Graded Krull–Schmidt for finite-dimensional graded modules

Let $X$ be a nonzero finite-dimensional graded left $A$-module. Its grading has finite support. Give ${\mathrm{End}}_k(X)$ the grading by degree shift: a homogeneous endomorphism of degree $r$ sends $X_d$ into $X_{d+r}$. Because the support of $X$ is finite, every $k$-linear endomorphism is a finite sum of such homogeneous maps, and composition adds degrees. The action map $\rho :A\to {\mathrm{End}}_k(X)$ sends $A_r$ into degree $r$; hence its image $B=\rho (A)={⨁}_r\rho (A_r)$ is a graded subalgebra. It is finite-dimensional as a subspace of ${\mathrm{End}}_k(X)$, and its identity is $1_B=1_X$. By [F1], $X$ is a graded $B$-module. Since $A\twoheadrightarrow B$, the graded $A$-submodules and graded $B$-submodules of $X$ coincide, as do their degree-zero endomorphism rings.

Existence follows by strong induction on $d={\dim }_{k}M$. If $d=0$, the empty sum is the required decomposition. If $M$ is nonzero and graded-indecomposable, it is already a one-term decomposition. Otherwise write $M=U\oplus V$ with nonzero graded submodules $U,V$. Each is a proper subspace of $M$, so [F4] gives ${\dim }_{k}U<d$ and ${\dim }_{k}V<d$. Induction decomposes $U$ and $V$ into finite sums of nonzero graded-indecomposables; combining those sums decomposes $M$.

If $X$ is also graded-indecomposable, it remains graded-indecomposable as a $B$-module. Apply [F3] to the finite-dimensional graded algebra $B$ and the module $X$ from step 1.1. It follows that the nonunits of ${\mathrm{End}}_{A,0}(X)={\mathrm{End}}_{B,0}(X)$ form a proper two-sided ideal.

Prove uniqueness by strong induction on $d={\dim }_{k}M$. When $M=0$, both decompositions are empty. For nonzero $M$, assume uniqueness in every smaller dimension and write $M=X\oplus C=Y_1\oplus \cdots \oplus Y_s$, where all summands are nonzero graded-indecomposables. Let ${\iota }_X,{\pi }_X$ and ${\iota }_j,{\pi }_j$ be the degree-zero inclusions and projections for these finite biproducts. Define $e_j={\pi }_X{\iota }_j{\pi }_j{\iota }_X\in {\mathrm{End}}_{A,0}(X)$. The identity ${\sum }_j{\iota }_j{\pi }_j=1_M$ gives ${\sum }_j{e}_j=1_X$. By step 2.1 the nonunits form a proper ideal, so at least one $e_j$ is invertible.

Fix such a $j$, and put $a={\pi }_j{\iota }_X:X\to Y_j$ and $b={\pi }_X{\iota }_j:Y_j\to X$. Then $ba=e_j$ is invertible. The degree-zero map $s:=a{e}_j^{-1}$ satisfies $bs=1_X$, so $Y_j=s(X)\oplus \ker b$: for each $y\in Y_j$, $y=s(b(y))+(y-s(b(y)))$, and the second term is in $\ker b$, while $s(X)\cap \ker b=0$. By [F2], the image and kernel are graded submodules. Since $s(X)\mathrm{\ne }0$ and $Y_j$ is graded-indecomposable, $\ker b=0$, so $s$ is bijective. Its inverse is $A$-linear, and it is degree-zero: for homogeneous $y\in Y_{j,d}$, write $s^{-1}(y)={\sum }_e{x}_e$ with $x_e\in X_e$; the direct grading of $Y_j$ and injectivity of $s$ force $x_e=0$ for $e\mathrm{\ne }d$. Thus $s$ is a degree-zero isomorphism $X\cong Y_j$.

Write $M=X\oplus C=Y_j\oplus D$, where $D$ is the sum of the other $Y$-summands. The projection ${\pi }_D{\mathrm{\mid }}_C:C\to D$ is a degree-zero isomorphism. Its kernel is zero because $C\cap Y_j=0$: if $c\in C\cap Y_j$, then ${\pi }_X(c)=0$ and the isomorphism $b={\pi }_X{\mathrm{\mid }}_{Y_j}$ from step 4.1 forces $c=0$. For any $z\in D$, choose the unique $y\in Y_j$ with ${\pi }_X(y)={\pi }_X(z)$; then $z-y\in C$ and ${\pi }_D(z-y)=z$, proving surjectivity. The inverse is degree-zero by the argument in step 4.1. Since $X,Y_j$ are nonzero, $C,D$ are proper subspaces of $M$, so [F4] gives ${\dim }_{k}C,{\dim }_{k}D<d$.

The decompositions of $C$ and $D$ into the remaining indecomposable summands have the same multiset by the induction hypothesis in step 3.1 and the degree-zero isomorphism in step 5.1. Adding $X\cong Y_j$ from step 4.1 proves uniqueness for $M$. Step 1.2 proves existence, so every finite-dimensional graded module has a finite decomposition unique up to permutation and degree-zero graded isomorphism.