# Let A &#x2286;<!-- ⊆ --> B be rings, B integral over A ; let <mi mathvari

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?
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bap1287dg
$\begin{array}{ccc}A& \stackrel{}{\to }& B\\ ↓& & ↓& \\ {A}_{\mathfrak{p}}& \stackrel{}{\to }& {B}_{\mathfrak{p}}\end{array}$
Going the right way, n is contracted to p by assumption. Hence the same holds for the left way. But the map $\mathrm{Spec}{A}_{\mathfrak{p}}\to \mathrm{Spec}A$ is well known to be injective and the sole pre-image of $\mathfrak{p}$ is $\mathfrak{m}$. Thus $\mathfrak{n}$ is contracted to m by the map ${A}_{\mathfrak{p}}\to {B}_{\mathfrak{p}}$. Of course the same arguments works for ${\mathfrak{n}}^{\prime }$.
For your second question, note that $B$ is an $A$-module, so this is just the usual notation ${M}_{\mathfrak{p}}$, when $M$ is an $A$-module and p a prime of $A$.
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