# 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.

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 $$\operatorname{swan}=p-1$$.

• @Macadam In this case we can just define the Swan conductor as the order of vanishing of the derivative minus $e-1$, which is $0$ in the case of tame ramification and $>0$ in the case of wild ramification. Jul 28 '20 at 22:21
• Any reference on Riemann-Hurwitz in characteristic $p$ should explain what you need here, because it's mostly the same ideas. Jul 28 '20 at 22:39
• @Macadam The derivative of $x^2 (x-3)^2 = 2 x (x-3)^2 + 2 x^2 (x-3) = 2x (x-3) (2x-3)$. There is a third ramification point $3/2$ which also has $e=2$. Jul 29 '20 at 11:59
• @Macadam If $\Psi$ is a well-defined morphism on $\mathbb P^1_{\mathcal O_K}$ then $\overline{\psi}$ cannot be constant (using that $\psi$ is finite, thus surjective.) For instance because $\overline{\psi}$ is a map from a projective variety, hence proper, so its image is closed. If the points are not in $\mathcal O_K$ (i.e. reduce mod $\pi$ to $\infty$) you get factors like $(cX -1)$ for $c \in \mathcal O_K$ and the same argument works. Alternately, you can extend to a larger field, which gives you enough freedom to change the variables so that none of the points reduce to $\infty$. Jul 29 '20 at 20:45
• @Macadam It is contradictory with the map being surjective in characteristic $0$ and constant in characteristic $p$ (then, what is the image? Is it closed?) The factors disappearing is OK, it just means the ramification points are going off to infinity. The field enlargement is to make a sufficient linear change of variables (rational linear transformation in $X$) that the points don't go off to $\infty$. Jul 30 '20 at 0:04