Shift-orbit bases for graded simple and projective classes

Every finite-dimensional graded module $M$ has finite length. If $M=0$, the empty chain is a composition series. If $M\mathrm{\ne }0$, the dimensions of its proper graded submodules form a nonempty subset of $\{0,1,\dots ,{\dim }_{k}M-1\}$; choose a proper graded submodule $N$ of maximal dimension. The quotient $M\mathrm{/}N$ is nonzero. Any proper nonzero graded submodule of $M\mathrm{/}N$ would lift to a proper graded submodule strictly containing $N$, contrary to maximality, so $M\mathrm{/}N$ is simple. Since ${\dim }_{k}N<{\dim }_{k}M$, induction gives a composition series of $N$; appending $M\mathrm{/}N$ gives one for $M$. This uses only a maximum in a finite set of dimensions and one submodule at a time.

By graded Krull–Schmidt [F5], write the regular graded module as a finite direct sum $A\cong {⨁}_{j=1}^m{Q}_j$ of nonzero graded-indecomposable modules; if $A=0$, take $m=0$. Each $Q_j$ is a direct summand of the shifted free module $A\{0\}$, so [F8] makes it a finite graded projective. If $A=0$, every unital left $A$-module is zero, so both groups are zero and the empty bases prove the theorem. Henceforth assume $A\mathrm{\ne }0$.

Let $Q$ be a nonzero finite-dimensional graded-indecomposable projective and $q:Q\twoheadrightarrow S$ a degree-zero epimorphism to a graded-simple module. Put $K=\ker q$. If $N\le Q$ is graded and $K+N=Q$, then $q{\mathrm{\mid }}_N:N\twoheadrightarrow S$ is epic. Projectivity [F9] lifts $q$ through $q{\mathrm{\mid }}_N$ to a degree-zero map $g:Q\to N$. After inclusion $N\hookrightarrow Q$, let $f$ be the resulting endomorphism. Then $qf=q$, and induction gives $q{f}^n=q$ for every $n\ge 1$, so $f$ is not nilpotent. By graded Fitting [F7], $f$ is invertible. Since $\mathrm{im}f\subseteq N$, this forces $N=Q$. Thus $K$ is superfluous among graded submodules and $q$ is a finite graded projective cover.

Every finite graded projective cover $p:P\twoheadrightarrow S$ of a graded-simple module is indecomposable. Indeed, if $P=U\oplus V$ with both summands nonzero, at least one restriction of $p$ is nonzero and hence surjective; its summand $U$ then satisfies $U+\ker p=P$, contradicting superfluity of $\ker p$. Moreover, $S$ is the unique graded-simple quotient of $P$ up to isomorphism. If $q:P\twoheadrightarrow T$ is another such quotient and $L=\ker q$, a nonzero graded image $q(\ker p)$ must be all of $T$ by simplicity. That would give $L+\ker p=P$, contradicting superfluity. Hence $q(\ker p)=0$, so $q$ factors through $P\mathrm{/}\ker p\cong S$; the induced nonzero map $S\to T$ is an isomorphism.

Every finite-dimensional graded module has finite support: if its dimension is $n$ and it had $n+1$ distinct nonzero homogeneous components, one nonzero vector from each would be linearly independent. If $M\mathrm{\ne }0$ and $M\cong M\{r\}$ by a degree-zero isomorphism, then $\mathrm{supp}(M)=\mathrm{supp}(M)+r$. Taking the maximum of this finite nonempty set gives $\max \mathrm{supp}(M)=\max \mathrm{supp}(M)+r$, hence $r=0$. Thus the shift action is free on the isomorphism classes of nonzero graded simples and nonzero indecomposable projectives; no lower or upper bound on the grading of $A$ is used.

The category of finite-dimensional graded modules is abelian: [F10] makes kernels, cokernels and finite biproducts degreewise, so these objects remain finite-dimensional and the full subcategory inherits the abelian structure. It is essentially small as well. For each finite-support dimension vector on $\mathbb{Z}$, fix the standard graded $k$-space with those component dimensions; the possible $A$-actions on it form a set of families of linear maps satisfying the module identities. Every finite-dimensional graded module is isomorphic to one of these models by choosing bases for its finitely many nonzero homogeneous components. The family of all such standard models is a set, and this object-by-object argument makes no simultaneous choice across an arbitrary family.

By step 1.6, the category of finite-dimensional graded modules is the essentially small abelian category used to define $G_0^{\mathrm{g}\mathrm{r}}(A)$ in [F1]. By step 1.1 all its objects have finite length, so [F4] says that its graded-simple isomorphism classes form a $\mathbb{Z}$-basis of $G_0^{\mathrm{g}\mathrm{r}}(A)$.

If $S$ is graded-simple, take a nonzero homogeneous $s\in S_r$. The graded submodule $As$ is nonzero, hence is $S$. The map $A\{r\}\to S$, $a\mapsto as$, is degree-zero because $1_A$ has degree $r$ in $A\{r\}$ and the action preserves degree; it is surjective. Decomposing $A\{r\}\cong {⨁}_{j=1}^m{Q}_j\{r\}$, at least one restriction to a summand is nonzero and therefore surjective onto the simple module $S$.

Each $Q_j$ has a graded-simple quotient: a composition series from step 1.1 has a simple final factor. By step 1.3 this quotient map is a projective cover, and by step 1.4 its simple quotient is unique up to isomorphism; denote that isomorphism class by $S_j$. For any graded-simple $S$, step 2.2 gives a surjection from some $Q_j\{r\}$ to $S$. Since shift is invertible [F16], both $Q_j\{r\}$ and $S_j\{r\}$ retain indecomposability and simplicity, respectively; $Q_j\{r\}$ is finite graded projective by [F8]. The shift of the cover $Q_j\twoheadrightarrow S_j$ is a cover of $S_j\{r\}$: [F10] preserves the epimorphism, and shifting back by $\{-r\}$ preserves the superfluity condition. By steps 1.3–1.4, the quotient $S$ is isomorphic to $S_j\{r\}$. Thus the finite list $S_1,\dots ,S_m$ meets every graded-simple shift orbit.

Every nonzero finite-dimensional graded-indecomposable projective $Q$ has a graded-simple quotient by step 1.1; step 1.3 makes that quotient map a projective cover. Existence and uniqueness of covers [F6] therefore identify $Q$ with the cover $P(S)$ of its simple quotient. Conversely, step 1.4 shows each $P(S)$ is indecomposable. Shifting a cover gives a cover of the shifted simple, as in step 3.1, so $P(S\{r\})\cong P(S)\{r\}$ by [F6]. If $P(S)\{r\}\cong P(T)$, that projective has simple quotients $S\{r\}$ and $T$; uniqueness from step 1.4 gives $S\{r\}\cong T$. Hence projective indecomposable shift orbits are in bijection with graded-simple shift orbits, and there are finitely many.

By steps 2.1 and 3.1, the $\mathbb{Z}$-basis of $G_0^{\mathrm{g}\mathrm{r}}(A)$ is partitioned into finitely many free shift orbits. Choose one simple $S_i$ from each orbit. By [F3], $v^r[S_i]=[S_i\{r\}]$, so the Laurent monomials $v^r$ map bijectively to the distinct $\mathbb{Z}$-basis classes in that orbit. The orbit spans therefore form a direct sum of copies of $\mathbb{Z}[v,v^{-1}]$, with basis $[S_i]$. This proves the stated finite Laurent basis for $G_0^{\mathrm{g}\mathrm{r}}(A)$.

By step 1.6, the set $\mathcal{I}$ of isomorphism classes of nonzero finite-dimensional graded-indecomposable projectives is a set. By graded Krull–Schmidt [F5], each finite graded projective $P$ has a unique finite decomposition into nonzero graded-indecomposable summands $Q_j$. By [F8], $P$ is a degree-zero summand of a finite sum of shifts of $A$; each $Q_j$, being a summand of $P$, is also a summand of that finite sum by transitivity of direct summands. Thus [F8] makes every $Q_j$ a finite graded projective. Sending $P$ to its multiplicity vector in $\mathcal{I}$ is additive under direct sum. By the split-group presentation [F14], it descends to a homomorphism $K_0^{\mathrm{g}\mathrm{r}}(A)\to \mathbb{Z}[\mathcal{I}]$. The map from the free abelian group [F15] sending each basis vector to its projective class is inverse: one composite fixes each indecomposable basis vector, while the other sends $[P]$ to the sum of its indecomposable classes, which equals $[P]$ by the split relation. Thus $\mathcal{I}$ is a $\mathbb{Z}$-basis of $K_0^{\mathrm{g}\mathrm{r}}(A)$. By steps 4.1 and 1.5, this basis is partitioned into finitely many free shift orbits represented by the covers $P_i$ of the chosen $S_i$. Using [F3] as in step 4.2 shows that $[P_i]$ is a finite $\mathbb{Z}[v,v^{-1}]$-basis of $K_0^{\mathrm{g}\mathrm{r}}(A)$. The cover classes are unique up to isomorphism by [F6], so the result is independent of the chosen covers.