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 $\mathfrak{P}$ be a non-zero prime ideal of $B$ such that $\sigma(\mathfrak{P}) = \mathfrak{P}$. Let $\mathfrak{p} = \mathfrak{P} \cap A$. Suppose $\mathfrak{p}B \neq \mathfrak{P}$.
My question: Is $\mathfrak{p}B = \mathfrak{P}^l$?