Over the $p$-adic field, intersecting a linear subspace with the vectors with entries in $\mathbb{Z}_p$

Question

Let $p$ be a prime number. Let $V$ be a linear subspace of $\mathbb{Q}_p^n$ of dimension $d$. Is $V \cap \mathbb{Z}_p^n$ a free $\mathbb{Z}_p$-module of rank $d$? Is every $\mathbb{Z}_p$-basis for $V\cap\mathbb{Z}_p^n$ a $\mathbb{Q}_p$-basis for $V$?

Answer 1

Write $\tilde{V} = V \cap \mathbb{Z}_p^n$ and let $v_1, \ldots, v_n \in \tilde{V}$ be linearly independent over $\mathbb{Z}_p$ ($n \geq 1$ any). Then the $v_i \in V$ are linearly independent over $\mathbb{Q}_p$ as well, for if there is a non-trivial relation $0 = \sum_{i = 1}^n a_i v_i$ with $a_i \in \mathbb{Q}_p$, you can multiply with a large enough power of $p$ to transform this into a relation over $\mathbb{Z}_p$ where no non-trivial such relations hold. On the other hand, if you start out with $w_1, \ldots, w_n \in V$ linearly independent over $\mathbb{Q}_p$, you may again scale with $p^r$, some $r \gg 0$ to assure that $w_i \in \tilde{V}$, and then the $w_i$ are obviously linearly independent over $\mathbb{Z}_p$.

In particular, this shows that $d$ is the maximal size of a linearly independent subset of $\tilde{V}$ over $\mathbb{Z}_p$, so since $\mathbb{Z}_p$ is a PID and $\tilde{V}$ is torsion-free, we conclude that $\tilde{V}$ is free of rank $d$, and so the answer to both of your questions is yes.