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](https://math.stackexchange.com/questions/3769654/ramification-point-and-reduction) in mathstackexchange and tell me in a comment.



Ramification and reduction
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 $\operatorname{swan}=p-1$.


