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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 $\mathbb{Z}$?

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    To answer your doubt on the hypothesis: the set $S$ is assumed to be finite by your hypothesis but if it is not finite, such a $v_m'$ may not exist.2011-10-16

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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.

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    Yes, I think I got too stuck in one idea. Anyway, thanks for your patience and explanations, I appreciate it very much.2011-10-18