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Let $M$ and $N$ be graded $R$-modules (with $R$ a graded ring). $\varphi:M\rightarrow N$ is a homogeneous homomorphism of degree $i$ if $\varphi(M_n)\subset N_{n+i}$. Denote by $\mathrm{Hom}_i(M,N)$ the group of homogeneous homomorphisms of degree $i$. We define $^*\mathrm{Hom}_R(M,N)=\bigoplus_{i\in\mathbb{Z}}\mathrm{Hom}_i(M,N)$. This is a (graded) $R$-submodule of $\mathrm{Hom}_R(M,N)$.

How can I prove that these two modules are equal if $M$ is finite? And do you know a counterexample if $M$ is not finite?

Wesley Farrel
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  • For the counterexample I have no idea. For the first question I believe to have an answer but I'm not sure. Here it is: $M=<x_1,\ldots,x_n>R$, take $\varphi\in\mathrm{Hom}_R(M,N)$. Then $\varphi(x_i)=n{i1}+\cdots+n_{im_{i}}$ where $n_{ij}\in N_j$ (for semplicity I took all the homogeneous elements in $N$ of positive degree). If we adjoin zero to the decomposition of $\varphi(x_i)$ we can suppose that they are of the form $\varphi(x_i)=n_{i1}+\cdots+n_{im}$. If we define $\varphi_j(x_i)=n_{ij}$ then $\varphi(x_i)=\sum^m_{j=1}\varphi_j(x_i)$ and so $\varphi=\sum^m_{j=1}\varphi_j$. – Wesley Farrel Feb 02 '11 at 16:28
  • Well, maybe I have the conterexample: take $R=\mathbb{Z}$ and $M=\mathbb{Z}[x]$. As an $R$-module $M$ is not finite. Take $N=M$ and $id\in\mathrm{Hom}_R(M,M)$. If $id=\varphi_1+\ldots+\varphi_n$ with $\varphi_i$ homogeneous of degree $i$, then for all $j$ we have $x^j=\varphi_1(x^j)+\cdots+\varphi_n(x^j)$, if this equality holds for all $j$ we have a contradiction. – Wesley Farrel Feb 02 '11 at 16:49
  • In the previous comment I'm assuming $\mathbb{Z}$ graded with the trivial grading. – Wesley Farrel Feb 02 '11 at 16:54
  • In my first comment I'm not sure that the homomorphisms $\varphi_j$ are homogeneous of some degree. – Wesley Farrel Feb 02 '11 at 16:58
  • Wesley, the identity map of every graded module is a morphism of graded modules, of degree zero. – Mariano Suárez-Álvarez Feb 02 '11 at 17:02
  • Yes...I forgot the grade 0, I believe that also my proof of the first question is wrong, because $\varphi_j$ maps all $M$ in $N_j$. – Wesley Farrel Feb 02 '11 at 17:07
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    maybe $\varphi\in\mathrm{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z}[x])$ such that $\varphi(x^i)=x^{2i}$ works – Wesley Farrel Feb 03 '11 at 16:26

1 Answers1

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I will just take a guess that you mean $M$ is finitely generated, not finite.

Pick a set of homogeneous generators $g_1, \ldots, g_n$ of $M$ such that $g_i \in M_{m_i}$. For any $g_i$, $\varphi(g_i) \in \bigoplus_{j=1}^{k_m} N_{n_{i,j}}$, i.e., degrees seen by applying $\varphi$ to $g_i$ are $n_{i,j} - m_i$. Since $\varphi$ is completely determined by $g_i$, we know that $\varphi \in \sum_{i,j} \hbox{Hom}_{n_{i,j} - m_i}(M, N)$. (I use "sum" because there might be duplicates.)

Tunococ
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