Prove that $\mathrm{rank}(A)=\dim(\mathbb{Q}\otimes_{\mathbb{Z}}A)$ Call a subset $X$ of an abelian group $A$ independent if, whenever $\Sigma m_ix_i=0$, where $m_i\in \mathbb{Z} $ and almost all $m_i =0$, then $m_i = 0$ for all $i$. Define $\mathrm{rank}(A)$ to be the number of elements in a maximal independent subset of $A$.

Prove that $\mathrm{rank}(A)=\dim(\mathbb{Q}\otimes_{\mathbb{Z}}A)$ and conclude that every two maximal independent subsets of $A$ have the same number of elements.

My attempt: $\mathrm{rank}(A)$ is the free part (non-torsion) of $A$. So, is that $\mathbb{Q}\otimes_{\mathbb{Z}}A$ kill the torsion part of $A$? And why.

A priori, the general element of $\Bbb Q\otimes A$ is a rational linear combination of elements of the form $q\otimes a$ with $q\in\Bbb Q$ and $a\in A$. As $$q\otimes a=q\cdot 1\otimes a$$ and $$\frac nm\cdot 1\otimes a+\frac rs\cdot 1\otimes b=\frac 1{ms}\cdot 1\otimes(nsa+rmb)$$
we can write each element of $\Bbb Q\otimes A$ more specifically in the form $q\cdot 1\otimes a$ with $q\in \Bbb Q$ and $a\in A$.

If $\{x_i\}_{i\in I}$ are independent in $A$, then $\{1\otimes x_i\}_{i\in I}$ are linearly independent in $\Bbb Q\otimes A$. Indeed, if $\sum q_i\cdot 1\otimes x_i=0$ (with almost all $q_i=0$), then with $N$ as common denominator of all $q_i$, we have $n_i:=Nq_i\in\Bbb Z$ and $$\begin{align}0&=N\sum (q_i\cdot 1\otimes x_i)\\&=\sum(n_i\cdot 1\otimes x_i)\\&=\sum 1\otimes n_ix_i\\&=1\otimes\sum n_ix_i\end{align}$$
and conclude that $M\cdot \sum n_ix_i=0$ holds in $A$ for some integer $M$. Then with $m_i:=Mn_i$, $\sum m_ix_i=0$ and so all $m_i$ are $=0$ and also all $q_i=0$, as was to be shown.

Conversely, let $\{\alpha_i\}_{i\in I}$ with $\alpha_i\in\Bbb Q\otimes A$ be linearly independent. As seen above, we can write $\alpha_i=q_i\cdot 1\otimes a_i$ with $a_i\in A$ and $q_i\in \Bbb Q$. Of course, $q_i\ne 0$ for our linearly independent family. If we multiply each $\alpha_i$ with a non-zero reatioal, we still have a linearly independent family. Hence we may assume wlog $\alpha_i=1\otimes a_i$.
Then the $\{a_i\}_{i\in I}$ are independent. Indeed, if $\sum m_ia_i=0$ (with almost all $m_i=0$), then 
$$\begin{align}0&=1\otimes 0
\\&=1\otimes \sum m_ia_i\\
&=\sum 1\otimes m_ia_i\\
&=\sum m_i\cdot 1\otimes a_i\\&=\sum m_i\alpha_i  \end{align}$$
and hence all $m_i=0$, as was to be shown.

Specifically, if $a\in A$ is a torsion element and $ma=0$ for some non-zero integer $m$, then 
$$1\otimes a=\frac 1m\otimes ma=\frac1m\otimes 0=0. $$
