On the relative discriminant of a cyclic extension of an algebraic number field whose relative degree is a prime number Let $K$ be a cyclic extension of an algebraic number field $k$ whose relative degree is a prime number $l$. Hasse wrote(see below) in his "Bericht" that the relative discriminant of $K/k$ is of the form $\mathfrak f^{l-1}$, where $\mathfrak f$ is an integral ideal of $k$. How do you prove this? Bericht, p.22 > Es sei also $k$ ein beliebiger Grundkörper und $K$ ein relativ-zyklischer Körper vom Primzahlgrade $l$ über $k$. Man kann zunächst zeigen, daß die Relativdiskriminante $\mathfrak d$ von $K$ nach $k$ die Form $\mathfrak d = \mathfrak f^{l-1}$ hat, wo $\frak f$ ein ganzes Ideal aus $k$ ist. By the way, I'm not conversant with German. I just decoded this using Google.

This follows from _Hilbert's Different Formula_. If $P'|P$ are primes of $K|k$, then the different exponent $d(P'|P)$ is $$ d(P'|P)=\sum_{i\ge0}\left(\operatorname{ord} G_i(P'|P)-1\right), $$ where $G_i(P'|P)$ are the higher ramification groups. Those are always subgroups of the Galois group $G=C_\ell$. So each and every term on the r.h.s. is either $\ell-1$ or $0$. Therefore all the different exponents will be multiples of $\ell-1$.

The relative discriminant is the norm of the different of $K/k$, so after taking the norm, we get an ideal of $k$ raised to power $\ell-1$.