In this question someone help me to understand why there is not always a model for a morphism $\mathbb{P}^1_K\to\mathbb{P}^1_K$. He says to me that the existence of model is equivalent of good reduction. Now I would like to understand this sentence ie take the "good" definitions and find a proof.
Let $K$ be a local field, of integer $\mathcal{O}$ and uniformizer $\pi$. Take $\varphi:\mathbb{P}^1_K\to\mathbb{P}^1_K$ a finite morphism and $f\in K(t)$ its associated rationnal function.
I propose the following definition for model: a model of $\varphi$ is an extension of $\varphi$ to $\mathbb{P}^1_\mathcal{O}$ that is a morphism $\Phi:\mathbb{P}^1_\mathcal{O}\to\mathbb{P}^1_\mathcal{O}$ with $\varphi$ the extention of scalar of $\Phi$ to $K$ that is $\varphi=\Phi\times_\mathcal{O}\text{Id}_K$ ie the following diagram commute, $\require{AMScd}$ \begin{CD} \mathbb{P}^1_K @>{\varphi}>> \mathbb{P}^1_K\\ @VVV @VVV\\ \mathbb{P}^1_\mathcal{O} @>{\Phi}>> \mathbb{P}^1_\mathcal{O} \end{CD}
I'm not sure of the "good" definition for good reduction, I propose: as $f\in K(t)\setminus K$, $K=\text{Frac}(\mathcal{O})$ and $\mathcal{O}$ is factorial, then one can write $f=P/Q$ with $P,Q\in\mathcal{O}[t]$ coprimes (no common irreducible divisor). Then one says that $\varphi$ has a good reduction if $\overline{P}$ and $\overline{Q}$ are coprimes in $k[t]$ and the degree of $P$ and $Q$ are preserve under reduction (I take this because $\pi t^2+1$ and $\pi t+1$ reducing to coprimly but I don't want to tell that this is a good reduction (not finit morphism), but I don't need it in the proof below...)
Question: with these definitions do we have that the existence of model is equivalent to good reduction?
I have elements of proof:
I think that $P,Q\in\mathcal{O}[t]$ will define a model if $(P,Q)=\mathcal{O}[t]$ that is $(P,Q)\cap\mathcal{O}=(1)$.
In general $(P,Q)\cap\mathcal{O}=(\pi^q)$ with $q\in\mathbb{N}$. So there is a model if $q=0$.
If there is a model then $q=0$ ie $1\in(P,Q)$ so reducing the Bezout relation one has $1\in(\overline{P},\overline{Q})$ so $\overline{P}$ and $\overline{Q}$ are coprimes (but how to prove that degree is preserved, or that the reduction stay a finite morphism??)
If $P$, $Q$ are coprimes then $(\overline{P},\overline{Q})=1$. Here I have a problem: I'd like to show that $(P,Q)\cap\mathcal{O}=(\pi^q)$ with $q>0$ is not possible. If $(\overline{P},\overline{Q})\cap k=(\overline{\pi^q})$ the conclusion comes shortly, but I have only $(\overline{P},\overline{Q})\cap k\supseteq(\overline{\pi^q})$
Thnaks for your help!