Grothendieck Groups and Graded Cartan Pairings

Let $F_{\mathcal{C}}$ be the free abelian group on $\mathrm{Iso}(\mathcal{C})$ and let $R_{\mathcal{C}}$ be the subgroup generated by $e_Y-e_X-e_Z$ for the short exact sequences of [F1]. By [F3], $\chi$ extends uniquely to $\tilde{\chi }\text{\hspace{0.17em}⁣}:F_{\mathcal{C}}\to H$. For each generator of $R_{\mathcal{C}}$, $\tilde{\chi }(e_Y-e_X-e_Z)=\chi (Y)-\chi (X)-\chi (Z)=0$, so $R_{\mathcal{C}}\subseteq \ker \tilde{\chi }$. Since $H$ is abelian, $R_{\mathcal{C}}$ is normal; [F4] gives a unique homomorphism from $F_{\mathcal{C}}\mathrm{/}{R}_{\mathcal{C}}=G_0(\mathcal{C})$ with the required values. The zero sequence gives $\chi (0)=\chi (0)+\chi (0)$, hence $\chi (0)=0$.

Let $F_{\mathcal{D}}$ be the free abelian group on $\mathrm{Iso}(\mathcal{D})$ and let $R_{\mathcal{D}}$ be generated by $e_{X\oplus Y}-e_X-e_Y$. By [F3], $\psi$ extends uniquely to $\tilde{\psi }\text{\hspace{0.17em}⁣}:F_{\mathcal{D}}\to H$, and the additivity hypothesis makes every generator of $R_{\mathcal{D}}$ lie in its kernel. The subgroup $R_{\mathcal{D}}$ is normal because $F_{\mathcal{D}}$ is abelian, so [F4] gives a unique homomorphism $F_{\mathcal{D}}\mathrm{/}{R}_{\mathcal{D}}=K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}(\mathcal{D})\to H$. Taking $X=Y=0$ gives $\psi (0)=0$.

Let $F\text{\hspace{0.17em}⁣}:\mathcal{C}\to {\mathcal{C}}^{\mathrm{\prime }}$ be exact. If $f$ is an isomorphism, applying [F9] to $f{f}^{-1}=1$ and $f^{-1}f=1$ shows $Ff$ is an isomorphism, so $[X]\mapsto [FX]$ is a function on $\mathrm{Iso}(\mathcal{C})$. In a short exact sequence, [F5] identifies $i$ as $\ker p$ and $p$ as $\mathrm{coker}i$. Exactness [F6] preserves these finite limit and colimit diagrams, so $0\to FX\to FY\to FZ\to 0$ is short exact in ${\mathcal{C}}^{\mathrm{\prime }}$. Thus the image class function is additive, and step 1.1 gives the induced homomorphism $G_0(F)\text{\hspace{0.17em}⁣}:G_0(\mathcal{C})\to G_0({\mathcal{C}}^{\mathrm{\prime }})$ with $G_0(F)([X])=[FX]$.

Let $F\text{\hspace{0.17em}⁣}:\mathcal{D}\to {\mathcal{D}}^{\mathrm{\prime }}$ be additive. If $f$ is an isomorphism, applying [F9] to $f{f}^{-1}=1$ and $f^{-1}f=1$ shows $Ff$ is an isomorphism, so $[X]\mapsto [FX]$ is defined on isomorphism classes. For a biproduct $X\oplus Y$, let $i_X,i_Y$ and $p_X,p_Y$ be its inclusions and projections. The equations $p_a{i}_b={\delta }_{ab}$ and $i_X{p}_X+i_Y{p}_Y=1$ characterize this biproduct. By [F7], $F$ preserves zero morphisms and addition; by [F9], it preserves identities and composition. The four image maps therefore exhibit $F(X\oplus Y)$ as a biproduct of $FX$ and $FY$, so $F(X\oplus Y)\cong FX\oplus FY$. The image class function is additive, and step 1.2 gives the induced homomorphism $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}(F)\text{\hspace{0.17em}⁣}:K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}(\mathcal{D})\to K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}({\mathcal{D}}^{\mathrm{\prime }})$ with $K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}(F)([X])=[FX]$.

For identity functors the formulas in steps 2.1–2.2 fix every generator. For composable exact functors $F,G$, and separately for composable additive functors, the composite is again of the same type. On every generator, $[X]\mapsto [FX]\mapsto [G(FX)]=[(GF)X]$; hence the induced homomorphism of $GF$ is the composite of the induced homomorphisms of $F$ and $G$. Equality on generators proves identity and composition laws for both assignments.

If $\eta \text{\hspace{0.17em}⁣}:F\Rightarrow G$ is a natural isomorphism, [F10] supplies a natural inverse, so each component ${\eta }_X\text{\hspace{0.17em}⁣}:FX\to GX$ is an isomorphism. Thus $[FX]=[GX]$ in the target isomorphism-class set, and the induced homomorphisms agree on every generator of the corresponding $G_0$ or split-$K_0$ group. They are therefore equal.

Since $\mathcal{C}$ is essentially small, $\mathrm{Iso}(\mathcal{C})$ is a set by [F1]. Simplicity is invariant under isomorphism, so $\mathrm{Simp}(\mathcal{C})\subseteq \mathrm{Iso}(\mathcal{C})$ is a set. Put $F:=\mathbb{Z}[\mathrm{Simp}(\mathcal{C})]$. The universal property of the free abelian group defines $\mathrm{\Phi }:F\to G_0(\mathcal{C})$ by $e_{[S]}\mapsto [S]$.

For $M\in \mathcal{C}$, choose a composition series $0=M_0<M_1<\cdots <M_n=M$. Define $[M:S]$ to be the number of indices $i$ for which $M_i\mathrm{/}{M}_{i-1}\cong S$. By [F4], this number is independent of the chosen series and of the representative of $[M]$. Only finitely many factors occur, so $\mu ([M]):={\sum }_{[S]\in \mathrm{Simp}(\mathcal{C})}[M:S]e_{[S]}$ is a well-defined finite-support element of $F$. Since this value is unique, no composition series is selected simultaneously for all isomorphism classes.

The class function $[M]\mapsto \mu ([M])$ is additive on short exact sequences. Indeed, take $0\to X\stackrel{i}{\to }Y\stackrel{p}{\to }Z\to 0$ and composition series $0=X_0<\cdots <X_r=X$ and $0=Z_0<\cdots <Z_s=Z$. Regard the first series as a series in $Y$ through the kernel isomorphism $i:X\cong \ker p$. For each $j$, let $Y_j$ be the inverse-image subobject of $Z_j$ under $p$, formed by the pullback of $p$ along $Z_j\hookrightarrow Z$. Then $Y_0=i(X)$ and $Y_s=Y$. The projection $Y_j\to Z_j$ is epic by [F7]; composing it with the quotient epimorphism $Z_j\to Z_j\mathrm{/}{Z}_{j-1}$ gives an epimorphism with kernel $Y_{j-1}$. The kernel assertion follows from the pullback universal property [F6]: a map into $Y_j$ is killed by the composite precisely when its $Z_j$-component factors through $Z_{j-1}$, which is precisely the defining property of $Y_{j-1}$. By [F8] this composite is a cokernel of $Y_{j-1}\hookrightarrow Y_j$, and by [F9] it therefore identifies $Y_j\mathrm{/}{Y}_{j-1}\cong Z_j\mathrm{/}{Z}_{j-1}$. These quotients are simple. Thus the chain $0=i(X_0)<\cdots <i(X_r)=Y_0<Y_1<\cdots <Y_s=Y$ is a composition series of $Y$, with the factors of the chosen $X$-series followed by those of the chosen $Z$-series.

The spliced series in step 1.3 has, for each simple $S$, exactly $[X:S]+[Z:S]$ factors isomorphic to $S$. Jordan-Hölder [F4] identifies these counts with the counts from any composition series of $Y$. Therefore $[Y:S]=[X:S]+[Z:S]$ for every simple $S$, so $\mu ([Y])=\mu ([X])+\mu ([Z])$. If $X=0$ or $Z=0$, the corresponding series is empty and the same argument gives the endpoint identity. Only two finite composition series are chosen for this one sequence; no arbitrary-index choice is used.

By steps 1.2 and 2.1, $\mu$ is an additive class function on $\mathrm{Iso}(\mathcal{C})$. The universal property [F10] gives a unique homomorphism $\bar{\mu }:G_0(\mathcal{C})\to F$ with $\bar{\mu }([M])=\mu ([M])$.

If $S$ is simple, then $0<S$ is a composition series with sole factor $S$ by [F5]. Hence $\bar{\mu }\mathrm{\Phi }(e_{[S]})=e_{[S]}$ for every basis vector, so $\bar{\mu }\mathrm{\Phi }=1_F$.

For a composition series $0=M_0<\cdots <M_n=M$, each short exact sequence $0\to M_{i-1}\to M_i\to M_i\mathrm{/}{M}_{i-1}\to 0$ gives $[M_i]=[M_{i-1}]+[M_i\mathrm{/}{M}_{i-1}]$ in $G_0(\mathcal{C})$ by [F1]. Since $[M_0]=[0]=0$, telescoping gives $[M]={\sum }_{i=1}^n[M_i\mathrm{/}{M}_{i-1}]=\mathrm{\Phi }\bar{\mu }([M])$. The classes $[M]$ generate $G_0(\mathcal{C})$, so $\mathrm{\Phi }\bar{\mu }=1_{G_0(\mathcal{C})}$.

Steps 4.1 and 4.2 show that $\mathrm{\Phi }$ and $\bar{\mu }$ are inverse isomorphisms. If $\mathrm{Simp}(\mathcal{C})$ is empty, every nonzero object would have a nonempty composition series with a simple first factor; thus every object is zero up to isomorphism, $F=0$, and the same inverse identities give $G_0(\mathcal{C})=0$. For $M=0$, the empty composition series gives $\mu ([0])=0$, consistent with $[0]=0$ from [F1]. The theorem is not an iff statement.

By induction on dimension, the regular left module $A$ has a finite composition series $0=A_0⊊A_1⊊\cdots ⊊A_n=A$: for a nonzero finite-dimensional module, choose a proper submodule of maximal dimension and continue with it. For any simple left module $S$, choose $0\mathrm{\ne }s\in S$; the map $A\to S$, $a\mapsto as$, is nonzero and hence surjective. If $j$ is the least index with the image of $A_j$ nonzero, then $A_{j-1}$ maps to zero and $A_j$ maps onto $S$. Thus the induced map $A_j\mathrm{/}{A}_{j-1}\to S$ is an isomorphism of simple modules. Consequently every simple is finite-dimensional and every simple isomorphism class occurs among the finitely many factors of this fixed series.

Let $P\mathrm{\ne }0$ be a finite-dimensional indecomposable projective. Among its proper submodules choose one, $K$, of maximal $k$-dimension; such a submodule exists because $0<P$ and the possible dimensions are finite. Then $P\mathrm{/}K$ is nonzero and has no proper nonzero submodule, so it is simple by [F5]. Let $p\text{\hspace{0.17em}⁣}:P\twoheadrightarrow P\mathrm{/}K$ be the quotient map.

Let $S_1,\dots ,S_t$ represent those finitely many isomorphism classes. For each $S_i$, choose a projective cover $q_i\text{\hspace{0.17em}⁣}:Q_i\twoheadrightarrow S_i$ using [F2]. These sources are finite-dimensional: choose a finite $k$-basis of $S_i$ and lift its vectors to $Q_i$. The submodule $Q_i^0$ generated by those lifts is finite-dimensional, since it is an image of a finite direct sum of copies of $A$, and $q_i(Q_i^0)=S_i$. Hence $Q_i=Q_i^0+\ker q_i$; [F3] gives $Q_i=Q_i^0$.

Fix $i$ and write $K_i=\ker q_i$. If $Q_i=U\oplus V$ with both summands nonzero, at least one of $q_i(U)$ or $q_i(V)$ is nonzero; by simplicity of $S_i$, that image is all of $S_i$. Say it is $q_i(U)$. Then $U+K_i=Q_i$, so superfluity of $K_i$ forces $U=Q_i$, contradicting $V\mathrm{\ne }0$. Thus each $Q_i$ is indecomposable.

Suppose $p\text{\hspace{0.17em}⁣}:Q\twoheadrightarrow S$ and $r\text{\hspace{0.17em}⁣}:Q\twoheadrightarrow T$ are surjections to simple modules, with superfluous kernels $K$ and $L$. If $r(K)\mathrm{\ne }0$, simplicity gives $r(K)=T$; then for each $x\in Q$ some $y\in K$ has $r(y)=r(x)$, whence $x-y\in L$ and $K+L=Q$. Superfluity of $L$ would give $K=Q$, impossible since $p$ is surjective onto the nonzero module $S$. Therefore $K\subseteq L$; interchanging $p$ and $r$ gives $L\subseteq K$. The equal kernels identify $S\cong Q\mathrm{/}K\cong T$. In particular, the $Q_i$ are pairwise nonisomorphic: transporting a cover map across any proposed isomorphism would give two such simple quotients of one source.

Suppose $N\le P$ and $N+K=P$. Then $p{\mathrm{\mid }}_N\text{\hspace{0.17em}⁣}:N\twoheadrightarrow P\mathrm{/}K$ is surjective. By projectivity [F4], it lifts $p$ to a map $h\text{\hspace{0.17em}⁣}:P\to N$; after inclusion into $P$, this gives $f\in {\mathrm{End}}_A(P)$ with $pf=p$, hence $p{f}^m=p$ for every $m\ge 1$. Regard $A$ and $P$ as concentrated in degree zero. Every submodule of $P$ is then graded, so its ordinary indecomposability makes it graded-indecomposable, and $f$ is degree-zero. By [F7], $f$ is invertible or nilpotent. Nilpotence is impossible because $p\mathrm{\ne }0$ and $p{f}^m=p$ for every $m$; therefore $f$ is invertible. Since $f(P)\subseteq N$, this forces $N=P$. Thus $K$ is superfluous and $p$ is a projective cover by [F3]. If $P\mathrm{/}K\cong S_i$, compose $p$ with such an isomorphism; uniqueness in [F2] identifies $P$ with $Q_i$ over $S_i$. Therefore every nonzero indecomposable finite-dimensional projective is isomorphic to exactly one $Q_i$.

By [F6], any finite-dimensional projective $Q$ is a finite direct sum of indecomposable modules. Each summand remains projective: precompose a map from the summand with the projection from $Q$, lift through the given surjection using projectivity of $Q$, and restrict the lift to the summand. Each summand is finite-dimensional, so step 5.1 identifies it with one of the $Q_i$. Step 4.1 makes those types distinct, and uniqueness in [F6] makes the multiplicities $m_i(Q)$ uniquely determined. The zero projective has the empty sum.

Let ${\mathbb{Z}}^{(t)}$ be the free abelian group with basis $e_1,\dots ,e_t$, and define $\mathrm{\Phi }(e_i)=[Q_i]$. Step 6.1 makes $\mathrm{\Phi }$ surjective. Send the free generator of the split group for each isomorphism class $[Q]$ to ${\sum }_i{m}_i(Q)e_i$; uniqueness and additivity of the multiplicities under direct sum, from [F6], make this assignment respect each relation $[Q\oplus R]=[Q]+[R]$ from [F1]. It therefore descends to a map $\mathrm{\Psi }\text{\hspace{0.17em}⁣}:K_0^{\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{t}}(\mathrm{proj}A)\to {\mathbb{Z}}^{(t)}$. The two maps are inverse: $\mathrm{\Psi }\mathrm{\Phi }(e_i)=e_i$, and the decomposition in step 6.1 plus [F1] gives $\mathrm{\Phi }\mathrm{\Psi }([Q])=[Q]$ for every generator. Hence the classes $[Q_i]$ form a free abelian basis. No step asserts that the Cartan map to $G_0(A)$ is invertible.

By [F4], ${\mathrm{GrMod}}_0(A)$ is abelian and its degreewise kernels, cokernels, and finite biproducts have the stated graded structures. For a map between finite-dimensional modules, its kernel is a subspace of the finite-dimensional domain and its cokernel is a quotient of the finite-dimensional codomain; [F8] gives finite-dimensionality of kernels and images. Thus the full subcategory ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}^{\mathrm{g}\mathrm{r}}(A)$ is closed under kernels, cokernels, images, and finite biproducts. The ambient coimage and image of any map therefore remain in this full subcategory, as does their canonical isomorphism; hence [F10] makes it abelian. It is essentially small: each object has finite support and a finite homogeneous basis; for each finite-support dimension vector, the graded $k$-space with those dimensions has a fixed standard model, and its $A$-actions are a set of families of linear maps satisfying the algebra-action identities. Every object is isomorphic to one of these set-many models. This objectwise coordinate argument selects no basis simultaneously from an arbitrary family. Hence [F1] defines $G_0^{\mathrm{g}\mathrm{r}}(A)$.

Every finite-dimensional graded module has a finite homogeneous basis and so is finitely generated. Conversely, a finite graded projective has finitely many homogeneous generators by [F6]; sending the homogeneous generator of each matching shift $A\{r\}$ to a generator of $P$ gives a degree-zero surjection from a finite sum of shifts of $A$, so $P$ is finite-dimensional. The category ${\mathcal{P}}_{\mathrm{f}\mathrm{d}}^{\mathrm{g}\mathrm{r}}(A)$ is essentially small as a full subcategory of ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}^{\mathrm{g}\mathrm{r}}(A)$. It inherits preadditive hom groups from that abelian category. By [F5], the zero module and a finite direct sum of finite graded projectives are again finite graded projective; their module biproducts therefore make this an additive category by [F9]. Thus [F2] defines $K_0^{\mathrm{g}\mathrm{r}}(A)$.

Write $A^{\mathrm{♭}}$ for the underlying ungraded algebra given the trivial grading, and regard an ungraded module as an $A^{\mathrm{♭}}$-module concentrated in degree zero. This identifies the ungraded module category with a full subcategory of ${\mathrm{GrMod}}_0(A^{\mathrm{♭}})$ that is closed under kernels, cokernels, images, and finite biproducts by [F4]. Its finite-dimensional subcategory remains closed because kernels and images are subspaces and cokernels are quotients of finite-dimensional vector spaces. The ambient coimage-to-image isomorphisms remain in this full subcategory, so it is abelian by [F10]; it is essentially small by the same finite-dimensional action-matrix argument as in step 1.1. Therefore [F1] defines $G_0(A)$.

For each $r\in \mathbb{Z}$, shift sends a finite-dimensional graded module to a finite-dimensional graded module and leaves degree-zero maps degree-zero. It is exact because degreewise kernels and cokernels in [F4] are merely reindexed, and its inverse is shift by $-r$ by [F3]. Thus shift is an exact autoequivalence of ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}^{\mathrm{g}\mathrm{r}}(A)$, and [F7] induces an automorphism ${\sigma }_r$ of $G_0^{\mathrm{g}\mathrm{r}}(A)$ with ${\sigma }_r([M])=[M\{r\}]$.

If $P$ is finite graded projective, [F5] exhibits it as a summand of a finite direct sum of shifts $A\{s\}$. Shifting this splitting by $r$ exhibits $P\{r\}$ as a summand of a finite direct sum of shifts $A\{s+r\}$, so [F5] shows $P\{r\}$ is finite graded projective. The shift functor and its inverse are additive on this category; [F7] therefore induces inverse automorphisms ${\tau }_r$ and ${\tau }_{-r}$ on $K_0^{\mathrm{g}\mathrm{r}}(A)$, with ${\tau }_r([P])=[P\{r\}]$.

The class function $[P]\mapsto [P]$ from $\mathrm{Iso}({\mathcal{P}}_{\mathrm{f}\mathrm{d}}^{\mathrm{g}\mathrm{r}}(A))$ to $G_0^{\mathrm{g}\mathrm{r}}(A)$ is additive: the split sequence $0\to P\to P\oplus Q\to Q\to 0$ gives $[P\oplus Q]=[P]+[Q]$ by [F1]. Hence [F7] gives a unique homomorphism $c_A^{\mathrm{g}\mathrm{r}}$ with $c_A^{\mathrm{g}\mathrm{r}}([P])=[P]$.

The identities ${\sigma }_0={\tau }_0=1$, ${\sigma }_r{\sigma }_s={\sigma }_{r+s}$, and ${\tau }_r{\tau }_s={\tau }_{r+s}$ follow on generators from $(M\{s\})\{r\}=M\{r+s\}$ and $(P\{s\})\{r\}=P\{r+s\}$ in [F3]. Therefore $v^r\cdot x:={\sigma }_r(x)$ on $G_0^{\mathrm{g}\mathrm{r}}(A)$ and $v^r\cdot y:={\tau }_r(y)$ on $K_0^{\mathrm{g}\mathrm{r}}(A)$ extend by finite integer linear combinations to $\mathbb{Z}[v,v^{-1}]$-module structures.

On every projective-class generator, $c_A^{\mathrm{g}\mathrm{r}}(v[P])=c_A^{\mathrm{g}\mathrm{r}}([P\{1\}])=[P\{1\}]=v[P]=v{c}_A^{\mathrm{g}\mathrm{r}}([P])$ by [F3]. Since these generators generate $K_0^{\mathrm{g}\mathrm{r}}(A)$ as an abelian group and $v$ is invertible, $c_A^{\mathrm{g}\mathrm{r}}$ commutes with every Laurent polynomial action and is $\mathbb{Z}[v,v^{-1}]$-linear.

The ungraded class function $[P]\mapsto [P]$ from finite-dimensional projective left $A$-modules to $G_0(A)$ is additive by the split short exact sequence $0\to P\to P\oplus Q\to Q\to 0$ in ${\mathcal{M}}_{\mathrm{f}\mathrm{d}}(A)$. By [F7] it factors uniquely through the split group $K_0(A)$, giving $c_A([P])=[P]$.

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.

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.