Doubt about higher ramification groups

Question

Let $L/K$ be a Galois extension of number fields with Galois group $G$. Let $O_K$ and $O_L$ be the ring of algebraic integers of $K$ and $L$ respectively. Let $P\subseteq O_K$ be a prime. Let $Q\subseteq O_L$ be a prime lying over $P$.

The $n$-th ramification group is defined as $$E_n(Q|P)=\lbrace \sigma\in G:\sigma(x)\equiv x\text{ mod } Q^{n+1}\text{ for all } x\in O_L\rbrace$$ In particular, $n=0$ gives the inertia group. How to prove the following:

Let $\pi$ be an element in $O_L$ such that $\pi\in Q-Q^2$.

Let $\sigma\in E_0$ be such that $\sigma(\pi)\equiv\pi\text{ mod }Q^2$. $\implies$ $\sigma\in E_1$.

Answer 1

You can find what you want in Serre's Local Fields.

Sketch: WLOG $L/K$ is totally ramified - i.e., replace $K$ with $K^{E_0}$, $P$ with $Q \cap K^{E_0},$ and $G$ with $E_0,$ so that $[L:K] = |E_0| = |G|$, and $Q$ is the only prime above $P$. Further, we may localise, so that $Q$ and $P$ are the only (non-zero) primes of ${\cal O}_L$ and ${\cal O}_K$. At that point, the rings are DVRs, $\pi$ generates ${\cal O}_L$ over ${\cal O}_K$, and we are done...

Precise reference - see Serre's Local Fields, Chapter IV, Section 2, Proposition 2 for a variant, with reference esp. to Chapter I, Section 6, Proposition 18 to show that $\pi$ generates ${\cal O}_L$.

Edit: As you imply in the comments below, one can get by with less (esp. without using localization, or $\pi$ generating the localization of ${\cal O}_L$) than the above.

• First of all, as Jyrki Lahtonen pointed out above, ${\cal O}_L/Q\simeq Q/Q^2$ over ${\cal O}_L$. An argument for this is: since $\pi\in Q\setminus Q^2$ (in particular $Q/Q^2 \ne 0$!), we have that $(\pi) = Q I$, for $I$ an ideal relatively prime to $Q$. Now, by the Chinese Remainder theorem, there exists $x\in {\cal O}_L$ such that $$x \equiv 1 \pmod Q,\text{ and }x\equiv 0 \pmod I.$$ Hence, for an arbitrary $z\in Q$, we have $$zx \equiv z \pmod {Q^2}\text { and } zx \equiv 0 \pmod {QI}.$$ But $(\pi) =QI$, so the last congruence means that $zx = u\pi$, for some $u\in {\cal O}_L$. Thus, $$ z \equiv u\pi \pmod {Q^2},$$ and $\pi$ generates $Q/Q^2$. Therefore, as multiplication by an element of $Q$ kills $Q/Q^2$, $Q/Q^2$ is a one-dimensional ${\cal O}_L/Q$ vector space.

• Next, for neatness, with the Galois simplification from the sketch above of the argument that basically follows that in Serre's book (not the localisation portion), we can assume that the canonical inclusion $${\cal O}_K/P \to {\cal O}_L/Q$$ is an isomorphism - in fact, is an equality.

Therefore, from the preceding and the filtration $$ {\cal O}_L \supset Q \supset 0 \pmod {Q^2},$$ we see that we can write any element of ${\cal O}_L/Q^2$ in the form $a + \pi b \pmod {Q^2}$, where $a$ and $b$ are elements of ${\cal O}_K$ - in particular $a$ and $b$ are fixed by Galois. Thus, if $\sigma$ fixes $\pi \pmod {Q^2}$, it acts trivially on ${\cal O}_L/Q^2$.

And we are done...

Answer 2

Further elaborating on peter a g's answer, if $ S/R $ is an extension of discrete valuation rings, $ \pi $ is an uniformizer of $ S $ and $ \mu $ generates the residue class field of $ S $ over that of $ R $, then we have $ S = R[\pi, \mu] $. This is because we have $ S = R[\pi, \mu] + \mathfrak p S $ where $ \mathfrak p $ is the maximal ideal of $ R $, and so we have $ S/R[\pi, \mu] = \mathfrak p (S/R[\pi, \mu]) $ as $ R $-modules. Since $ J(R) = \mathfrak p $, Nakayama kicks in, and the factor module is trivial, i.e $ S = R[\pi, \mu] $. When the extension $ S/R $ is totally ramified, the residue class field extension is trivial, and therefore we have $ S = R[\pi] $.