Exercise 5.28

Question

Let $A$ be an integral domain, $K$ its field of fractions. Show that the following are equivalent:
  \begin{enum}
    \item $A$ is a valuation ring of $K$;
    \item If $\mathfrak{a},\mathfrak{b}$ are any two ideals of $A$, then either $\mathfrak{a} \subseteq \mathfrak{b}$ or $\mathfrak{b} \subseteq \mathfrak{a}$.
  \end{enum}
  Deduce that if $A$ is a valuation ring and $\mathfrak{p}$ is a prime ideal of $A$, then $A_{\mathfrak{p}}$ and $A/\mathfrak{p}$ are valuation rings of their fields of fractions.

Amit Awasthi · Accepted Answer

\begin{proof}
  $(1) \Rightarrow (2)$. Consider two ideals $\mathfrak{a},\mathfrak{b} \subseteq A$. Suppose, without loss of generality, that there exists $x \in \mathfrak{a} \setminus \mathfrak{b}$, and let $0 
e y \in \mathfrak{b}$. Then, $x/y 
otin A$, for otherwise $x \in \mathfrak{b}$ since $\mathfrak{b}$ is an ideal, and so $y/x \in A$, and so $y \in \mathfrak{a}$. Thus, $\mathfrak{b} \subseteq \mathfrak{a}$.
  \par $(2) \Rightarrow (1)$. Suppose there exist $a,b \in A$ such that $b 
e 0$ and $a/b 
otin K$; in particular, this implies $a 
e 0$. Let $\mathfrak{a} = (a),\mathfrak{b} = (b)$. If $\mathfrak{a} \subseteq \mathfrak{b}$, then there exists $c \in A$ such that $a = bc$, i.e., $a/b = c \in A$, a contradiction. Thus, $\mathfrak{b} \subseteq \mathfrak{a}$, and so there exists $c \in A$ such that $b = ac$, i.e., $b/a = c \in A$, and so $A$ is a valuation ring of $K$.
  \par Now suppose $\mathfrak{p} \in \operatorname{Spec}(A)$. Any two ideals in $A_{\mathfrak{p}}$ (resp.~$A/\mathfrak{p}$) are of the form $\mathfrak{a}_{\mathfrak{p}},\mathfrak{b}_{\mathfrak{p}}$ (resp.~$\mathfrak{a}/\mathfrak{p},\mathfrak{b}/\mathfrak{p}$), where $\mathfrak{a},\mathfrak{b}$ are ideals of $A$. Since $A$ is a valuation ring, without loss of generality $\mathfrak{b} \subseteq \mathfrak{a}$, and so $\mathfrak{b}_{\mathfrak{p}} \subseteq \mathfrak{a}_{\mathfrak{p}}$ (resp.~$\mathfrak{b}/\mathfrak{p} \subseteq \mathfrak{a}/\mathfrak{p}$). Thus, $A_{\mathfrak{p}}$ (resp.~$A/\mathfrak{p}$) is a valuation ring of its field of fractions.
\end{proof}

Exercise 5.28

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Exercise 5.28

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