Let $L/K$ be a finite extension of $p$-adic fields and let $K'/K$ be the maximal unramified subextension of $L/K$. Then $L/K'$ is totally ramified, and if $\alpha \in L$ is a uniformizer then the minimal polynomial $\varphi(x) \in K'[x]$ of $\alpha$ over $K'$ will be Eisenstein. The ramification polygon $P$ of $L$ is essentially the Newton polygon of the ramification polynomial $$\rho(x)=\varphi(\alpha x + \alpha)/(\alpha^n)\in L[x]$$ of $L/K$. Write $\rho(x)=x^n+\sum_{i=0}^{n-1}\rho_ix^i$. Then the ramification polygon of $L/K$ is the lower convex hull of the set of points $(-i,v_L(\rho_i))$, where we ignore all points with $\rho_i=0$. Thus the ramification polygon is the reflection of the Newton polygon of $\rho(x)$ through the vertical axis. This is independent of the choice of uniformizer $\alpha$ for $L$.
If $I/K$ is a family of extensions then the ramification polygon of $I/K$ is defined to be the ramification polygon of any $L/K\in I/K$. The slopes of the segments of $P$ are the rams of $L/K$ and $I/K$.
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- Last edited by Kevin Keating on 2025-05-14 14:59:46
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