Let $F/K$ be an extension of number fields and let $\mathcal{O}_F$ and $\mathcal{O}_K$ be their respective rings of integers. The ring of integers of a number field is a Dedekind domain, and these enjoy the property that every ideal $\A$ factors uniquely as a finite product of prime ideals (see the entry fractional ideal). Let $\p$ be a prime ideal of $\intK$ . Then $\p \intF$ is an ideal of $\intF$ . Let us assume that the prime ideal factorization of $\p \intF$ into primes of $\intF$ is as follows: \begin{eqnarray} \label{eq1} \p \intF=\prod_{i=1}^r {\P_i}^{e_i} \end{eqnarray}We say that the primes $\P_i$ lie above $\p$ and $\P_i|\p$ (divides). The exponent $e_i$ (commonly denoted as $e(\P_i|\p)$ ) is the ramification index of $\P_i$ over $\p$ . Notice that for each prime ideal $\P_i$ , the quotient ring $\intF/\P_i$ is a finite field extension of the finite field $\intK/\p$ (also called the residue field). The degree of this extension is called the inertial degree of $\P_i$ over $\p$ and it is usually denoted by: $$f(\P_i|\p)=[\intF/\P_i:\intK/\p].$$

Notice that as it is pointed out in the entry ``inertial degree'', the ramification index and the inertial degree are related by the formula: \begin{eqnarray} \label{eq2} \sum_{i=1}^r e(\P_i|\p)f(\P_i|\p)=[F:K] \end{eqnarray}where $r$ is the number of prime ideals lying above $\p$ (as in Eq. ()). See the theorem below for an improvement of Eq. () in the case when $F/K$ is Galois.

When the extension $F/K$ is a Galois extension then Eq. () is quite more simple: