Let $v_1, v_2,\dots, v_n$ be $n$ vectors in $\mathbb{Q}^m$ (defined as a Vector Space over the field $\mathbb{Q}$) having all integer entries. Let $p$ be a prime number and let $\bar{v}_1, \bar{v}_2,\dots, \bar{v}_n$ be the $n$ vectors in $\mathbb{F}_p^m$ obtained by reducing the entries of $v_i$ modulo $p$.
Given that $\bar{v}_1,\bar{v}_2,\dots,\bar{v}_n$ span $\mathbb{F}_p^m$ (defined as a Vector Space over the field $\mathbb{F}_p$), prove that $v_1, v_2, \dots, v_n$ span $\mathbb{Q}^m$.
Here, $\mathbb{F}_p$ denotes the field $\mathbb{Z}/p\mathbb{Z}$.
I don't know how to proceed with this one. Please help.
Thanks in advance.
Of course you must have $n\ge m$. Without loss of generality, we may assume that $\bar{v}_1,\ldots,\bar{v}_m$ span $\mathbb{F}_p^m$.
By contradiction, if $v_1,\ldots,v_m$ does not span $\mathbb{Q}^m$, then there exist rational numbers $a_1,\ldots,a_m$ such that $\sum_i a_iv_i=0$. Up to multiplying by some rational factor, we may actually assume that each $a_i$ is an integer, and that some $a_i$ is not divisible by $p$. Now, by reducing the linear equation $\sum_i a_iv_i=0$ modulo $p$, we obtain a contradiction.
Edit: Here's a less elementary proof. Consider the $\mathbb{Z}$-module $M$ generated by the $v_i$. As it is torsion-free and of finite type, it is a free module of some finite rank $r$. By reducing it modulo $p$, we see that $r=m$; in particular $\mathbb{Q}\cdot M=\mathbb{Q}^m$.