Let $K$ be an algebraic number field, $\mathcal{O}_K$ its ring of integers, $\alpha\in\mathcal{O}_K$ such that $K=\mathbb{Q}(\alpha)$, $A:=\mathbb{Z}[\alpha]$, $\mathfrak{p}$ a non-zero prime ideal (so a maximal ideal) of $A$, and $\mathfrak{f}:=\{x\in\mathcal{O}_K\mid x\mathcal{O}_K\subseteq A\}$ the conductor.
If $\mathfrak{p}+\mathfrak{f}=A$ then, by extension/contraction, $\mathfrak{p}\mathcal{O}_K$ is a prime ideal of $\mathcal{O}_K$. My question is:
In case $\mathfrak{f}\subseteq\mathfrak{p}$, can $\mathfrak{p}\mathcal{O}_K$ be prime?
Best Answer
Sometimes yes, sometimes no.
For a number field $K \not= \mathbf Q$ and positive integer $c$, the order $\mathbf Z + c\mathcal O_K$ of $K$ has conductor $\mathfrak f = c\mathcal O_K$ (that is not true if $K = \mathbf Q$!), and if $c = p$ is a prime number then $p\mathcal O_K$ has index $p$ in $\mathbf Z + p\mathcal O_K$: see Example 2.1 here. An ideal of a ring with prime index is a prime ideal, so for each prime number $p$ the order $A = \mathbf Z + p\mathcal O_K$ in $K$ has a conductor $\mathfrak f = p\mathcal O_K$ that is a prime ideal of $A$. Whether or not $p\mathcal O_K$ is a prime ideal in $\mathcal O_K$ of course varies; it's the main question of algebraic number theory, isn't it? :)
Taking $K = \mathbf Q(i)$, $p\mathcal O_K$ is a prime ideal in $\mathcal O_K$ if $p \equiv 3 \bmod 4$ and it is not a prime ideal in $\mathcal O_K$ if $p \not\equiv 3 \bmod 4$.