🪴 Quartz 3

Search IconIcon to open search

ramified extensions

Last updated January 27, 2022

Definition

Suppose

The extension $K/k$ is called

  1. unramified iff $e=1$
  2. tamely ramified iff $\text{char}(k_\mathfrak{p})\nmid e$
  3. totally ramified iff $f=1$

Properties

  1. TFAE:
    1. $K/k$ is tamely ramified
    2. $Tr_{K/k}(\mathcal{O}_K)=\mathcal{O}$
    3. $v_\mathfrak{B}(\mathfrak{D}_{K/k})=e-1$
  2. (totally ramified extensions)
    1. Suppose $K=k_P$ where $P\in\mathcal{O}[T]$ is an Eisenstein polynomial. Then $K/k$ is totally ramified. Also, if $\Pi\in K$ and $P(\Pi)=0$, then $v_\mathfrak{B}(\Pi)=1$.
    2. Conversely, Suppose $K/k$ is totally ramified, $\Pi\in\mathcal{O}K$, $v\mathfrak{B}(\Pi)=1$. Then $P_\Pi$ is Eisenstein and $K=k(\Pi)$ and $\mathcal{O}_K=\mathcal{O}[\Pi]$.
  3. (unramified extensions)
    1. Suppose $K=k_P$, $P\in\mathcal{O}[T]$ is monic of degree $n$, such that $\overline{P}\in k_\mathfrak{p}[T]$ (the residue mod $\mathfrak{p}$) is irreducible and separable. Then $K/k$ is unramified and $K_\mathfrak{B}=(k_\mathfrak{p})_{\overline{P}}$.
    2. Suppose $K/k$ is unramified. Then there exists $x\in\mathcal{O}K$ such that $K\mathfrak{B}=k_\mathfrak{p}(\overline{x})$. If this is the case then $K=k(x)$, $\mathcal{O}_K=\mathcal{O} $, $\overline{P_x}\in k_\mathfrak{p}[T]$ is irreducible and separable.
  4. Suppose $K/k$ is unramified, $K'/k$ any finite separable extension. Then the natural map $\sum_{K/k}^{K'/k}\to\sum_{K_\mathfrak{B}/k_\mathfrak{p}}^{K'_{\mathfrak{B}'}/k_\mathfrak{p}}$ is a one-to-one corespondence. If $K_\mathfrak{B}/k_\mathfrak{p}\simeq K'_\mathfrak{B}/k_\mathfrak{p}$ and $K'/k$ is unramified then $K/k\simeq K'/k$.
  5. (maximal unramified subextension)
    1. There exists a unique subfield $k\subset K_0\subset K$ which is unramified and contains any other unramified subfield. Any subfield $K\subset K_1\subset K_0$ is unramified. $(K_0)_{\mathfrak{B}0}=K\mathfrak{B}$.
    2. (inertia group) If $K/k$ is normal, so is $K_0/k$. Suppose $G=\text{Gal}(K/k)$. Then $K_0/k$ corresponds to the subgroup $G_0\subset G$ defined as $G_0\coloneqq{g\in G\mid\forall x\in\mathcal{O}K,\ v\mathfrak{B}(g(x)-x)>0}$.
    3. The group $G_0$ is called the inertia group. There exists a decreasing chain of subgroups $G\supset G_0\supset G_1\supset\dots$ which are called ramification groups.

Interactive Graph