Possibility of choice of normal basis $\{ \sigma_1w \cdots \sigma_nw\}$ such that $w\in \mathcal{O}_L$ (valuation ring of local field)? I'm reading Neukirch, Algebraic number theory, p.318, proof of the class field axiom for local fields and some question arises.
Let $L|K$ be a finite (possibly cyclic) Galois extension of local fields of degree $n$. Let $\sigma_1, \cdots ,\sigma_n$ be the elements of the Galois group $G$. Then by the normal basis theorem, there exists an element $w \in L$ such that $\sigma_1w,\cdots \sigma_nw$ form a basis of $L$ over $K$.
Proof of the normal basis theorem is as follows (Serge Lang, Algebra, p.312) :

Then my question is, can we choose such $w$ satisfying $w \in \mathcal{O}_L := \{a\in L | v_L(a) \ge 0 \}$ (the valuation ring of $L$ with the discrete valuation normalized by $v_L(L^{*})=\mathbb{Z}$ )?
My frist attempt is, since by the Neukirch's book, p.121, for each $x \in L$, $x\in \mathcal{O}_L$ or $ x^{-1} \in \mathcal{O}_L$. So for such above $w$, if we show that  $\sigma_1(w^{-1}), \cdots ,\sigma_n(w^{-1})$ also form a basis of $L$ over $K$, then we are done.
And to show this, by imitaing the proof of the Normal basis theorem, I'm trying to show that
$\operatorname{det}(\sigma_{i}^{-1}\sigma_j(w^{-1}))\neq 0$. And I'm stuck at this point.
If furthurmore, $L|K$ is cyclic extension, then is it possible that $\sigma_1(w^{-1}), \cdots ,\sigma_n(w^{-1})$ form a basis?
This question originates from following proof of the Neukirch's book (p.317, (1.1) Theorem) :


Why $\alpha \in \mathcal{O}_L$ is true?
Can anyone helps?
 A: It is simpler to scale using a uniformizer than to use the inverse. This argument assumes that the valuation on $K$ is chosen to be the unique one extending the valuation on $k$. Otherwise, none of this really matters.
More concretely, once we have such a non-zero $w \in K$, we pick a uniformizer $\varpi$ of $O_k$ and  a positive integer $r$ large enough such that $v_K(\varpi^rw) \geq 0$. Since $\varpi $ is fixed by all the $\sigma_i$, the determinant of your matrix will just be scaled by a large power of $\varpi$ ( if I remember linear algebra correctly, by $\varpi^{nr}$) and therefore is still non-zero.
This is a very common trick btw.
Alternate proof that works over any finite extension of number fields and local fields. In the following, we interpret $O_k$ as the ring of integers or the valuation ring as appropriate.
Find the minimal polynomial for $w$ over $k$ and clear denominators so that all coefficients are in $O_k$. Say $$\sum a_n w^n = 0 \text{ with } a_n \in O_k$$
Then multiply both sides by $a_n^{n-1}$ and rearrange to get a polynomial equation $$(a_n w)^n + a_{n-1}(a_nw)^{n-1} + \cdots + a_0a_n^{n-1} = 0 $$ which we can read as a monic polynomial relation satisfied by $a_nw$. This implies that $a_nw$ is integral over $O_k$ and therefore lives in $O_K$.
