Minimal polynomial has coefficients in the valuation ring? Question
Let $K$ be a complete nonarchimedean valued field with valuation ring $\mathcal{O}$ and let $L/K$ be a finite extension.
Let $\alpha$ be an element of $L$ and $f\in K[x]$ its minimal polynomial.
Is is true that $f\in\mathcal{O}[x]$?
Comments
First of all, I am asking this, because of a proof I saw, where the author said "...let $f$ be the minimal polynomial of $\alpha$ and let $\overline{f}=f\pmod{\kappa}$, where $\kappa$ is the residue field of $K$."
He is implicitly using the fact that he can send the coefficients of $f$ to the residue field, and we can only do that if the coefficientes are in $\mathcal{O}$.
I strongly suspect that this is not true in general. I know that if $f$ is monic and the independent term is in $\mathcal{O}$ then indeed $f\in\mathcal{O}[x]$, but only with the information that $f$ is the minimal of some element, I cannot deduce that.
I think that the author meant the following:
"Take the minimal polynomial $f$, clean denominators, and now you can send it to $\kappa[x]$, even though $f$ is not monic anymore".
Is this correct?
If so, after sending $f$ to $\kappa[x]$, it feels like we could perfectly make it monic again, right?
Sorry if this question is kind of annoying or obvious, but Neukirch's style is yet kind of hard for me, making a lot of implicit assumptions and I want to understand things correctly.
Thank you in advance for explanations.

Let $\alpha \in L$ with minimal polynomial $f \in K[X]$ (assuming to be monic).  Then $f \in \mathcal O_K[X]$ if and only if $\alpha \in \mathcal O_L$.
Pf: $\Leftarrow$: We have $\mathcal O_K = \mathcal O_L \cap K$, and the coefficients of $f$ are symmetric functions of $\alpha$ and its conjugates over $K$.  Since $\alpha \in \mathcal O_L$ (by definition $\mathcal O_L$ is the set of things in $L$ which are integral over $\mathcal O_K$), so do all its conjugates.  So the coefficients of $f$ are in both $\mathcal O_L$ and $K$.
$\Rightarrow$: $\alpha$ is a root of a monic polynomial with coefficients in $\mathcal O_K$, so $\alpha$ is in $\mathcal O_L$.
This is true more generally when you replace $\mathcal O_K$ by an integrally closed ring.
