The vertex modules S_i and their prime quotients

The action is a well-defined ring homomorphism on the path ring. Let $R=\mathbb{Z}{\mathrm{\Gamma }}_m$ and define $\phi :R\to {\mathrm{End}}_{\mathbb{Z}}(\mathbb{Z})$ on the free basis of paths by $\phi (p)={\delta }_{p,e_i}{\mathrm{i}\mathrm{d}}_{\mathbb{Z}}$, the Kronecker symbol being $1$ exactly when $p$ is the vertex path $e_i=(i)$. This is additive, and it is multiplicative: for composable paths $p,q$ the product $pq$ is the concatenated path when the endpoints match and $0$ otherwise, and $pq=e_i$ happens exactly when $p=q=e_i$, because path length is additive over concatenation and a product of two paths is a vertex path only if both factors are vertex paths; comparing with $\phi (p)\phi (q)={\delta }_{p,e_i}{\delta }_{q,e_i}\mathrm{i}\mathrm{d}$ gives $\phi (pq)={\delta }_{pq,e_i}\mathrm{i}\mathrm{d}=\phi (p)\phi (q)$, while for a noncomposable pair both sides are $0$. Hence $\phi$ is a ring homomorphism, and $\phi ({\sum }_j(j))={\mathrm{i}\mathrm{d}}_{\mathbb{Z}}$ shows that it is unital, so it makes $\mathbb{Z}$ a left $R$-module by [F5]; in this action the vertex idempotents act by ${\delta }_{j,i}\mathrm{i}\mathrm{d}$ and every positive-length path acts by $0$.

Descent to $A_m$ and degree-zero grading. Every generator of the ideal $I_m$ displayed in [F1] is a length-two path or a difference of two length-two paths, hence is sent to zero by $\phi$; since $I_m$ is the two-sided ideal generated by these elements and $\phi$ is multiplicative, $\phi (I_m)=0$, and by [F5] the homomorphism factors through the quotient $A_m=\mathbb{Z}{\mathrm{\Gamma }}_m\mathrm{/}{I}_m$, making $\mathbb{Z}$ a left $A_m$-module on which the class of a path acts by ${\delta }_{p,e_i}\mathrm{i}\mathrm{d}$; in particular $e_i$ acts as the identity and every other vertex idempotent and every positive-length path class acts as zero. This action is graded in the sense of [L3]: $A_m$ is spanned over $\mathbb{Z}$ by vertex classes and positive-length path classes by [F1], the positive-length classes act by $0$ and the vertex classes lie in degree $0$ with $A_{m,0}(S_i{)}_0\subseteq (S_i{)}_0$ and $A_{m,d}(S_i{)}_0=0\subseteq (S_i{)}_d$ for $d\mathrm{\ne }0$. The element $1$ generates $S_i$ over $A_m$ because $e_i\cdot 1=1$, so $S_i$ is a finitely generated graded left $A_m$-module, and $S_i\mathrm{\ne }0$ because $1\mathrm{\ne }0$.

The prime quotient and its exactness. By [L3] the submodule $p{S}_i=A_{m}p$ is graded with $(p{S}_i{)}_0=p\mathbb{Z}$ and $(p{S}_i{)}_d=0$ for $d\mathrm{\ne }0$, and the quotient $S_i\mathrm{/}p{S}_i$ is a graded left $A_m$-module with $(S_i\mathrm{/}p{S}_i{)}_0=\mathbb{Z}\mathrm{/}p\mathbb{Z}$, zero in other degrees, and the induced action in which $e_i$ is the identity, every other vertex idempotent and every positive-length path is zero. By [L4] the sequence $0\to S_i\stackrel{p}{\to }{S}_i\to S_i\mathrm{/}p{S}_i\to 0$ is exact if and only if it is exact degreewise, and degreewise it is the sequence $0\to \mathbb{Z}\stackrel{p}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}p\mathbb{Z}\to 0$ in degree $0$ together with the sequence $0\to 0\to 0\to 0$ in every other degree. The latter is exact. For the former, multiplication by $p$ is injective because $pa=0$ with $p\mathrm{\ne }0$ forces $a=0$ by [F6]; its image is $p\mathbb{Z}=\ker (\mathbb{Z}\to \mathbb{Z}\mathrm{/}p\mathbb{Z})$; and the quotient map is surjective.

Conclusion. For every $0\le i\le m$ the assignment displayed in the Definition descends to a well-defined action of $A_m$, making $S_i$ a finitely generated nonzero graded left $A_m$-module with $e_i$ acting as the identity, every other vertex idempotent acting as zero and every positive-length path acting as zero (steps 1.1 and 1.2); for every prime $p$ the quotient $S_i\mathrm{/}p{S}_i$ is the $\mathbb{Z}\mathrm{/}p$ vertex module and the sequence $0\to S_i\stackrel{p}{\to }{S}_i\to S_i\mathrm{/}p{S}_i\to 0$ is degreewise exact (step 1.3); and the internal shifts $S_i\{r\}$ are the shifts of $A_m\text{-mod}$ (Definition).