Let $K/\mathbb{Q}$ be a number field and suppose a prime $p\in\mathbb{Z}$ factors in $\mathcal{O}_K$ as $\prod_{i=1}^r \mathfrak{p}_i^{e_i}$. From algebraic number theory, we have the identity $$ [K:\mathbb{Q}]=\sum_{i=1}^r e_if_i. $$ If $v_{ \mathfrak{p}_i}$ is the $ \mathfrak{p}_i$-adic valuation on $K$ and $ K_{\mathfrak{p}_i}$ the corresponding extension of $\mathbb{Q}_p$, is it true that $$ [K_{ \mathfrak{p}_i}:\mathbb{Q}_p]=e_if_i? $$ I know a statement like this exists for local fields - i.e., if $L/K$ is an extension of local fields then $[L:K]=ef$ - but I'm wondering how these 'local' and 'global' statements interact... Any references would be welcome.
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Yes your "natural intuition" is right here. $e,f$ can be defined using statements about the the decomposition group (and the inertia group), and the natural map $D_{\mathfrak{p}_i/p} \rightarrow \textrm{Gal}(K_{\mathfrak{p}_i}/\mathbb{Q}_p)$ is an isomorphism. (In general you can pass to a Galois closure and use the tower laws for $e$ and $f$.)
I learnt this from Jack Thorne a few years ago. Here's the notes: https://www.dpmms.cam.ac.uk/~jat58/antm2015/notes.pdf. The specific statement you are looking for is Proposition 6.14.3.
