On the common divisor of discriminants of integers

Abstract

Let $F$ be an algebraic number field of a finite degree, and let $K$ be an extension of $F$ of a finite degree. Denote by $\delta(K/F)$ the greatest common divisor of the discriminants of integers of $K$ with respect to $K/F$. Then, $\delta(K/F)$ is divisible by the discriminant $d(K/F)$ of $K/F$. Let $\mathfrak{p}$ be an arbitrary prime ideal of $F$, let $\mathfrak{p}=\mathfrak{q}_{1}^{e_1} \mathfrak{q}_{2}^{e_2}\cdots \mathfrak{q}_{g}^{e_g}$ be the decomposition of $\mathfrak{p}$ in $K$ into primes, and let $f_i$ be the degree of $\mathfrak{q}_i$. The set of indices $\{1, 2, \ldots, g\}$ is then divided into the union of maximal subsets $I$ such that $f_i = f_j$ whenever $i$ and $j$ belong to a common $I$. We write $f_I$ instead of $f_i$ for $i \in I$, and denote by $g_I$ the number of elements in $I$. Put on the other hand $c(I)= \sum_{d|f_{I}}\mu(f_{I}/d)N\mathfrak{p}^{d}$, where $\mu$ is the Möbius function. Then, $\mathfrak{p}$ divides $\delta(K/F)d(K/F)^{-1}$ if and only if there exists an $I$ such that $c(I) \lt f_{I}g_{I}$.