Let $A$ be a graded ring. Note that the grading of $A$ may not be $\mathbb{N}$, for example, the grading of $A$ could be $\mathbb{Z}^n$. Actually, my question comes from the paper of Tamafumi's On Equivariant Vector Bundles On An Almost Homogeneous Variety, Proposition 3.4. And I translate this proposition to modern language:
Let $A = \mathbb{C}[\sigma^{\vee} \cap \mathbb{Z}^n]$. If $M$ is a finitely generated $\mathbb{Z}^n$-graded $A$-projective module of rank $r$, then there exists $u_1,u_2,\dots,u_r$ in $\mathbb{Z}^n$ such that \begin{eqnarray*} M \simeq A(-u_1) \oplus A(-u_2)\oplus \dots \oplus A(-u_r) \end{eqnarray*} as $\mathbb{Z}^n$-graded $A$-module. In particular, $M$ is an $A$-free module.
(page 71) He says since $\operatorname{Hom}(\widetilde{M},\widetilde{F})$ is $T$-linearized vector bundle, $\operatorname{Hom}(M,F)$ is a $\mathbb{Z}^n$-graded A-module.
My Questions: I don't know what this statement means. I know why we need to show that $\operatorname{Hom}(M,F)$ is a $\mathbb{Z}^n$-graded $A$-module, but I don't understand his reason. What is the "grading" of "$\operatorname{Hom}(M,F)$"? What does "$\operatorname{Hom}(M,F)$" mean? "$\operatorname{Hom}_A(M,F)$", "$\operatorname{Hom}_{\mathbb{Z}}(M,F)$" or what? I think it is just a purely algebraic question, so why do we need to use a "vector bundle"? I feel uncomfortable about this question.