TITLE: Showing a subgroup of $\mathbb{R}^n$ is a free abelian group on $\leq n$ generators
QUESTION [8 upvotes]: I'm a bit stuck on Exercise III.5 of Lang's Algebra. (Page 166.)

Let $A$ be an additive subgroup of Euclidean space $\mathbb R^n$, and assume that in every bounded region of space, there is only a finite number of elements of $A$. Show that $A$ is a free abelian group on $\leq n$ generators. 
[Hint: Induction on the maximal number of linearly independent elements of $A$ over $\mathbb R$. Let $v_1, \ldots, v_m$ be a maximal set of such elements, and let $A_0$ be the subgroup of $A$ contained in the $\mathbb R$-space generated by $v_1, \ldots, v_{m-1}$. By induction, one may assume that any element of $A_0$ is a linear integral combination of $v_1, \ldots, v_{m-1}$. Let $S$ be the subset of elements $v \in A$ of the form $v= a_1 v_1 + \cdots + a_m v_m$ with real coefficients $a_i$ satisfying 
  $$
\begin{eqnarray*}
0 \leq a_i < 1, &\ \ \text{if } i=1,2,\ldots, m-1
\\
0 \leq a_m \leq 1 .&
\end{eqnarray*}
$$
  If $v'_m$ is an element of $S$ with the smallest $a_m \neq 0$, show that $\{ v_1, \ldots, v_{m-1}, v'_m \}$ is a basis of $A$ over $\mathbb Z$.]

Following the hint, I suppose $c_1v_1+\cdots+c_{m-1}v_{m-1}+c_mv'_m=0$ for $c_i\in\mathbb{Z}$. Then
$$
c_1v_1+\cdots+c_{m-1}v_{m-1}+c_m(a_1v_1+\cdots+a_mv_m)=0
$$
implies
$$
(c_1+a_mc_1)v_1+\cdots+(c_{m-1}+c_ma_{m-1})v_{m-1}+c_ma_mv_m=0.
$$
But $v_1,\dots,v_m$ are linearly independent, so $c_ma_m=0$, thus $c_m=0$ since $a_m\neq 0$. Then since $c_i+c_ma_i=0$ for all other $i$, $c_i=0$ for $i=1,\dots,m-1$, and the vectors are linearly independent over $\mathbb{Z}$.
I'm trying to show $\{v_1,\dots,v_{m-1},v'_m\}$ also span $A$ over $\mathbb{Z}$. Since $v_1,\dots,v_m$ is a maximal linearly independent set, I think I can write any $x\in A$ as
$$
x=c_1v_1+\cdots+c_mv_m,\quad c_i\in\mathbb{R}.
$$
I realized
$$
x=(\lfloor c_1\rfloor v_1+\cdots+\lfloor c_m\rfloor v_m)+(c'_1v_1+\cdots+c'_mv_m)
$$
where $\lfloor \cdot\rfloor$ is the floor function, and $0\leq c'_i<1$. So the second summand in parentheses is in $S$, and the first summand is a linear integral combination of $v_1,\dots,v_m$. I'm not sure if this observation leads anywhere, and I'm not sure where the fact that $A$ has only finitely many elements in every bounded region of space comes in. What's the right way to proceed? Thanks.
Added: With Arturo Magidin's help, $ka_m=c'_m$ for some positive integer $k$. Thus
$$
x=(\lfloor c_1\rfloor v_1+\cdots+\lfloor c_m\rfloor v_m)+k((c'_1/k)v_1+\cdots+a_mv_m),
$$
so taking $v'_m=(c'_1/k)v_1+\cdots+a_mv_m$, $\{v_1,\dots,v_m,v'_m\}$ is a spanning set of $A$ over $\mathbb{Z}$. How can I show from this that $\{v_1,\dots,v_{m-1},v'_m\}$ spans $A$ over $\mathbb{Z}$?

REPLY [6 votes]: You used the fact that $A$ has only finitely many elements in any bounded region of space implicitly when you asserted that the set $S$ would contain an element $v'_m$ with smallest $a_m\neq 0$; you are using the fact that the set $S$ is contained in the ball of radius $\lVert v_1\rVert+\cdots+\lVert v_m\rVert$, hence $S$ is finite, so you can pick that element with "smallest $a_m\neq 0$". Otherwise, since the coefficients are real numbers, there might not be any element with "smallest $v'_m$.
You are doing well. Now, since $c'_1v_1+\cdots +c'_mv_m\in S$, then $c'_m\geq a_m$, where $a_m$ is the coefficient of $v_m$ in the expression for $v'_m$. 
Now, there is a smallest positive integer $k$ such that $ka_m \leq c'_m$. You may want to show that $ka_m$ will actually be equal to $c'_m$.