The <prime ideal factorization> of $pB$ is
$$
pB=\prod_{q\mid p}q^{e_{q/p}},
$$
which defines the <ramification index of a prime ideal> $e_{q/p}=v_q(pB)$. The <residue-field degree> is
$$
f_{q/p}=[B/q:R/p].
$$
Localizing at $p$ makes $B_p$ a free $R_p$-module of rank $[L:K]$, so $B/pB$ has dimension $[L:K]$ over the <residue field> $R/p$. The <Chinese remainder theorem> and the filtration by powers of each $q$ decompose this vector space into $e_{q/p}$ successive quotients isomorphic to $B/q$, each of dimension $f_{q/p}$. Therefore the <fundamental identity for prime decomposition> gives
$$
\boxed{[L:K]=\sum_{q\mid p}e_{q/p}f_{q/p}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 1 / i / Solution

Let $B$ be the <integral closure> of a <Dedekind domain> $R$ in a finite separable extension $L/K$ of degree $n$. For a nonzero <prime ideal> $\mathfrak p$ of $R$, write
$$
\mathfrak pB=\prod_{\mathfrak q\mid\mathfrak p}\mathfrak q^{e_{\mathfrak q/\mathfrak p}}
$$
and let $f_{\mathfrak q/\mathfrak p}=[B/\mathfrak q:R/\mathfrak p]$. Then
$$
n=\sum_{\mathfrak q\mid\mathfrak p}e_{\mathfrak q/\mathfrak p}f_{\mathfrak q/\mathfrak p}.
$$

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