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.