There's a small theorem in algebra that I've been toying with. Let $R$ be integrally closed in its quotient field, and $E$ a splitting extension of the quotient field, and $S$ be the integral closure of $R$ in $E$. Then for a maximal ideal $p$ of $R$, there are only finitely many maximal ideals of $S$ whose intersection with $R$ is precisely $p$.
I was trying to formulate a small example. Let $f(X)=X^3+X+1$, and $E\supseteq\mathbb{Q}$ a splitting extension for $f$. So taking $R=\mathbb{Z}$ in this case, let $S$ be the integral closure of $\mathbb{Z}$ in $E$. I choose say $(3)$ as a maximal ideal of $\mathbb{Z}$. So there must be finitely many maximal ideals $q$ of $S$ such that $q\cap\mathbb{Z}=(3)$.
This sparked my curiosity, if we know that there are only finitely many such maximal ideals $q$, how many are there actually, at least in this case? I couldn't think of a way to actually enumerate them all to be able to say "There are 9 such maximal ideals!," or something along those lines. Thank you kindly.