Homological and internal shifts on K_0(C_m)

The homological shift is negation on $K_0$. By [L2] the triangle $X\to 0\to X[1]\stackrel{1}{\to }X[1]$ is distinguished for every object $X$ of $C_m$, so its relation $[0]=[X]+[X[1]]$ holds in $K_0(C_m)$ by [L3]; with $[0]=0$ from [L3] this gives $[X[1]]=-[X]$, and taking $X$ to be any object shows the stated identity for all of $K_0(C_m)$.

$\{1\}$ is an automorphism of $C_m$. By [F4] the internal shift acts termwise, fixes every differential, is the identity on morphisms shifted termwise, and satisfies $\{1\}\{-1\}=\{0\}=\mathrm{i}\mathrm{d}$; a bounded complex of finite graded projective modules is carried to a bounded complex of finite graded projective modules of the same supports, so $\{1\}$ is an automorphism of $C_m$ with inverse $\{-1\}$.

$\{1\}$ commutes with the homological shift. By [L1] and [F4] both $(X[1])\{1\}$ and $(X\{1\})[1]$ have $n$-th term $X^{n+1}\{1\}$, and both carry the differential $-d_X$, since neither the shift of complexes nor the internal shift changes $d_X$ except for the sign $(-1{)}^1$ of the homological shift; hence $(X[1])\{1\}=(X\{1\})[1]$ on the nose for every object, and similarly on morphisms.

$\{1\}$ commutes with cones. Let $f:X\to Y$ be a chain map in $C_m$. The complex $\mathrm{Cone}(f)\{1\}$ has $n$-th term $(Y^n\oplus X^{n+1})\{1\}=Y^n\{1\}\oplus X^{n+1}\{1\}$ with differential $d(y,x)=(d_{Y}y+fx,-d_{X}x)$, while $\mathrm{Cone}(f\{1\})$ has the same $n$-th term and differential $d(y^{\mathrm{\prime }},x^{\mathrm{\prime }})=(d_{Y\{1\}}{y}^{\mathrm{\prime }}+(f\{1\})x^{\mathrm{\prime }},-d_{X\{1\}}{x}^{\mathrm{\prime }})=(d_Y{y}^{\mathrm{\prime }}+f{x}^{\mathrm{\prime }},-d_X{x}^{\mathrm{\prime }})$ by [F4]; the identity on the two summands is therefore an isomorphism of complexes $\mathrm{Cone}(f)\{1\}\to \mathrm{Cone}(f\{1\})$, natural in $f$.

$\{1\}$ preserves distinguished triangles. By [L1] every distinguished triangle of $C_m$ is a cone triangle or an isomorphic image of one; by steps 2.1 and 2.2 the internal shift carries the cone triangle on $f$ to a triangle isomorphic to the cone triangle on $f\{1\}$, namely $X\{1\}\stackrel{f\{1\}}{\to }Y\{1\}\to \mathrm{Cone}(f\{1\})\to X\{1\}[1]$ via the isomorphisms of steps 2.1 and 2.2, and isomorphic images of distinguished triangles are distinguished by the first axiom of the triangulated structure. Hence $\{1\}$ is an exact automorphism of $C_m$.

The induced automorphism $q$ of $K_0(C_m)$. Put $q([X]):=[X\{1\}]$ on generators. This is well defined because isomorphic objects of $C_m$ have isomorphic shifts, the resulting map on the free abelian group on the classes is additive by construction, and it respects the defining relations, because a distinguished triangle is carried by step 3.1 to a distinguished triangle and the relation attached to the image is the image of the relation; hence $q$ is a well-defined endomorphism of $K_0(C_m)$. Applying the same construction to $\{-1\}$, which is also an automorphism by step 1.2, gives an endomorphism with $q{q}^{-1}=\mathrm{i}\mathrm{d}=q^{-1}q$ on generators by $\{1\}\{-1\}=\{0\}$, so $q$ is an automorphism of the abelian group $K_0(C_m)$.

Powers of $q$ realise all internal shifts. For $r\ge 0$ one has $q^r([X])=[X\{r\}]$ by induction on $r$, the case $r=0$ being $[X\{0\}]=[X]$ and the inductive step using $X\{r+1\}=(X\{r\})\{1\}$ and step 4.1; for $r<0$ one has $q^r=(q^{-1}{)}^{-r}$ and $X\{r\}$ is obtained by applying $\{-1\}$ exactly $-r$ times to $X$, so induction on $-r$ with the inverse automorphism gives $q^r([X])=[X\{r\}]$ for every $r\in \mathbb{Z}$.

The group-ring module structure. Let $G$ be the infinite cyclic group with generator $g$, so that $\mathbb{Z}[G]$ has basis $[g^r]$ with $r\in \mathbb{Z}$ and $[g^r][g^s]=[g^{r+s}]$ by [L5]. Define $\rho$ on basis elements by $\rho ([g^r]):=q^r$ and extend $\mathbb{Z}$-linearly; this is a well-defined $\mathbb{Z}$-linear map from the free group $\mathbb{Z}[G]$, and it is multiplicative on basis elements because $\rho ([g^r][g^s])=\rho ([g^{r+s}])=q^{r+s}=q^r{q}^s=\rho ([g^r])\rho ([g^s])$, hence multiplicative on all of $\mathbb{Z}[G]$ by distributivity, and it is unital because $\rho ([g^0])=q^0=\mathrm{i}\mathrm{d}$. Therefore $\rho :\mathbb{Z}[G]\to \mathrm{End}(K_0(C_m))$ is a unital ring homomorphism and $K_0(C_m)$ is a module over the group ring of the infinite cyclic group generated by $q$, whose elements are the Laurent polynomials $\mathbb{Z}[q,q^{-1}]$; under this structure the generator acts by $q$, that is $q\cdot [X]=[X\{1\}]$.

Conclusion. The homological shift acts by $[X[1]]=-[X]$ by step 1.1, so the homological shift induces negation on $K_0(C_m)$; the internal shift $\{1\}$ is an exact automorphism of $C_m$ commuting with $[1]$ and with cones by steps 2.1 and 2.2, it preserves distinguished triangles by step 3.1, and it induces the automorphism $q$ of $K_0(C_m)$ of step 4.1 with inverse induced by $\{-1\}$; the powers of $q$ realise all internal shifts by step 5.1, and $\mathbb{Z}[G]$ acts through step 6.1. Consequently $K_0(C_m)$ is a module over the group ring generated by $q$ with $[X\{r\}]=q^r[X]$ and $[X[1]]=-[X]$, the two shifts being distinguished exactly by the sign $-1$ and the invertible parameter $q$. No skeleton is used, no representative of an isomorphism class is selected, and no choice principle is invoked.