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.
Best Answer
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.