Injective group homomorphism from $\mathbb Z^n$ into $\mathbb{R}^m$ What is a nice way to prove that an injective grouphom $$\mathbb{Z}^n\rightarrow \mathbb{R}^m$$
implies $m\geq n$?
 A: Given an injective group homomorphism, we obtain that the image of $\mathbb{Z}^n$ in $\mathbb{R}^m$ is a finitely generated module of rank $n$. This is possible for every $m$. In fact, we can always define $\phi:\mathbb{Z}^n\to \mathbb{R}$ as $\phi(\vec{x})=x_1+x_2\alpha+\dots+x_n\alpha^{n-1}$, where $\alpha$ is a algebraic number of degree $n$ over $\mathbb{Q}$. Composing this with the immersion of $\mathbb{R}$ into $\mathbb{R}^m$ we obtain a homomorphism of groups for $m,n$ general.
However, if we either 


*

*consider injective homomorphisms from $\mathbb{Z}^n$ to $\mathbb{Q}^m$ 

*restrict our consideration to homomorphism such that the image of $\mathbb{Z}^n$ is discrete in $\mathbb{R}^m$
the requested inequality $m\ge n$ holds.
Proof:


*

*Let us write $\{v_1\dots, v_n\}$ as the image in $\mathbb{Q}^m$ of the vectors $(1,\dots,0),(0,1,\dots,0)$ and so on. We want to prove that $\{v_1,\dots,v_n\}$ are linearly independent over $\mathbb{Q}$. From there the assertion follows. To prove that they are l.i., let $$\sum a_i v_i=0.$$ 
Multiplying by the MCD of the denominators of $a_i$ we obtain $$\sum b_i v_i=0.$$ Since $b_i\in \mathbb{Z}$, from the injectivity of the homomorphism follows that $\forall i, b_i=0$, and the claim is proved.

*Let $\{v_1,\dots, v_n\}$ be defined as before with the obvious modification that $v_i\in \mathbb{R}^m$.
If $m<n$, this set of vectors is linearly dependent, and we can write
$$\sum_{l=1}^{n-1}a_lv_l=v_n\ \ (a_l\in\mathbb{R},\text{but they are not all in } \mathbb{Q}).$$
Thanks to the simultaneous version of Dirichlet's approximation theorem we can find integers $p_i,q$ such thar $|2qa_i-p_i|\le\frac{1}{q^\frac{1}{n-1}}<\varepsilon$. Thus we can write
$$0<||\sum_{l=1}^{n-1}p_l v_l-qv_n||=||\sum_{l=1}^{n-1} (p_l-qa_l)v_l||\le K\varepsilon.$$
Note that the first linear combination has all integer coefficients, and is thus in the image of $\mathbb{Z}^n$. Letting $\varepsilon \to 0^+$ we obtain that $0$ is a cluster point for the image, which is not discrete.
