I've been working through some of the problems of Marcus' Number Fields and am stuck on a problem relating to ramification groups.
Let $K$ be a number field, $L$ is a normal extension of $K$, $G$ is Galois Group of $L / K$, $R$ the ring of integers of $K$, $S$ the ring of integers of $L$, $P$ a prime of $R$, $Q$ a prime of $S$ lying over $Q$. $E$ is the inertia group of $G$, e.g. $E = \{\sigma \in G
|\ \sigma(\alpha) \equiv \alpha\ \text{mod}\ Q \}$.
Take $\pi \in Q – Q^2$. The exercise claims for each $\sigma$ there is some $\alpha \in S$ so that $\sigma(\pi) \equiv \alpha\pi\ \text{mod}\ Q^2$, and $\alpha$ is in fact determined uniquely mod $Q^2$. (The goal of the question is ultimately to embed $E / V_1$ into the multiplicative group $(S / Q)^{*}$.)
The hint the book provides is to write $\pi S = QI$ and use the Chinese Remainder Theorem to find an $x$ so that $x \equiv \sigma(\pi)\ \text{mod}\ Q^2$ and $x \equiv 0\ \text{mod}\ I$.
Taking such an $x$, since $\pi S$ is a principal ideal this gives that for any $q \in Q$, $qx = s\pi$ for some $s \in S$. It's not obvious how to isolate $x$ as taking quotients isn't generally allowed in the ring of integers.
I'm not sure how to proceed from here. Other things I've tried:
- Since $\sigma$ is in the inertia group, $\pi, \sigma(\pi), \sigma^{-1}(\pi) \in Q$.
- $p \in Q$ which shows that $x = (s/p)\pi$; however it's not obvious $s / p$ is an algebraic integer.
- Fiddling a bit with Sage, if $\sigma(\pi) / \pi$ is an algebraic integer then that could serve as $\alpha$, however it is not obvious it always should be one.
I've done some reading online and it looks like this situation is much easier in the local case as $S$ is a PID. I'm looking to understand the global case as that's more in the spirit of the original text. Thanks for any assistance you can provide.
Best Answer
Okay, I think I have something that works.
Claim: $x \in Q$. Proof of claim: since $\sigma \in E$ (the inertia group), $\pi \sigma(\pi) \in Q^2$. As $\sigma(\pi) - x \in Q^2$, $\pi(\sigma(\pi) - x) \in Q^2$, so $\pi x \in Q^2$; because $\pi \not\in Q^2$, $x \in Q$.
Since $QI = (\pi)$ and $\pi \not \in Q^2$, $Q$ and $I$ are coprime and so by a theorem of commutative ideal theory $Q \cap I = QI$; as $x \in Q, I$, $x \in QI = (\pi)$ and so $x = \alpha \pi$ for some $\alpha \in S$.
Therefore $\sigma(\pi) \equiv \alpha \pi \pmod{Q^2}$, the required result.