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Let $A$ be a domain with $\max(A)$ a finite set. It is known that if $\mathfrak{m}\in \max(A)$ then $A_{\mathfrak{m}}$ is a PID.

Show that $A$ is also a PID.

Here is what I was thinking. Suppose that $A$ is a local ring first. It is enough to show that every $\mathfrak{p}\in \text{spec}(A)$ is a principal ideal. We know, by hypothesis, $\mathfrak{p}'$ is a principal ideal where $\mathfrak{p}'$ denotes the image of $\mathfrak{p}$ under the localization map $A\to A_{\mathfrak{m}}$. Thus, we can write $\mathfrak{p}' = (\frac{a}{1})A_{\mathfrak{m}}$ for some $a\in A$. We claim that $\mathfrak{p} = (a)A$.

Let $x\in \mathfrak{p}$, since $\frac{x}{1}\in \mathfrak{p}'$ we can write $dx = ay$ for some $y\in A$ and $d\in A\setminus \{m\}$. Since $A$ is a local ring, and $d$ is not in the maximal ideal, $d$ is a unit, therefore $x = d^{-1}ay$. This shows that $\mathfrak{p}$ is generated by $a$.

What if $A$ is not a local ring? If we pick a prime ideal $\mathfrak{p}$ then we at least know there is a finite list of maximal ideals $\mathfrak{m}_i$ for $1\leq i\leq n$ which contain $\mathfrak{p}$. As before let us denote by $\mathfrak{p}_i$ to be the image of $\mathfrak{p}$ under the localization. We can write $\mathfrak{p}_i = ( \frac{a_i}{1}) A_{\mathfrak{m}_i}$ for some $a_i\in A$.

Therefore, for every $x\in \mathfrak{p}$, and every $1\leq i\leq n$, there exists $y_i\in A$ and $d_i \in A\setminus \{m_i\}$ such that $d_i x = a_i y_i$.

Now we have two issues. Even if we can somehow show that each $d_i$ is a unit then it will only show that $\mathfrak{p} = (a_1,a_2,...,a_n)$. The issue of "simplifying" this ideal all the way down to the form $(a)$ for some $a\in A$ is essentially asking for a gcd.

It feels like we need to use the Chinese Remainder Theorem here somehow but it is not clear to me how to apply it.

user26857
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Nicolas Bourbaki
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  • I don't understand your first point: There is nothing to prove, if $A$ is local, since $A = A_{\mathfrak{m}}$ for its maximal ideal $\mathfrak{m}$. – Ulli Sep 07 '20 at 11:37
  • BTW, a domain such that each of its localisations is PID, hence DVR, is called almost Dedekind domain, see, for instance here. The above statement is Theorem 3 in this paper. – Ulli Sep 07 '20 at 12:04
  • I am not assuming that $A$ is a Dedekind domain. – Nicolas Bourbaki Sep 08 '20 at 00:03
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    With these prerequisites $A$ easily turns out to be Dedekind. See my answer below. – Ulli Sep 08 '20 at 05:11

1 Answers1

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Let $\mathfrak{m}_1, \dots, \mathfrak{m}_n$ be all maximal ideals of $A$.
$A$ is noetherian: Let $\mathfrak{a}$ be an ideal of $A$. For each $1 \le i \le n$ pick $a_i \in \mathfrak{a}$ such that $\mathfrak{a} A_{\mathfrak{m}_i} = a_i A_{\mathfrak{m}_i}$. Then $\mathfrak{a} = (a_1, ..., a_n)$.
Clearly, each non-zero prime ideal is maximal in $A$, hence dim $(A) \le 1$.
Moreover, $A = A_{\mathfrak{m}_1} \cap \dots\cap A_{\mathfrak{m}_n}$ is integrally closed, hence $A$ is a Dedekind domain.
A semilocal Dedekind domain is PID (which, as I guess, is not so trivially to prove), see, for instance, here.

Ulli
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  • Why is each non-zero prime ideal maximal? – Nicolas Bourbaki Sep 08 '20 at 00:07
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    Let $\mathfrak{p}$ be a non-zero prime ideal in $A$. Pick a maximal ideal $\mathfrak{m}$ containing $\mathfrak{p}$. Then $\mathfrak{p} A_\mathfrak{m} = \mathfrak{m} A_\mathfrak{m}$, since $A_\mathfrak{m}$ is PID. Hence $\mathfrak{p} = \mathfrak{m}$. – Ulli Sep 08 '20 at 05:32
  • Why is $\mathfrak{a} = (a_1,...,a_n)$? – Nicolas Bourbaki Sep 11 '20 at 20:46
  • Assume $x \in \mathfrak{a} \setminus (a_1, \dots, a_n)$. Then ${s \in A: sx \in (a_1, \dots, a_n)} \subseteq \mathfrak{m_i}$ for some $i$. It follows $x \in \mathfrak{a} A_{\mathfrak{m_i}} \setminus (a_1, \dots, a_n) A_{\mathfrak{m_i}}$, which contradicts $\mathfrak{a} A_{\mathfrak{m}i} = a_i A{\mathfrak{m}_i}$. – Ulli Sep 13 '20 at 06:18