Q:

For $A_j$ abelian, $\text{rank } (\oplus_{j \in J} A_j )= \sum_{j \in J}\text{rank } (A_j)$

Let $A_j$ be abelian for $ j \in J$.
Prove: $\text{rank } (\oplus_{j \in J} A_j )= \sum_{j \in J}\text{rank }  A_j$

My reasoning is this:
Let $A = \text{rank } (\oplus_{j \in J} A_j )$, where $a \in A \Rightarrow a = (a_j)_{j\in J}$.  Let $F$ be the subgroup generated by a maximally independent subset.  Then $F$ is a subset of $A$ and therefore, for $a^{(i)} \in F$,
$\sum_i m_i a^{(i)} = 0 \Rightarrow \sum m_i (a^{(i)}_j)_{j \in J} = 0 \Rightarrow \sum (m_ia^{(i)}_{j \in J}) = 0 \Rightarrow \sum m_i a^{(i)}_j = 0, \forall j \in J$ and $m_i = 0, \forall i$.
But then doesn't this mean that $(a_1, 0, 0, \dots, 0), \dots(\dots), \dots, (a_{|F_{A_{J_1}}|}, 0, 0, \dots, 0)$ is in $F$? And therefore you can repeat this $n_1 \times n_2 \times \dots $ times and instead of it being equal to a sum of the ranks it's equal to a product of the ranks?
What am I misunderstanding here?

A:

I don't understand your argument, maybe reading the following proof will help.
Take $B_j\subseteq A_j$ a maximal linearly independent subset. Consider $B = \bigsqcup_{j\in J} i_j(B_j)$ where $i_j\colon A_j \to A$ is the canonical inclusion. We claim that $B$ is a maximal linearly independent subset. 
First, we check that $B$ is a linearly independent subset. Indeed, the projection in any coordinate $j$ of a linear relation in $B$ will yield a linear relation in $B_i$.
Second, for maximality, if $B' \supsetneq B$, then there exists some $j$ so that $\pi_j(B'\setminus B)\neq 0$, where $\pi_j\colon A\to A_j$ is the standard projection. However, this contradicts maximality of $B_j$.
I guess that your problem is that you might be thinking that projecting a maximal linearly independent subset of $A$ will yield a maximal linearly independent subset of $A_j$. This is incorrect, since the projection has kernel! 

