I don't know how to prove that a regular domain is a Krull domain.
If we say that $\mathcal{P}=\{p\in \operatorname{Spec}(A)\mid \operatorname{ht}(p)=1 \}$ a Krull domain is a domain s.t.:
- $A_p$ is a DVR for any $p\in\mathcal{P}$;
- $A=\bigcap_{p\in\mathcal{P}}A_p\subseteq \operatorname{Frac}(A)=k$;
- $\forall\ a\in A$ the cardinality of $\{p\in\mathcal{P}\mid a\in p\}\}$ is finite.
The first and the third condition are okay for regular rings. In a intuitive way also the second is okay but I have a doubt:
I take $\frac{a}{s}\in k\ s.t.\ \frac{a}{s}\in A_q\cap A_p$, this means that $\exists\ \frac{a_1}{s_1}\in A_p$, $\frac{a_2}{s_2}\in A_q$ with $\frac{a_1}{s_1}=\frac{a_2}{s_2}=\frac{a}{s}\in k$. How can I conclude that there exists $a_3\in A$ $s_3\in A\setminus(p\cup q)$ $s.t.\ \frac{a}{s}=\frac{a_3}{s_3}$ ?
Because I don't know if $s_1\in A\setminus p$ or $s_2\in A\setminus q$.
thank you.
I find a solution:
$A_p\subseteq\{\frac{a}{b}\in k\mid \frac{b}{a} A\cap A\not\subseteq p\}\cup\{0\}$. One cn check that it is an equality, but the inequality is immediate, because if $\frac{a}{b}\in A_p\setminus\{0\}$ we have that $\frac{b}{a}a=b\in A\setminus p$.
So $A\subseteq\bigcap_{p\in \mathcal{P}} A_p \subseteq \bigcap_{p\in \mathcal{P}}\{\frac{a}{b}\in k\mid \frac{b}{a} A\cap A\not\subseteq p\}\cup\{0\} = \{\frac{a}{b}\in k\mid b A\subseteq\frac{b}{a} A\cap A\not\subseteq p\ \forall p\in\mathcal{P}\}\cup\{0\} \subseteq\{\frac{a}{b}\in k\mid b A \not\in p\ \forall p\in\mathcal{P}\}\cup\{0\}\subseteq A$
Where the last inequality holds because if $b\in A\setminus\bigcup_{p\in \mathcal{P}}p$ the ideal $b A$ could be not contained in a prime ideal, otherwise (Krull's Hauptidealsatz) there is a $p\in\mathcal{P}$ s.t. $b\in p$ that is absurd, so $bA=A$ and $b$ is invertible in A.