Directness of a sum of modules I have  found two different definitions and I wonder if they are equivalent:
Choose $(M_i)_{i\in I}$ a family of submodules of an $R$-Module $M$.


*

*The sum of the $M_i$ over all $i\in I$ is derect.

*For every finite subset $F\subset I$ the sum of all $M_i$ with $i\in F$ is direct.


So, is this 'the same'? What is the idea of the proof and/or any references?
 A: For $J \subseteq I$, let $M_J$ be defined as the direct sum
$$\bigoplus\limits_{i \in J} M_i$$
By definition, this is the $R$-submodule of the cartesian product $\prod\limits_{i \in J}M_i$ consisting of all elements $(m_i)$ such that $m_i = 0$ for all but finitely many $i \in J$.
Since I'm not sure of your definition, let me use the use the following definition: The sum of the $M_i : i \in J$ is direct if and only if the $R$-module homomorphism $\phi_J: M_J \rightarrow M$ given by 
$$(m_i) \mapsto \sum\limits_i m_i$$
is injective (equivalently, is an isomorphism onto its image $\sum\limits_i M_i$, the submodule generated by the $M_i$).
Assume that for each finite subset $F \subseteq I$, the sum of the $M_i : i \in F$ is direct, i.e. $\phi_F$ is injective.  Let $m = (m_i) \in M_I$, and suppose that $\phi_I(m) = 0$.  We want to show that $m$ is zero, i.e. all the $m_i$ are zero.  Already, we know that all the $m_i$ are zero except for those $i$ in a finite set $F = \{i_1, ... , i_r\}$.
Let $$n = (m_{i_1}, ... , m_{i_r})$$ which is an element of $M_F$.  Our hypothesis is that 
$$0 = \phi_I(m) = \sum\limits_i m_i = \sum\limits_{j=1}^r m_{i_j} = \phi_J(n)$$
and since $\phi_J$ is injective, this means that all the $m_{i_j}$ are zero, as required.
A: The sum of the submodules is direct if the obvious homomorphism
$$
\sigma_I\colon\bigoplus_{i\in I}M_i\to \sum_{i\in I}M_i
$$
is an isomorphism or, in other words, an element $x=\sum_{i\in I}M_i$ can be written in a unique way as
$$
x=\sum_{i\in I}x_i \quad\text{with}\quad x_i\in M_i,\text{ for } i\in I
$$
Note that $\sigma_I$ is surjective by construction, so directness is the same as saying that $\sigma_I$ is injective.
Suppose the sum is direct and consider the diagram
$$\require{AMScd}
\begin{CD}
\bigoplus_{i\in F}M_i @>\sigma_F>> \sum_{i\in F}M_i \\
@VVV @VVV \\
\bigoplus_{i\in I}M_i @>\sigma_I>> \sum_{i\in I}M_i
\end{CD}
$$
Since the vertical arrows are injective and $\sigma_I$ is an isomorphism, also $\sigma_F$ is injective.
Conversely, if the sum is not direct, take a nonzero element $z$ in the kernel of $\sigma_I$; then this element belongs to the image of some injection
$$
\bigoplus_{i\in F}M_i\to \bigoplus_{i\in I}M_i
$$
so also the sum restricted to $F$ is not direct. Say that
$$
z=(x_i)_{i\in I}
$$
and consider $F=\{i\in I:x_i\ne0\}$, which is a finite set. Then $z'=(x_i)_{i\in F}$ is sent to $z$ by the canonical embedding. Since $\sigma_I(z)=0$, also $\sigma_F(z')=0$.
