A question of the uniqueness of the prime ideal lying over in an integral extension (Corollary 5.9 in Atiyah's Commutative Algebra)

Question

I am reading Atiyah's Commutative Algebra. I can understand all the proof of Corollary 5.9 except the sentence $\mathfrak{n}^c=\mathfrak{n}'^c=\mathfrak{m}$.

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

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 $\mathfrak{m}$ 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?

Corollary 5.9. Let $A\subseteq B$ be rings, $B$ integral over $A$; let $\mathfrak{q}, \mathfrak{q}'$ be prime ideals of $B$ such that $\mathfrak{q}\subseteq\mathfrak{q}'$ and $\mathfrak{q}^c=\mathfrak{q}'^c=\mathfrak{p}$ say. Then $\mathfrak{q}=\mathfrak{q}'$.

Proof. By (5.6), $B_{\mathfrak{p}}$ is integral over $A_{\mathfrak{p}}$. Let $\mathfrak{m}$ be the extension of $\mathfrak{p}$ in $A_{\mathfrak{p}}$ and let $\mathfrak{n}, \mathfrak{n}'$ be the extensions of $\mathfrak{q}, \mathfrak{q}'$ respectively in $B_{\mathfrak{p}}$. Then $\mathfrak{m}$ is the maximal ideal of $A_{\mathfrak{p}}$; $\mathfrak{n}\subseteq \mathfrak{n}'$, and $\mathfrak{n}^c=\mathfrak{n}'^c=\mathfrak{m}$. By (5.8) it follows that $\mathfrak{n}, \mathfrak{n}'$ are maximal, hence $\mathfrak{n}=\mathfrak{n}'$, hence by (3.11)(iv) $\mathfrak{q}=\mathfrak{q}'$.

Corollary 5.6. Let $A\subseteq B$ be rings, $B$ integral over $A$.
i) If $\mathfrak{b}$ is an ideal of $B$ and $\mathfrak{a}=\mathfrak{b}^c=A\cap \mathfrak{b}$, then $B/\mathfrak{b}$ is integral over $A/\mathfrak{a}$.
ii) If $S$ is a multiplicatively colsed subset of $A$, then $S^{-1}B$ is integral over $S^{-1}A$.

Corollary 5.7. Let $A\subseteq B$ be integral domains, $B$ integral over $A$. Then $B$ is a field if and only if $A$ is a field.

Corollary 5.8. Let $A\subseteq B$ be rings, $B$ integral over $A$; let $\mathfrak{q}$ be a prime ideal of $B$ and let $\mathfrak{p}=\mathfrak{q}^c=\mathfrak{q}\cap A$. Then $\mathfrak{q}$ is maximal if and only if $\mathfrak{p}$ is maximal.

Answer 1

To answer your first question, look at the commutative diagram

$$\require{AMScd} \begin{CD} A @>{}>> B\\ @VVV @VVV \\ A_\mathfrak p @>{}>> B_\mathfrak p \end{CD}$$

Going the right way, $\mathfrak n$ is contracted to $\mathfrak p$ by assumption. Hence the same holds for the left way. But the map $\operatorname{Spec} A_\mathfrak p \to \operatorname{Spec} A$ is well known to be injective and the sole pre-image of $\mathfrak p$ is $\mathfrak m$. Thus $\mathfrak n$ is contracted to $\mathfrak m$ by the map $A_\mathfrak p \to B_\mathfrak p$. Of course the same arguments works for $\mathfrak n'$.

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 $\mathfrak p$ a prime of $A$.

Answer 2

Since $\mathfrak{p}\subseteq \mathfrak{q}$, we have $\mathfrak{m}=\mathfrak{p}^{e}\subseteq \mathfrak{q}^{e}\cap A_{\mathfrak{p}}=\mathfrak{n}\cap A_{\mathfrak{p}}$.

Since $\mathfrak{q}\cap A=\mathfrak{p}$, we have $\mathfrak{q}\cap (A-\mathfrak{p})=\emptyset$. By Lemma III.4.9.(iii) in Hungerford's Algebra, $\mathfrak{n}=\mathfrak{q}^{e}$ is prime in $B_{\mathfrak{q}}$. It follows that $\mathfrak{n}\cap A_{\mathfrak{p}}$ is prime in $A_{\mathfrak{p}}$ ($\mathfrak{n}\cap A_{\mathfrak{p}}$ is the inverse image of $\mathfrak{n}$ under the inclusive mapping from $A_{\mathfrak{p}}$ to $B_{\mathfrak{q}}$) and $\mathfrak{n}\cap A_{\mathfrak{p}}\neq A_{\mathfrak{p}}$.

Since $\mathfrak{m}$ is maximal in $A_{\mathfrak{p}}$ and $\mathfrak{m}\subseteq \mathfrak{n}\cap A_{\mathfrak{p}}\neq A_{\mathfrak{p}}$, we have $\mathfrak{n}\cap A_{\mathfrak{p}}=\mathfrak{m}$.