Indecomposable projective classes form a basis of split K0

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.