Ramification and reduction Let $K$ a local field ($K$ finit extension of $\mathbb{Q}_p$), $\mathcal{O}_K$ the integer of $K$ and $k$ the residue field of $\mathcal{O}_K$.
Let $\psi:\mathbb{P}^1_K\to\mathbb{P}^1_K$ a finit separable morphism, $\widetilde{\psi}=\Psi:\mathbb{P}^1_{\mathcal{O}_K}\to\mathbb{P}^1_{\mathcal{O}_K}$ a model of $\psi$ that is $\Psi$ is the extension of scalar of $\psi$ ie $\Psi=\psi\times_{\mathcal{O}_K}\text{Id}_K$.
$$ \require{AMScd}
\begin{CD}
\mathbb{P}^1_K @>{\psi}>> \mathbb{P}^1_K\\
@VV{\alpha}V @VVV \\
\mathbb{P}^1_{\mathcal{O}_K} @>{\Psi}>> \mathbb{P}^1_{\mathcal{O}_K}
\end{CD} $$
Let $\overline{\Psi}=\Psi\times_{\mathcal{O}_K}\text{Id}_k$ the reduction of $\Psi$.
$$ \require{AMScd}
\begin{CD}
\mathbb{P}^1_k @>{\overline{\Psi}}>> \mathbb{P}^1_k\\
@VV{i}V @VVV \\
\mathbb{P}^1_{\mathcal{O}_K} @>{\Psi}>> \mathbb{P}^1_{\mathcal{O}_K}
\end{CD} $$
If the branching points (ie ramification points)  $P_1,\ldots,P_n$ of $\psi$ are $K$-rationnals, as $\mathbb{P}^1_{\mathcal{O}_K}(\mathcal{O}_K)=\mathbb{P}^1_K(K)$ (by mutliplication of denominators) one can take their reductions $\overline{P_1},\ldots,\overline{P_n}\in\mathbb{P}^1_k(k)$.
Question: I'd like to prove that if the ramification indices of the $P_i$ are resp. $e_i$, they are the same for the $\overline{P_i}$ and if there are ``coalescence'' then the ramification indices of the resulting ramification point $\overline{Q}$ is the sum of the indices $e_i$ for which $\overline{P_i}=\overline{Q}$. I don't have the beginning of an explanation of that, if it's true...
I guess that we shouldn't have wild ramification so the sums of $e_i$ of point that collapse in the same point shouldn't be nul in $k$.
I guess that a general reference for that is SGA1 (Exposé X) but for the moment it's to difficult for me... If someone has a simpler reference for my specific case I'l take it! Thanks!
If you find this question to easy for mathoverflow feel free to answer here in mathstackexchange and tell me in a comment.
 A: In your setting, you can just do everything concretely using the derivative.
The correct statement is for $\overline{Q}$ in $\mathbb P^1_k$,
$$e(\overline{Q}) + \operatorname{swan}(\overline{Q}) = 1 + \sum_{\substack{ i \in \{1,\dots n \} \\ \overline{P}_i = \overline{Q} }} (e_i - 1).$$
This is under your assumptions except that we need to assume $\overline{\Psi}$ is separable.
To prove this, first we can assume by a change of variables that $\overline{\Psi}(\overline{Q}) \neq \infty$. Then express $\widetilde{\psi}$ as a rational function $f$ in $\mathbb Z_p[X]$, without a pole at $\overline{Q}$, and thus without a pole at any of the $P_i$ that reduce to $Q$. Now consider its derivative $\frac{df}{dx}$.
In characteristic zero, this function vanishes exactly at the ramification points $P_1,\dots, P_n$, and its order of vanishing at $P_i$ is $e_i-1$.
In characteristic $p$, its order of vanishing at a point is $e$ plus the Swan conductor minus $1$.
Now we just need to know that the order of vanishing of $\frac{df}{dx}$ at $\overline{Q}$ is the sum of its orders of vanishings at $P_i$ for all the $P_i$ that reduce to $Q$. This follows from factoring the numerator of $\frac{df}{dx}$ into linear factors, and noting that the order of vanishing is the number of linear factors that vanish at a point.
We can't rule out wild ramification here, as the example $x^p-x$ (for $K = \mathbb Q ( p^{1/(p-1)})$) shows. In that case we have one point $\infty$ with $e=p$ and $p-1$ points (the $p-1$st roots of $p^{-1}$) with $e=2$, that all reduce to $\infty$, and in the reduction, $\infty$ has $e=p$ and $\operatorname{swan}=p-1$.
