Totally ramified extension of global field in terms of local field case

class-field-theorylocal-fieldramification

Let $K$ be a number field of class number $1$(in other words, it's ring of integers $O_K$ is PID.
Let $p$ be a nonzero prime ideal of $O_K$.

$L/K$ is an extension with $n=[L:K]$, we say that $L$ is totally ramified at $p$ if $p\mathcal{O}_L=q^n$ for some prime $q$ of $L$.

On the other hand, $L_p/K_p$ is tottally ramified if $L$ and $K$ have the same residue field.

Then, what is the relation between '$L/K$ is totally ramified at $p$' and '$L_p/K_p$ is totally ramified as local field' ?

I think latter does not imply former in general, but if I add some condition (like $[L:K]＝[L_p:K_p]$), can't we say something like the latter implies the former ?
Thank you for your help.

Best Answer

Nothing you're describing depends on the class number. The general situation is as follows (see, for example, Milne's Algebraic Number Theory, Thm. 3.34; Neukirch's Algebraic Number Theory, Prop. I.8.2; or Serre's Local Fields, Prop. I.10).

Let $A$ be a Dedekind domain with field of fractions $K$, let $L$ be a finite separable extension of $K$ with $n = [L : K]$, and let $B$ be the integral closure of $A$ in $L$. Then $B$ is a Dedekind domain. For each prime $P$ of $A$ and each prime $Q$ of $B$ dividing $PB$, let $e_Q$ be the ramification index, i.e., the largest integer such that $Q^{e_Q}$ divides $PB$, and let $f_Q = [B/Q : A/P]$ be the residue degree (also called the inertia degree). Then $$n = \sum_{Q \mid P} e_Q f_Q.$$ In particular, the following two notions of "totally ramified at $P$" are equivalent:

  • $PB = Q^n$ for a prime $Q$ of $B$.
  • There is a unique prime $Q$ of $B$ dividing $PB$, and the residue field $B/Q$ is a degree 1 extension of $A/P$ (i.e. the residue fields are canonically isomorphic).

In the case of an extension of local fields, there's only one prime in total, so the uniqueness part of the second condition is automatic.

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