Fundamental identity for ramification and inertia

Remarks

Proof idea: count elements of $\mathcal{O}_L/\mathfrak{p}\mathcal{O}_L$. It is a vector space over $\mathcal{O}_K/\mathfrak{p}$ of dimension $n$, by localising at $\mathfrak{p}$, where $\mathcal{O}_L$ becomes free of rank $n$. By the Chinese remainder theorem it is the product of the rings $\mathcal{O}_L/\mathfrak{P}_i^{e_i}$, and each has dimension $e_i f_i$, using the filtration by powers of $\mathfrak{P}_i$. In the Galois case, the Galois group acts transitively on the $\mathfrak{P}_i$, and conjugate primes have the same $e$ and $f$.

Example: in $\Q(\sqrt{-5})/\Q$, with $n = 2$: the prime $2$ has $e = 2, f = 1, g = 1$; the prime $3$ has $e = 1, f = 1, g = 2$; the prime $11$ is inert since $-5$ is not a square modulo $11$, with $e = 1, f = 2, g = 1$. In each case $efg = 2$.

Going deeper

For a student

The identity limits the possible splitting types and is used constantly in computations. In a Galois extension, $e f$ is the order of the decomposition group of any prime over $\mathfrak{p}$, and $e$ is the order of its inertia group. For a cyclic extension of prime degree $\ell$, the identity leaves only three cases: split completely ($g = \ell$), inert ($f = \ell$) or totally ramified ($e = \ell$).