Lattice of a finite dimensional $\mathbb{Q}_p$-vector space Let $V$ be a finite dimensional vector space over the field $\mathbb{Q}_p$ of $p$-adic numbers. Let $T$ be a $\mathbb{Z}_p$-submodule of $V$ which is free of finite rank and generates $V$ over $\mathbb{Q}_p$. 
Why can $V$ be identified with $T\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$? 
Thanks
 A: As $T$ is of finite rank free $\mathbb{Z}_p$ module, suppose $\{x_1,x_2,...,x_n\}$ be a basis of $T$ and that also generates $V$ over $\mathbb{Q}_p$. So we can think $T=\mathbb{Z}_p^r$ and $V=\mathbb{Q}_p^r$. Hence obviously, $V=\mathbb{Q}_p^r\cong\mathbb{Z}_p^r\otimes_{\mathbb{Z}_p}\mathbb{Q}_p=T\otimes_{\mathbb{Z}_p}\mathbb{Q}_p$.
A: The assumption that the $\mathbf{Z}_p$-submodule $T$ of $V$ spans $V$ over $\mathbf{Q}_p$ is equivalent to surjectivity of the canonical map $T\otimes_{\mathbf{Z}_p}\mathbf{Q}_p\rightarrow V$ (this map is canonical in that it requires no choice of bases, and is given by $t\otimes\lambda\mapsto\lambda t$). The map is always injective, because every element of the source is of the form $t\otimes \lambda$ with $t\in T$ and $\lambda\in\mathbf{Q}_p$ (because of common denominators), and this maps to $\lambda t$, which is zero precisely when $\lambda=0$ or $t=0$ (since $V$ is a vector space over $\mathbf{Q}_p$), and either of these implies $t\otimes\lambda=0$.
So, in summary, a $\mathbf{Z}_p$-submodule $T$ of $V$, finitely generated or not, free or not, spans $V$ if and only if the natural $\mathbf{Q}_p$-linear map $T\otimes_{\mathbf{Z}_p}\mathbf{Q}_p\rightarrow V$ is surjective (equivalently, an isomorphism). 
