Let $A\subseteq B$ be rings, $B$ integral over $A$; let $\mathfrak{q},{\mathfrak{q}}^{\prime}$ be prime ideals of $B$ such that $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime}$ and ${\mathfrak{q}}^{c}={\mathfrak{q}}^{\prime c}=\mathfrak{p}$ say. Then $\mathfrak{q}={\mathfrak{q}}^{\prime}$.

Question 1. Why ${\mathfrak{n}}^{c}={\mathfrak{n}}^{\prime c}=\mathfrak{m}$?

My attempt: Since $\mathfrak{p}\subseteq \mathfrak{q}$, we have $\mathfrak{m}={S}^{-1}\mathfrak{p}\subseteq {S}^{-1}\mathfrak{q}=\mathfrak{n}\subseteq {B}_{\mathfrak{p}}$. But m is maximal in ${A}_{\mathfrak{p}}$, which is not necessarily maximal in ${B}_{\mathfrak{p}}$. I can't get $\mathfrak{m}=\mathfrak{n}$ by this.

Question 2. When we use that notation ${A}_{\mathfrak{p}}$, which means the localization ${S}^{-1}A$ of $A$ at the prime ideal $\mathfrak{p}$ of $A$. But in this corollary, $\mathfrak{p}$ doesn't necessarily be a prime ideal of $B$. Why can he write ${B}_{\mathfrak{p}}$? Should we write ${S}^{-1}B$ rigorously?

Question 1. Why ${\mathfrak{n}}^{c}={\mathfrak{n}}^{\prime c}=\mathfrak{m}$?

My attempt: Since $\mathfrak{p}\subseteq \mathfrak{q}$, we have $\mathfrak{m}={S}^{-1}\mathfrak{p}\subseteq {S}^{-1}\mathfrak{q}=\mathfrak{n}\subseteq {B}_{\mathfrak{p}}$. But m is maximal in ${A}_{\mathfrak{p}}$, which is not necessarily maximal in ${B}_{\mathfrak{p}}$. I can't get $\mathfrak{m}=\mathfrak{n}$ by this.

Question 2. When we use that notation ${A}_{\mathfrak{p}}$, which means the localization ${S}^{-1}A$ of $A$ at the prime ideal $\mathfrak{p}$ of $A$. But in this corollary, $\mathfrak{p}$ doesn't necessarily be a prime ideal of $B$. Why can he write ${B}_{\mathfrak{p}}$? Should we write ${S}^{-1}B$ rigorously?