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If $M,N$ are graded modules over a graded ring $R$, then $\operatorname{Hom}_{R}(M,N)$ is also a graded module and how ?

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If by $\text{Hom}_R$ you mean graded homomorphisms (those that preserve the grading), then no. However, there is a "graded Hom" where the $i^{th}$ graded component consists of homomorphisms which raise degree by $i$, and the zeroth graded component of the graded Hom is the ordinary Hom. The general keyword here is internal Hom.

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Is $Hom_R(M,N)$ true for all ordinary $R$-module homomorphism (doesn't require preserve grade)? I can't understand you answer, but I think your answer is correct and useful. Could you write down details about your answer? Thanks a lot. –  Peter Hu Jun 6 '12 at 10:10
    
@Peter: I don't understand what you mean by "[i]s $\text{Hom}_R(M, N)$ true". –  Qiaochu Yuan Jun 6 '12 at 14:46
    
Dear @Qiaochu Yuan I think I need to arise a new question because there only allow 5xx characters... –  Peter Hu Jun 6 '12 at 15:04
    
Dear @Qiaochu Yuan my question post in math.stackexchange.com/questions/154715/… Thank you very much –  Peter Hu Jun 6 '12 at 15:24
    
My main question is that I don't understand the meaning of $i^{th}$ graded component consists of homomorphisms which raise degree by $i$. Could you tell me? Thank you very much. –  Peter Hu Jun 6 '12 at 15:42
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