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Let $M$, $N$ be $R$-graded modules, say: $$M= \bigoplus_{i\in\mathbb Z} M_{i}, N= \bigoplus_{j\in\mathbb Z} N_{j}.$$
Then $$\operatorname{Hom}(M,N) \cong \prod_{i\in\mathbb Z} \bigoplus_{j\in\mathbb Z}\operatorname{Hom}(M_{i},N_{j})$$ and if $\phi:M\to N $ then it can be decomposed as $$\phi= \prod_{i \in Z} \bigoplus_{j\in Z}\phi_{j}^{i}$$ with $\phi^{i}_{j}:M_{i}\to N_{j}$

I have tried a lot to prove it and to define $\phi^{i}_{j}$ but without any success :(

Any response will be greatly appreciated.

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Let $\imath_i: M_i \to M$ the inclusion and $\pi_j: N \to N_j$ to projection. Then $\phi_j^i = \pi_j \circ \phi \circ \imath_i$. –  martini Mar 27 '12 at 19:34

2 Answers 2

The formula is incorrect. Take for instance $\mathbb{Z}$ as ground ring, and $S$ to be the set of all sequences of integers indexed by integers with only finitely many nonzero terms: $$S=\bigoplus_{i\in\mathbb{Z}} \mathbb{Z}$$ consider now $M$ with $M_0=S$ and all other $M_i=0$. There is a natural map $\psi:M\rightarrow S$ which sends every sequence in $M_0=S$ to the sum of its terms in the corresponding degrees: $$\psi(s=(\dots,0,s_{-N},\dots,s_{N},0,\dots))=\sigma=(\sigma_i)_{i\in\mathbb{Z}}$$ where for each $i\in\mathbb{Z}, \sigma_i=s_i$... Doesn't this look silly? Have I gone to such lengths to define the identity morphism? $\psi$ is indeed the identity morphism between the two modules, but it is absolutely not the identity morphism between the two graded modules: by construction, $M$ is concentrated in degree $0$ while $S$ has a copy of the integers in every rank. In any case, $\psi$ is by no means a finite sum of morphisms from $M_0$ to different degrees of the graded group $S$.

What this means, is that in general you don't have that a morphism into a direct sum decomposes as a sum of morphisms into each component of the target space. However, if you assume that for all $i$ the group $M_i$ is finitely generated as a $R$-module, then the formula is true!

For completeness sake, you should remember that $$\mathrm{Hom}_R(\bigoplus_{i\in I} M_i, N)\simeq\prod_{i\in I}\mathrm{Hom}_R(M_i,N)$$ and $$\mathrm{Hom}_R(M,\prod_{i\in I} N_i)\simeq \prod_{i\in I}\mathrm{Hom}_R(M,N_i)$$ and whenever $M$ is finitely generated as a $R$-module, then $$\mathrm{Hom}_R(M,\bigoplus_{i\in I} N_i)\simeq\bigoplus_{i\in I}\mathrm{Hom}_R(M,N_i), $$ while generally theright hand side may be a strict subset of the left hand side.

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I think the graded module structure just falls naturally $$\phi(R_{i}M_{j})=R_{i}(\phi(M_{j}))=R_{i}(\bigoplus \phi_{k}[M_{j}])$$ with $\phi_{k}$ be the projection of $\phi$ onto $N_{k}$. Of course you may just define $\phi^{i}_{j}=\pi \circ \phi \circ i$ with $\pi$ be the projection $N\rightarrow N_{j}$, $i$ be the inclusion $M_{i}\rightarrow M$, etc.

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I think your answer might be incorrect :S –  Olivier Bégassat May 8 '12 at 21:20
    
@OlivierBégassat: I am not bothered with this question anymore. I do not mind if it is correct or incorrect. –  Kerry May 8 '12 at 23:24

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