finite extensions of discrete valuation fields: A method to find a basis

Question

Suppose that $L|K$ is a finite extension of discrete valuation fields. Namely $w$ is a discrete valuation on $L$ extending a valuation $v$ on $K$. Now consider the respective rings of integers $\mathcal O_L$ and $\mathcal O_K$; I don't understand the following fact used, without any explanations, in Neukirch's book to prove Proposition II.6.8:

If $\{x_1,\ldots,x_m\}\subset\mathcal O_L$ is a basis of $\mathcal O_L$ as $\mathcal O_K$-module, then $\{x_1,\ldots,x_m\}$ is a basis of $L$ as $K$-vector space.

Answer 1

Take an element $y\in L$. There is an element $a\in\mathscr{O}_K$, not zero, for which $ay\in\mathscr{O}_L$, hence $ay=\sum_i a_ix_i$ for some $a_i\in\mathscr{O}_K$. Now multiply both sides by $a^{-1}$ to get $y$ as a $K$-linear combination of the $x_i$. They are also $K$-linearly independent by a similar scaling-to-$\mathscr{O}_L$ argument.

Answer 2

$\mathcal O_L$ is a free $\mathcal O_K$-module of rank $m$, and $\{x_1,\dots,x_m\}\subset\mathcal O_L$ is a basis. Then this is a basis for $L=S^{-1}\mathcal O_L$ over $K=S^{-1}\mathcal O_K$, where $S=\mathcal O_K\setminus\{0\}$.

(This is a standard fact: if $F$ is a free $R$-module and $\{x_1,\dots,x_m\}\subset F$ is a basis over $R$, then $S^{-1}F$ is a free $S^{-1}R$-module and $\{\frac{x_1}1,\dots,\frac{x_m}1\}\subset S^{-1}F$ is a basis over $S^{-1}R$.)