Definition of tame ramification from the ramification groups

Let $R$ be a d.v.r., $K$ its field of fractions, $L/K$ a separable finite field extension and $S$ the integral closure of $R$ in $L$. Let $p$ be the prime ideal of $R$ and assume that $K$ is complete in the valuation topology so that likewise be $L$. Thus we can define $\mathfrak P$ to be the unique prime ideal of S lying above $p$.

How do you show that the following two definition of tame ramification coincide?

$(1)\operatorname{char} K \not |\,\, e(L/K) \\(2)\, G_1 = \{1\}$

where $G_1$ is the first ramification group, i.e. $\{\sigma \in \Gamma(L/K)\,|\,\sigma x\equiv x\pmod {\mathfrak P^2}\quad\forall x\in S\}$. (Okay, I need to assume the extension is actually Galois for this second definition to work)

• You should call $S$ the integral closure of $R$ in $L$, not the algebraic closure of $R$ in $L$. Have you looked in books that discuss higher ramification groups? – KCd Jan 17 '14 at 1:28
• @KCd oops, that slipped from me. I had this question from reading the following article math.ucsb.edu/~cindytsy/talks/… in which both are mentioned to be equivalent but no explanation is given. Otherwise, I'm following Froehlich's ANT – Rodrigo Jan 17 '14 at 1:30
• Your question is a standard result about higher ramification groups. The index $[G_0:G_1]$ is a power of $p$ and the indices $[G_i:G_{i+1}]$ are relatively prime to $p$ (when the residue field has characteristic $p$). See books on local fields that discuss higher ramification groups (e.g., the book by Fesenko and Vostokov). – KCd Jan 17 '14 at 1:35
• @KCd: Dear Keith, Do you have things backwards (i.e. tame inertia is prime-to-$p$ and wild is $p$-power)? Or do I? Cheers, – Matt E Jan 17 '14 at 3:36
• Oops, of course it'd be hard for $G_1$ to be trivial in the tamely ratified case if the later indices were not powers of $p$. Yes, I had my description exactly backwards. – KCd Jan 17 '14 at 4:24