Ramification index

Let $L\mathrm{/}K$ be a finite extension of number fields and let $\mathfrak{p}$ be a nonzero prime of ${\mathcal{O}}_K$. In the factorisation $\mathfrak{p}{\mathcal{O}}_L={\prod }_{\mathfrak{P}\mid \mathfrak{p}}{\mathfrak{P}}^{e(\mathfrak{P}\mathrm{/}\mathfrak{p})},$ the positive exponent $e(\mathfrak{P}\mathrm{/}\mathfrak{p})$ is the ramification index of $\mathfrak{P}$ over $\mathfrak{p}$.

Here $\mathfrak{P}\mid \mathfrak{p}$ means contraction to $\mathfrak{p}$, as in Primes above and residue degree. The factorisation exists and its exponents are unique by Integral ideal factorisation in a number field, in ZF. For completeness, the extended ideal is proper: a finite integral basis of ${\mathcal{O}}_L$ over $\mathbb{Z}$ also generates it over ${\mathcal{O}}_K$ (The ring of integers has rank the degree). If $\mathfrak{p}{\mathcal{O}}_L={\mathcal{O}}_L$, these generators satisfy $v=Av$ with all entries of $A$ in $\mathfrak{p}$. The adjugate identity makes $\det (I-A)$ annihilate ${\mathcal{O}}_L$, hence $1$; this contradicts $\det (I-A)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{p})$. Every prime factor contains $\mathfrak{p}{\mathcal{O}}_L$, so its contraction contains $\mathfrak{p}$ and equals it by maximality. Conversely, a prime above $\mathfrak{p}$ contains that finite product, hence contains one of its prime factors and equals it. Here nonzero primes are maximal because their quotient rings are finite domains. Thus the displayed product indexes exactly the primes above $\mathfrak{p}$, using only finite algebra.