If $M,N$ are graded modules over a graded ring $R$, then is $\operatorname{Hom}_{R}(M,N)$ also a graded module and how?
Is the module of homomorphisms between graded modules also a graded module?
2
$\begingroup$
modules
graded-modules
1 Answers
4
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.
-
0Is $Hom_R(M,N)$ 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. – 2012-06-06
-
0@Peter: I don't understand what you mean by "[i]s $\text{Hom}_R(M, N)$ true". – 2012-06-06
-
0Dear @Qiaochu Yuan I think I need to arise a new question because there only allow 5xx characters... – 2012-06-06
-
0Dear @Qiaochu Yuan my question post in http://math.stackexchange.com/questions/154715/if-m-and-n-are-graded-modules-what-is-the-graded-structure-on-homm-n Thank you very much – 2012-06-06
-
0My 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. – 2012-06-06
-
0@Peter: if $M, N$ have graded components $M_j, N_j$, then the $i^{th}$ graded component of the graded Hom consists of sequences of morphisms $d_j : M_j \to N_{j+i}$ (respecting the action of $R$). An easier setting in which to understand internal Homs is the category of representations of a group $G$; here we have the ordinary Hom (intertwining operators) but the internal Hom is the space of all linear maps between two representations $V, W$ and it is canonically also a representation of $G$ (if $V$ is finite-dimensional it is just $V^{\ast} \otimes W$). The ordinary Hom is the invariant... – 2012-06-06
-
0...subspace of the internal Hom. Taking $G = \mathbb{Z}$ gives us $\mathbb{Z}$-graded vector spaces and then one can see the direct relevance to this situation. – 2012-06-06
-
0If the domain is f.g. then there is a natural grading on the hom, so in that case the answer is yes, actually. If the domain is free of infinite rank, for example, there is no grading giving the obviou degree to the elements one wants. – 2015-10-16