Prime ideal factorization

Every nonzero ideal of the ring of integers of a number field factors uniquely as a finite product of nonzero prime ideals. Thus each $x\in K^{\times }$ has integer valuations ${\mathrm{ord}}_{\mathfrak{p}}(x)$, almost all zero, determined by the factorization of its principal ideal.

If $\mathfrak{p}$ is a prime ideal of ${\mathcal{O}}_K$ above a rational prime $p$, its ramification index is the exponent $e_{\mathfrak{p}}$ of $\mathfrak{p}$ in the factorization of $(p)$. Equivalently, for the normalized ideal valuation, $e_{\mathfrak{p}}=v_{\mathfrak{p}}(p)$.

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 $R$, writeand let $f_{\mathfrak{q}/\mathfrak{p}}=[B/\mathfrak{q}:R/\mathfrak{p}]$. Then