mathrm{id}}_P$, so $\phi$ or $\psi$ is an isomorphism by the two-term case of step 5.1. If $\phi$ is an isomorphism then $i_Q$ is a split monomorphism and an epimorphism, hence an isomorphism $Q\cong P$, contradicting the strictness $\mathrm{\ell }(Q)<\mathrm{\ell }(P)$ for the proper inclusion $Q⊊P$ from step 1.1; the same argument applies to $\psi$. Hence proper subobjects of $P$ have proper sum. Now choose a proper subobject $M\subseteq P$ of maximal length, which exists by step 1.1. For any proper $Q$ the sum $M+Q$ is proper, so $\mathrm{\ell }(M+Q)\le \mathrm{\ell }(M)$, while $M\subseteq M+Q$ gives $\mathrm{\ell }(M)\le \mathrm{\ell }(M+Q)$; hence $M=M+Q$ and $Q\subseteq M$. Therefore $M$ is the unique maximal proper subobject $J(P)$, and $P\mathrm{/}J(P)$ is simple, since a proper subobject of the quotient pulls back to a proper subobject of $P$ contained in $J(P)$.

Keep the notation of step 6.1, put $B={⨁}_{j\ge 2}{Y}_j$, and use the isomorphism $b_1:X_1\to Y_1$ to define $\mathrm{\Theta }=i_{X_1}{b}_1^{-1}{p}_{Y_1}+i_B{p}_B\in \mathrm{End}(X)$. Relative to $X=Y_1\oplus B$, its matrix is $\left(\begin{array}{cc}1& 0\\ w& 1\end{array}\right)$, where $w=p_B{i}_{X_1}{b}_1^{-1}:Y_1\to B$, because $p_{Y_1}{i}_{X_1}=b_1$. Thus $\mathrm{\Theta }$ is invertible with inverse $\left(\begin{array}{cc}1& 0\\ -w& 1\end{array}\right)$, sends $Y_1$ onto $X_1$, and fixes $B$. Therefore $X=X_1\oplus B$. Taking the quotient by $X_1$ in this decomposition and in $X=X_1\oplus A$ gives $A\cong X\mathrm{/}{X}_1\cong B$, establishing cancellation.

Let $X=X_1\oplus \cdots \oplus X_n=Y_1\oplus \cdots \oplus Y_m$ with all $X_i,Y_j$ indecomposable, and induct on $n$. For $n=0$ we have $X=0$, so $m=0$ because the $Y_j$ are nonzero. For $n\ge 1$, steps 6.1 and 7.1 applied with $A={⨁}_{i\ge 2}{X}_i$ provide $j_0$ with $X_1\cong Y_{j_0}$ and ${⨁}_{i\ge 2}{X}_i\cong {⨁}_{j\mathrm{\ne }{j}_0}{Y}_j$; the left-hand side is a sum of $n-1$ indecomposables and the right-hand side of $m-1$, so the induction hypothesis gives $n-1=m-1$ and a bijection matching the remaining factors up to isomorphism, and $X_1\cong Y_{j_0}$ completes the correspondence.

Collecting the results: step 1.2 gives the finite decomposition into indecomposables, step 5.1 the local endomorphism ring together with the finite-sum criterion, step 3.1 the dichotomy isomorphism-or-nilpotent, step 8.1 the uniqueness up to isomorphism and permutation, and step 6.2 the unique maximal proper subobject $J(P)$ of an indecomposable projective $P$ with simple quotient. Moreover the canonical epimorphism $\pi :P\to P\mathrm{/}J(P)$ is essential: if $N\subseteq P$ satisfies $N+J(P)=P$ and $N$ were proper, then $N\subseteq J(P)$ by step 6.2 and $P=N+J(P)=J(P)$, a contradiction; hence $N=P$. With $P$ projective, $(P,\pi )$ is a projective cover of the simple object $P\mathrm{/}J(P)$ in the sense of [F3].