So there's a statement in Lang that I would like to understand better.
It's contained in his proof of the following statement:
Every finite abelian p-group is isomorphic to a product of cyclic p-groups. If it is of type $(p^{r_1},...,p^{r_2})$ with $$r_1\geq r_2\geq\cdots\geq r_s>1$$*then the sequence of integers $(r_1,...,r_s)$ is uniquely determined.*
So what I would understand is this part of his proof:
We want to establish $$A=A_1\oplus\cdots\oplus A_s$$ where each $A_i$ are cyclic subgroups of order given in the statement of the problem
Conversely, suppose that $m_1,...,m_s$ are integers $\geq0$ such that $$0=m_1x_1+\cdots+m_sx_s$$Since a_i has period $p^{r_i}$(i=1,...,s), we may suppose that $m_i<p^r{_i}$. Putting a bar onthis equation yields $$0=m_2\overline{x_2}+\cdots+m_s\overline{x_s}$$Thus $m_i=0$ for all $i=2,...,s$. But then$m_1=0$. From this it follows at once that $(A_1+\dots+A_i)\cap A_{i+1}=0$
Question: I just want to make sure I understand the last statement of the proof correctly. So if the only way we can write 0 is if the all coefficients $m_i=0$ , then it must follow that $(A_1+\cdots+ A_i)\cap A_{i+1}=0$, why is that?
Suppose $\;x\in\left(A_1+\ldots+A_i\right)\cap A_{i+1}\;$ , then
$$\begin{cases}x\in A_1+\ldots + A_i\implies& x=m_1a_1+\ldots +m_ia_i\\{}\\x\in A_{i+1}\implies& x=m_{i+1}a_{i+1}\end{cases}$$
and from both equations above we get
$$m_1a_1+\ldots +m_ia_i-m_{i+1}a_{i+1}=0\iff m_k=0\;\;\forall\;k=1,2,...,i+1$$
and in particular, $\;x=m_{i+1}a_{i+1}=0\cdot a_{i+1}=0\;$