Definitions: Let $A$ be a commutative and unitary ring and $M$ be an $A–$module. We will call the submodule the sum of a family of submodules $\displaystyle\{N_i\}_{i∈I}$ of $M$:
$\displaystyle \sum _{i\in I}N_i=\left <\bigcup _{i\in I}N_i\right >$
Comments before the statement: If we have a family of $A–$modules $\{M_i\}_{i\in I}$ , the Cartesian product $\prod_{i\in I}M_i$ is an $A–$module with the operations defined by: $(x_i)_{i\in I}+(y_i)_{i\in I}=(x_i+y_i)_{i\in I}\ \ , \ \ r.(x_i)_{i\in I}=(r.x_i)_{i\in I}$.
Likewise, maps $\displaystyle\iota_i:M_i \rightarrow \prod_{i\in I}M_i$ given by:
$\iota_i(m)(j)=m$ if $j=i$ and $\iota_i(m)(j)=0$ if $j\not =i$ are monomorphismof modules. But by identifying each module $M_i$ with its image it is no longer true that $\displaystyle \sum _{i\in I}M_i=\prod_{i\in I}M_i$ but $\displaystyle \sum _{i\in I}M_i=\{f\in \prod_{i\in I}M_i|\{i\in I|f(i)\not = 0\}$ is finite $ \}$
Show:
$\displaystyle \sum _{i\in I}M_i=\{f\in \prod_{i\in I}M_i|\{i\in I|f(i)\not = 0\}$ is finite $ \}$
I have tried to prove it without success. Is there any way to prove it following typical reasoning? I mean that it assumes that an element $x$ belongs to the set on the left, then it also belongs to the set on the right, and vice versa.
This is what I have tried:
If $f\in \{f\in \prod_{i\in I}M_i|\{i\in I|f(i)\not = 0\}$ is finite $ \}$. Then $\exists n\in\mathbb N\ \exists J\in \mathscr PI \ \forall i\in I (f(i)\in M_i\ \land |J|=n\ \land (f(i)\not=0 \longleftrightarrow i\in J)$ . So $\exists \alpha_f \exists n\in\mathbb N\ \exists J\in \mathscr PI \ \forall i\in I(f(i)\in M_i\ \land |J|=n\ \land\alpha_f=\displaystyle \sum _{i\in J}f(i))$. Then $\alpha_f \in \displaystyle \sum _{i\in J}M_i\subset \sum _{i\in I}M_i$. In this way, an element from the set on the right has been "identified" with an element from the set on the left. Now suppose that $\alpha \in \displaystyle \sum _{i\in I}M_i$. Then $\exists n\in\mathbb N\ \exists J\in \mathscr PI \ \forall i\in I (f_i\in M_i\ \land |J|=n\ \land (f_i\not = 0 \longleftrightarrow i\in J) \land (\alpha =\displaystyle \sum _{i\in J}f_i))$. So $\displaystyle \exists f_\alpha \in \prod_{i\in I}M_i\ \exists n\in\mathbb N\ \exists J\in \mathscr PI \ \forall i\in I(f_\alpha(i)=f_i\in M_i\land |J|=n\ \land (f_\alpha(i)\not =0\longleftrightarrow i\in J))$. Then $f_\alpha \in \{f\in \prod_{i\in I}M_i|\{i\in I|f(i)\not = 0\}$ is finite $ \}$. In this way, an element from the set on the left has been "identified" with an element from the set on the right.
Let $\{M_\lambda\}_{\lambda\in\Lambda}$ be a family of $R$-modules over a ring $R$, with $\Lambda$ a (non necessarily finite) set. Set $M:=\prod_{\lambda\in\Lambda}M_\lambda$ and, if $m\in M$, let $m_\lambda$ be its (image via the) projection on $M_\lambda$.
Define the map $i_\lambda:M_\lambda\to M$ as follows: for $n\in M_\lambda$, set $i_\lambda(n)_\lambda=n$ and $i_\lambda(n)_\mu=0$ if $\mu\neq \lambda$. Clearly $i_\lambda$ is injective, so we can view $M_\lambda$ as a submodule of $M$.
Let $S\subset M$ be the submodule of elements $m\in M$ such that $m_\lambda\neq 0$ only for finitely many $\lambda$. Then $M_\lambda\subset S$ for all $\lambda$, so that $\sum_{\lambda\in\Lambda}M_\lambda\subset S$. Conversely, take $m\in S$, and let $\lambda_1,\dots,\lambda_n\in\Lambda$ be the (finitely many) elements with $m_{\lambda_1},\dots,m_{\lambda_n}\neq 0$. Then $m=m_{\lambda_1}+\cdots+m_{\lambda_n}$, so that $$m\in \sum_{i=1}^nM_{\lambda_i}\subset\sum_{\lambda\in\Lambda}M_\lambda .$$