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Let $K$ be an algebraic number field. Let $A$ be the ring of integers in $K$. Let $L$ be a cyclic extension of $K$ of degree $l$, where $l$ is a prime number. Let $B$ be the ring of integers in $L$. Let $G$ be the Galois group of $L/K$. Let $\sigma$ be a generator of $G$. Let $\mathfrak{D}_{L/K}$ be the relative different of L/K.

My question: Is the following proposition true? If yes, how would you prove this?

Proposition Let $\mathfrak{P}$ be a prime ideal of $B$. Let $\mathfrak{p} = \mathfrak{P} \cap A$. Then $\mathfrak{P}$ divides $\mathfrak{D}_{L/K}$ if and only if $\sigma(\mathfrak{P}) = \mathfrak{P}$ and $\mathfrak{p}B \neq \mathfrak{P}$.

Related question: Selfconjugate prime ideal of a cyclic extension of an algebraic number field of prime degree.

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General theory about splitting of primes shows that either $\mathfrak p B$ is prime (inert case), that $\mathfrak p B$ splits as a product of $l$ distinct primes which are acted on simply transitively by $G$ (split case), or that $\mathfrak p$ is the $l$th power of a single prime ideal in $B$ (ramified case).

By assumption $\mathfrak P$ divides $\mathfrak p$, and it is fixed by $\sigma$, so we are not in the split case. The assumption that $\mathfrak p B \neq \mathfrak P$ shows that we are not in the inert case either. Thus we are in the ramified case.

General ramification theory implies that the ramified primes (i.e. the $\mathfrak P$ which are $l$th roots of the $\mathfrak p$ that are ramified) are precisely the primes dividing $\mathfrak D_{L/K}$. This gives the proposition. [It also gives an answer to the linked question.]

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