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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.

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2 Answers 2

up vote 4 down vote accepted

If $A$ is a (commutative) $\newcommand\ZZ{\mathbb Z}\ZZ^n$-graded ring and $M$ and $N$ are $\mathbb Z^n$-graded $A$-modules, we can consider the $A$-module $\hom_A(M,N)$ of all $A$-linear maps. For each $g\in\ZZ^n$ we can look at the subset $\hom_A(M,N)_g$ of all $A$-linear maps $f:M\to N$ such that $f(M_h)\subseteq N_{h+g}$ for all $h\in\ZZ^n$: we call the elements of $\hom_A(M,N)_g$ the homogeneous $A$-linear maps of degree $g$.

It is easy to see that the sum $\hom_A(M,N)_{\text{homog}}=\bigoplus_{g\in\ZZ^n}\hom_A(M,N)_g$ is a direct sum, giving a $A$-submodule of $\hom_A(M,N)$.

If $M$ is a finitely generated $A$-module, then we have $\hom_A(M,N)_{\text{homog}}=\hom_A(M,N)$. In general, though, $\hom_A(M,N)_{\text{homog}}$ is a proper submodule of $\hom_A(M,N)$.

In what you are reading, it is most likely that $\hom(M,N)$ denotes $\hom_A(M,N)$.

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Thank you very much! I guess I understand your argument, and I need to go bed to sleep now. (Taiwan times). I will check some details you tell to me. I really appreciate your help. –  Peter Hu Jun 6 '12 at 16:52
4  
Don't dream of graded modules! –  Mariano Suárez-Alvarez Jun 6 '12 at 16:58
    
I try to write down the proof in the following note: dropbox.com/s/ptcx1cx58x8io0u/Tamafumi2.pdf Does I checked correctly? Thanks you! –  Peter Hu Jun 7 '12 at 12:10

I dont know if this will help but, Let $A$ be a graded ring, $N$ a graded $A$ module, and $M$ an $A$-module. The submodules $Hom_A(M,N_n)$ of $Hom_A(M,N)$ define on $Hom_A(M, N)$ a graded module structure.

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Please ignore this, it is just a comment and not an answer. I am new to this and accidentally put it here, instead of just making it as a comment. –  messi Jun 6 '12 at 16:46
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But your answer also help me to think this problem! Thank you very much! –  Peter Hu Jun 6 '12 at 16:54

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