Suppose that $A$ is a Dedekind domain with field of fractions $K$. Let $L$ be a finite separable extension of $K$, and $B$ be the integral closure of $A$ in $L$. We know that $B$ is also a Dedekind domain.
Let $\mathfrak{p}$ be a prime ideal of $A$. Suppose that $x\in B$ has $Nm_{L/K}(x)\in \mathfrak{p}$, is it necessary that $x$ is contained in some prime ideal of $B$ that lies above $\mathfrak{p}$? Stated in another way, if $x$ is not in any prime ideal of $B$ that lies above $\mathfrak{p}$, is it possible that we still can have $\dfrac{Nm_{L/K}(x)}{x}\in (\mathfrak{p})$ ($(\mathfrak{p})$ is the ideal generated by $\mathfrak{p}$ in $B$)?
If $x$ is not in any prime of $B$ above $\mathfrak{p}$ then it is a unit of the ring $B/\mathfrak{p} B$.
This is a $[L:K]$-dimensional $A/\mathfrak{p}$-vector space, the multiplication by $x$ is an invertible endomorphism, and $N_{L/K}(x)$ is its determinant, which is non-zero, ie. $N_{L/K}(x)\not \in \mathfrak{p}$.